John-c-hull-options-solutions-manual.pptx

SOLUTIONS MANUAL

Options, Futures, and Other Derivatives Eighth Edition Global Edition

John C. Hull

Maple Financial Group Professor of Derivatives and Risk Management Joseph L. Rotman School of Management University of Toronto

Pearson Educatlon

Acquisitions Editor: Tessa O’Bricn Senior Acquisitions Editor, International: Laura Dent Editorial Project Manager: Amy Foley Associate Production Project Manager: Alison Eusden Senior Manufacturing Buyer: Carol Melville Cover Image: Image Source/Alairy Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companics around the world Visit us on the World Wide Web at. www.pearsoned .co.uk CPearson Education Limited 2012 The right of John C. Hull to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1955. Pearson Education is not responsible for the content o1 third party intcrnet sitcs. All rights reserved. Permission is hereby given for the material in this publication to be rcproduccd for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any forin or by any means, electronic, mechanical, photocopying, recording, or otherwise without either the prior written perinission of the Publishers or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd., Saffron House, ó- 10 Kirby Street, London EC IN 8TS. This book may not be lent, resold, hired out or otherwise disposed of by way of made in any form of binding or cover other than that in which is it is published, without the prior content of the Publishers. ISBN 13: 97f›-0-273-75911-9 ISBN 10: 0-273-75911-6 10 9 8 7 6 5 4 3 2 1 15 14 13 12 11 Printed and bound by Courier Westford in The United States ot America.

Contents Preface Chapter 1 Cfapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter fi Chapter 7 Chapter 8 Chapter 9 Chaptes 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27

Introduction Mechanics of Futures Markets Hedglng Strategies Using Futures Interest Rates Determination of Forward and Futures Prices Interest Rate Futures Swaps Securitization and the Credit Crisis of 2007 Mechanics of Options Markets Properties of Stock Options Trading Strategies Involving Options Binomial Trees Wiener Processes and Ito’s Lemma Thc Black-Scholes-Merton Model Employee Stock Options Options on Stock Indices and Currencies Options on Futures Greek Letters Volatility Smilcs Basic Numerical Procedures Value at Risk Estimating Volatilities and Correlations Credit Risk Credit Derivatives Exotic Options More on Models and Numerical Procedures Martingales and Measures lntercst Rate Derivatives: The Standard Market Models Convexity, Timing and Quanto Adjustments Interest Rate Derivatives: Models of the Short Rate Interest Rate Derivatives: HIM and LMM Swaps Revisited Energy and Commodity Derivatives Rea1 Options

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1 10 1 fi 22 30 38 45 54 57 65 72 79 87 94 107 110 1 19 127 139 146 162 1 67 173 183 190 195 209 216 223 229 239 244 247 250

CHAPTER 1 Introduction Problem 1.1 Whcit is the difference between a long forward yo.sition and o .short [omrard position?

When a trader enters into a Jong forward contract, she is agreeing to Sri)' fhe underlying asset for a certain price at a certain time in the future. When a trader enters into a short forward contract, shc is agreeing to se// the underlying asset for a certain price at a certain time in the fbturc. Problem 1.2. ExJ›lain carefully the difference between hedging, spe‹uilation, anJ arbitrage. A trader is hedging when she has an exposure to the price of an asset and takes a position in a derivative to offset the exptisure. In a .yy›ect/ation the trader has no exposure to otlsct. She is betting on the future movements in the price of the asset. Arbitrage involves taking a position in tw'o or more different markets to lock in a profit. Problem 1.3. Tal is the Ji erence between entering into a long fori»ard contract n›lien the/orward J›rice /.s .S56 and taking a long position in a call option with a 5’trike price oJ $50.’" In the first case the trader is obligatcd to buy thc asset for $50. (The trader does not have a choice.) In the second case the trader has an option to buy the asset for $50. (The trader does not hate to exercise the option.) Problem 1.4. £xy/nin careJiillv the difference lleW’een .selling a call option and l›uying a pm optian. Selling a call option involves giving someone else the right to buy an asset from you. It gives you a payoff or — max(N—t K,0) — min(K S , D) Buying a put option involves buying an option from someone else. It gives a payotl‘ ot mex(K Nt, 0) In both cases the potential payoff is K — S . When you write a call option, the payoff is negativc or zero. (This is because the counterparty chooses whether to exercise.) Wheil you buy a put option, thc payoff is zero or positive. (This is because you choose whether to exercise.) Problem 1.5. An in»e.stor enters into a short forward contract to sell 100,000 British pounds jor US dollars at an exchange rate of 1.4ñ00 k'S dollar.s jeer J›ound. Herr much does the investor gain or lose i[the exchange rate ct the end o/ the contract /.s (a) 1.3 90P and IN) 1.4200? (a) The investor is obligated to sell pounds for 1.4000 when they are worth 1.3900. Thc gain is (l .4000-1.3900) ^ 100,000 — S1,000. 1 /20 12 Pearson Education

(b) The investor is obligated to sell pounds for 1.4000 when they are worth 1.4200. The loss is (1.4200- 1.4000)^ 100.000 = $2,000 Problem 1.6. A traJer enters into u short c'otton futures ‹’ontru‹ t nhen the}utw-es price is 50 cents per pounJ. The contract is]or the deliver’ of 50,000 pounJs. Hey inns h Joes the traJer gain or la.ye i[the cotton price at the end of the contrac't is (a) 48.2h cents per pound, tb) 51.30 cen/.y per pound? (a) The trader sells ftir 50 cents per pound something that is worth 45.20 cents per pound. Gain = (50.5000 — $0.4820) x 50, 000 = $900 . (b) The trader sells for 50 cents per pound something that is pound. Loss ($0.51 30 — $0.5000) x 50, 000 = 5650 .

orth 51.30 cents per

Problem 1.7. Suf›po.se that you irrite a put contract with a strike price o[$40 and an expiration date in three month.s. The current .stock prices i.s $41 and the contract i.s on 100 share.f. What have vow committed your.se/[to.'• How much could you gain or lo.se? You have sold a put option. You have agreed to buy 100 shares for 540 per share if the party on the other side of the contract chooses to exercise the right to sell for this price. The option will be exercised only when the price or stock is below $40. Suppose, for example, that the option is exercised when the price is $30. You have to buy at 540 shares that are z'orth S30; you losc S10 per share, or $1,000 in total. 11’the option is exercised v.hen thc price is $20, you lose $20 per share, or $2,000 in total. Thc v orst that can happcn is that thc price o1 the stock declines to almost zero during the three-month pcriod. This highly unlikely event would cost you $4,000. In return for the possible future losses, you receive the price of the option from the purchaser. Problem 1.8. What is the difference hetsreen the over-the-counter market and the exchange-traded market? What are the bid and offer quotes of a market maker in the o› er-the-counter market'? The over-the-counter market is a telephone- and computer-linked network of financial institutions, fund managers, and corporate treasurers where two participants can enter into any mutually acceptable contract. An exchange-traded market is a market organized by an cxchange where traders either meet physically or communicate electronically and the contracts that can be traded have been defined by the exchangc. Whcn a market maker quotes a bid and an otlcr, thc bid is thc price at which thc market maker is prepared to buy and the offer is thc price at which thc markct makcr is prcparcd to scll. Problem 1.9. Yoa would like to .specii late on a ri.se in the price ofa cci-lain .stock. The current .stock price i.s $29, and a three-month call ii’ith a strike o[$30 co.sts S2.90. You /inv'e 5J,800 to invest. Ident f' tit o alteryutiye .strategies, othe inve lving an iiive.ffmeftf in the stocL and the other

involt ing inve.stment in tltc• option. What at-e the potential guin.s anJ los,sc,s from each!! One strategy would be to buy 200 shares. Another would be to buy 2,000 options. If the share price does v cll thc sccond strategy will give rise to grcatcr gains. For cxample, if the share 2 '201 2 Pearson Education

price goes up to $40 you gain [2, 000 x (54—0$30)—] $5, 800 = $14, 200 from the second strategy and only 200 x ($4—0 $29) = $2, 200 from the iirst strategy. However, if the share price does badly, the second strategy gives greater losses. For example, it the share price goes down to $25, the first strategy leads to a loss of 200 (529 — 525) = $s00. whcreas the second strategy leads to a loss of the » hole $5,800 investment. This example shows that options contain built in leverage. Problem 1.10. Suppose you ow n 5,ONO shares that are worth $25 each. How can yut option.s lie u.sed to provide you with insurance against a derline• in the values of your holding ovc•r the nyxt four oionths*

You could buy 50 put option contracts (each on 100 shares) with a strike price ot $25 and an expiration date in four months. II at the end o1 four months the stock price proves to be less than $25, you can exercise the options and sell the shares tor $25 cach. Problem I.11. When first i.s.sued, a .stock provide.s fund.s for a company’. I.s the .same true of an e.xchangetraded stock option'? Di.scuss. An exchange-traded stock option provides no funds for the ctiinpany. It is a security sold by one investor to another. The company is not involved. Sy contrast, a stock u hen it is first issued is sold by the company to investors and does provide funds for the company. Problem 1.12. Explain why a futures contra‹’t ccin be tised for er/her spe‹’ulation or heéfging. If an investor has an exposure to the price of an asset, he or she can hcdgc with futures contracts. If the investor will gain when the price decreases and lose when the price increases, a long futures position will hedge the risk. If the in›estor will lose when the pricc decreases and gain when the price increases. a short futures position will hedge the risk. Thus either a long or a short futures position can be entered into for hedging purposes. If the investor has no exposure to the price of the underlying asset. entering into a futurcs contract is speculation. If the investor takes a long position, he or she gains when the asset’s pricc increases and loses when it decreases. If the investor takes a short position, he or she loses when the asset’s price increases and gains when it decreases. Problem 1.13. Suppose that a March call option to buy a share Jor $50 ‹’osts $2.50 and is field until starch. Under what circumstances will the holder o[the option make u pro]it? C’ndei- ›i hat circum.stance.s will the option he exerci.red° Draw a diagram showing how the profit on u long go.si/ier7 in the option deg›end.s on thy .y/ock /›rice at the maturin oJ the option. The holder of the option will gain if the price of the stock is above $52.50 in March. (This ignores the time value of money.) The option »'ill be exercised if the price of the stock is above $50.00 in March. The profit as a function of the stock price is shown in Figure S1.1.

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St oc k Price

45

Figure SI.1

50

55

60

Profit from long position in Problem 1.13

Problem 1.14. Suppose that u June put option to se// a share for $60 co.st.s $4 and i.s lielJ until June. UnJer what circumstances ill the seller of the option (i.e., the party' with a short position) make u profit? Un‹1ei‘ yvhat cin:umstances will the option he exercised." bra iv a diagram showing how the pro]it]rom a short position in the option depend.s on the stock price al flue muturin of the oytion.

The seller of the option w ill lose money if the price of the stock is below 556.00 in June. (This ignores the time value of money.) The option will be exercised if the price of the stock is below 560.00 in June. The profit as a function of the stock price is shown in Figure SI.2.

4 St ock Price 60

Figure 51.2

7 0

65

Profit from short position in Problem 1.14

Problem 1.15. It is .kay anJ a lraJer writes a Septeml›er call option with a .strike pricr o] $20. The 5’to‹’k 4 ?2012 Pearson Education

price i.y 518, end the option price is $2. Describe the inve.stor ’.s cash flows if the option is held until $e ›tember am' the s tock price is $25 at this time. The trader has an inflow of $2 in May and an outflow o1 $5 in September. The $2 is the cash received from the sale of the option. The $5 is the rcsult or the option being exercised. The investor has to buy the stock for $25 in September and sell it to the purchaser of the option for S2fl. Problem 1.16. A trader writes a December put oplion vvifh u strike pt-ice of $3t1. The J›rice of the option is $4. L“nJer hat ‹'ircuinytance› Jors the traJer make a gaits." The trader makes a gain if thc price of the stock is above $26 at the time of exercise. (This ignores the time value of money.) Problem 1.17. A company knows that it i.s due to receive a certain amount o] a foreign ctirrency in four monlhs. V"hut type o[option contract i.s appropriate for heJging? A long position in a four-month put option can proc ide insurance against the cxchange rate falling below the strike price. lt ensures that the foreign currency can be sold for at least the strike piicc. Problem 1.18. A US company expec'ts to have to pay I million Canadian dollars in .six months. Explain heir the exchange rate risk can be liedgeJ using (a) a]o arJ contract,’ (b) an option. The company could enter into a long forward contract to buy 1 mil lion Canadian dollars in six months. This w’ould have the effect of locking in an exchange rate equal to the current for ard exchange rate. Alternatively the company could buy a call option giving it the right (but not the obligation) to purchase 1 million Canadian dollars at a certain exchange rate in six months. TO is v ould provide insurance against a strong Canadian dollar in six months while still allowing the company to benefit from a weak Canadian dollar at that time. Problem 1.19. A trader enters into a short fort ard contras i on 10t1 million yen. The forward exchange rate is $0.00RT per yen. How mu‹’h Joes the li-ader gain or lo.se if the exchange rate at the end of //ie contract i.y (a) $0.0074 per i en, tb) $0.0091 per yen'? a) The trader sells 100 million yen for $0.0080 per yen when the exchange rate is $0.0074 per yen. The gain is 100 x 0.0006 millions of dollars or $60,000. b) The trader sells 100 million yen for $0.0050 per yen when the exchange rate is $0.0091 per yen. The loss is 1 00 x 0.0011 millions of dollars or $110,000. Problem 1.20. The Chreap o BoarJ of Tra Je ofers a fulures contract on long-term Treasur) honds. Characterize the reveller-s likely to use this contract. Most investors z ill use thc contract because they want to do one of the following: a) Hedge an exposure to long-tcrm intcrcst rates. 5 2012 Pearson Education

b) Speculate on thc future direction of long-term interest rates. c)Arbitrage between the spot and futures markets for Treasury bonds. This contract is discussed in Chapter ñ. Problem 1.21. “Options and fixtures are zeto-sum games.” What do you think is meant lay this statement? The statement means that the gain (loss) to the party z'ith the short position is equal to the loss (gain) to the party with the long position. In aggregate, the net gain to all parties is zero. i'roblem 1.22. Describe the profit from the following portfolio: a long Jorw ard contract on an a.s.yes cmd a long European put option on //ie asset with the same maturity' as the forward contract unJ a striM price that is equal to the forward j›rice oj the asset at the time the yor¡folio is set up. The terminal value of the long forward contract is: where 6 is the price of the asset at maturity and I, is the delivery’ price, which is the same as the forward price of’ the asset at the time the portfolio is set up). The terminal value of the put option is: max (N0 zS ,ñ)

The terminal value of the portfolio is therefore

This is the same as the terminal value ot’ a European call option with the samc maturity as the forward contract and a strike price equal to F . This result is illustrated in the Figure S1.3. The profit equals the terminal value of the call option less the amount paid for the put option. (It does not cost anything to enter into the forward contract. 15

__ 10

.* 20

__ 30

40

-10

- - - Put Total

20

Figure SI.3

Profit from portfolio in Problem 1.22

Problem 1.23. In the 1980s, Bankers Trust developed indeX currency' OF!ion notes (ICO.U.s). These are bonds 6

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in which the amount received bv the holder at maturity varies with a foreign exchange rate. One example was r/i trade with the Long Term Credit Bank of Japan. The ICON specified that /the yen—U S. dollar exchange rate, S , is greater than 169 yen per dollar at maturity (in 1995), the holder of the bond receive.s $1,0P6. If it is less than 169 yen per dollar, the amount received by the holder of the bond i.s 1, 00—0

max 0, 1, 000

SI

—1

When the exchange rate is belo84.5, nothing is received hy the holder at maturity. Show that thi.s ICON is a combination of a regular bond and two option.s. Suppose that the yen exchange rate (yen per dollar) at maturity of the ICON is S . The payoft’ from the ICON is 1, 000 1, 000 — 1, 169 — 000 0 1 if

When 84.5 S,

if

S, > 169

if 84.5 ñ ST 169 3 < 84.5

Si

169 the payoff can be written 169, 000 2, 000 — The payoff from an ICON is the payoiT from: (a) A regular bond (b) A short position in call options to buy 169,000 yen with an exercise price of 1/169 (c)A long position in call options to buy 169,000 yen with an exercise price of 1/84.5 This is demonstrated by the following table, which shows the terminal value of the various componcnts of the position

Sz > 169 Sz < 84.5

Bond 1000

1000

Short Call.s 0

169, 000(/ —

Whole position 1000

Long Calls 0

6

169, 000( )

2000 — 1 69. 000

—

0

Problem 1.24. On July 1, 2011, a company enters into a forward contract to buy 10 million Japanese yen on January I, 2012. On September 1, 2011, it enters into a%rward contract to sell 10 million Japanese yen on January 1, 2012. Describe the payoff from thi.s .strategy. Suppose that the forward price for the contract entered into on July 1, 2011 is N, and that the forward price for the contract entered into on September 1, 2011 is Hzwith both ' and f 2 being measured as dollars per yen. If the value of one Japanese yen (measured in US dollars) 7 â20l2 Pearson Education

is S, on January 1, 2012, then the value of the first contract (in millions of dollars) at that time is 10(S, — £} ) while the value of the second contract tper yen sold) at that time is: 10(F_, — S ) The total payolT from the two ctintracts is therefore I 0(S — , ) * 10(F2 Nt ) = 10(I, — , ) Thus if the for ard price for delivery on January 1, 2012 increased between July 1, 2011 and September 1, 2011 thc company will makc a profit. (Note that the yen/USD exchange rate is usually cxprcssed as the number of yen per USD not as the number of USD per yen) Problem 1.25. Suppo.se that USD-.sterling spot and forward exchange rates are as follow.s: 1.45R0 1.4556 1.4518 What opportunitie.s are open to an arbitrageur in the following situation.f? (a) A 180-day European call option to buv £1 %r $1.42 cost.s 2 cent. . (b) N 99-day' European put option to sell £1 for $1.49 cost.s 2 cent.s.

ta) The arbitrageur buys a l80-day call option and takes a short position in a 180-day forward contract. If St is the terminal spot rate, the profit from the call option is max( S — 1 .42, 0) — 0.02

The profit from the short forward contract is 1.4518 — S z

The profit from the strategy is therefore max( .S;. — 1 .42, 0) — 0.02 + 1 .4518 — 5

max( S — 1.42, 0) + 1.4318 — S z

This is 1.431S— S, 0.118

when Si °1.42 when 5'¿>1.42 This shows that the profit is always positive. The time value o1 money has been ignored in these calculations. However, when it is taken into account the strategy is still likcly to be profitable in all circumstances. (We would require an extrcmely high interest rate for $0.0118 interest to be required on an outlay of $0.02 over a l80-day period.)

(b) The trader buys 90-day put options and takes a long position in a 90 day forward contract. If S is the terminal spot rate, the profit from the put option is max( 1. 49 S, , (i) — 0.02

The profit from the long forward contract is Sr-1.4556 Thc profit trom this strategy is therefore â'20l2 Pearson Education

max(1 .49 - .S, . 0) — 0.02 * 3, — 1 .4556 rriax( 1 .49 - 3t , 0) S z — 1.4756

This is S —1.4756 when St> l .49 0.0144 when St<1.49 The profit is therefore always positive. Again, the time value of money has been ignored but is unlikcly to affect the overall profitability of the strategy. (We would require interest rates to be extremely high for $0.0144 interest to be required on an outlay of $0.02 over a 90-day period.)

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CHAPTER 2 Mechanics of Futures Markets

The open inte•rc•.st of a futures contract at a particular time is the total number of’ long positions outstanding. (Equivalently. it is the total numbcr o1’ short positions outstanding.) The trading ro/time during a cenain period o1 time is thc number of contracts traded during this period. Problem 2.2. What i.s ltte difference l›etv'een a local and a futures comrni.y.tion merchant'? A futttres comrui.s.slow merchant trades on behalf of a client and charges a commission. A loco/ trades on his or her own behalf. Problem 2.3. Suppose that you enter into a short Jiitures ‹’entrap I to sell July silver for $1.7.20 per ounce. The size oj!the conli-acl is 5,000 ounces. The initial marz«in is $4,000, and the maintenance mark in is $3,000. What ‹ harize in the fixtures pri‹’e v’ill lead to a margin call? What happens if von do not meet the margin call? There w'ill be a margin call » hen $1.000 has been lost frtim the margin account. This will occur when the price of silver increases by 1,000/5,000 = $0.20. The price of silver must therefore rise to S17.40 per ounce for there tti be a margin call. If the margin call is not met, your broker closes out your position. Problem 2.4. Suppose lhut in Neyfe/nfier +°072 u cc›inyo/n lames o long position in u ‹ extract on way 2013

Guide oil Jtitures. It closes out ils po›'ition in 'Vtai ‹ li 2013. The futures price (per barrel) is $68.30 w lien it enfers’ into the contru‹’t, S70.5.0 when it closes out iis position, and $69.10 at the en‹J uj De‹'eniber 2012. One’‹:onIra‹I is fut like ‹Jeli i'e ' o J,0iJ6 barrels. fJ 7/at i.s the company ’s total pro]it? When is i/ realized? How is it ta.red i[it is (a) a hedger and (h) a sf›eculator? Assume that the compare ha.s a December 31 yc•ar-enJ. The total profi I is (S70.50 $fi8.30) x I .0titJ = $2,2fltl. Ol this ($69.1—0 $68.30) x 1. 000 or 5800 is realized on a day-by-day basis between September 201.2 and December 31, 2012. A further (570.50 — $69.10) x 1, 000 = $1. 400 is realized on a day-by-day basis between January 1, 2013, and March 20.13. A hedger ’ould be taxed on the whole profit of $2,200 in 20.13. A speculator would be taxcd on $800 in 20.12 and S1,400 in 20.13. Problem 2.5. What doe.s a stop order to sell at 12 mean? When might it he it.red? 4’hat does a limit order to .sell at 52 mean? When might it he u.sed? A crop order- to sell at $2 is an order to sel1 at the best available price once a price of S2 or l tl 02012 Pearson Education

less is reachcd. It could be used to limit the losses from an existing long position. A limit order to sell at $2 is an order to sell at a price of $2 ter more. It could be used to instruct a broker that a short position should be taken, pro›'iding it can be done at a price more favorable than $2. Problem 2.6. What is the di ffyrence betw'een the operation o[the mai gin accounts adiiiin istered hy a clearing house and tliose a‹fminis'fered fir a hroker? The margin account administered by the clearing house is marked to market daily, and the clearing house member is required to bring the account back up to the prescribed level daily. The margin account administered by the broker is also marked to market daily. However, the account does not have to be brought up to the initial margin level on a daily basis. It has to be brought up to the initial margin level whcn the balance in the account falls below the maintenance margin level. The maintenancc margin is usually about 75% of the initial margin. Problem 2.7. What diJ/erence.s exist in the ›vuv prices are quoted in the foreign exchange futures market, the foreign exchange .spof market, anJ thc foreign exc'hange forward market? In futures markets, prices are quoted as the number o1 US dollars per unit of foreign currency. Spot and forward rates are quoted in this way for the British pound, euro. Australian doliar, and New Zealand dollar. For other major currencies, spot and forward rates are quoted as the number of units of foreign currency per US dollar. Problem 2.8. The party with a short position in a futures contract sometimes has options as to the precis‹• asset that ivi// he delivereJr, here delivery will take place, w›hen delivery ›•'ill take place, unJ so on. Do the.se options increase or decrea.se the futures price? Explain your reasoning. Thcse options make the contract less attractive to the party with the long position and morc attractive to the party with the short position. They therefore tend to reducc the futures price. Problem 2.9. What are the mosl important a.spects of the Jesigil of a ne futures contract? The most important aspects of the design o1 a new futures contract are the specification of the underlying asset, the size of the contract, the delivery airangcments, and the deliv cry months. Problem 2.10. Explain hoii’ margins prote‹’t investors against the possibility of default. A margin is a sum of money deposited by an investor with his or her broker. It acts as a guarantee that the investor can cover any losses on the futures contract. Thc balance in the margin account is adjusted daily to reñect gains and losses on the futures contract. If losses are abo\‘e a certain level, the investor is required to deposit a further margin. This system makes it unlikely that the investor will default. A similar system of margins makes it unlikely that the in 'estor’s broker will default on the contract it has with the clearing house member and unlikely that the clearing house inembcr will default with the clearing house. 11 '2012 Pearson Education

Problem 2.1 J. A traJer buys twin July futw es contract.s on frozen orange juice. Each contract i.s %r the Jelii'ery o] 15,000 poutiJs. The current htm-es price is 160 c'en/s per pound, the initial margin is $6,000 per contract, and the mainte•nance maigrn is $4,5t18 her contract. Whal price change would leod to a margin call.’* kinder ii hat circumstance.s could 52,000 he withdrawn from the margin account’? There is a margin call it more than $1,500 is lost on one contract. This happens i1 thc fixtures price of frozen orange juice falls by more than 10 cents to below 150 cents per pound. $2,000 can be withdrawn from the margin account if there is a gain on one contract of $1,000. This will happen if the futures price rises by 6.67 cents to 166.G7 cents per pound. Problem 2.12. Show that, i f the future.s j›rice ofa commoJil y is greater than the spot jar-ice during the delivery J›eriod, then thc•rr i.s an arhitruge opporttiniH . Does an arbitrage opportunity exist if the natures J -ice is /e.ss thati the spot price’? Exploin 5 our answer. lf the futures price is greater than the spot price during the delivery period, an arbitrageur buys the asset, shorts a futures contract, and makes delivery for an immediate profit. If thc futures price is less than the spot price during the delivery period, there is no similar perfect arbitrage strategy. An arbitrageur can take a long futures position but cannot force immediate delivery of the asset. Thc dccision on when delivery w’ill be made is made by the pany v ith the short position. Ncvertheless companies interested in acquiring the asset may find it attracti v e to enter into a long futures contract and wait for delivery to be made. Problem 2.13. Explain the difference hetv ecu a market-if-toiicheJ orJer and a stop order. A market-if-touched order is executed at the best available price after a trade occurs at a specified price or at a price more favorable than thc specified price. A stop order is executed at the best available price after there is a bid or oticr at the specified price or at a price less far orable than the specified price. Problem 2.14. Explain i»hat a ›top-limit order to sell at 0.30 with a limit of2ñ.10 mean›'. A stop-limit order to scll at 20.30 with a limit of 20.10 means that as soon as there is a bid at 20.30 the contract should be sold providing this can be done at 20.10 or a higher price. Problem 2.15. At the end of one day a clearing house memher is long 100 contracts, and the .sei//emenr price i.s $50,000 per contract. The original margin is $2,000 per contract. On the follo n ing Jay the memher l›ecome.s resf›onsible for‘ clearing an aJJitional 20 long contract.s, entereJ into at a price of $51,ONO per rotilracl. The settlement pri‹’e at the end o/ this day is $50,200. He» much Joes the member- have to aJH to its margin account with the exchange c'le.aring house? The clearing house mcmber is requircd tti provide 20 x 52, 000 = $40. 000 as initial margin for the ncu contracts. There is a gain tif (50,200 50,000) x 100 $20,000 on the existing contracts. There is also a loss of (5 1, 00—0 50, 200) x 20 $1 6. 000 on the new contracts. The 12 9201 2 Pearson Education

member must therefore add 40, 000 — 20, 000 + 16, 000 = $36, 000 to the margin account. Problem 2.16. in July 1, 2012. a Japanese company enters into a [orwarJ contract to huy $1 million ith yen on January 1, 2013. On September 1, 2012, it enters into a forward contract to sell 81 million on January I, 2ñ13. Dec crihe the yrofit or‘ los.s the company will make in dollar.f as a fz/ nction of! the forward e.rchange. rates on .July 1, 2012 and September I, 2012. Suppose I, and N, are the forward exchange rates for the contracts entered into .July 1, 2012 and September 1, 2012, and N is the spot rate on January 1, 2013. (All exchangc rates arc measurcd as yen per dollar). The payoff from the first contract is (3 — F) million yen and the payoff from the second contract is (I— N) million yen. The total payoff is therefore (S — I ) + (I,— d) = ( — F,) million yen. Problem 2.17. The jomvard price on ltte St+i.s.f francQtr delivery in 45 Jays is quoted as 1.1ñ00. The}titures price for a contract that ii ill he delivereJ in 45 day.s is 0.9000. Explain these two quotes. Which is more favorable for an ins'eslor ‹inling to sell Swiss franc.s? The 1.1000 forward quotc is the number o1 Swiss francs per dollar. The 0.9000 futures quote is thc number of dollars per Swiss franc. When quoted in the same way as the futures price the forward price is1 1.1000 — 0.9091 . The Swiss franc is therefore more valuable in the forward market than in the futurcs market. The forward market is therefore more attractive for an investor wanting to sell Swiss francs. Problem 2.18. Suppose you rall your broker ated issue instructions to .sell one Jtily /tog.y contract. Describe w›hat happens. Live hog futures are traded on the Chicago Mercantile Exchange. The broker will request some initial margin. The order will be relayed by telephone to your broker’s trading desk on the Roor oI the exchange (or to the trading dcsk ot another broker). It will then be sent by messenger to a commission brokcr who will execute the trade according to your instructions. Confirmation of the trade eventually reaches you. It there are adverse movemcnts in the matures price your broker may contaei you to reqiicst additional margin. Problem 2.19. '’Spec’u:lation in futures markets i.‹pure gambling. It is not in the puhlic interr 'I to allow' speculators to trcide on ci]uiures exchan pe.” Discus.f this view'point. Speculators are important market participants because they add I iquidity to the market. However, contracts must be useful ftir hedging as well as spcculation. This is because regulators generally tinly approve contracts when they are likely to be of interest to hedgers as well as speculators. Problem 2.20. Live cattle [uture› lradc »ith June, Augu 't, October, December, Februarv, and April 13 92012 Pearson Education

maturities. Why Jo you think that the oj›en intere.st for the June ‹’ontract i.s le.ss than that for fhe ñii sf contract in Table 2.2? Normally, the shorter the maturity of a contract is, the higher the open interest. However, traders tend to close out their positions in the month immediately before the maturity month. This means that the open interest for the closest maturity month can be less than that for the next closest maturity month Problem 2.21. What do you think w'ould happen iJ an exchange .s/armed /radiag a contract in whi‹’h the quality of flue underlying asset was incompletely sJ›ecified!" Thc contract would not be a success. Parties with short positions would hold their contracts until delivery and then deliver the cheapest form of the asset. This might well be viewed by the party with the long position as garbage! Once news o1 the quality problem became widely known no one would be prepared to buy the contract. This shows that futures contracts are feasible only when therc are rigorous standards within an industry for defining the quality of the asset. Many futures contracts have in practice failed because of the problem of defining quality. Problem 2.22. “When a futurcs.s eontrcict i fruéfeJ on the floor of the exchange, it may' be the case that the open interest increase•s by one, s tays the same, or decrea.se.s hy one. ’ Explain this statement. II both sides or the transaction are entering into a new contract, the open interest increases by one. If both sides of thc transaction are closing out existing positions, thc open interest decreases by one. It one party is entering into a new contract while the other party is closing out an existing position, the open interest stays the same. Problem 2.23. Suppose that on Octol›er 24, 2t)12, a company’ sells one April 2013 live-cattle futures contracts. It closes out it.s position on January 21, 2013. The hityres price (per pounJ) is 91.20 cents when it enters into the contract, 88.30 cents when it closes out its position, and RR.Rh cents at the end of December 2012. One contract is for the delivery of 40,000 pounds of ca/cfc. What i.s the total profit.’" How is it ta.xed if the comI›at’ t y is (a) u heifer and (b) a .speculator? A.s.crime that the company has a December 3 I ’ear enJ. The total profit is 40, 000 x (0.912—0 = $1,160 If the company is a hedger this is all taxed in 2013. If it is a speculator 40, 000 x (0.9120 — 0.5880) = $960 is taxed in 2012 and 40, 000 (0.8580 — 0.8830) = $200 is taxed in 2013.

0.8530)

Problem 2.24. A cattle farmer expects to have 12a,000 pounds of live• cattle to sell in three month.s. The livecattle futures contract truJeJ by the CMI Grouy is for the ‹delivery of 40,000 pounds of cof//e. How c un the farmer u.se the contract]or hedging? From the [armyr’s viewpoint, what are the pros and cons of hedging'! 14

The farmer can short 3 contracts that have 3 months to maturity. If the price of cattle falls, the gain on the futures contract will of1’set the loss on the sale of the cattle. If the price of cattle rises, the gain on the sale of the cattlc will be offset by the loss on the natures contract. Using futurcs contracts to hedge has the adsantagc that it can at no cost reduce risk to almost zero. Its disadvantage is that the farmer no longer gains from favorable movements in cattle prices. Problem 2.25. It i.s July 2011. A mining company Isa.s ju.st discovereJ u small Jeposit o[gold. It vNll take six month.s to construct the mine. The gold n'ill then he extras tel c'n u more or /e3,s continuous basis for one year. Futures con/roc/s on gold a+ e available ii ith Jeli ver5: /nou//is every h› o months from Ange› 1 2011 to December z!0l z°. Each contrae’t is]or lhe ile•/i»ci of 100 ount e.. Discuss how the mining company mighl use future.s inarket.s [or lieJging. The mining company can estlmate its production on a month by month basis. It can then short fixtures contracts to lock in the price rcceived for the gold. For example. i1’a total of 3.000 ounces are expected to be produced in Scptember 2011 and October 2011, the price received for this production can be hedged by shorting 30 October 2011 contracts.

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CHAPTER 3 Hedging Strategies Using Futures Problem 3.1. k"ndei what circum.stancc•s are (o) a short heJge anJ (h) a long l1edp‹• appropriate? A short hedge is apprtipriate when a company owns an asset and expects to sell that asset in the future. lt can also be used when the company does not currently own the asset but expects to dti so at some time in the future. A long heJge is appropriate when a company knows it will have to purchase an assct in the future. It can also be used to offset the risk from an existing short position. Problem 3.2. Explain what i› meant b» Isa.sis risk v'hen fixtures contracts are useJ for hedging. Basis risk arises from the hedger’s uncertainty as to the difference betw’een the spot price and futures price at the expiration of the hedge. Problem 3.3. ExJ›l‹iin s‹hat i.s meant lay u per fe•ct hedge. Does o perfect hedge alat c9›s lead to a better outcome than an imperfect h‹•Jge? Explain your on.River. AI›erf‹•ct lieJpe iS one that completely eliminates the hedger’s risk. A perfect hcdgc does not al ays lead to a

better outcomc than an imperfect hedge. lt just leads to a morc certain outcome. C?onsider a company that hcdgcs its exposure to the price of an asset. Suppose the asset’s price movements prove to be favorable to the company. A perfect hcdge totally neutralizes the company’s gain from these favorable price movements. An imperfect hedge, which only partially neutralizes thc gains, might u elf give a better outcomc. Problem 3.4. L“nJer what circumstance.s doe. a minimum-variant e hedge portfolio lead to no hedging at A minimum ›‹iriance hedge leads to no hedging when the coefficient or correlation between the futures price changes and changes in the price of the asset being hedged is zero. Problem3.5. Give three reason,s ii h the treasurer ofa company might not heJge the comJ›any’.s exposure to a particular risk.

11 the company’s compctitors are not hedging, the treasurer might feel that the company will experience lcss risk if it does not hedge. (See Table 3.1.) The shareholders might not want the company to hcdgc because the risks are hedged within their portfolios. lf there is a loss on the hcdgc and a gain from the company’s exposure to the underlying asset, the treasurer might feel that he or she w ill have difficulty justifying the hedging to other executives within the organization. ‹cD2012 Pearson Education

Problem 3.6.

The optimal hedge ratio is 0.65

0.8x 0.81

0.642

This means that the size of the futures position should be G4 2'?o of the size of the company’s exposure in a three-month hedge. Problem 3.7. A company has a $28 million porI]olio »ith u beta oJ 1.2. It ’ould like tG H3’€ $iltUl“€S conti acts on the S&P 500 to heJge its risk. The index futures is ‹ ui’rently standing at 1080, anal each conlru‹’t is]or Jelivei y of 5250 times the index. V'hat is the heJge that minim i-es risk? What should the c'ompuny Jo il it ii’ants to reduce the beta o[tl1e portfolio to H.6? The formula for the number of contracts that should be shurted gis es 2f1. 1)()(IQ ‘ 1 1080 x "" = Rounding to the nearest whole number, S9 contracts should be shorted. To reduce the beta to 230 88.9 0.6, half of this position, or a short position in 44 contracts, is required.

Problem 3.8. In the ‹’own futures confror‘/, the jollo›ring Heliverv months are a› ailable. .larch, iUav, Jul y, September, anJ December. State the contract that .should lie u.red for he‹lging alien the expiration of the hedge is in ay June, h) July, and c j .January A good rule of thumb is to choose a futures contract that has u delivery month us close as possible to, but later than, the month containing the expiratitin of the hedge. The contracts that shtiuld be used are tliereftire (a) .tuly (b) September (c) March Problem 3.9. Does a per}e‹’t hedge ulwuvs suc'c'eeJ in locking in the ‹’urren t spot prim e of an asset for a future transaction? Explain voui ans»ei’. No. Consider, for example, the rise of a forward contr‹ict to hedge a known cush inflow in a foreign currency. The lonvard contract locks in the forw urd exchange rat—e which is in general different from the spot exchange rate. Problem 3.10.

The basis is the amount by which the spot price exceeds the futures price. A shurt liedger is 17 20.12 Pearson E‹1u cation

long the asset and short futures contracts. Thc ›'a1ue of his or her position theietorc improves as the basis increases. Similarly, it orsens as the basis decreases. Problem 3.11. Imagine you are thr lrea3’urer of a JaJ›ane.se compan v exporting electt-onic equipment to the UniteJ States. Di 'cues hon you v ould de.sign a foreign exchange heJ„•ing .strateg) and the arguments you wotilJ use to .sell the .strate › to ’our Jello» executive.s. Thc simple answer to this question is that the treasurer should 1. Estimate the company’s fumre cash flows in Japanese yen and U.S. dollars 2. Enter into forw'ard and futures contracts to lock in the exchange rate for the U.S. dollar cash flows. However, this is not the whole story. As the gold jewelry example in Table 3.1 shows, the company should examine whether the magnitudcs of the foreign cash fiows depend on the exchange rate. For example, wi]l the company be able to raise the price of its product in U.S. dollars if the yen appreciates'? If the company can do so, its ftireign exchange exposurc may be quite low. The key estimates required are those showing the overall effect on the company’s profitability of changes in the exchange rate at various times in the future. Once these estimates have been produccd the company can choose between using futures and options to hedge its risk. The results of the analysis should be prcsented carefully to other executives. lt should be explained that a hedge does not ensure that profits will be higher. lt means that profit will be more certain. When futures/forwards are used both the downside and upside are eliminated. With options a premium is paid to climinate only the downside. Problem 3.12. ffuppose lhal iti E.sample 3.2 o[Section .1.3 the company Jecides to u.ye a hedge ratio of 0.8. Hoiv does the ‹Incision affect the ii’ay in which the heJge is implemented and the result? If the hedge ratio is O.S. the company takes a long position in 16 December oi Ifutures contracts on June S when the futures price is Sfi8.00. It closes out its FOSition on Novcmber 10. The spot price and tiitures price at this time are 570.00 and $69.10. The gain on the futures position is (69.1—0 6S.00) x 1 6, 000 = 17. 600 The eflcctive cost of the oil is thereforc 20, 000 x 70 17, 600 = 1, 352, 400 or $fi9.12 per barrel. (This compares with $65.90 pcr barrel when the company is fully hedged.) Problem 3.13. 'Ij'the minimum-»ariance hedgE rtltiG i3' CfllCUfOtCd a.s 1.0, thr heJge must he f›erfect.” Is this siutement true? E.xplain your answ'er. The statement is not true. The minimum variance hedge ratio is

It is 1.0 when p = 0.5 and cr, = 2w, . Since p < 1.0 the hedge is clearly not perfect. Problem 3.14. If there is no ba.sis risk, the minimum variant e hedge ratio is alwav.s 1.0.” Is this statement 1S 2012 Pearson Education

Thc statement is true. Using the notation in the text, if the hedge ratio is 1.0, thc hedger locks in a price off, b, . Since both , and b, are known this has a variance of zero and must be the best hedge. Problem 3.15 “For an asset where fixtures prices are u.suallv /e.s.s than .sf›ot J›rice.s, long hedge.s are likely to be particularlv attractive.“ Explain thi.s .statement. A company that knows it will purchase a commodity in the future is able to lock in a price close tti the futures price. This is likely to be particularly attractiv e when the futures price is less than the spot price. Problem 3.16. The stanJard Je v'iation o] rnonthly ‹’hanger in the spot pi’i‹ e of live ‹’attle is (in cents per pounJ) 1.2. The 5’landur‹1 Jeviation o[monthly chanz«es in the futures price of live cattle %r //ie c/oses/ contru‹:t is 1.4. The correlation hem een he /ii/iire.y price change.s and the .spot price ‹’han,qes is 0.7. It is now Octol›er 15. A hee[j›rodiicer i.s committed to f›urchasing 200,000 pounJs of live caltle on November 15. The J›roducer ant.s to u.se the December /n'ecattle futures contracts fo hedge it.s ri.wk. Each contract i.s for the delivery o/#9, 000 yooañf o/ cattle. Mat .ytrotegy .should the l›eef producer follow.'! The optimal hedge ratio is

The becl produccr requircs a long position in 200000 x 0.6 = 120, 000 lbs of cattle. The beef producer should therefore take a long position in 3 December contracts closing out the position on November 15. Problem 3.17. A corn farmer argue.s ”I do not u.se [utw es contract.s for hedging. My real risk is not the price o[corn. It is that my »liole crop gets v’ipeJ out lay the weather.”Discus's this i ieivpoint. Should the farmer estimate his or her- expected production o] corn anJ hedge to ti3 In lock in a price for expected proJurtion.’* lf weather creates a significant uncertainty about the volume of coin that ill be harvested, the farmer should not enter into short forward contracts to hedgc the pricc risk on his or her cxpccted production. The reason is as follou=s. Suppose that the weather is bad and the larmcr’s production is lower than expected. Other farmers are likely to have been affected similarly. Com production overall will be ltiw and as a consequence the price of ctirn will be relatively high. The farmer’s problems arising from the bad harvest will be made worse by losses on the short futures position. This problem emphasizes the importance of looking at the big picture when hedging. The farmer is correct to question whether hedging price risk while ignoring other risks is a good strategy. Problem 3.18. On July 1, an inve•stor holJs 50,000 shares oJ a certain stock. The market price is $30 per share. The. investor i› interesteJ in heJging against movements in the market over the next 2012 Pearson Education

month anal HeciJes to it.se the .$’eptenil›e•r Mini S&P 500 futures contract. The index is currently 1,500 and owe contract is [or deli ver oJ 350 time.s the index. The Ucta of the stock i.s /. 3. IJhat .strate .should the in vestor]olloii ? 2’nder what circumstances w'ill it be prohtal›le? A shtirt posititin in

5

0 30 = 26

’ 50 x 1, 500 contracts is required. It will bc profitable if the stock outperforms the markct in the sense that its return is greater than that predicted by the capital asset pricing model. Problem 3.19. SuJ›po.se that in Tahle 3.5 the compan v Jecides In ii›’e a hertz e ratio of 1.5. How do‹•s the dcci.tion affect the 'ay the hc•‹lge is imple i‹•nted anJ the result? If the company uses a hedge ratio ot 1.5 in Table 3.5 it would at each stage short 1.50 ctintracts. The gain from the futures contracts would be I .50 x 1. 70 — $2.55 pei barrel and the company would be $0.85 per barrel better off.

Siipposc that you enter into a short futures contract to hedge thc SñlC Of tin asset in six months. II the price of the asset rises sharply during the six months, the futures price will also rise and you may get margin calls. The margin calls will lead to cash outflows. Eventually the cash oiilflows u ill be tiffset by the extra amount you gct when you sell the asset, but there is a mismatch in the tinting of the cash outflow s and inflow s. Your cash outflow s occur earlier than j‹nn cash inflows. A similar situation could arise if ytiti used a long position in a titures contract to hedge the purchase of an assct at a lu ture time und the asset's price fell sharply. An extreme example ot what we arc talking about here is provided by Metallgcscllschaft (see Business finagshct ñ.2J.

Proiilem 3.21. An airline e. e‹:utive has argued.’ “ There is no yoint in our- us rug oil future.s. There is Just as that it will be greater than thi.s J›rice.” Dist•iiss thr e.vec'iltive ’s viewpoint. It may well be true that there is just as much chancc that the price of oil in the future will be above the failures price as that it ill be below the futures price. This means that the use o1 a futures ctintract ftir speculation would bc like bctting on whether a coin comes up heads or tails. But it might make sense for thc airline to use futures for hedging rather than speculation. The futures contract then has the effect tif reducing risks. It can bc argued that an airs ine should not exposc its shareholders to risks associated with the future price of oil when there are contracts availablc to hedge the risks.

20.2 Pearson Education

Goldman Sachs can borrow 1 ounce of gold and sell it for $1200. It in› ests the $1,200 at 5% so that it becomes S1,260 at the end of the year. lt must pay the lease rate of 1 .5% trim S1,200. This is $18 and leaves it with $1,242. It follows that if it agrees to buy the gold for less than $1,242 in one year it will make a profit. Problem 3.23. The expee'teH return on the S&P 500 is 12%v anJ the risk-free rate is 5%‹. N hat is he expel feel return on the investment with a beta of éa) 0.2, (bj 0.5, anJ (‹’) 1.4? u) 0.05 * 0.2 (0.12 — 0.05) = 0.064 or 6.4% b) 0.05 * 0.5 x (0.1—2 0.05) = 0.0b5 or b.5"% c) 0.05 * 1.4 x (0.12 — 0.05) = 0.148 or 14.8%

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CHAPTER 4 Interest Rates Problem 4.1. A bank quotes y’ou an interest rate of 14°X per annum wiih quarterly compounding. What is the equivalent rate wilh (a) continuous compounding and tb) annual compounding? (a) The rate with continuous compounding is + ' 14 4ln 1 = 0.1376 4 or 13.76% per annum. (b) The rate w ith annual compounding is 1+

0.14

— 1= 0.1475

4

or I 4.75* o per annum.

Problem 4.2. What is meant bv LIBOR anJ LIBID. Which is higher'? LIBOR is thc London InterBank Offered Rate. It is calculated daily by thc British Bankers Association and is the rate a AA-rated bank requires on deposits it placcs with other banks. LIBID is the London IntcrBank Bid rate. It is thc ratc a bank is prepared to pay on deposits from other AA-ratcd banks. LIBOR is grcatcr than LIBID. Problem 4.3. The s L-month and one-year zero rate.s are hoth 10% per annum. For a hond that hue a life o[ /8 month,s and pan'.f a coitfioH Hf R‘No fi€t‘ ctHtl tlitt ( with semiannual payments and one having just heert tradej, the j’iOld i.S I F.4‘No fCt- annum. What i.s the l›ond’.s price." What i,s the 18month zero rate.* All rates are quote•d i‹'ith Se•miannUal comFoiinding. Suppose the bond has a face value of $100. I ts price is obtained by discounting the cash flows at 10.4%. The price is 1

1.052 +

4

+

4 4 1.052’

104

= 9f›.74

104

— 1.052 1.05 1.05' If the 18-month zero rate is , we must hay e Problem 4.4. An iin ester rec'eives $1,100 in one year in return %r an investment o(S 1,000 now. Calculate ltte percentage return per annum Frith a) annual comyotinding, h) .seniiannual compounding, c) monthly’ comj›ounding and d1 continues.s compounding. (1 + A / 96.74 2)’

(a) With annual compounding the return is 22 ‹ ?2O1 2 Pearson Education

110()

or 10% per annum.

— 1= 0.1 1000

(b) With semi-annual compounding the return is fi * here 2 i.e.,

= 1100

R 1 1 = 1.0488 so that fi = 0.0976 . The percentage— return is therefore 9.760/» pcr annum. (c) With monthly compounding the return is k whcrc 1000 1

R

12

i.e.

2

= 1100

I+ 1 00797

= I.

.

12

so that fi = 0.0957compounding . The percentage returnisisr therefore (d) \Vith continuous the return where: 9.57% per annum. 1000c’ = 1100 i.e., so that fi = In 1.1 = 0.0953 . The percentage return is therefore 9.53% per annum. Problem 4.5. Suppose that zero interest rates w'ith continues.s compounding are as follows. Maturit y (monk/i.sJ 3 6

Rate (%‹› per annum) 8.0 b.2

9 12

8.5

5.4

15 8.6 18 8.7 Call ulate jorw’arJ intere•st rules/or the .seemed, third, [ourth, fifth, anJ sixth quarters. The forward rates with continuous compounding are as follows to Qtr 2 Qtr 3 Qtr 4 Qtr 6

9.2% 23 2012 Pearson Education

Problem 4.6. Assuming that zero rates are as in Problem 4.5, what i.s the value o[an FRA that enahle.s the holder to earn 9.5%/or- a three-month period.starting in one year on a principal of $1,ONO, ti00'? The intere.st rate is expres.sed worth quartcrly comfaounding. The form ard rate is 9.0*» z ith continuous compounding or 9.102*o Wlth quarterly compounding. From equation (4.9), the value of the FRA is therefore [1, 000, 000 x 0.25 x (0.095 — 0.09102)]e*’""' " = 893.56 or $593.56. Problem 4.7. The term structure oJ interest rates is upward sloping. Put ltte Jollov ing in orJer o/ magnitude.(a) The five-year zero rate (b) The yield on a five-year coupon-bearing bonJ (c) The %rward rate corresponding to the perioJ between 4.75 anJ 5 years in the future What is the answer to this question w'hen the term structure o/ interest rarer is dow'n w'ard .sloj›ing? When the term structure is upward sloping, c > u > ñ . When it is downward sloping, Problem 4.8. What does duration tell you ahout the sensitivity ma bond portfolio to interest i-ate.s? What are the limitations of the duration measure'? Duration provides information about the effect of a small paral let shift in the yield curve on the value of a bond ponYol io. The percentage decrease in the value of the portfolio equals the duration of the portfolio multiplied by the amount by which interest rates are increased in the small parallel shift. The duration measure has the following limitation. It applies only to parallel shifts in the yield curve that are small. Problem 4.9. What rate oj interest with continuous compounding is equivalent to 15%› per annum with monthly compounding? The rate of interest is R where:

i.e., fi = 121n

0.15

1 • 12

= 0.1491

The rate of interest is therefore 14.91% per annum. Problem 4.10. A deposit account pays 12’Xo per annum with c.ontinuous compounding, but interest is acltially 24 2012 Pearson Education

paid quarterly. How much intere.st will he paid each quarter on a $10,000 deJ›osit? The equivalent rate of interest z'ith quarterly compounding is s where 4 R

4(e'" —1) = 0.1218

The amount of interest paid each quarter is therefore: 0.1218 10.000 • = 304.55 4 o $304,55. Problem 4.11. SuJ1po.se that 6-month, 12-month, 18-month, !4month, and 30-month z•cro rn/e.y are 4%, 4.2oXu, 4.4%, 4.6%, and 4.8% yer annum v'ith continues.s compounding re.spectively. E.climate the ca.sh price ofa hond with a [ace value o[ 100 that iv ill mature in 3t) more/i,i anal;›ay.s a EN ttfloH O f 4“Xo yer an niiin .semianiiiiu// '.

The bond pays $2 in é*. 1.2, 1.8. and 24 months, and $102 in 3tl months. The cash price is 2e ' "" + 2e “ ' 2e " "‘ ' ’ 2e “" 102c " '““ ' = 98.04 Problem 4.12. A three-year l›ond j›rovide.s a corepo H i9 i¥ o .$€irtitlllHttolly and ha. O Ea.sh j1t‘ice o[ 104. What

is the hond’s yield) The bond pays $4 in 6. 1.2, 1.8, 24. and .30 months, and $104 in 36 months. The bond yield is the value of i that solvcs 4e ' ' + 4e ’ ' * 4e ’ * 4e " * 4e “* 104e "‘ = 104 Using the Solver or Coal Seek tool in Excel = 0.06407 or 6.407'?o. Problem 4.13. Suppose that the 6-month, 12-month, 18-month, anJ 24-month zero rates are SP, 6%, 6.5°X, and 7°X‹› respectively. What is the twin-year par »ield? x 2 . Also Using the notation in the text, m(l0(-i = 2 ,10c I = e0 'sc94) ” ’ = 0.8694 7.072 A = e "’”“ + e ' ”' ' e ' ””' ’3.6935 e " '° " = 3.6935 To The verify that this is correct we calculate the value formula in the text gives the par yield as ot a bond that pays a coupon of 7.fJ72% per year (that is 3.53ti5 every six months). The value is 3.53Ge ' "“' ' 3.5365c ’ " ' ’ + 3.536e°' '"’ ' ’ * 103.536c ' '°” ' = 1.00 verifying that 7.072o/ » is the par yield. Problem 4.14. Suppose thut zero interest rate ' ritli continuotis compoiin‹ling are as follow›.25 C 2012 Pearson Education

MattiriR ( year’s)

Rate t°X per annym)

5

4.2 4.5

The forward rates with continuous compounding are as follows: Year 2. 4.00 s Year 3: 5.1@ « Year 4: 5.7% Year 5: 5.7% Problem 4.15. Use the rate in Pt oblem 4. 14 to value an FRA wtiere yoH u’ill pOy 5’Xo jot‘ the third year on $1 million. The forw'ard rate is 5.1*Zo with continuous compounding or e’ — '“ I — 5.232 % with annual compounding. The 3-year interest rate is 3.7% with continuous compounding. From equation (4.1 0), the value of the FRA is therefore [1, 000, 000 (0.05232 — 0.05) x l]e*' "”' = 2, t)78.85 or S2,07S.85. Problem 4.16. A I 0-V€tlF, N’Xo Coupon bond currently sells Jar $90. A 10-year, 4% couyon bonJ currently se•lls Jor $d0. Whal is the 10-year zero rate.! (Hint. Consider taking a long po›ilion in twin of lhe 4% coupon bonJs and a .Thor-I position in one of the 8% common bonJs.) Taking a long position in two ot the 4% coupon bonds and a short position in one of the 8%

coupon bonds leads tu the following Ash

.1 0.100 w0 —20.0357 x 50 = —70 I ñln Year 10 : 20—070 100 100 or 3.57%» per annum. because the coupons cancel out. $100 in 10 years time is equivalent to $70 today. The 10year rate, 4.17. p , (continuously compounded) is therefore giv en by Problem 100 =is70c''" Explain carefully’ why liqiiidit y preference them3 consistent 'ilh the observation that the The ratc is term structure of intere.st rate.s tend,s to he upward sloping more often than it i.s downward sloping.

If long-tern rates were simply a reflection of expected future short-term rates, we would expect the tcrm structure tti be dow nward sloping as often as it is upward sloping. (This is 26 2012 Pearson Education

based on thc assumption that half of the time invcstors expect rates to increase and half o1 the time investors expect rates to decrease). Liquidity preference theory argues that long term rates are high relative to expected future short-term rates. This means that the term structure should be upward sloping more often than it is dow’nward slopins

The par yield is thc yield on a coupon-bearing bond. The zero rate is the yield on a zcrocoupon bond. When the yield curv e is up ard sloping, the yield on an fi' -year couponbearing bond is less than the yield on an i\' -year zcro-coupon btnad. This is because the coupons are discounted at a lower rate than the .\' -year rate and drag the yield down below' this rate. Similarly, z'hen the yield curve is downward sloping, the yield on an -ycar coupon bearing bond is higher than the yield on an :\' -year zero-coupon bond. Problem 4.19. Why are L.S. Trea.erm j.' rates sign fiicantl; lower than other rates that are close to ri.sk free? There are three reasons (see Business Snapshtit 4.1). 1. Treasury bil Is and Treasury bonds must be purchased by financial institutions to fulfi11 a variety of regulatory requirements. This increases demand for these Treasury instruments driving the price up and the yield down. 2. The amount o1 capital a bank is required to hold to support an investment in Treasury bills and bonds is substantially smaller than the capital rerJiiired to support a similar investment in other very-low-risk instruments. 3. In the United States, Treasury insti uinents are given a far tirable tax treatment compared with most other fixed-income investments because they are not taxed at thc state level. Problem 4.20. A repo is a contract where an investment dealer who owns securities agrees to sell them t‹i another company not and buy them back later at a slightly higher price. The other company is providing a loan to the investment desler. This loan involves very little credit risk. lf the borrower does not honor the agreement. the lending company simply keeps the securities. If the lending company does not keep to its side of the agreement, the original owner of the securities kceps the cash. Problem 4.21. Explain ›vhy an h’RA i.‹ equi valrnt to the exchange ma floating rate of interest for a fixed rate of intere.st." A FRA is an agreemcnt that a certain speci fled interest rate, ñ; , will apply to a certain principal, /. , for a certain specified future time period. Suppose that the rate obscrved in the market for the future time period at the beginning ot the timc period proves to be fi„ . If the FRA is an agreement that fi, will apply when the principal is invested, the holder of the FRA can borrow the principal ut fi„ and then invcst it at R, . The net cash flow at the end o1 27 . 2012 Pearson Education

the period is then an inflo ' of R L and an outflow of R, L . If the FRA is an agreemcnt that fi wi |1 apply when the principal is borrowed, the holder of the FRA can invest the borrowed principal at fi„ . The net cash flow‘ at the end of the period is then an inflow of fi„L and an outflow ot fi,£ . In either case we see that the FRA involves the exchange of a fixcd rate ot interest on the principal of £ for a floating rate of interest on the principal. Problem 4.22. A five-year honJ the end of each year.

ills a 5 ield of 11 X‹› (continuous/ compounded) pays an 8% couJ›on at

a) What i.s the bond’s yrice'? b) but is the hond '.f function'' r) 2’se lhe duration to calculate the effect on the bond's price ofa L).2‘No de‹’rea› e in its yield. d) Re‹-alculate the hond 's price on the basi.s ma 10.8% per annum yield anJ verijy lhat the result is in agreement with soul- an.swer to fc). c) The bond’s price is d) The boncl’s duration is ' 86 o

8

i'

2 S ' '" + 3 x 8e*' ’ '

4 8r• ’ '

5 x 108e i’ '

c) Since, z'ith the notation in the chapter AF = —BD the effect on the bond’s price of a 0.2% dccrease in its yield is 86.80 x 4.256 x 0.(i02 0.74 The bond’s price should increase from 86.50 to 87.54. d) With a 10.8% yicld the bond’s price is Se'' '“ + 8e ' “ ’ ’ ”" ' + 8e '’“’ + I I)Ye '°’ 87.54 This is consistent with the ansz'er in (c).

8e =

Problem 1.23. The cas hV›rire.s ofsix-month anal one-5'ear Treasu r; bills ure 94.11 and 89.0. A 1.5-year hond that mill pay cou)›ons of S4 every six rnon/Jzs currently sells for 94.84. A two-yeur bond that will puv coupon.s of $5 every six months currently sells Jâr $97.12. Calculate the six-riionth, one-›’enr, 1.5-ye•ar, ancl h 'o-year -em rates. The I›-month Treasury bill provides a rcturn of 6 94 = 6.383% in six months. This is 2 x 6.353 = 12. 766 % per annum with senaiannual compounding or 21n(1.06383) = 1 2.38* per annum w ith continuous ctiinpounding. The 1 2-month rate is 11 89 = 1 2.3 60* with annual compounding or In(1 .1236) = I 1.6ñ*‹ with continuous compounding. For the 1 - year bond we must have

4e ' '”“’’° * 4e ' ' " " + I 04e ‘ " = 94.84 where fi is the 1 - year zero rate. It follows that 3.76 + 3.56 * 104e ' ‘“ = 94.84

e°' ’* = 0.8415 A — 0.115 or 11.5%. For the 2year bond we must have 5e*’ "”" ‘ + 5e ' ' '"°" * 5e ' "”' ' + l 05e '* = 97.12 where s is the 2-year zero rate. It follows that e “2

— 0.7977

R— — 0.113

or I 1.3% . Problem 4.24. “An interest rate swap u’here six-month LIBOR i.s exchanged for a)ixed rate 5°Xu on a principal of $100 million for five year.y i.s a portfolio of nine FRAS.” Explain thi.s .statement. Each exchange of payments is an FRA where interest at 5% is exchanged for interest at LIBOR on a principal of $100 million. Interest rate swaps are discussed further in Chapter 7.

29 ' 2012 Pearson Education

CHAPTER 5 Determination of Forward and Futures Prices Problem 5.1. Explain v hat happens when an inve.stor shorts a certain share. The investor’s broker borrows the shares from another client’s account and sells them in the usual way. To close out the position, the investor must purchase the shares. The broker then replaces them in the account of the client from whom they were borrowcd. The party with the short position must remit to the broker dividends and other inctime paid on the shares. The broker transfers these funds to the account of the client irom whom the shares were borrowed. Occasionally the broker runs out of places from which to borrow thc shares. The investor is then short squeezed and has to close out the position immediately. Problem 5.2. What i› the difference between they fomvard prire and the value o,/ a jot-w ard contract The forward price of an asset today is the price at z'hich you w ould agree to buy or scll the asset at a fiiture time. The value ot a forward contract is zero when you first enter into it. As time passes the underlying asset price changes and the value of thc contract may become positive or negative. Problem 5.3. Suppo,se that you enter into a .sin-monlh forn›ard contract on a non-dividenJ-yaying ,stock lien the stock price i $3 II and the risk-/ree• intere.st rate trilli continuoti.s compounding) i.s The forward price is 30c' ' ' = $31.86 Problem 5.4. A stock iijdzy ‹’urrentlv stand.s at 350. Thr risk-free interest rate is S% per annum (with continuou.s compounding) and the dividend yield on the index is 4“Xoyer annum. What should the_fu ture.s price for u fow -month con/i o‹ / be'! Thc fixtures price is 350c

""

’ = $354.7

Problem 5.5. Explain carefiillv whv the/u/iire.s pt-ice of gold can he calc'ulule‹l front it.s .spot price and other obsrrv'able variables ivherea,s the fixtures price of copj›er cannol. Gold is an investment asset. If the futures price is too high, investors z ilf find it profitable to incrcase their holdings o1 gOld and short iñtures contracts. If the futures price is too low, they will find it profitable to decrease their holdings of gold and go long in the futures market, Clipper is a consumption asset. If the futures price is too high, a strategy of buy coppcr and short futures works. How ever, because ink estors do not in general hold the asset, the strategy of sell copper and buy futures is not available to them. There is therefore an upper bound, but ‹â201 2 Pearson Education

no lower bound, to the fixtures price. Problem 5.6. F.xplaii1 carefull the meaning o[the term.s coiwenience yield anJ cost of carry. Wha! is the relation,ship he•nveen future.s Price, spot J›rire, convenience yield, and co.st of carry? €?oni enience yield measures the extent to which there are benefits obtained from owncrship of the physical asset that are not obtained by owners of long futures ctintracts. The cost o/ c u p ' is the intercst cost plcs storagc cost lcss the income earned. The futures price, , and spot price, N, , are i elated by hcrc c is the cost of carry, ,. is the convenience yield, and T is the time to maturity of the futures contract. Problem 5.7. Explain iiliy a foreign currenc›’ can be treateJ as an nsse/ providing a known yield. A ftireign currency provides a known interest rate, but the interest is received in the foreign currency. The value in the dtimestic currency tif the income provided by the foreign currency is therefore knt»vn as a percentage of the value of the foreign currency. This means that the income has the properties of a know'n yield. Problem 5.8. Is the futures prim e oj a sloch inde. gtx•aler llian or less than the exfiectcd future value o[the The futures pricc of a stock index is always less than the expected future value of the index. This follow s from Section 5.14 and the fact that the indcx has positive systematic risk. For an alteniative argument, let /, be the expecie‹1 return required by investors on the index so

Problem S.9. A one-year long [orv ard c.ontrcic.t oti ‹i non-Jim idend-pay ing .stock is entereJ into hen the ›'to‹’l‹ prim e is 540 anJ the ri,sk-free rate of intere.st i.s 10% per annum ii’ith continuous a) Whal arr they fori+'ard price and the initial value of the foin›’ard contract? b) Six months later, lhr pierce of thr stock is $45 and they risk-free intere.st rate is still 10%. What are the foini'arH pri‹’‹ and lh‹• valiir oJ the form ard contract." a) The ftirw‹ird price. F„ , is given by equation (5.1) as: F 44.21

40e’’ =

or- $44.21. The initial value of the forward contract is zero. b) The delivery price v in the contract is $44.21. The value of the contract, f , after six months is given by equation (5.5) as: / — 45 — 44.2 le ' ' = 2.95 20.1.2 Pearson Education

i.e., it is 52.95. The forward price is: 45e’ ' ‘ or $47.31.

47.31

Problem 5.10. The risk-free rate of intere.st i.y 7%‹ jeer annum with continuGH3’ conipoundin g, and the

lividenJ yield on a stocL inde i.s 3.2% per annum. The current value of the inJe. i ' 150. What is the si. -month futures price'? Using equation (5.3) the six month futures price is 1 $ be'" ’° ' "”"’ — 1.5.2. S8

or $152.55 . Problem 5.11. Assume chef the risk-free intere.st rate i.s 9% Jeer annum with continuous compounding and that the JiviJend yie•ld on u stock index varie.s throughout the ) ear. In Februam , stay, August, anJ.tovemb‹r, dividends are paid at a rate of 5% per aniuis. In other months, dividends are puiJ ri/ o roll’ Gf2"o peer annum. Suppose that the value of the index on July 31 i.s 1,300. What is the ji.itures price Jor a contract drliverahle on December 31 of the same year? The futures contract lasts for live months. The dividend yield is 20c for three of the months and 5% for two of the months. Thc average dividend yield is therefore (3 x 2 -F 2 x 5) = 3.2 The futures price is therefore 1300c "’ "'" """ = 1, 33.1, b0 oz $1331.80.

Problem 5.12. Suppose that the risk-f-ee intere.st rate i.s 10% her annum with continuous compounding anJ that the Ji vidend yield on a stock inde.x i.s 4% per annum. The inclex is stanJing at 400, anJ the futures price%r a contract deliverable in four month.s i.s 4ñ5. What arbitrage opportunities Joes Hiis create'! The theoretical fixtures price is 400c " "’ “""‘ '" = 405.OS The actual futures price is only 405. This shows that the index futures price is too low relative to the index. The correct arbitrage strategy is (a) Buy fixtures contracts (b) Short the shares underlying the index. Problem 5.13. Estimate the difference between short-term in Iere5 I rates in Me•.Pico and the L'nited States on Mai' 26, 2010 front the information in Table 5.4. The settlement prices for the futures contracts are to Sept: 0.76375 Dec: 0.75625 32

The Dccember price is about 0.9Sfi» below the September price. This suggests that the shortterm interest rate in the Mexicti exceeded short-term interest rate in the United Statcs by about 0.9b o per three months or about 3.920 d per year. Problem 5.14. The two-month intere.st rutes in Sii'r/zerfond and lhe nited States an ]%i Ol'td 5’No fler anntlnj, respe.‹’lively, B itli continuous cony›aiinJing. The .yyn/yiric.e o] the Swi.s.s/ranc is $0.Rñññ. The future.s jet-ice for a ‹ ontract d‹•li v'eruble in Pro month› is $0.R I 0t1. What arbitrage opportunitie.s don this create'? The theoretical futures price is 0.s 000e '

' '“ =

0.s040 The actual tutures price is too high. This suggests that an arbitrageur should buy Swlss francs and shtirt Swiss francs futures.

The present value of the storage costs for nine nitinths are 0.06 * 0.06e ' '"” ' "“' = 0.1.76 or $0.176. The futures price is from equation (5.11) given by I, where F, — — (15.000 * 0. I 75)e" '" "’ = 1.6.36 i.e., it is S16.3b per ounce.

0.06e '

Problem 5.16. Supf›ose thal F cinJ F, are tw'o]titure.s contracts on the .some commoJiiy » ith time.s to

If an investor could makc a riskless prollt by’ (a) Taking a long position in a fiitures contract u hich manures at time /, (b) Taking a short position in a futurcs contract which matures at time i, When the first futures contract matures, the asset is purchased for using funds borrowed at rate r. It is thcn held rinti1 time at which point it is exchanged for f 2 under the second contract. The costs o1 the funds borrowed and accumulated interest at time /; is f,e”' ' ' . A positive profit ot' _ g, i , is then realized at time /_ . This type of arbitrage opportunity cannot exist for long. Hence: I, e”-“'

fi'20l 2 Pearson Educe tion

Problem 5.17. When a know it Jiilure ‹ ash outflow in a foreign currenc is heHgeJ bv u company ti 'ing u forward contract, there is no foreign exchange risk. When it is heJged using fixtures contract.s, the daily settlement pro‹’ess does leave the company e.xposeJ to some risk. Explain the tiature of thi.s ri.wk. In particular, com.wider w hether the com;›an is better off using a futiire,s contract or a forward contract i hcn a) The »alue o[the foreign curi-etic5'/alls raJ›idly during the life of the conn-act b) The value o] thr foreign currency rises rapidly during the life of the contract c) The volue of the fore•ign cut-rency ff-st ri.Yes cmd them [alls hack to its initial value d)The »alue oj the foreign currency first falls and then rises back to its initial i alue Assuine thot the jorwurJ price rquuls Ihr futures price. In ttital the gain or loss under a futures contract is equal to the gain or loss under the corresponding form ard contract. However the timing of‘ the cash tions is different. When the time value of money is taken into account a futurcs contract may prove to bc more valuable or less valuable than a forward contract. Of course the company docs not know in advance which wit 1 wtirk out better. The long forward contract pro›'ides a per1’ect hedge. The long futures contract provides a slightly imperfect hedge. e) In this case the forward contract would lead to a slightly better tiutcoine. The company will make a loss on its hedge. If the hedge is with a forw'ard contract the whtile of the loss will be realized at the end. lt it is v•ith a futures contract the loss will be realized day by day throughout the contract. On a present value basis the fomier is preferable. b) In this case the futures contract would lead to a slightly better outcome. The company »'ill make a gain on thc hcdgc. It thc hedge is with a forward contract the gain will be realized at the end. If it is ith a futures contract the yain will be realized day by day throughout the life of the contract. On a present value basis thc latter is pretérablc. c) In this case the fixtures contract would lcad to a slightly better outcome. This is because it w’ould involve positive cash flow› s early and ncgative cash flows later. d) In this case the form art contract u ould lead to a slightly better outcome. This is because. in the case tif the futures contract, the early cash flow s w ould be negative and the later cash flow would be positive. Problem 5.18. It is sometimes argued th6tt a forw'cfKd e.xchange rate is an unbiased predictor of/uture eychunge mtce. 7’nJer ii’hut circumstances is this so.'" From the discussion in Section 5.14 of the text, the for ard exchange rate is an unbiased predictor of thc tuturc exchange rate 1'hcn the exchange rate has no systcmatic risk. To have no systematic risk thc cxchangc iatc must bc uncorrclatcd ith thc rcturn on the market. Problem 5.19. Short that the growth rate in an index [iinire.s J›rice equals the exce.s.f rettirn o/ the portfolio underlying the index o»er the i-i.sk-[ree rate. A.s.sume that the ri.sk-See inter-est rate and the Jivid‹•nJ yir.IJ are c.on,stant. Suppose that F is thc futurcs price at time zero for a contract maturing at time r and , is thc liitures price for the same contract at time . lt follo» s that F„ = S „e” ”’

F, — S,e "" "

whcrc .S, and S, arc the spot price at times zero and /, , r is the risk-free rate, and y is the dividend yield. These equations imply that F, S, _,,. , Define the excess return of the portfolio undcrlying the indcx over the risk-free rate as x , The total return is r + .r and the return realized in the form of capital s ains is r + — y . It follows which is the required result. Problem 5.20. Show that equation (5.3) is truer b)’ conside•ring an inv'esImenl in ltte asset ‹:ombineJ›i'ilh a short position in a Ltures contract. Assume lhal all income}t-oin fhe ass ct is rein vesled in //ie asset. Use an argument similar to flu/ in footnotes 2 anJ 4 anal explain in detail what an arbitrageur w'ould do iJ equation (5.3) JiJ not hold. Supposc we buy :Y units of the asset and invest the income from the asset in the asset. The income from the asset causes our holding in the asset to grow' at a continuously compounded rate y . By time T our holding has grown to .\?" units of the asset. Analogoiisly to footnotes 2 and 4 ot Chapter 5, we therefore buy \ units of the asset at time zero at a cost of S, per unit and enter into a for 'ard contract to sell .\'e unit for I, per unit at time T . This generates the following cash flow s: Time 0: — Y.S, Because thcre is no uncertainty about these cash fltiws, the present value of the time T inflow must equal the time zero outflow when we discount at the risk-free rate. This means that Y.Y„ — ( .Y£ e" )e ”

F, = S,e" '""

This is equation (5.3). IfF„ > s„ ' "' , an arbitrageur should borrow money at rate r and buy .\' units of the asset. At the same time the arbitrageur should enter into a fonvard contract to sell L’e'" units of the assct at time T . As income is received, it is reins ested in the asset. At time T the loan is repaid and the arbitrageur wakes a profit of .v(F r” S,e ) at time T II F„ < s, e” " , an arbitrageur should short .\' units of the asset investing the proceeds at rate r . At the same time the arbitrageur should enter into a forward contract to buy .\e•’ units of the asset at time T . When income is paid on thc asset, the arbitrageur owes moncy on thc short position. The investor meets this obligation from the cash proceeds tif shorting liirther units. The result is that the number of units shorted grows at rate ‹ to Ye•’ . The cumulative short position is closed out at time T and the arbitrageur makes a profit tif .U(S e"’ Fze“’ ) .

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Problem 5.21. Ex¡›lain care[ull) what is meant h;’ the exJ›ected price ma cmmmcdi on a particulor/uture date. Suppose that the fftHT€.s ¡›t iCt: o Crude oil decline.s ›i ith the maturity of the contract at the rates of 2“Xoyer year-. Assume that .speculator.s tend to he short crude oil futures anJ hedgei-s lenJed to b‹• long cruder oilJiitw-es. N hat does the Ke ries and His argt1itt‹•l1t fii7yJv about the e. pcrted future pricr. of oil'? To understand the meaning of‘thc cxpcctcd future pricc ot a commodity. supposc that thcrc are \ different possible prices at a particular future time: P. Pp,, . P„. Define q, as the (subjectiv e) probability thc price being P («ith q + q ... + q, = 1 ). The cxpected fixture

Different people may have different expected future prices for the c‹Hn1rodity. The expected future price in the market can be thought of as an average of the opinions of different market participants. Ot course, in practice the acmal price of the commodity at the future time may provc to be higher or lower than the expected price. Keynes and Hicks argue that speculators on ay erage make money from commodity futures trading and hedgers on average lose money from commodity tutures trading. II speculators tend to have shon positions in crude oil futures, the Keynes and Hicks argument implies that futures prices overstate cxpected future spot prices. If crude oil futures priccs decline at 20o per year the Keynes and Hicks argument therefore implies an even faster decline for the expected price of crude oil in this case.

Whcn thc geometric average of the price relati› es is used, the changes in thc value of the index do not correspond to changes in the value ot a portfolio that is tradcd. Equation (5.8) is therefore no longer corrcct. Thc changcs in the valuc ol the portfolio are monitorcd by an index calculatcd from thc arithmctic averagc o1’ thc prices of the stocks in the portfolio. Since the geometric av'cragc of‘a set o1‘numbers is always less than the arithmetic average, equation (5.8) overstates the futures price. It is rumored that at one time (prior tti 1958), equatitin (5.5) did hold for the Value Line Index. A major Wall Street finn was the first tti recognize that this represcnted a trading opportunity . It made a financial killing by buying the stocks underlying the index and shooting the futures. Problem 5.23. A L! S. company is interested in using lhe jiiiures ronlracl› lr‹iJeJ by flue C HE Group to he‹lge i/6 Nffsff«/ion dollui c. posui-e. De]inc r as tlse interest male (all maturities) on lhr U.S. dollar anJ r,. as the interest rate (all ni‹ituritie›) on flue Australian dollcii. Assume that r unJ

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exposure at time i {T > t). La) Show that the optimal hedge ratio is

(b) Show that, when t i.s one doz y, the up/inns/ /teéfge ruf/c' /s tlltttost exa‹’tly Sz I F where S, is the current spot price of the currenc5 nnd F i.y the current futui es price oj the currency for the contract maturing at lime T (c) Show that the company can take account of the Jail settlement o[future.f contract.s for a hedge //ia/ la.ate longer than one kay by adjusting the hedge ratio .so that it always equals the spot price of I/ie currenr v divided by the fixtures price o/ the currency. (a) The relationship bet een the futures price I, and the spot price I, at timc / is Suppose that the hcdgc ratio is ñ . The price obtained with hedging is li(F — F,) S where Ut is the initial futures price. This is hFz

S, — hS,e'"!'’ "’!''

If h — — e’*' ""' ", this reduces to hF and a zero variance hedge is obtained. (b) When i is onc day, ii is approximately e'” "” = So /F o . The appropriate hedge ratio is therefore I, / / \ (c) When a fumres contract is used for hedging, the price movements in each day should in thcory be hedged separately. This is because the daily settlement means that a matures contract is closed out and rev ritten at the end or‘each day. From (b) the correct hedge ratio at any given time is, therefore, S/F \ here S is the spot price and F is the iiiturcs price. Suppose there is an exposure to .V units o1 the foreign currency and .u units of the foreign currency underlic onc matures contract. With a hedge ratio of 1 we should trade .S'f M contracts. With a hedge ratio of SQF wc should trade

contracts. In other words z e should calculatc the number of contracts that should be truded as the dollar value or our cxposure divided by the dollar value of one futures contract. (This is not the samc as the dollar value of our exposure divided by the dollar value of the assets underlying one futures ctintract.) Since a futures contract is settled daily, e should in theory rebaluncc our hedge daily so that the outstanding number of fiiturcs contracts is always (S \' )/(¿ i/ ) . This is known as tailing thc hcdgc. (Sec Chapter 3.)

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CHAPTER 6 Interest Rate Futures Problem 6.1. A ChS. Treastii5 bonJ pays H 7’Xo Coupon on January 7 anJ Jul 7. How much interest accrued y›er $100 ofprincipal to the bonJ holJer betw ecu July 7, 2ñ11 and Augirst 9, POH ? Ho woulJ your answer be Ji erent if it were a corporate bond? There are 33 calcndar days between July 7, 2011 and August 9, 2011. There are 184 calendar days between July 7, 2011 and January 7, 2011. The interest earned per $100 of principal is therefore 3.5 x 33, 1 b4 = 30.6277 . For a corporate bond u e assume 32 days beñveen July 7 and August 9, 2011 and 180 days bctu'een July 7, 2011 and January 7, 2011. The interest earned is 3.5 x 32 , 1 80 = 30.6222 . Problem 6.2. It is Jantiary 9, 2013. The price of a Treasury hond willi a I2’Xu coupon that matures on October 12, 2020, is quoteJ as 102-07. What is the c'ash pt ic'e? There arc 89 days bed ecu October 12, 2012, and January 9, 2013. There are 182 days betwccn October 12, 2012, and April 12, 2013. Thc cash price of the bond is obtained by adding the accrued interest to the quuted price. The quoted price is 102or 1 02.21 575. The cash price is therefore 152 Problem 6.3. flow i.s the conversion factor o[a hond calculated h the C..WF. Group'^ //o

is it a,sed.'

The conversion factor for a bond is equal to the quoted price the bond 1'ould have per dollar of principal on the first day of the delivery month on the assumption that the interest rate for all maturities equals fi% per annum (z'ith semiannual compounding). The bond maturity and the times to the coupon payment dates are rounded down to the nearest thrcc months for thc purposes of the calculation. The conversion factor dellnes ho much an in›'cstor 'ith a short bond matures contract receii es when bonds are deli›'ered. It thc conversion factor is 1.2345 the amount investor receives is calculatcd by multiplying 1.2345 by the most recent futures pricc and adding accrued intcrcst. Problem 6.4. A 6urodo/for Ju/ures yi-ice cñorige.y fi-om 96.76 to 96.R2. What i.s the gain or lo.s.s to an inve.stor who is long two contract.s? The Eurodollar futures price has increased by Ii basis points. The ini'estor makes a gain per contract of 2i x 6 = $1i0 or $300 in total. Pzoblem6.5. What i.s the pur ›o.se o/ the convexitv adJ'ustment made to F.w-odollar f/tures rat •s!• Why is the con vexity am.stment neces,sai y.* .'201 2 Pearson Education

Suppose that a Eurodollar futures quote is 95.00. This gives a fixtures rale of 5% for the threemonth period covered by the contract. Thc convexity adjustment is the amount by which futures rate has to be reduced to give an cstimate of the form ard rate for the period. The convexity adjustment is necessary bccause a) the futures contriict is settled daily and b) the futures contract expires at the beginning of the three months. Both of these lead to the futures rate being greater than the forward rate. Problem 6.6. The 350-Jay LIBOR rate /.s 3° with continuous ‹ ompounding and the lorv'ard rate calculated from a Eurodollar fixtures c.onlract that matures in 350 days i.s 3.2% with ‹ ontinuous compounding. E.stimate the 440-Juv zero rate. From equation (6.4) the rate is 3.2 x 90 + 3 x 350 440

or 3.0409%.

3.0409

Problem 6.7. It is January 30. You are managing a bond porfolio worth $6 million. The Juration of the portfolio in six months will he ‹5.2 year.s. The September Treasury bonJ futures pri‹’e is currently 108-15, and fhe cheapest-to-deliver bond whiff h‹ive a Juration of 7.6 years in September. How .should you h••dge against changes in interest rates over the next six month.s? The value of a contract is 108 }2 x 1, 0(i0 — S108, 468.75. The number of contracts that should be shorted is 6 000 000 x 8.2 = 59.7 108, 468.75 7.6 Rounding to the nearest whole number, 6tl contracts should bc shorted. The position should be closed out at the end of.luly. Problem 6.8. The yrice of a 90-day Treasury bill i› fueled os /6.00. What continuou.sly compounded return (on an uclual/365 basis) Joes an investor earn on the Treasury bill for the 90-day perioJ? The cash pricc o1 thc Treasury bill is

90 100 —360 x 10 = 597.50 The annualized continuously compoundcd rcturn is

365 ln I+ 2.5 90 97.5

= I 0.27 %

Problem 6.9. It is Muy 5, 2011. The quoted yrice o/a govei-nnic•nt honJ w’ilh ct 1 E’NoCoupon that matures on Jtilv 27, 2014, is 110-17. What i.s the ca.sh price'" The number of days between January 27, 2011 and May 5, 2011 is 98. The number of days betwccn January 27, 2011 and July 27, 2011 is 181. The accrued interest is therefore ?^.2012 Pearson Education

6 • 98 - 3.2486

The quoted price is I 10.5312. The cash pricc is therefore 110.53 12 * 3.2486 = 113.7798 or $113.78 . Problem B‹ lid Conversion Factor 6.10. 1.2131 Suppose that the Treasury hond fixtures Jirice i.s 01- 12. Which o/ the/olloi+ ing 1. four 5 792honds i.s cheapest to deliver. I. 1149 JS-?) 4 J44-02 1. 4026 The cheapest-to-deliver bond is the one for which Quoted Price — futures Price Conversion Factor is least. Calculating this factor for each of the 4 bonds we get Bond 1 : 125.15625 —101.375 x 1.2131 = 2.178 Bond 2 : 142.46875 —101.375 x 1.3792 = 2.652 Bond 3 : 115.96575 — 101.375 x 1.1149 = 2.946 Bond 4 : 1 44.06250 — 101.375 x 1.4026 = I .574 Bond 4 is therefore the cheapest to deliver. Problem 6.11. It is July 30, 2013. The cheapest-to-Jeliver bonJ in u September 2013 Treasury bond futures contract is a 13°X coupon bonJ, anJ cIe/ivery is expecteJ to be maJe on September 30, 2013. Coupon payments on the bond are made on February 4 anJ August 4 each year. The term .sNcmre i.s flat, and the rate of interest v ith semiannual compounding is 12% per annum. The conver.tion factor for the hond is 1.5. The current quoted bond price is $110. Calculate the quoted future.s price/or the contract. There are 176 days between FebruaW 4 and July 30 and 1 8 I days between February 4 and August 4. The cash price of the bond is, therefore: 110 + 1 76 x 6.5 = 116.32 181 The rate of interest with continuous compounding is 21n 1 .06 = 0.11 65 or 11.65% per annum. A coupon of 6.5 will be received in 5 days (= 0.01366 years) time. The present value of the coupon is 6.5e '^*“ " = 6.490 The futures contract lasts for 62 days (= 0.1694 years). The cash futures price if the contract were written on the 13% bond would be (1 1 6.32 — 6.490)e’ '” 4‘° "” = 1 12.02 At delivery there are 57 days of accrued interest. The quoted futures price if the contract were written on the 13% bond would therefore be

57 112.0—2 6.5 x184 = 110.01 Taking the conversion factor into account the quoted fixtures price should be: 110.01 73.34

Problem 6.12. An investor i.s looking [or arbitrage op¡›ortunities iii the Treasury Fund/iifures mai-ket. What comj›lication.s are created hy the [act that the party ii’ith a short position con choosed to deliver any hond with a matiirih o/ other 15 years:! If the bond to be delivered and the time of delivery v ere known, arbitragc would be straightforward. When the futures price is too high, the arbitragcur buys bonds and shorts an equivalent number of bond futures contracts. When the futures price is too low. the arbitrageur shorts bonds and goes long an cquivalcnt number of bond futures contracts. Uncertainty as to which bond will be dclivcred introduccs complications. The bond that appears cheapest-to-delivcr nov may not in fact be cheapest-to-deliver at maturity. In the casc whcrc thc fixtures price is too high. this is not a major problem since the party with the short position (i.c., the arbitrageur) determines w hich bond is to be delivered. In the case z'here the futures price is too low, the arbitrageur’s posititin is far more difficult since he ter she docs not know' which bond to short; it is unlikely that a profit can be fucked in for all possible outcomes. Prnhlem 6.J3. .Suppose that tic nine-inonlh SIB 0 £ intel CS I i ¢Ilfi IS â"o fi!ñI LJIJI1 fllTt ‹init [he siX-mGti ih V I D O O

intei-esi role’ i.s 7. fi"Xo ›er annum (froth ills ‹ie/fzu//fiñi.5 and can///›ucz‹a' co//Ipuundiny). Estimate he iliree-month Eui-odollui-futures f›rice quote Koi a conlrucl multiring in six months. The fom'ard interest rate for thc timc period betw'een months 6 and 9 is 9% per annum with continuous compounding. This is because 9% per annum for three months when combincd w ith 7 - % per annum for six months gives an average interest rate of S% per annum for the nine-month period. With quarterly compounding the forward interest rate is 4(e’ ” ’ — 1) = 0.09102 or 9.102%. This assumes that the day count is actual/actual. With a day count of actual/3f›0 the rate is 9. 102 x 360 ‘ 365 = 8.977 . The three-month Eurodollar quote for a contract maturing in six months is therefore 1 00 — 8.9 7 7 = 91.02 Problem 6.14. Suppose that the 300-Jan LIBOR Piero rate i.y 4’X‹ and Eurodollar quotes for contracts maturing in 300, 398 anJ 489 days are 95.83, 95.62, and 95.4R. Ca/ct//a/e 398-dny and 489day LIBOR zero rates. A.f.Hume no difference l›eRveen foovard ated future.s rate.s [or the purposes of your calculation.s. The form ard rates calculated form the first tu o Eurodollar futures are 4.17'Z« and 4.38%. These are expressed with an actual/3f›0 day count and quarterly compounding. With continuous compounding and an actual/365 day count they are 4|

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(365/90)1n( 1*0.0417/4)=4.2060%› and (365/90)ln(1+0.0438/4)= 4.41 b7%. It follows from equation (6.4) that the 395 day rate is (4•300*4.2060•98)/39S=4.0507 or 4.05L7 o. ThC 459 day rate is (4.0507^39S*4,41fi7• 9 I )/489=4. 11 85 or 4.1 ISS* . We are assuming that the first futures rate applies to 95 days rather than the usual 91 days. The third futures quote is not needed. Problem 6.15. Suyy›o.se that a bon‹1 pot tf‹›lio ith a duration of 12 year.s i.s hedged u.sing a future.s contract in i+hicJi the imderlyinq as.set ha.y a durafion of four year.f. What i.s likel j’ to lie the impact on rñe hedge o[the fact that the 12-year rate i.s /e.s.s volatile than the fow -year rate? Duration-based hedging procedures assume parallel sh ifts in the yield curve. Since the I 2year rate tends to move by less than the 4-year rate, the portfolio manager may find that lie or she is over-hedged. Problem 6.16. Suppose that it is February 20 anJ a trea›’iirer realizes that on July 17 the company ›i ill have to issue $5 million of ‹’ommercial paper v ith a inaturi3 of 180 Jaya. If the paper were issueJ today, the c'ompanv would realize 54,820,000. (In othe r »orJs, the company would receive $4,820,000 for its paper anJ have to reJeem it at $o“, 000,000 in 180 days’ time.) The Septemher EuroJollar future.s price is quoted a.s 92.00. How .should the trea.surer hedge the compan›' ’.s exJ›ostire? The company treasurer can hedge the ctinipany’s exptisure by shorting Eurodollar futures contracts. The E rirodollar futures position leads to a profit if rates rise and a loss if they fall. The duration of the commercial paper is hv'ice that tif the Eurodollar deposit underlying the Eurodollar futures contract. The contract price of a Eurodollar futures contract is 980,000. The number of contracts that should be shorted is, therefore, 4,820, 000g —29 980, 84 000 Rounding to the nearest whole number 10 contracts should be shortcd. Problem 6.17. On August 1 u portfolio manager has a honJ porrfi›lio worth $10 million. The duration of the should the portfolio manager imnluni r the portfolio again.st changes in interest rates over thc next tw'o month.s? The treasurer should short Treasury bond futures contract. If bond prices go doc n, this futures position will provide offsetting gains. The number of contracts that should be shorted 10, 000, 000 x 7.1 $$ 91, 375 x 8.8 Rounding to the nearest whole number 88 contracts should be shorted. 42 2012 Pearson Educatiota

Problem 6.18. How can the portfolio manager change the duration of the portfolio to 3.0 years in Problem 6. 7/.• The answer in Problem 6.17 1s designed to reduce the duration to zero. To reduce the duration from 7.1 to 3.0 instead of from 7.1 to 0, the treasurer should short 4.1 .30 = 7.1 50.99 or 51 contracts. Problem 6.19. Benveen October 30, 2012, and November 1, 2012, you have a choice between owning a U.S. government bond paying a 12% coupon and a U.S. corporate hond paying a 12% coupon. Consider carefully the day count conventions discus,sed in this chapter and decide which of the Evo bonds Mott w'ould prefer to ow'n. Ignore the risk of default. You would prefer to own the Treasury bond. Under the 30/360 day count convention there is one day between October 30 and November 1. Under the actual/actual (in period) day count convention, there are two days. Therefore you would earn approximately tw'ice as much interest by holding the Treasury bond. Problem 6.20. Crypo.se that a Eurodollar futures quote is R8 for a contrat’t maturing in 60 days. What is the LIBOR forward rate for the 60- to 150-day period? Ignore the difference between futures and forw'ards for tlic purpo.ses o[thi.s question. The Eurodollar futures contract price of 85 means that the Eurodollar futures rate is 12% per annum with quarterly compounding. This is the forward rate ftir the 60- to 150-day period with quarterly compounding and an actual/360 day count convention. Problem 6.21. The three-month EuroJollur futures price for a contract muluring in six years is quoted as 95.20. The stanJarJ Jeviation of the change in the shorl-ierm interest rate in one vear i.s /. I‘No. Estimate the %rw ard LIBOR interest rate for the perioJ between 6.00 and 6.25 veors in the future. Using the notation of Section 6.3, o = 0.011, i, = 6, and i, — 6.25. The convexity adjustment iS

1 0.011’ x 6 x 6.25 = 2 0.002269 or about 23 basis points. The futures rate is 4.8% with quarterly compounding and an actual/3fi0 day count. This becomes 4. h x 365 / 360 = 4. b67* with an actual/actual day count. It is 4ln(1 + .04567 /' 4) — 4.84% with continuous compounding. The forward rate is therefore 4.6—4 0.23 = 4.61 % with continuous compounding. Problem 6.22. Explain why the forwarJ interest rate is less than the corresponJing futures intet-est rate calculated from a Eurodollar futures contract. 43 ?°?2012 Pearson Education

Suppose that the contracts apply to the interest rate betwecn times T, and T,_ . There are two reasons for a difference between the forward rate and the futures rate. The first is that the natures contract is settled daily whereas the for ard contract is settled once at time z' . The second is that without daily settlcment a futures contract would be settlcd at time T not T, . Both reasons tend to make the futures rate greater than the forward rate.

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CHAPTER 7 Swaps Problem 7.1. Companies A and B have been offered the}ollow'ing rates per annum run n $20 million fiveyear loan: Floating Rate

Fi. ed Rate Company A Company B

5.0% 6.4%

LIBOR 0. IN o

LIBOR+0.6%

Company A requires a fioating-rate loan,- company B requires a fixed-rate loan. Design a sw'ap that will net a bank, acting a.s intermediary, 0.1% per annum ynd that will appear equally attractive to both companies. A has an apparent comparative advantage in fixed-rate markets but wants to borrow floating. B has an apparent comparative advantage in floating-rate markets but wants to borrow fixed. This provides the basis for the swap. There is a 1.4% per annum differential between the fixed rates offered to the two companies and a 0.5% per annum differential between the floating rates offered to the two companies. The total gain to all parties from the swap is therefore 1.4 — 0.5 = 0.9 % per annum. Because the bank gets 0.1% per annum of this gain, the swap should make each of A and B 0.4% per annum better off. This means that it should lead to A borrowing at LIBOR —0.3 % and to B borrowing at 6 OOH. The appropriate arrangement is therefore as shown in Figure S7.1. 5.3%

Company A

—

5.4%

Company B

Financial Institution LIBOR

Figure S7.1

LI BOR0.6%

Swap for Problem 7.1

Problem 7.2. Company X wishes to horrow L’.S. dollars at a JixeJ rate of interest. Company Y wishes to borrow Japanese yen at a fixed rate of interest. The amounts required by the two companies are roughly the same at the current exchange rate. The companies have been quoted the following interest rates, which have heen adju.sted %r the impact o] taxes:

45 201 2 Pearson Education

Dollar.s 5.0% 6.5%»

Companv X Company Y

9.6% 10.th»

Design a sii’ap ihai will net a hank, acting a.s infermediai3, 50 basis poinls per annum. Make the swap equally attractive to Ihe two companie.s and ensure that all foreign exchange risk is ass umeJ by the bank. X has a comparative adsantage in yen markets brit * ants to borrow dollars. Y has a comparative advantage in dollar markets but wants to borro ' yen. This provides the basis for the swap. There is a 1.5% per annum differential betwecn the yen rates and a 0.4% per annum differential betw een the dollar rates. The total gain to all parties from the sw’ap is therefore 1— . 5 0.4 = 1.1 % per annum. The bank requires 0.5% per annum, leaving 0.30o per annum for each of X and Y. The swap should lead to X borrowing dollars at 9.6 — 0.3 = 9.3 % per annum and to Y borrowing yen at 6.5 — 0.3 — 6.2 % per annum. The appropriate arrangement is therefore as shown in Figure S7.2. AU foreign exchange risk is borne by the bank.

h'en 5*v

Financial

Company

X }

Company Dollars 1 0’.;

Do II ars 9.3’, «

Dollars 1 0°, i

Figure S7.2 Swap for Problem 7.2 Problem 7.3. A $100 million intere.at rate s 'ap has a remain ing li]e oj 10 months. Under the term.s of the swap, .six-fllonth LIBOR is i3xchOH@€d fOi‘ 7’Xo per annum f‹ expounded.semiannually). The average of the bid—offer rate heing exchangeJ for six-month LIBOR in swaps of all maturities is currently 5’No yer annum with I ontinuous compounding. The .six-month LIBOR rate was 4.6’o per annum tw o months ago. What is the ‹’urrent value of the .sv'ap to the partv paying fioating!! What is its value to the partv payinff fixeJ? In four months 53.5 million ( — 0.5 0.07 SI 00 million) will be receix’ed and $2.3 million ( = 0.5 x 0.046 x $100 million) will be paid. (We ignore day count issues.) In 10 months $3.5 million will be rcccived. and the LIBOR rate prevailing in four months’ time will be paid. The value of thc fixed-rate bond underlying the swap is 3 .5e°’’”’ " 103.5e ””" '" = $102.7 1 S million The value of the floating-rate bond underlying the sw ap is (100 + 2.3)e ' ' = $100.609 million The value of the swap to the party paying floating is Sl 02.718 $100.609 = $2.109 million. The value of the swap to the party paying fixed is —$2.109 million. These results can also bc dcrived by decomposing the swap into forward contracts. Consider the party paying floating. The first forward contract involves paying $2.3 million and receiving $3.5 million in four months. lt has a value of 1.2e '”“ " = $1.180 million. To value the second forward contract, we note that the forward interest rate is 5% per annum with contlnuous compounding, or 5.063fio pcr annum with semiannual compounding. The value of the forw'ard contract is 46 i^'2012 Pearson Education

100 x (0.07 0.5 — 0.05063 x 0.5)c " " ” = 50.929 million The total value of the forward ctintracts is therefore $1.180 $0.929 = $2.109 million. Problem 7.4. EN plain H hat a swap rate is. What is the rel‹itions hip he•tiveeii .stray› rate.s and par › ields.’" A sv ap rate for a particuliir maturity is the avcrage of the bid and offer fixed rates that a market maker is prepared tti exchange for LIBOR in a standard plain vanilla swap with that maturity. The swap rate for a particular maturity is the LIBOR/swap par yield for that maturity. Problem 7.5. A currency .swop has a remaining life of 15 months. lt involve.s e.Jehan ging interest ett I 0“Xo Otl £2.0 million for interest at 6% on 53O million once a year. The term stru‹’ture oJ iaferes/ rate.s in both rite k"ilitecl Kingdom and the United States is currently flat, anal i f the sw'af› w'e•re negotiated t o d p thein/eres/ i-ri/e’6 exchanged would he 4’No in do Htm › anal 7% in › tcrling. All

interest ra/e,s are quoted with annual conipotineling. The curt-ent exchange rate LHollar› fler pound ste•rling) is 1. R500. what i.s the value of the so ap to the park'paying .sterling? V"liut is the vultie oJ thc .swap to the j›ar5 yaw ing Jollai's? Thc swap involves exchanging the sterling interest of 20 x 0.10 or £2 million for the dollar interest of 30 x u.06 = 1.s million. The principal amounts are also exchanged at the end of the life of the swap. The v'a1ue of the sterling bond iinderlyin s the s ap is 2 22 = 22.152 million pounds ) “ (1.07)’ (1.07 The value of the dollar bond underlying the sv ap is I .8 3 I .8 $32.061 million (1.04)' “ (1.04)’ The value of the su-ap to the party paying sterling is therefore 32.061 — (22.1 82 1 .85) = —58.97a› million The walue of the sz ap to the party paying dollars is $8.976 million. The results can also be obtained by viewing the sw ap as a portfolio o1 forward contracts. The continuously compounded interest rates in sterling and dollars are 6.766"% per annum and 3.922% per annum. The 3-month and 15-month ftirward exchangc rates are 1. b5e" "^' """’ ' = 1 . 83G9 and 1.S 5e " ' " ” " " ° = I.7554 . The values of the two for\s ard contracts corresponding to the exchange of interest for the party paying sterling are therefore (1.—8 2 z 1.8369)c°’ "’ 2'“’" = —$1. 855 million and (1.8 — 2 x 1.7S54)e ' '” 22 ' ” = —S1.6 86 million The value of the forward contract corresponding to the exchange of principals is (3 —0 20 x 1 .7d54)e°" ”’ 2' ' " = —$5.43 5 mjI lion The total value ot the swap is —SI . 855 $1.6 b—6 55.435 —SS.9 75 million. Problem 7.6. Explain the difference heU›een the credit riwk anal the market risk tit a financial conlra‹'t. Credit risk arises from the possibility of a default by the countcrparty. Market i isk arises from movements in market ›'ariables such as intei est rates and exchange rates. A coinpl ication is 47 0â20l2 Pearson Edricatitin

that the credit risk in a swap is contingent on the values o1 market variables. A company’s position in a swap has credit risk only when the value of the swap to the company is positive. Problem 7.7. borrowing at six-month LIBOR ylus 150 busis yoints and.swapf1ing LIBOR for 3.7°X. He goes on to .say that tlii.s n’as possible because his ‹’ompany In.s a c.omparative rdvon/oge in the floating-rate market. V’hat has the trras urer overlooked? The rate is not truly fixed because, if thc company’s credit rating declines, it will not be able to roll over its Vitiating rate borrowings at LIBOR plus 150 basis points. Thc clfcctive fixed borrowing rate then increases. Suppose for example that the treasurer’s spread over LIBOR increases from 150 basis points to 200 basis points. The borrowing rate increases from 5.2% to 5.7"%. Problem 7.8. E. pl‹iin liy a bank is suh feet to credit risL whe•n if enters into twin off.setting swap contracts. At the start ot’ thc swap, both contracts have a value o1 approximately zero. As time passes, it is likely that thc swap values will change, so that one swap has a positive value to the bank and thc othcr has a negative value to the bank. If the counterparty on the other side of the poSltivev£lliie sw ap defaults, the bank still has to honor its ctintract with the other counterparty. lt is liable to lose an amount equal to the positive value of the swap. Prohlem 7.9. Coinpanie s X and Y hav'e been ojJereJ the %llowing rate.s peer umum on a 85 million 10-year invy.stment.’ Floating Rate.

LIBOR LIBOR

Company X Company’ Y

L’miJ›anv X rcquires a JixeJ-mite investment, coinJ›any Y require's a floating-rate rinse.stment. Die.sign a s›vap that ii ill net a bank, acting a.s intermediai y, 0.2% per annum and will appear The spread between the interest rates offered to X and Y is 0.8% per annum on fixed rate investments and 0.0% per annum t›n floating rate investmcnts. This means that the total apparent benefit to all parties from the swap is 0.S%pcrannum. Of this 0.2% per annum will go to the bank. This lead es 0 30» per aiuium for each of X and Y. In other words, company X should be able to get a fixed-rate return ot 8.3fio per annum while company Y should be able to get a floating-rate return LIBOR + 0.Who per annum. The required s ap is shown in Figure S7.3. The bank earns 0.2%, company X earns 8.3o/«, and company Y cams LIBOR * 0.3%.

48 D20 12 l'earson Education

RAN K

LIB(JR

C ompan} LIBtJR

I igurc S7.3 Sw'ap for Problem 7.9

At the end of year 3 the financial institiititin was due to receive 5500,000 ( = 0.5 1 0 % ot $10 million) and pay $450,000-( 0.5 x 9 0o o1 $10 million). Thc immediate loss is therefore $50,t1tl0. To value the remaining swap we assume than foovai d i ates are real ized. All for iv ard rates are b"% per annum. The remaining cash flow's are therefore valued tin the assumption that the floating payment is 0.5 x 0.08 10. 000, 000 — $400, 000 and the net payment that would be reccivcd is 500, 000 — 400, 000 $1.00, ONO. The total cost or default is therefore the cost of foregoing thc following cash how s: 3 year: 3.5 year: 4 year: 4.5 year: 5 year:

550,000 $100,000 $100,000 $100,000 $100,000

Discounting these cash flow's to year 3 at 4fio per six months we obtain the cost of the default as $413,000.

US dollars ( floating rate) Canadian dollars (fi xcd rate)

.4 LIBOR*0.5°0

B LIBOR+ 1 .0*

5.t1%’o

N.5°fi

As.mine that A n ants to borroii C’.S. Jollars at a jloating rate oJ internet ‹iilJ B w'ants to l›orro›+ Canadian Jollars at a hxeJ i’ate of internet. A financial iilstitution is pluilning to arrange a .sway and requires a 5H-hmsis-point .fyi-eaif. If the s wap is equallv uttracli ve lv A and B, ›rliat rate.s of intere.st ill A and B end up pai ing? Company A has a comparative advantage in the Cmidiun dtillur fixed-rate market. Company B has a comparative advantage in the U.S. dollar floating-rate market. (This may be because of their tax positions.) However, company A wants to borrow in the U.S. dollar fl‹iating-rate market and company B ants to borro iii the Canadian dollar fixed-rate market. This gis es 49 ‘2012 Pearson Ldu cation

rise to thc swap opportunity. The differential bctwccn the U.S. dollar floating rates is 0.5% per annum, and the dirferential betw'een the Canadian dollar fixed rates is 1.5'?o per annum. The difference betwecn the differentials is 1% pcr annum. The total potential gain to all parties from the swap is therefore 1% per annum, or 100 basis points. If‘thc financial intermediary requires 50 basis ptiints, each ofA and B can be made 25 basis points bcttcr off. Thus a swap can be designed so that it provides A with U.S. dollars at LIBOR 0.25% per annum, and B with Canadian dollars at 6.2b o per annum. The swap is shown in Figure S7.4. C$: 6.25°/

Financial Institution

A

B

Figure S7.4 Su ap for Problem 7.11 Principal payments fio ' in the opposite direction to thc arrows at the start of the life of the swap and in the same direction as the arrow s at the end of the lile of’ thc swap. The financial institution ould be exposed to some foreign exchange risk which could bc hedged using forward contracts. Problem 7.12. A financial in.stitution has entei-eJ into u IN-year currency swap with company Y. L’nJer the /erm,i o/the .swap, the hnancial institution recci ves interest at 3’to per annum in SwissJonc.s cmd pay’.s inleres I at 8%‹› per annam in 2'.S. dollars. Inte•res I p‹n'ment.s are exchanged once a year. The princiJ›al amounts are 7 million dollars anJ 10 million francs. SupJ›o.se that co/iya/iJ' X declare.v banL+iipt‹’v at the enJ ofveur 6, t hen the exchange rate i.‹ 8.80 per ]runc. What i.s the co.s/ to the financial institution.‘" As 'rune that, at the end of yea+ 6, the in/eres t rate i.s 3‘No fler annum in S i.is dance and 8% per annum in L.S. Jollars for all maturities. All ititere.st rate.s are quoted vNth annual coinpounJing. When interest rates are c‹rrnpounded annually 1* r where F, is the T -ycar forward rate, S, is the spot rate, r is the dtiirestic risk-free rate, and s se ra, .

'

the end o

Spot: 1 year forw-ard: 2 year forward: 3 year forward: 4 year forw'ard:

II.f18 and r — — 0.03. thc spot and forward exchange carea e

’y

"" 0.8000 0.8388 0.5796 0.9223 0.9fi70

The value or the swap at the time of the default can be calculated on the assumption that for ard rates are realized. The cash flow s lost as a result of the default are thcrclore as follows. 50 20.12 Pearson Education

7

5G0,000 560.000 560,000 560,000

3t10,000 300,000 300,000 300,000

0.8000 0.5358 7.6 0.9223

240,000 251.600 263,900 27G,70tl

10

7,560,000

10,300.000

0.9670

9,960.100

-320.000 *-308,400 -296.100 -283.300 2,400.100

Discounting the numbers in the final column to the end of year ti at 8%› per annum, the cost of the default is $679,800. Note that, if this were the only contract entered into by company Y, it would make no sense for the company to default at the end ot year six as the exchange of payments at that time has a positive value tti company Y. In practice, company Y is likely to be defaulting and declaring bankruptcy for reasons unrelated to this particular contract and payments on the contract are likely to stop whcn bankruptcy is cleclarcd. Problem 7.13. A/ter it hedges its foreign e.re/iaHge risk using forward conii-ac/s, is the financial institution’.s average .sj›read in Figure 7.11 likelv io be greater than or /e.s.r than 2.0 ha.f i.s points? f.. pl‹iin 'our an.s 'er. The financial institution will ha›'e to buy 1.1% of the AU f9 principal in the forw arm market for each year of the life of the swap. Since AUD interest rates are higher than dollnr interest rates, AUD is at a disctiunt in forward markets. This means that the AU D purchased for year 2 is less expensive than that purchased for year 1; thc AUD purchased for year 3 is less cxpensive than that purchased ftir year 2; and so on. Thisw orks in favor ot the financial institution and means that its spread inci-eases w ith time. Thc spread is always above 20 basis points. Problem 7.14. "Companies »ith high credit risks are the ones that cannot acces.s kixeJ-rate marI:rls in an intere.st rate swap.” .Assume thal lhis slalem)1t i.s true. Rio you think it increas es or Jec.rea,ses the ri.sk o/ a financial institution’s s›‹ap poi-tfolio? A.s.Hume that companies are mo› I likel v' to default when interest rates are high. Consider a plain-vanilla intcrest rate s ap inv olving two companies X and Y. We supposc that X is paying fixecl and receii ing floating while Y is paying floating and receiving fixed. The quote suggests that company X w ill usually bc less creditworthy than company Y. (Company X might be a BBB-rated ctimpany that has dilticulty in accessing fixed-rate markets directly; company Y might be a AAA-rated company that has no difficulty accessing fixed or floating rate markets.) Presumably company X wants lixcd-rate fends and company Y wants floatingrate funds. The financial institution will realize a loss if company Y defaults iv hen rates are high or if ctimpany X defaults hen rates are low. These events are relatii ely unlikely since (a) Y is unlikely to default in any circumstances and (b) defaults are less likely to happen when rates are low’. I or the purposes of illustration, suppose that the probabilities tif various evcnts are

Default by Y:

0.00 1

5.1

8.'20 12 Pearson Education

default by X. Rates high when default ticcurs: Rates low z.hen default occurs:

0.010 0.7 0.3

Thc probability of a loss is tLtltd 1 x (1.7

0.0 10 x 0.3 = 0.003 7

If the roles of X and Y in the sw ap had been rev ersed the probability of a loss ould bc 0.OF 1 0.3 + 0.010 0.7 = 0.0073 Assuming companies are more likely to ‹tc1‹aiilt hen interest rates are high, the above argument show's that the observation in quotes has the effect o1 dccrcasing the risk of a linancial institution’s su ap portfolio. It is » orth notins that the assumption that defaults are more likcly u hen interest rates are high is ‹open to question. The assumption is motivatcd by the thought that high interest rates often lead to financial difficulties for corporations. However, tlierc is oltcn a time lag beta een interest rates heing high and the resultant default. When the default actually happens interest rates may be relatively low.

In an interest-rate swap a financial institution’s cxposurc dcpends on the difference beñveen a fixed-rate of interest and a floating-rate of intcrcst. lt has no cxposure to the notional principal. In a loan the whole principal can be lost.

The thank is paying a floating-rate on thc deposits and receiving a fixed-rate on the 1o‹ins. It can offset its risk by entering into interest rate s1'aps ( ith other financial institutions or corporations) in w hich it contracts to pay fixcd and rcccivc tloatinp.

The floating payments can be valued in currency A by (i) assuming that the foot ard rates are realized, and (ii) discountins the resulting cash fiows at appropriate currency A discount rates. Suppose that the value is I', . The fixed paymcnts can be valued in currency B by disctiunting them at the appropriate currency B discount rates. Suppose that the value is I . lf (j is the current ex change rate (number tif units of currency A per unit of crirrcncy B), the 'alue of the s ap in currcncy A is P—, I . A lternatively. it is k / Q —VBin currency B.

i2012 Sea son Education

The two-year swap rate is 5.4%. This means that a two-year LI holt bond paying a semiannual coupon at the rate of 5.4% per annum sells tor par. It @_ is thc h o-year LIBOR zero rate 2.7e*’ °‘ ‘ + 2.7c*' ’”' + 2.7e*' ’ " ‘ * 1 02.7e ' — 100 Solving this givcs A, = 0.05342 . The 2.5-year sw'ap rate is assumed to be 5.5%›. This means that a 2.5-year LIBOR bond paying a semiannual coupon at the rate of 5.5"/o per annuiai sells for par. lf R is the 2.5-year LIBOR zero rate 2.75e ' ’

* 2.75e ' “ ' * 2.75e ’^" 5 * 2.73e ’ " ' ' * 02.75e*“ ’ 2 = 100

Solving this gives R,. ,. — — 0.05442 . The 3-year swap rate is 5.6%i. This means that a 3-year LI BOR bond paying a semiannual coupon at the rate o1 5.600 per annum sells ior par. 11 fi, is the three-year LIBOR zero rate 2.Se ' " ' ‘ -I- 2.Se ' “ ' * 2.Se*’ "" ‘ + 2 8e 0 0'*"‘2.0 + 2. 8e

"""“'

-F102.8e “"’ = 100 Solving this gives fi, = 0.05544. The zero rates for maturities 2.0, 2.5, and 3.tl years are therefore 5.342*/o, 5.442"%, and 5.544 o, respectively.

53 ›2012 Pearson Education

Chapter 8 Securitization and the Credit Crisis of 2007 Problem 8.1 Whal us ltte role of G..'X'’ 1970s .'!

(Finnie :lf«e) in the mortgvEe-hackeJ s ecuritie› market of the

GNMA guaranteed qualifying mortgages against default and created securities that were sold to investors. Problem fL2 An ABS is a set tif branches created frtiin a portfolio of mortgages or other assets. An ABS CDO is an ABS created from particular tranches (e.g. the BBB tranches) of a number of different ABSs. Problem 8.3 The mezzanine tranche o1 an ABS or ABS CDO is a tranche that is in the middle as far as seniority goes. It is ranks below lhe senior tranchcs and therefore absorbs losses before they do. It ranks above the et uity tranche (so that the equity tranche absorbs losses before it does).

The waterfal I defines how the cash flows front the underlying assets are allocated to the tranches. 1 n a typical arrangement, cash flows are fi rst used to pay the senior ti an ches their promised remrn. The cash fiows ( if any) that are lefi over are used to provide the mezzanine tranches with their promised returns. The cash flows (if any) that are left o› er are then used to provide the equity tranches with their promised returns. Any residual cash tiows are used to pay do n the principal on die senior branches. Problem 8.5 blind are the nuns hers in Table 8.1 for a loss rate o[a) 1s!%‹› and h) 15%?

17.9% 48.7%

Problem 8.6 Idol is a suhprime mortgage? A subpriine mortgage is a mortgage w here the risk of default is higher than normal. This may be because the borrow er has a poor credit history or the ratio of the loan to value is his h or both. 54 20.12 Pearson Education

Problem 8.7 Why do you think the increase in house prices Jarring the 2000 to 2007 period is referreJ to The increase in the price of houses was caused by unsustainable factors. Problem 8.8. Why did mortgage lenders Jreqiientlv not check on information provided by potential borro›v'ers on mortgage application forms during the 2t!if to 20LI 7 period'! Subprime moitgages were frequently securitized. The only information that z as retained during the securitization process was the applicant’s FICO score and the loan-to-value ratio of the mortgage. Problem 8.9. How were the ri.yLy in ABS CDO.f mi.sjudged hy the market? Investors underestimated how high the default correlations between mortgages would be in stressed market conditions. Investors also did not always realize that the tranches underlying ABS CDOs were usually quite thin so that they were either totally wiped out ter untouched. There was an unfortunate tendency to assume that a tranche with a particular rating could be considered to be the same as a bond with that rating. This assumption is not valid for the reasons just mentioncd. Problem 8.10. What i.s meant by the term “agencv ‹ nets "? HeJiJ agency costs pluy a role in the credit cri.sis? “Agency costs” is a tens used to describe the costs in a situation where the interests ot‘two parties are not perfectly aligned. There v’ere potcntial agency costs between a) the originators of mortgages and investors and b) employees of banks who earned bonuses and the banks themselves. Problem 8.11. How is an ABS CDO createJ? What was the inofivo//on /o cremate ABS CDOS .’• Typicallj an ABS CDO is created from the BBB-rated tranches of an ABS. This is because it is difficult to find investors in a dircct way tor the BBB-rated tranches of an ABS. Problem fi.12. Ex Plain the imJ›act of an increa.se in default correlation on the risks of the senior tran‹’hr oj an A RS. What i.s it.s impact on the ri.sks n/rite equity tranche? As default correlation increases, the senior tranche tif a CDO becomes more risky because it is more likely to suffer losses. lt therefore becomes less valuable. As default ctirrelation increases, the equity tranche becomes more valuable. To understand why this is so, note that in the limit vhen there is perfect correlation there is a high probability that there will be no dclaults and the equity tranche ill suffer no losses. 55 2012 Pearson Education

Problem 8.13. Explain why the AAA -rated tranche o/ an ABS CDO is more risky than the AAA-rated tranche of an ABS. A moderately high default rate will wipe out the tranches underlying the ABS CDO so that the AAA-rated tranche of the ABS CDO is also wiped out. A moderately high default rate will at worst ipe out only part of the AAA-rated tranche of an ABS. Problem 8.14. Explain why the end-o/-year honus is sometime.s referred to as “.short-term compensation. ” The end-of-year bonus usually reflects performance during the year. This type of compensation tends to lead traders and other employees of banks to focus on their next bonus and therefore have a short-term time horizon for their decision making.

56 tâ>20l 2 Pearson Education

CHAPTER 9 Mechanics of Options Markets Problem 9.1. An investor buys u European put on a sltare for $3. The stock price is $42 and the strike J›rice is 840. UnJer what circumstances does the investor make a profit'? L’ncler u'hat ctrl umstances will the option be exercised? Dram' a Jiagram showing the variation of the investor’s profit with the stoc'k price at the maturity of the option. The investtir makes a profit if the price of the stock on the expiration date is less than $37. In these circumstances the gain from exercising the option is greater than $3. The option will be exercised if the stock price is less than $40 at the maturity of the option. The variation or the investor’s profit with the stock price in Figure 59.1.

6

Pr

4

2

e lS) 2

40

30

4S

50

-4

Figure 59.1

Investor’s profit in Problem 9.1

Problem 9.2. An investor sells u European c'ull on u share for $4. The slock price is $47 anJ the strike price i.y 850. UnJer what circumstances Joes the inv'estor make u profit.! Under what circum.stance.s will the option be exercised? Draw a Jiagram showing the variation oj the inve.sfor ’.s profit with the stock price at the maluri of the option. The investor makes a profit if the price of the stock is below $54 on the expiration datc. It the stock price is below $50, the option will not be exercised, and the investor maLes a profit ot S4. If the stock price is between $50 and $54, the option is exercised and the investor makes a profit between $0 and $4. The variation of the investor’s profit z'ith the stock price is as shown in Figure S9.2. 57 ifi20l2 Pearson Education

6

Profit (S)

4

tock Price ($) -2 40

50

S

60

-4

-6

Figure S9.2

Investor’s profit in Problem 9.2

Problem 9.3. An investor sells a European call option with ›hike pri‹’e of n anJ maturity T and buys a put with the same strike price and maturity. Describe the inve.ftor’s yen.sition. The payoff to the investor is — max ‹S, — K,0) + max (c— S , 0) This is K — S in all circumstances. The investor’s position is the same as a short position in a forward contract with delivery price x . Problem 9.4. Explain » h9› margin.s are required n’hen clients write options buI not when the y buy options. When an investor buys an option, cash must be paid up front. There is no possibility ot future liabilities and therefore no need for a margin acctiunt. When an investor sells an option, there arc potential future liabilities. To protect against the risk of a default, margins are required. Problem 9.5. A .stock option is on a February, May, Augu I, anJ No vember cycle. What option.s trade on (a) April 1 anJ (b) May 30? On April 1 options trade with expiration months of April, May, August, and November. On May 30 options trade with expiration months of’ June, July, August, and November. Problem 9.6. A company declare.s a 2-for-1 stock .sy/if. Explain how the terms change for a call option with o sir/ke price of $60. C?201 2 Pearson Education

The strike price is reduced to $30, and the option gives the holder the right to purchase twice as many sharcs. Problem 9.7. “Employee .stock oj›tioiis' issued bv a company are Jifferent from regular exc'hange-Iraded call option.s on the company s stock because they can affect the capital structure of the company. ' Explain thi› statement. The exercise of employec stock options usually leads to new shares bcing issued by the company and sold to the employee. This changes the amount of equity in the capital structure. When a regular exchange-traded option is exercised no ncw shares are issued and the company’s capital structure is not affected. Problem 9.8. A corporate treasurer is deg igning a hedging pi‘ogrum imolving foreign ‹ urrency option.s. What ure the pro.y and cont o] using (a) the S’ASDAQ Oil and th) the over-the-counter market for trading? The NASDAQ OMX otters options with standard strike prices and times to maturity. Options in the over-the-counter market have the advantage that they can be tailored to meet the precise needs of the treasurer. Their disad›'antage is that they expose the treasurer to some credit risk. Exchanges organize their trading so that there is virtually no credit risk. Problem 9.9. Su 'pose that a Eurojlean call option to huv a shares for $100.00 co.Arr .SS. 90 aa‹f is helJ until maturity. L"nder n'hat ci‘ctrmsfrrnc'es will the holder- o} the option make a proJit? Under what cirt iimstan‹ es will the option br exercised? Draw' ci Jiagrain illustrating how' the pro]it from a long position in the option depeiirJ.s on the stock pri‹!e at maturi3 of the oplion. Ignoring the time value of money, the holder of thc option will make a profit if the stock price at maturity of the option is greater than $105. This is because the payoff to the holder of the option is, in these circumstances, greater than the $5 paid for the option. The option will be exercised if the stock price at maturity is grcater than $1 00. Note that if the stock price is between $1(IO and $105 the option is exercised, but the holder of the option takes a loss overall. The profit from a lting position is as show’n in Figure S9.3.

80

Figure S9.3

4 0

Profit from long position in Problem 9.'7 59 â20l 2 Pearson Lducation

Problem 9.10. .Suppose that a European put option to se/l u share %r $60 costs $8 and is held until maturin’. Uncler »liat circumstunce•s w'il! the seller of the option (the party north the shorl position) make a profit? Under what circum.stan‹'es ›i'ill the option be exercised.'" Draw u diagram illustrating how the profit Jrom a short 'osition in the option Jepends on the stock price at maturin of the option. Ignoring thc time value of money, the seller of the option will make a profit if the stock price at maturity is greater than $52.00. This is because the cost to the scller of the option is in these circumstances less than thc price received for the option. The option will be exercised if thc stock price at maturity is less than $60.00. Note that if the stock price is between $52.00 and $60.00 the sellcr of the option makes a profit even though the option is exercised. The profit mom the short position is as shown in Figure S9.4.

45

Figure S9.4

ProJit from short position in Problem 9.10

Problem 9.11. De.scribe the ter-in inal value of the [ollowing portfolio.’ a newly entered-into long%rw'ard contract on an a,s,set anJ a long position in a European put option on the a,s.set ’ith lhe same matiiriH as the fori»ard con/i eel and a .y/rike price that i.s equal to the forward price of they asset at the time the port%lio is set up. Show that lhe European put option ha.s the same value os a European ‹’all option with the same .strike pri‹:r anal maturih. The terminal value of the long forward contract is: ›‹'heie Sz i s the price of the ;isset ct mamrity and F is the foovard price of the asset at the time the portfolio is set up. (The delivery price in the torw‹ird contact is alsti F .) The terminal v alue of the put option is: 60 âz 2012 l'earson Education

The terminal value of thc portfolio is therefore No fnax ( F, — N,, 0)

= max (0. S — Fz ] This is thc same as the terminal value of a European call option with the same maturity as the forward contract and an exercise price equal to . This result is illustrated in the Figure S9.5.

1O

’ * x Price ” *

0

.^ Asset ------ -

--- --- For •‹ard - - - PutTotal

-10 -15

•“*

—20

Figure S9.5

Profit from portfolio in Problem 9.11

We have show'n that the forward contract plus the put is worth the same as a call with the same strike price and time to maturity as the put. The forward contract is worth zero at the time the portfolio is set up. It follows that the put is v orth the same as the call at the time the portfolio is set up. Problem 9.12. A trader buys a ‹ all oytion with a strike prie'e of $45 and a put oytion w'ilh a strike price of $40. Both options have the .same maturity. The call co.st.s $3 and they ptil costs $4. Draw a diagi-am showing the variation o] the lraJer’s profit with the asset price, Figure S9.6 shows the \’ariation o1 thc trader’s position with the asset price. We can divide the alternative asset prices into three ranges: a) When the asset price less than $40, the put option provides a payoff of 4—0 S, and the call option provides no payot1\ The options cost $7 and so the total profit is 33 — S . b) When the asset price is between $40 and $45, neither option provides a payoff. There is a net loss of $7. c) When the asset price greater than $45, the call option provides a payolT of S — 45 and the put option provides no payoff. Taking into account the 57 cost of the options, the total profit is I — 52. The trader makes a profit (ignoring the time value of money) i1 thc stock pricc is less than 61

$33 or greater than 552. This type of trading strategy is known as a strangle and is discussed in Chapter 11.

CuII

15 10

,,•**’

.•‘"*’’ , •’“"‘’“

0 2

— Total

Asset Price

.;10.-...- ..s40.-....-'

60

-1 0

Figure S9.6

Profit from trading strategy in Problem 9.12

Problem 9.13. Explain why an American option is alwuys worth at least as much as u EUrOF•an option on the same asset ›viIli the some .ytrike pi-it’e and exerci.se date. The holder of an American option has all the same rights as the holder of a European option and more. It must therefore be n orth at least as much. If it were not, an arbitrageur could short the European option and take a long position in the American option. Problem 9.14. Explain ii’hy an American option is a/ivay.s worth at leost as much as its intrinsic value. The holder of an American option has the right to excrcise it immediately. The American option must therefore be worth at least as much as its intrinsic value. If it were not an ai bitrageur could lock in a sure profit by buying the option and exercising it immediately. Problem 9.15. Lxplain carefully the difference hehveen writing a put option and buying a call option. Writing a put gives a payoff of min(S — K, 0) . Buying a call gives a payoff of max(S — K, 0) . In both cases the potential payoff is S, — K . The ditTerence is that for a writtcn put the counterparty chooses n'hether you get the payoff(and will allow you to get it only when it is ncgative to you). For a long call you decide z'hether you get the payoff(and you choose to get it when it is positiv-e to you.) Problem 9.16. The trea.surer o] a corporation is tr; ing to choc.se hetween options and forward contracts to hedge the corporation’s foreign exchange risk. Discu.ss the advantages and disadvantages of 62 â20l 2 Pearson Ectucaiion

Forward contracts lock in the exchange rate that mill apply to a particular transaction in the torture. Options provide insurance that the exchange rate will not be worse than some level. The advantage of a forw'ard contract is that uncertainty is eliminated as far as possible. The disadvantage is that the outcome w'ith hedging can be significantly u’orse than the outcome with no hedging. This disadvantage is not as marked with options. However, unlike forward contracts, options ini’olve an up-front cost. Problem 9.17. G"onsiHei an exchange-traded ca// opfron contract to buy 500 .shares with a strike price of $40 an‹1 maturii)' iit jour iTiontllS. -r /Oif7 f o r ' t h el e r mr o [ t h e optie n ‹'ontrar.t change hen

there i› a) A I H“Xos IOCk dividend

b) A 1t1“>ocash dis idend c) A 4-foi--1 stock.split d) The option contriict becomes one to buy 509 x I . 1 = 550 shares with an exercise price 40/1.1 = 36.36. e) There is no effect. The terms of an options contract arc not nonrally adjusted for cash dividends. f) Thc option contract becomes one to buy 500 x 4 — 2, 000 shares with an exercise price of 40/4 — $10 . Problem 9.18. “If mcs t of the call option.f on a slo‹’k are in they mon‹n , it i.s likel5’ that the .stock price hue risen rapiJl›’ in the last few months. ” Discu.s.s this ›latement. Thc exchange has ccrtain rules govcrning when trading in a new option is initiated. These mean that the option is close-to-the-money when it is first traded. If all call options are in the money it is therefore likely that the stock price has increased since trading in the option began. Problem 9.19. What i› the e/[ect o] an unexpected ca›h Jivideitd on (a) a call option price and (1) a put option price? An uncxpected cash dividend would rcduce the stock price on the ex-dividend date. This stock price reduction would not be anticipated by option holders. As a result there w ould be a reduction in the value of a cal1 option and an increasc the i’alue of a put option. (Note that the terms of an option are adjusted for cash dividends only in exceptional circumstances.) Problem 9.20. Option.s nn General .motors stock are on a Varch, .time, $cytember, and Dec ember cycle. What options trade on (a) Varch 1, (1›) June 30, end éc August 5? a) March, April, June and September b) July, August, September, December c) August, September, December, March. 63 ’.?*2012 Pearson £¢t\lcation

Longer dated options may also trade. Problem 9.21. Explain why the markc•t maker’s bfJ-offer pr€od repre.sent.s a real cost to options investor.s. A “fair” price for the option can reasonably be assumed to be half way betw'een the bid and the offer price quoted by a market maker. An investor typically buys at the market maker’s offer and sells at the market maker’s bid. Each timc he or she does this there is a hiddcn cost equal to half the bid-offer spread. Problem 9.22. A United States inve.stor writes five nakeJ call o¡›tion contracts. The option price i.s .S3.50, the strike price is $6a.00, and the stock price is $57.00. W/taf i.s the initial margin requirement? The t» o calculations are necessary to determine the initial margin. Thc first gives 500 x (3.5 -F0.2 x 57 — 3) = 5, 950 The second gives 500 x (3.5 * 0.1 x 5 7) = 4, 600 The initial margin is the greater of these, or $5,950. Part of this can be provided by the initial amount of 500 x 3.5 = $1, 750 received for the options.

64 r(.2012 Pearson Education

CHAPTER 10 Properties of Stock Options Problem 10.1. List the six]actais afje‹ tiring stock oj›tion yrices. The six factors aflccting stock option prices are the stock price, strike price, risk-free interest rate, volatility, time to maturity, and dividends. Problem 10.2. What i.s a low er bound Jor the prim e oja four-month call option on a non-JiviJenJ-paving stock n'hen the stock firice is $28, the str ike price i.s $25, and the risk-]re.e interest rate is 8°Xa per annum? The low'er bound is 28 — 25e ”'“ ”'

— $3.66

Problem 10.3. What i.s a lower hounJ for the pri‹:e ofa one-month European put option on a non-JiviJenHpay’ing stoc.k when the stock pric'e is $12, the .strike price i,s $15, and ltte risk-free interest rate is 6’
65 '20 12 Pearson Education

An Amcrican call option can be exercised at any time. If it is cxcrcised its holder gets the intrinsic value. It follows that an American call option must be worth at least its intrinsic value. A Europcan call option can be worth less than its intrinsic value. Consider, for example. the situation where a stock is expected to provide a very high div idend during the life of an option. The price o1 the stock will decline as a result of the dividend. Bccause the European optitin can be exerciscd only after the dividend has been paid. its value may bc lcss than the intrinsic value today. Problem 10.7. The price of a non-diviJend saying .stock i.s 57 P anJ the price of a //iree•-monfh F.uropcan call option on the stock with a strike price of 820 is SI. The risk-li-ee rale is 4’X per annum. iffin/ i.s the price of u lhree-month Euro mean yiof oJ›tion with a strike pri‹:e o] $20’? In this case, c = 1, T — — 0.25 , .Y, = 19. K — 20, and i = 0.04 . From putwall parity — c + Ke —' S„ p = 1 -F 20c “‘ so that the European put price is S1.50.

— ‘ ’ 19 = 1.80

Problem 10.8. Explain why the arguments leaJing fci yu/—cal/yari/J/or F.uro mean option.s cannot be used to giv'e u similar re.sult [or Ameri‹’an options. When early exercise is not possible, wc can argue that two portfolios that are worth the same at time T must be worth the same at earlier timcs. When early exercise is possible, the argument falls down. Suppose that P S > C + Ke "" . This situation does not lead to an arbitragc opportunity. II we buy the call, short the put, and short the stock, we cannot be sure of the result becausc wc do not know when the put will be exercised. Problem 10.9. Whul is a lower hound for flue pri‹’e of a si.x-moi1lh call option on a non-dividend-paying stow k when thc stock price i.s .S80, the .y/rike price is $75, anJ the risk-free interest rate is fi% per annum’? The lower bound is Problem 10.10 What is a lo w'er bound for- the price ma tw’o-month Europ‹•an yiil oplion on a non-dividendpaying s/ocâ w'hen the stock price i.s S58, the strike price i3’$65, and the risk-frec interest rate i.y 5%i› yer annum.” The lower bound is 65e

” " — 58 = $6.46

02012 Pearson fiduco tion

Problem 10.11. A]oiir-month European call option on a JiviJenJ-paving stock is currently .selling for $5. The stow:k prim e is $64, the strike price is $60, and a dividend off 0.80 i.s expected in one month. The risk-free interest rate is 12%i per annum for all n1aturitie.s. Mat opportunitie.s are there for an arbitrageur? The present value of the stri ke price is 60c ' ' 2“ ” = $57.65 . The present value of the dividend is 0.S0e ' ' z" = 0.79 . Because 5 < 6—4 57.6—5 0.79 the condition in equation (I 0.8) is violated. An arbitrageur should buy the option and short the stock. This generates 64 — 5 = $59 . The arbitrageur invests SO.79 of this at 12% for one month to pay the dividend of $0.50 in onc month. Thc remalning 558.21 is invested for four months at 12%. Regardless of what happens a profit will materialize. lf the stock price declines bclow $60 in four months, the arbitrageur loses the $5 spent on the option but gains on the short position. The arbitraSe rir shorts when the stock price is $64, has to pay dividends with a prcscnt value of $0.79, and closes out the short position 'hen the stock pricc is $60 or less. Because $57.65 is the prescnt value of $ti0, the short position generates at least 6—4 57.C5 0.79 = $5,56 in present value terms. The present value of the arbitrageur’s gain is thcrcf’orc at least 5.56 — 5.00 $0.56 . If the stock price is abtive $60 at the expiration of the option, the option is exercised. The arbitrageur buys the stock for $60 in four months and closes out the short position. The present value of the $60 paid for the stock is $57.65 and as before the dividend has a present value of $0.79. The gain from the short position and the excrcisc of the option is thcrcforc exactly 6—4 57.65 0.79 = S5.56 . The arbitrrtgeur’s gain in present value terms is exactly 5.S—65.00 = $0.55 . Problem 10.12. A one-month European put option on a non-dividend-pawing stock i,s currently selling for $2.50 . The .yiociyrice i.s 547, the .strike yt-ice is $50, and the risk-free intere•st rate is 6"No per annum. What opportunities are there for an arbitrageur ? In this case the present value of the strike price is 50c“’"" " = 49.75 . Because 2.5 < 49.75 47.00 the condition in equation (I 0.5) is violated. An arbitrageur should borrow $49.50 at ti% for one month, buy the stock, and buy the put option. This generates a profit in al1 circumstances. If the stock price is abtw e $50 in one month. the option expires worthless, but the stock can be sold for at least S50. A sum of 550 received in one month has a present value of $49.75 today. The strategy therefore generates protit with a present value ot at least $0.25. If the stock price is below' 550 in one month the put option is exercised and the stock owned is sold for exactly $50 (or S49.75 in present value terms). The trading stratcgy therefore generates a profit of exactly $0.25 in present value terms. Problem 10.13. Gii’e an intuiti› e e. plancition of wliv the early exercise o[an American put hecome.s more attra‹’tive as the i isk-(ree rate increa.se.s and volatiliu decrea.se,s. Thc early exercise of an American put is attractive when the interest earned on the strike price is greater than the insurance element lost. W hen interest rates increase. the value of the interest earned on the strike price increases making early exercise more atuactive. When

volatility decreases. the insurance element is less valuablc. Again this makes early exercise more attractive. Problem 10.14. The price of a European en// lhal expires in six months and hos u strike price of $3a is $2. The unJerlying stock pri‹'e is 529, unJ a divide+iJ oJ $0.50 is c. pected in tw'o months unJ again in five months. The term structure is flat, w'ith all risk-free interns t rates being 10%. What i.s the price ofa European yut option that expires in six n7OH/h.s and has a strike price of Using the notation in the chapter. put-call parity [equation (10.10)] gives c * Ke '!' D — /› * St p— —c

Ke ' + D— „S

In this case 2 30e '' ” (0.5 e ° '” " + 0.5e '“ )— 29 = 2.5 l In other w ords the put price is $2.51. Problem 10.15. E.xpl‹iin the arbitrage opportunitics in Problem 10. 14 i(tlie Eurof›ean put price i.s 53. If the put price is $3.00, it is too high relative to the call price. An arbitrageur should buy the call, short the put and short the stock. This generates —2 3 29 $30 in cash z hich is inv cstcd at 10@ ». Regardless of what present 3—1 happens 30 —4 a profit 3—1 u’ith 30ca “°’ ” value of 3 . 0 — 0 2.51 = $0.49 is locked in. or If’ thc stock price is above 530 in six months, 1.00 the 4.00call —option 1 1.59is exercised, and the put option expircs orthless. The call option enables the stock to be bought for $30, or ?"^2012 Pearson Education 30c*’ "" ' = $28.54 in present value terms. The dividends on the short position cost

2, 41 < P < 3.00

Upper and lower bounds for the p icc ot an American put are therefore $2.41 and $3.00. Problem 10.17. Explain carefully the arbitrage opportunitie.s in Problem 10. 1 d if the American put price is greater than the calculated ur Fer houHd. If the American put price is greater than $3.Elf) an arbitrageur can sell the American put, short the stock, and buy the American call. This realizes at least 3 31 — 4 = $30 which can be invested at the risk-lrcc intcrest rate. At some stage during the 3-month period either the American put or the American call will be cxercised. The arbitrageur then pays $30, receives the stock and closes out the shtirt position. The cash flows to the arbitrageur are 30 at time zero and —$30 at somc fiiture time. These cash flo s have a positive present value. Problem I 0.18. Pro›’e the result in equation (I ñ.7). (Hint.- For- lh‹• Jirsl parl oj the relationship consiJer (a) a 1›ort[olio const.sting o[a F.urope•an call plus an amotinl of ‹ a›'h equal to x anal tb) a portfolio ronsi 'ting oJ on American put option plus one share.) As in thc text we use c and , to denote the European call and put option price, and C and r to denote the American call and put option prices. Because P _ p , it follows from put—call parity that P> r Ke '' —„S

and since ‹

i? , P

C + Ke “ „S

OF

For a further relationship between C? and p , consider Portfolio 1: One European call option plus an amount o1 cash cqual to x . Portfolio I: One American put option plus onc share. Hoth options have the same exercise price and cxpiration datc. Assume that the cash in portfolio I is invested at the risk-trcc interest rate. If the put option is not exercised early portfolio J is worth max (S , K) at time T . Portfolio I is worth max (S—, K)— K

A, O) Ke“ — — max ( S , Ke“

at this time. Portfoliti I is therefore worth more than portfolio J. Suppose next that the put option in portfolio J is exercised early, say, at time r . This means that porttolio J is worth K at timc r . However, even if the call option were worthless, portftiliti I would be worth Ke"' at time r . It follows that portfolio I is v orth at least as much as porttolio J in all circumstanccs. Hencc Since c = C ,

69 2012 Pearson Education

Combining this with the other inequality derived above for —€’ P , +'e obtain .St — Y < G — P < 5’t — Ke " Problem 10.19. Prove the result in equation (10.11). (Hint.' For ltte first part of the re/ofions/rig c'ons iHet‘ la) a portfolio ‹ onsisting of a European call plu.s an amount o] cash eq,iri/ /o la x OnJ (b) a portfolio con.listing o/ an American put option plus one shame.) As in the text v’e use c and to denote the Europcan call and put option price, and C’and to denote the American call and put option prices. The present waliie of the dividends wi11 be denoted by D . As shown in the answer to Problem 10.18, w hen thcre are no dividends 6’ — I S— Ke°'' Dividends reduce C and increase r . Hence this rclationship must also be true when there are dividends. For a further relationship between C and P . consider Portfolio I: one European call option plus an amount of cash equal to D • K Portfolio J: one American put option plus one share Both options have the same exercise price and expiration date. Assuine that the cash in portfolio I is invcstcd at the risk-free interest rate. If the put option is not exercised early, portfolio J is worth max ( Yt , K) D e"

at time T . Portfolio 1 is worth inn y (S — K, 0) + (D + K)e' = max (S , K)+ be' Ae"—' A at this time. Portfolio I is therefore worth more than portfolio J. Suppose next that the put option in portfolio I is exercised early, say, at time r . This means that portfolio J is worth at most KDe” at time r . How’ever, even if the call option were worthlcss, porttol io I would be u’orth (I + K)e' at time r . It follows that portfolio I is worth more lhan portfolio J in alI circumstances. Hence c E K> P +S Because C c

Problem 10.20. ConsiJer a jive-year c'all option on a non-JiviJenJ-pming .s/ocJ granted to emy›/uyees. The option can be exerciseJ at any time after the end o(the Qtr.st year. Unlike a regular e.TC/irinQ6'traJed ‹’all option, the employee stock oj›tion cannot he .sold. ti’/in/ is the likc•lv impac.I o] this An employee stock option may be exercised early because the employee needs cash or because he or she is uncertain about the company’s future prospects. Regular call options can bc sold in the market in either of these o situations, but 70employcc stock options cannot bc sold. In theory an employee can short the company’s stock Education as an altcrnati› e to excrcising. In practicc ‹ 2012 Pearson this is not usually encouraged and may e›'en bc illcgal tor senior managers.

The graphs can be produced from the first worksheet in DerivaGem. Select Equity as the Underlying Type. Select Black-Scholes as the Option Type. Input stock price as 50, volatility as 30%, risk-free rate as 5•Z , time to exercise as 1 year, and exercise price as 50. Leave the dividend table blank because we are assuming no dividends. Select the button corresponding to call. Do not select the implied volatility button. Hit the Enter key and click on calculate. DerivaGem will show the price of‘ the option as 7.15562248. Move to the Graph Results on the right hand side of the worksheet. Enter Option Price for the vertical axis and Asset price for the horizontal axis. Choose the minimum strike price value as 10 (software will not accept 0) and the maximum strike price value as 100. Hit Enter and click on Draw' Graph. This will produce Figure 10.la. Figures 10.1 c, 10.1e, 10.2a, and IU.2c can be produccd similarly by changing the horizontal axis. By selecting put instead of call and recalculating the rest of the figures can be produced. You are encouraged to experiment with this worksheet. Try different parameter values and different types of options.

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CHAPTER 11 Trading Strategies Involving Options Problem 11 .1. What is meant hy a protective put!" What position in call options is equivalent to a protective A protective put consists of a long position in a put option combined with a long position in the underlying shares. It is equivalent to a long position in a call option plus a certain amount of cash. This follows from putoall parity: p*S — — c + Ke ” D

Problem 11.2. Explain lwo ways in which a bear spre•ad can be cre•ate•d. A bear sprcad can be created using two call options with the samc maturity and diftcrcnt strike prices. The investor shorts thc call option with thc lower strikc price and buys thc call option with the higher strike price. A bcar spread can also bc crcatcd using two put options with the samc maturity and dillcrcnt strikc prices. In this casc, the investor shorts the put option with the lower strikc price and buys the put option with the higher strike price. Problem 11.3. When i.s i/ appropriate/or an inve.stor In purchase a butterfly .spread? A butterfly spread involves a position in options with three different strike prices (ñ„ K., and K, ). A butterfly spread should be purchased when thc investor considers that the price of the underlying stock is likely to stay close to the central strike price, Hz . Problem 11.4. Call options on a stock ore availablc w'ith strike prices o] $15, S17 -, and $20 and expiration dates in three month.s. Their prices are $4, $2, and $ — , ' , re.effectively. E. plain hon' the options ca» be useJ to create a butterfly spreaJ. Construct a iable show'ing how pro]it varies w'ith stock price%r the bulterfiv spreuJ. An investor can createStock a butterfly Price.spread Sz by buying Profit call options with strike prices of $15 and S < 15 $20 and selling two call options with strike prices of15) $17— — ' -. The initial investment is (S,. — 4 +— 2 x 2 = $ . The following table shows the o1- profit with the final stock (20variation — Sz) — price:

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Problem 11.5. What trading strategy create.s a rever.‹e calendar spread'^ A rcvcrse calendar spread is created by buying a short-maturity option and selling a longmaturity option, both with the same strike price. Problem 11.6. What is the difference betn›een a strangle anJ a straJJle? Both a straddle and a strangle are crcated by combining a long position in a call with a long position in a put. I n a straddle the tw o have the same strike price and expiration date. In a strangle they have different strike prices and the same expiration date. Problem 11.7. A cull opticn with a strike J -ice o($5 H co,sts $2. A put option viih a slike price o/ 545 costs $3. Explain how a strangle can he created from these tw'o oy//Oo6. Ref i› the pattern ol profits from the strangle? A strangle is created by buying both options. The pattern of profits is as follows: Stock Price. Si

Profit

(45 — N —) 5

Problem 11.8. Use put—‹’all pari5 to relate the initial investment for a bull spre•uJ ct-raleJ using t «//s to the inifia/ investment for a hull .spread created using puts. A bull spread using calls provides a profit pattern with the same general shape as a bull spread using puts (see Figures 11.2 and 11.3 in the text). Define J›, and c us the prices of put and call with strike price K, a14d p, And c, as the prices o1 a put and call with strike price c 2. Front put-call parity p S— — c, + K e "'

Hence: This shows that the initial investment when the spread is created froth puts is less than the initial invcstmcnt when it is created frtim cal Is by an aiiiount K, — K,)‹ ‘ . In fact as mentioned in the text the initial investment when the bull spread is created from puts is negative, while the initial investment when it is created from calls is positive. investment of (K—calls . K,)e ” overtothe put strategy. This earns interest The profit when are used create thc bull spread is higher thanofwhen puts are used by (K z — K,)(1 — e ") . This rcllects the fact that the call strategy irn olves an additional risk73 free '2012 Pearson Education

(K—,

K )e '’ (e"—’ I) = (K — K, )(1 —

'!’) .

Problem 11.9. Explain hey an aggres.five hear spread can he created using put options. An aggressi e bull sprcad using call options is discussed in the text. Both o1 thc options used have relatively high strike prices. Similarly, an aggressive bear sprcad can be created using put options. Both of the options should be out of the money (that is, they should have relatively low strike prices). The spread then costs very littlc to set up because both of the puts are worth close to zero. In most circumstances thc spread will provide zero payoff. However, there is a small chancc that the stock price will fall fast so that on expiration both options will be in the moncy. Thc spread then provides a payoff equal tti the difference between the twti strike prices, N—, K,. Problem 11.10. Suppose their put options on a stow k with strike pt ices $30 and $35 cG.sf $4 and S7, res pectively. How c’an the options lie useéf fo create (a) a hull .spread and (h) a hear spread? Construct a table that show’s the profit and j›ayof/ for hoth .spreads. A bull spread is created by buying the $30 put and selling the $35 put. This strategy gives rise to an initial cash inflow of S3. The outcome is as ft›1lows: Stock Price N, 35

30

5,—33

S, < 35

S, — 32 —2

A bear spread is created by selling the $30 put and buying the $35 put. This strategy costs $3 initially. The outcome is as ftillows: Stock Prire

2 Problem 11.11. U.se put—call parih to .shot ’that the cost of a biitterfly spread createJ jrom European puts is identical to the co.st ofa l›utterflv’.‹f:u-eaJ creafed om Eiiropr‹ui calls. Define c, , c, , and c, as the prices of calls z'ith strike prices K, , H z and K. . Ref ne y , p, and p as the prices of puts vith strike prices A, , K and K, . With the usual notation c, + K e '’ — — p, S c, + K e '' — — +

Hence

c, — * p, Because K,, — K, — — K, — K, , it follows that K,

c

2c, + (K, K, — 2A, )e "' = p, * p K — 2x,. — — 0 and

c, c— c — — p, p, — 2pz The cost of a butterfly spread created using European calls is therefore exactly the same as the cost of a butterfly spread created using European puts. Problem 11.12. A call with a strike price o/ $60 co.sfs $5. A put Smith the same strike pric'e and expiration date costs $4. Construct a table that shows the profit from o .straddle. For what range ofstock prices would lhe straddle leaJ to a loss'? A straddle is created by buying both the call and the put. This strategy costs $10. The T'royit Pricein the following table: Pav‹›ff profit/lossStock is shown S, > 60 S—z 60 S, 60 60 — S, This shows that the straddle will lead to a loss if the final stock price is between $50 and $70. Problem 11.13. Construct a tahle .showing the payoff from a bull spreaJ›+'hen puts w'ith strike prices K, anJ K, are used (K_ > K, ) . The bull spread is created by buying a put with strike price K, and selling a put with strike price K- . The payoff is calculated as follows: Pavofl from Long Ni‹J

Payoff from Short Put

0

0

Problem 11.14. An inve›'tor believes that there w'ill be a hig jump in a .stock price, but is unc'ertain as to the direction. Identifv s‘r\ d ‘erent strategies the investor can follow and explain the Jifferences among them. Possible strategies arc: Strangle Straddle Strip Strap Reverse calendar spread Reverse butterfly spread

75 fi2012 Pearson Education

The strategies all prtivide positive profits when there are large stock price moves. A strangle is less expensive than a straddle, but requires a bigger move in the stock price in order to provide a positive profit. Strips and straps arc more expensive than straddles but provide bigger profts in certain circunistanccs. A strip will provide a bigger profit when there is a large downward stock pricc move. A strap w ill provide a bigger profit when there is a large upward stock price movc. In the case of strangles, straddles, strips and straps, the profit increases as the sizc of the stock price movemcnt increases. By contrast in a revcrsc calendar spread and a reverse butterfly spread there is a maximum potential protit regardless of the size of the stock price movement. Problem 11.15. how can a [orward conti-act on a stock w'ilh a particular deliver y price and delivery date lie createJ from options.! Suppose that the delii’ery price is K and the dclivery date is T . The ftirward contract is created by buying a European call and selling a European put when both options have strike price K and exercise date r . This portfolio provides a payoff of S, — K undcr all circumstances where S is the stock price at time T . Suppose that I, is the forward price. If K — F , the forward contract that is created has zero valuc. This show s that the price of a call equals the price of a put u hen the strike price is . Problem 11.16. “A be. spread comprises four options. Two ‹’an be combined to create a long [orst ard posilion anJ Iii’o ‹ an be ‹ oinbined to create a short %rsvard j›o.Simon. ” Explain this A box sprcad is a bull spread created using calls and a bear spread created using puts. With the ntitatitin in the text it consists of a) a long call with strike K, , b) a short call with strike K, , c) a long put ith strike V. , and d) a slitirt put w'ith strike K, . a) and dJ give a long foovard contract v ith delivery price K, .' b) and c) givc a short forward contract with delivery price Kg . The to o foru-ard contracts taken together give the payoff orñ —2 K, . Problem 11.17. What is the restilt if tlte slrihe price of the put is higher than the .strikeVarice of the call in a sli-ungle.’" The result is shown in Figure S11.1. The profit pattern from a long position in a call and a put when the put has a highcr strike price than a caII is much the same as when the call has a higher strike pricc than the put. Both the initial investment and the tinal payoff are much highcr in thc first case

76 2012 Pearson liducation

Figure Sl 1.1

Profit Pattern in Problem 11 .17

Problem 11.18. One Australian dollar is currently »or!h $0.64. A one-year bullcrfly spread is set up using European ‹’all options with strike prices o}$0.60, S0.65, cuff $0.70. The risk-fee interest rates in the UniteJ States unJ Australia are 5%› anJ 4% respectively, and the volatiliiy of the exchange rate i› 75°o. Use lhe DerivuGem software fo calculate the cost oJ'seffing up the butterfly spreaJ position. Show that the cost is the same i fEuropean put options are use‹1 instead ofEuropean call options. To use DerivaGem select the first worksheet and choose Currency as the Underlying Type. Select Black--Scholes European as the Option Type. Input exchange rate as 0.64, volatility as 15%, risk-free rate as 5%, foreign risk-frcc interest rate as 4%, time to exercise as 1 year, and exercise price ds 0.60. Select the button corresponding to call. Do not select the implied volatility button. Hit the Enter key and click on calculate. DerivaGem will show the price of the option as 0.061 S. Change the exercise price to 0.65, hit Enter, and click on calculate again. DerivaGem will show the value of the option as 0.0352. Change the exercise price to 0.70, hit Enter, and click on calculate. DerivaGem will show the value of the option as 0.01S1. Now select the button corresponding to put and repeat the procedure. DerivaGem shows the values of puts with strike prices 0.60, 0.65, and 0.70 to be 0.0176, 0.0356, and 0.0690, respectively. The cost of setting up the butterfly spread when calls are used is therefore 0.061 8 + 0.01 fi 1 — 2 x 0.0352 = 0.0095 The cost of setting up the butterfly spread when puts are used is 0.01 76 + 0.0690 — 2x 0.0386 — 0.0094 Allowing for rounding errors these two are the same. Problem 11.19 An index provides a dividend vield of I°Xo and has a volatility of2F’Xo. the risk-free interest rate is 4%. How long does a principal-protected note, created as in Example 11. I, have to last for ii to be profitable to the bank.! Use DerivaGem. Assume that the investment in the index is initially $100. (This is a scaling factor that makes no diff’erence to thc result.) DerivaGem can be used to value an option on the index with the index level equal to 100, the volatility equal to 20%», the risk-free rate equal to 4%, the 77 2012 Pearson Education

dividend yield equal to 1%, and the exercise price equal to 100. For different times to maturity, T, we value a call option (using Black-Scholes European) and the amount available to buy the call option, z'hich is 100-1 00e '““’. Results are as follows: Time to maturitv, T 2 10

1i

5

Funds Available 3.92 7.69 15.13

Value of Option 9.32 13.79 23.40

32.97 35.60

33.34 34.91

This table shows that the answer is between 10 and 11 years. Continuing the calculations we find that if the life of the principal-protected note is 10.35 year or more, it is profitable for the bank. (Excel’s Solvcr can be used in conjunction with the DerivaGem functions to facilitate calculations.)

75 2012 Pearson Education

CHAPTER 12 Binomial Trees Problem 12.1. A stock j›rice is currently $40. lt is knoii’n flow or the enJ of one month it n’ill he either $42 or $38. The ri.wk-free intere.st i-ate is b’Xo per annum with t’on/inuou.s coinj›ounding. What is the value ofa one-month European call option w'ii/f a Strike price of $39." Consider a portfolio consisting o1 —1 : Call option A. Shares If the stock price rises to $42, the ptirtfolio is z'orth 42A $38, it is worth 38A . These are the same when

3 . II the stock price falls to

42A = 38A

3

or k — — 0.75 . The value o1 the portfolio in tone month is 28.5 for both stock prices. Its value today must be the present value of 28.5. or 25.5e ” " ""' 25.31 . This means that — / -F 40A 28.31 where f is the call price. Becausc a = 0.75 , the call price is 40 x 0.7—5 28.31 = $1.69 . As an alternative approach, e can calculate the prtibability, ,of an up movement in a risk-neutral world. This must satisfy: 42/a 38(1 — y›) = 40c° "" '^ " so that 4 p = 4be’""“" "’'—' 38 or = 0.50f›9 . The value of the option is then its expected payoff discounted at the riskfree rate: [3 x 0.5669 0 x 0.433 l]e ”" ”’’' — — I .69 or $1.69. This agrees with the previous calculation. Problem I 2.2. Exfi'lain tlte no-urhitraye anti I-ink-neutr‹iI aliintiun nyproacl1e to valuing a Lui’nyean option using a one-s'Iey binutniul tree. In the no-arbitrage approach, c set op a riskless portfolio consisting ot a position in the option and a position in the stock. By setting the retuni on the portfolio equal to the risk-free interest rate. we are able to value the option. Whcn we use risk-neutral valuation, we iirst choosc probabilities for the branches of the tree so that the expected return on thc stock equals thc risk-tiec interest rate. We then value the option by calculating its cxpccted payoft and discounting this expected paytiff at the risk-free interest rate. Problem 12.3. What i.s meant h)' the delta of a stock oplion? The delta of a stock option measures the scusitivity ot the option price to the price of the

Problem 12.4. A slow L pric’e is ‹’urrently $50. It is known that at the end o/.six months it will he either $45 or 555. Tye risk-free in/ere.sr rate i.s 76% Jeer annum with continuous rompounJing. What is the value of a .yi.v-mould European put motion n'itli a .str-ike prire oJ $50." C’tinsider a portfoliti c‹insistiiig of —1 : Put option A: Shares If the stock price rises to $ñ5, this is worth 55a . lt' the stock price falls to $4J, the portfolio is worth 45 A 5 . These are the same when 4.5 A— 5 = S 5A

or ñ = —0.50 . The value o1 the ponfolio in six months is —27.5 for both stock prices. Its value today must be the present value of —27.5 , or —2 7.5e '“ " = —26.16 . This means that + 50a = —2G. l G where / is the put price. Because A= —0.50 , the put price is $1.16. As an alternati›'c approach » e can calculate the probability, p , or an up movement in a risk-neutral orld. This must satisfy: 55a * 45(—1 y) = 50e’ "’ ‘ so that ION = 50e" “’ — 45 or p — — 0.7564 . The ›-a1ue of the option is then its expected payoft discounted at the risk-free i ate: [0 x 0.7564 5 x 0.2436]e ’ ' ' = 1.1 6 or $1.16. This agrees witli the previous calculation. Problem 12.5. A .stock price i.s currently $ 100. O ct- each of the next two six-monlh perioJs it is expec'teJ to go up hy I O“o or do›v n hY / f4‘o. The I ink-fI‘€€ intere•s I ratr is 8% per cuinum with continuous compounding. What i.s the values o[a one-year Europe‹in call option with a strike prim e o[ In this case u = 1.10, d — — 0.90 . Al = 0.5 , and r = 0.0s . so that e’"‘" 0.90 = 0.7041 I .10 — 0.90 The tree for stock price movements is shown in Figure S12.1. We can work back from the end tif the trce to the beginning, as indicated in the diagram, to give the valuc o1‘thc option as S9.61. The option value can also be calculated directly from equation (12.10): [0.7041" x 21* 2 x 0.7041 x 0.2959 x 0 + 0.2959' x 0]e°2" "" ‘ = 9.61 or S9.61 .

20.12 Pearson Education

110

21 21 99

61 81

Figure S12.1

Tree for Problem 12.5

Problem 12.6. For the situation considered in Problem 12.5, n›liat is the »alue ofa one-year European put opiion wiili a str-ike price of $100'! Verf that the European call and Liiropea• r• Frices Figure S12.2 shows how we can value the put option using the same tree as i n Problem 1 2.5. The value o1 thc option is $1.92. The option value can also be calculated directly front equation (12.10): e 2" "‘° ' [0.7041“ x 0 + 2 0.7041 x 0.2959 x 1 + 0.2959“ x 19] = 1.92 or $1.92. The stock price plus the put price is 100 1.92 = $101.92 . The present value of the strike price plus the call price is 100c 0"' 9.61 S1 01.92 . These are the same, vcrifying that put—call parity holds.

21 110

99

81 19 Figure S12.2

Tree for Problem 12.6

Problem 12.7. What are the for-mulas for u and d in term.s of volatility?

Sl fi'201 2 Pear.son Education

Problem 12.8. Consider the situation in u hich stock firice i ovemeni,s during the life of a European option are governeJ hy a two-step binomial lrre. Explain w'h ’ ff iS I'lOt fiOB'Sible• to sent up a position in the stock anJ the opiion that rem‹i/es ri 'kless for the trhole of the life of lhe option. The riskless portfolio consists of a short position in the option and a long position in n shares. Because changes during the lilc of the option, this riskless portfolio must also change. Problem 12.9. A stock price i.s currently ,SS fi. It i.s knon'n that at the end o/ two months it will be either $53 or $4b. The risk-free interest rate is 1O’fo yer annum with continuou.f compounding. What is the value of a live-month European call option with a strikeprice of $49’? Use no-arbitrage arguments. At the end of two months the value of the option will be either $4 (if the stock price is $53) or $0 (if the stock price is $48). Consider a portfolio consisting of: +A : shares —1 option

:

The value of the portfolio is either 4sA or 53a — 4 in two months. II 48A = 53A 4

i.e., where

is the valuc of thc option. Since the portfolio must earn the risk-free rate of interest h = 0.8 (0.8 x 50 — / ) e ’ " ' " = 38.4 the value of the portfolio is certain to be 38.4. For this value ofw the portfolio is i.e., therefore riskless. The current value of the portfolio is: = 2.23 0.8 x 50 — f The value of the option is therefore $2.23. This can also be calculated directly from equations (12.2) and (12.3). u = 1.06 , d = 0.96 so that N'

and

0 2 e0 1 ’'," — 096

= 0.5681 1.0fi — 0.96 / = ego " x 0,5681 x 4 = 2.23

Problem 12.10. A stock price is currentl y .$80. It i.s /mown that at the end of four months it will be either $75 or $65. The risk-free intere.st rate i.s 5%‹› per annum with continuous compounJing. What is the value ofa four-month European j›tit option with a strike price of $80? Use no-arbitrage arguments. At the end of four months the value of the option wit 1 be either $5 (if the stock price is $75) or $0 (if the stock price is SS5). Consider a portfolio consisting of: S2 tfi/2O 1 2 Pearson Education

— A

shares

option + (Note: The delta, of a put option is negative. We have constructed the portfolio so that it is 1 and +a shares so that the initial investment is * I option and shares rather than — i option positive.) The value of the portfolio is either —85A or —75A 5 in four months. If —85A = —75 A * 5

i.e., A = —0.5

the value of the portfolio is certain to bc 42.5. For this value of a the portfolio is therefore riskless. The current value or the portfolio is: 0.5 x 80 * ,/ where is the value of the option. Since the portfolio is riskless (0.5 80 f)r’”“’ " = 42.5 i.e., = 1.80 The x alue of the option is thereiore $1.80. This can also be calculated directly from equations (I 2.2) an‹j (12.3). u = 1 .0G25 , I = 0.9375 so that e° ‘ 4 ' — 0.9375 ' 1.062 (1.9375 = 0.6345 —1 t› = 0.3655 and —5 = e '"'“‘ " x 0.3655 x 5 = 1.80

Problem 12.11. A s rock price is ‹’urrently 540. It i.y knon›n that at the end of thi ee monlhs it »ill be either $45 or 535. The risk-See rate of intere.at north quarterly coinpoun‹ling is 8% per annum. Calculate thr value ofa three-month European y›«/ option on the stock wllh an exerc'ise pri‹’e of $40.

Veri ff' that no-arbitrage arguments and risk-neutral valuation arguments give the same ans 'ers. At the end of three months the value of the option is cithcr $5 (if the stock price is $35) or $0 (if the stock price is $45). Consider a portfolio consisting of: -A shares

:

+1 option (Note: The delta,w , o1 a put option is negative. We have constructed the portfolio so that it is + I option and —n sharcs rather than —i option and z shares so that the initial investment is positive.) Thc value of the portfolio is either —35A 5 or —45a . 11: —3 S A * 5 = —45A

the value of the portfolio is certain to be 22.5. For this value of a thc portfolio is therefore riskless. The current value of the portfolio is —40A+ / wherc / is the value of the option. Since the portfolio must carn the risk-free rate of intcrest 83 '°D20l2 Pearson Education

(40 x 0.5 * /) x 1.02 = 22.5 Hence

/ — 2.06

i.e., the value of the option is $2.06. This can also be calculatcd using risk-neutral valuation. Suppose that y› is the probability of an upward stock price movcment in a risk-neutral world. We must ha e 45p * 35(l — ) = 40 x 1 .02 10p — — 5.S = 0.5S The expected value oi’ the option in a risk-neutral world is: 0 x 0.5S * 5 x 0.42 = 2.10 This has a present value of 2.10 2 06 1.02 This is consistent w ith the no-arbitrage answer. Problem 12.12. A stow k price is ct rrentl ‹SS 0. Over each of the next tw'o lliree•-month period.s it is expected to go up by 6% ct- Jo tJ tt hi' 5‘Xo. The ri.sk-free interest rate is 5% per annum with continuou.s compounJing. What is the alue of a .six-month European c'all option wiih a strike price of A tree describing the behavior of the stock price is shown in Figure S12,3. The risk-neutral probability of an up move, p , is given by — ‘’" ' ‘' —0 1.0—6 5689 0.95 There is a payoff from the optitin of 5€a.l 8 — 51 5.1s for the highest final node (which corresponds to two up moves) zero in all other cases. The value of the option is therefore 5.1S x 0.5659’ e "" ’' = 1.635 This can also be calculated by working back through the tree as indicated in Figure S1 2.3. The value o1 the call option is the low er number at each node in the figure.

84 'c\20 1 2 Pearson Education

53 50 1.635

2.910

56.18 5.18 50.35

47.5 45.125

Figure S12.3 Tree for Problem 1 2.12 Problem 12.13. For the situation considered rn Prol›lem 12.12, n’liat is the value oJ a st. -monlh European put option with a strike price of $51? Serf that the European rall anJ Eur oyean pul pt-ices .coli.s 'yii/—cal/ynriq.. If the put option ii ere American, ii oulJ it ever be optimal to exercise it early’ at any o[the nodes on the n-ee."' The tree for valuing the put option is shown in Figure S12.4. We gct a payoff of 51 — 50.35 = 0.6i if the middle final node is reached and a payoff of 5 1 — 45. 125 = 5.87 if the lowest final node is reached. The value of the option is therefore (0.65 x 2 x 0.56S9 x 0.4311 5.575 0.4311“)e * — 1.376 This can also be calculated by working back through the tree as indicated in Figure S I 2.4. The value of the put plus the stock price is 1 .376 * 50 = 51.376 The value of the call plus the present value of the strike price is l .635 + 5 le 0 °’ " = 5 1.3 7G

This verifies that put—call parity holds To test whether it worth exercising the option early e comparc thc ›'a1ue calculatcd tor thc option at each node v ith the payoff from immediate exercisc. At nodc C the payo1”1 trom immediate exercise is 5 1 — 47 5 = 3.5 . Because this is greater than 2.8664. the option should be exercised at this node. The option should not be exercised at either node A or code D. 56.18

53 1.376

.277

47.5

50.35 0.65

2.86 45.125 5.875 F i g u r e

Problem 12.14. A stock J›rice is ‹’urrently S2J. It is know'n that at the end of tv'o month.s it will be either 523 or $2.7. The ri.sk-See interest i’ate iS 10‘Xo per annum w'ith continuous compounding. Suppose Sz is ltte. stock price at the end of two months. What is the ve/tie oJ a deri vati ve that pays off

At the end o1 two months the value of the derii ati› e will be eithcr 529 (i1 the stock price is 23) or 729 (it‘the stock price is 27). Consider a portfolio consisting of: shares A deri›’ative — The value of the portfolio is cither 2.7 1n 729 or 23—a 529 in tw'o months. If 27A — 729 = 23A — 5 2"? i.c., the value of the portfolio is certain to be 621. For this i alue tif n the portfolio is therefore riskless. The current › alue of’ the portfolio is: 50 x 25 — ./ where / is the value ot thc derivative. Since the portfolio must earn the risk-lrcc rate o1 interest

(50 x 25 — ( )e ' '““ '" = 621 i.e., / = 639.3 The value of the tiption is therefore $639.3. this can also be calculated directly from equations (12.2) and ( 12.3). u = 1.08, d — — 0.92 so e" " " 0.92 = 0.6050 1.08 — 0.92

and

/ = e ' ” '" (0.6050 729 + 0.3950 x 529) = 639.3 Problem 12.15. Calculate u , d, and p when a binomial ti-ee is constructed to value an option on a %reign ‹ urrcnc . The free .Cley› .yize i.s one month. the domestic interest rates iS 5“o per annum, the foreign intere•st rate i.s 8’o yer annum, ancl the volatility is 12% pet- oontirn. In this case

o = e’ ”’"" = 1.0352 0.9975 — 0.4553 0.9660 1.0352 — 0.9660 S6 20.1 ? Pearson Educalion

CHAPTER 13

Wiener Processes and Itô’s Lemma Problem 13.1. Mat would it mean to assert that the temperature at a certain place follow.s a Markov process? Do you think thai temperatures do, in [act, %llow a Markov process? Imagine that you have to forecast the future temperature from a) the current temperature, b) the history of the temperature in the last week, and c) a knowledge of seasonal averages and seasonal trcnds. If temperature followed a Markov process, the history of the temperature in the last week would be irrelevant. To answer the second part of the question you might like to consider the following scenario for the first week in May: 1. Monday to Thursday are warm days; today, Friday, is a very cold day. 2. Monday to Friday are all very cold days. What is your forecast for the weekend? If you are more pessimistic in the case o1’ the second scenario, temperatures do not follow' a Markov process. Problem 13.2. Can a trading rule based on the past histor y ofa stock ’s price ever produce returns that are consistently above average? Discuss. The first point to make is that any trading strategy can, just because of good luck, produce above average returns. The key question is whether a trading strategy consistently outperforms the market when adjustments are made for risk. It is certainly possible that a trading strategy could do this. However, when enough investors know about the strategy and trade on the basis of the strategy, the profit will disappear. As an illustration of this, consider a phenomenon known as the small firm effect. Portfolios of stocks in small fines appear to have outperformed portfolios of stocks in large firms when appropriate adjustments are made for risk. Research was published about this in the early 1950s and mutual funds were set up to take advantage of the phenomenon. There is some evidence that this has resulted in the phcnomcnon disappearing. Problem 13.3. A company ’s cush position, measured in million.s o[dollar.s, follows a generalized Wiener process with a Jrift rate of 0.5 per quarter and a variance rate of 4.0 per quarter. How high Joes the company’.s initial ca.sh position have to be for the company to have a less than 5°X‹› chance of a negative ca.sh po.sition by the end of one year.’" Suppose that the company’s initial cash position is x . The probability distribution of the cash position at the end of one year is p(i * 4 x 0.5, 4 x 4) = ‹p(x * 2.0,1 6) where q(m, v) is a normal probability distribution with mean m and variance v . The probability of a negative cash position at the end of one year is i+2.0 4 87 U201 2 Pearson Education

where ñ'(.r) is the cumulative probability that a standardized normal variable fwith mean zero and standard deviation I .0) is less than .r . From nomial distribution tables

when: . + 2.0

4 = —1.6449 i.e., when x = 4.5796 . The initial cash position must therefore be $4.58 million. Problem 13.4. Variahles X, anJ J 2 jâllow generalized Wiener proce.s.see with dri[t rates y anJ ¡5 and variances a," and o,"- . What proces.s doe.s X X _ follow ij: (a) The changes in X, anJ J 2 in any short interval of time ai-e uncorrelated? (b) There is a correlation p between the changes in X, anJ X,, in any shorl interval of time'? (a) Suppose that /i and lz equal ai and a2 initially. After a time period of length T ,I has the probability distribution ‹p(n, + p,T, a T) and A has a probability distribution

From the property of sums of independent normally distributed variables, I, + I, probability distribution

has the

i.e., 2

)T}’

This shows that A, + A, follows a generalized Wiener process with drift rate //, /i, and variance rate o-/ + cr2 . (b) In this case the change in the value o1‘I I I in a short interval of time dt has the probability distribution: ‹p (y, +//,)fi/,(w + cr, + 2yo-,cr,)A/ If

,

, cr, , cr2and p are all constant, arguments similar to those in Section

13.2 show that the change in a longer period of time T is

The variable, I, +J 2 , therefore follows a generalized Wiener process with drift rate y, y 2 and variance rate o, + n/ + 2pcr,n, 88 'i°02012 hourson Education

Problem 13.5. Consider a variable, S , that follows the process dS — — ¿i dt + cr dz For the first three years, -- 2 anda — — 3, Jùr the next three years, — —3 uuéfrr = 4. IQ the initial value of the variable is 5, what is the jlrohahility distribution o] the value of the ariable at the end o/ year six? The change in S during the first three years has the probability distributitin p(2 x 3, 9 3) = ‹p(6, 27) The change during the next three years has the probability distribution 9(3 x 3, 16 x 3) = 9(9. 48) The change during the six years is the sum ot a variable u ith probability distribution p(6, 27) and a v ariable with probability distributions(9, 48) . The probability distribution of the change is therefore (6 * 9, 27 + 48) = (15, 75) Since the initial value of the variable is 5, the probability distribution of the value of the variable at the end o1 year six is ‹p(20, 75) Problem 13.6. Suppose f/inf Ci is a fiinction o[a stock price, S ant time. Suppo.se that o and a are the i olcitiliiie.s of S and G. Show that when the expected return o] S inc'reases by 2a , the ironth rate of G increa.‹es 1›i’ñm , where 2. is a constant.

From 1tô’s lemma Also the drift of G is

z'here is the expected return on the stock. When p increases by Vcr„ , the drift of G increases by

89 2012 Pcarson Education

the expected increase in the stock price and the variability of the stock price are constant in absolute terms. For example, i1 the expected gro th rate is $5 per annum when the stock price is $25, it is also $5 per annum when it is $100. If the standard dcviation of weekly stock price movements is $1 when the price is $25, it is also $1 when thc price is S100. In:

the expected increase in the stock price is a constant proportion or the stock price while the variability is constant in absolute tems. ln:

the expected increase in the stock price is constant in absolute terms or the proportional stock price change is constant. The model.

hile the variability

is the most appropriate one since it is most realistic to assume that the expected percentage return and the i-ariability of the percentagc return in a short interval are constant. Problem 13.9. It has been suggested that the short-term interest rate, r , follows the stochastic proce.s.s

where a, 1› , and c are positive constants and ‹Iz i.s a Wirner process. De.scrihe the nature of this process. The drift rate is «(5 — r) . Thus, when the interest rate is above b the drift rate is negative and, when thc interest rate is below 6 , the drift rate is positive. The interest rate is thereforc continually pulled towards the level b . The rate at which it is pulled toward this level is o. A volatility equal to c is superimposed upon the “pull” or the drift. Suppose a — 0 4 , ñ = 0.1 and c = 0 15 and the current interest rate is 20?o per annum. The interest rate is pulled towards the level of 10% per annum. This can be rcgarded as a long run average. The current drift is —4 0Zo per annum so that the expected rate at the end of one year is about 16% per annum. (In fact it is slightly greater than this, because as the interest rate decreases, the “pull” decreases.) Superimposed upon the drift is a volatility of 15% per annum. Problem 13.10. Suf›pose that a stoc'k price, S , follows geonieti-ic Brownian motion with e. pecte‹1 returtt yrand volatility cr . its — yS dt + aS dWhat is the process followed by ihe variohle S" .! Short that S" also follows geometric Bros nian motion. 91 C8 2012 Pearson Education

II‘ G(S,t) - S" then ‹“›G / âi — — o , âG I ñ:S — — nS" ' , and â’Cr I âS’ — — n(n — I)S“ “" . Using 1U’s lemma:

This shows that G = S" follows geometric Brownian motion where the expected return is

2 gn - 2 n(n —i and the volatility is no . The stock price S has an expected return of value of I is She'” . The expected value of S" is

and the expected

Problem 13.11. Suppose that x i.s the yielJ to maturity with continuous compounJing on a zeroroi ›mibond tltat frays off $1 at time T . Assume that x ;tollow› the proces.s

where a, y, and s are yo.sitive constants and dz is a Wiener process. What is the process falloweJ by the hond price.‘! The process followed by B , the bond price, is from Ito’s lemma: ’“ d/ xdz B 0/

Since: the required partial derivatives are c*B

— — xB

gp , Dr- —(T — /)e ”"*’ ' — -(T — t) B

GB 2’B .

{T—I)' e°’'" °'' — (I— f) 2 B

Hence: J —

I' - t)

1 2

Problem 13.12 A siock whose price is 30 ha.s an expected return of 9%‹ and u volatility oj 20%. In Excel simulate the .s/ock price path over 5 years using monthly time steps and ranJoni sample.s Iron a normal dii/rfAf7fioo. Chart the simulated stock price path. By hitting 19 observe how the path changes a.s the ranJom sample change. 92 2012 Pearson Education

The process is Af = 0.09 x S x At + 0.20 x S x c x

Where At is the length of the time step (=1/12) and e is a random sample from a standard normal distribution.

93 2012 Pearson Education

CHAPTER 14 The Black-Scholes-Merton Model Problem 14.1. What due.s the Black—Schole.f—werton siock option pri‹’inq mod‹•l assume ahout the probabilih' distribution of the .stock price in one vcar? What does it assume ahout the continuum.‹1y’ e'oinpoundrd rare o[return on the .stock during the ye•ar'! The Black—Scholcs—Merton option pricing model assumes that thc probability distribution of the stock price in 1 year (or at any other fixture time) is ltignormal. It assumes that the continuously compounded rate of return on thc stock during the year is normal!) distributed. Problem 14.2. The volatiliF Ofa .sfockyric’e is 30’X per annum. What is the standard deviation of the percentage prices chaflQ€ fn OHe trading day? The standard deviation of the pcrcentage price changc in time fi is cry hcre m is the volatility. In this problem = 0.3 and. assuming 252 trading days in one year, Al = 1./ 25 2 = 0.004 so that v = 0.3 0 00 = 0.019 or 1.9%.

The price of an option or other deri›'ative when expressed in terms of the price o1 thc underlying stock is independent of risk preferences. Options therefore have the samc value in a risk-neutral w’orld as they do in the real world. We may therefore assume that the world is risk neutral for the purposes of valuing options. This simplifies the analysis. In a risk-neutral world all securities have an expected return cqual to risk-free interest rate. Also, in a risk- neutral world, the appropriate discount rate to use for expected future cash flow's is the risk- free interest rnte. Problem 14.4. Calculate the price o[a force-inonf/t Etiroyean J›ut option on o non-dividend-paying stock »'ith a .strike price ofs5t) when the current stork rice is $50, ///« riN-f‘ee• interest rate is I H No yer annum, and the volatility is 30°Xo per annum. In this case S, = 50 . K = 50 , r — — 0. 1 , cr = 0.3 , T — — 0.25 , and J The European put price is

or $2.37.

In(50/ 50) + (0.1* 0.09 / 2)0.25 0.3

0 2417

5 50 \"(—0.09 l7)e "' " — 50.Y(—0.24 17) I, — I, — 0.3 = 50 0.4634c ' ’ ’ 2’ — 50 x 0.4045 = 2.37 25 = 0.0917 94 t°^20 12 Pcarson Education

Problem 14.5. What different e does it make to your c'alculations in Problem 14.4 tea Jividend of $1.50 is c.rpecfed in hr'o month.s? In this casc we must subtract thc present value of thc dividend from the stock price before using Black—Scholes-Merton. Hence the appropriate value of S, is 5„ = 50 1 .50e ' ” ' = 48.52

As before K — — 50, r — — 0. l , o = 0.3, and d

— — 0.25 . I n this case

Jn(48.52 / 50) +(0. + 0.09 i 2)0.25 — 0.0414 0.3 25

c/, = d, — 0.3 0 2 = —0.1056 The European put price is "' ""—48.52 45.52 x\’(—0.0414) =50 50 5'(0.10S6)e x 0.5432e*’' "’ 0.4835 = or $3.03. 3.03 Problem 14.6. ml is implied 'o mlili ? how ca n it be calculated'? The implied volatility is the volatility that makes the Black—Scholes-Merton price of an option equal to its market price. It is calculated using an itcrative procedure. Problem 14.7. .4 stock price is currently .540. Assume that the expected return from the stock is 15% nod ifi i olalility is 25’X‹›. What is lhe probahili} distribution for the ratc o/rettn-n (with cnntinuotis compounding) earned o v'er a Rvo-vear perioJ? In this case = 0.15 and o = 0.25 . From equation (14.7) the probability distribution for the rate of return over a one-year period with continuous compounding is: ‹p 0.15

0.25’ 0.25 2

2

’

2

i.e.,

o(0.11875,0.031 25) The expected value of the return is 11.875%» per annum and the standard deviation is 17.7%° per annum. Problem 14.8. 1.stock price hns nn expecieJ return of 1 b“% and a volatility of 35’Zo. The current price is $3d. a) What i.s the pi-ohubility that a European ‹’all option on thr stoe'k with an exercise price of $40 and a nuilurih Jate in .six months will he exercised? b) WhO*!• !!›* I*"obubility that a European put option on the ›too k with thy same exercis e prire and matw-ity w'ill lie exercise‹1? 95 U20 1 2 Pearson Edtic aliol

a) The required probability is the probability of’ thc stock price being above $40 in six months time. Suppose that the stock price in six months is Sz 0.35 2

In N— ‹p(In 38 (0.1—6

)0.5, 0.35' x 2 0.5)

i.e.,

In St — p(3.687, 0.247’ ) probability is Since ln 40 = 3.659 , the required v 3 689 3 657 0.247 From normal distribution tables N(0.00b) — 0.5032 so that the required probability is 0.4968. b) In this case the required probability is thc probability of the stock price being less than $40 in six months time. It is 1 — 0.4968 = 0.5032

Prob1em14.9. Using the notation in the chapter, prov'e that a 95°» conJiJence interval for S, is between S,e'"°‘’ ’" ' "°”

and

S e'^ °" ”"" "°”

From equation (14.3): 2 95% confidence intervals for In S are therefore

and In S, + (y —

)T + 1 . 9 6 p T 2 9J%t confidence intervals for N, are therefore d i.e. In 3t +( -/ O

o' '' 2) F-1 .96 rr

e

an

(,fi m 2 d fi+I .R6 e’’ "

Problem 14.1tl. A portfolio manager announce.s that the average o{the returns realized in each of the last 10 years i.s 20.ohm per yonom. In what respect is this statement misleaJing? This problem relates to the material in Section 14..3. The statement is misleading in that a certain sum of money, say $1000, when invested for 10 years in the fund would have realized a return (with annual compounding) of less than 20*o per annum. The average of the returns realized in each year is always greater than the return per annum (with annual compounding) realized over 10 years. The first is an arithmetic average of the returns in each year; the second is a geometric average o1 these returns. 96 fi*2012 Pearson Education

Problem 1J.11. A.ssume that a non-Jividend-yaying stack has an expected return of p and a volatility o] a. An innovative financial in.stitution has just announced that it will trade a de•rivuti ve that pays of] a Jollar amount equal to ln Si at time T where S Jenotes the values o}!the stock price at time T

a) Use risL-neutral valid ation to ca fenlate the price of file derivative at time t in term o] the s tock price, S, at time I b) Confirm that yotir price svtisJies the differential equation (14. /d c)

At time i, the ex}iected value of In IT i s from equation (14.3) In S + (y — m‘"1 2){T— t) I n a risk-neutral world the expected value of In Si is therefore lit S + {r — a' 1 2) T — i) Using risk-neutral valuation the value of the derivative at time I is e°""°"[In S + {r — cr“"12)(T — t)] b) If then = r e ' The left-hand side o1 " the Black-Scholes-Merton differential equation is e ’'" " r ln N / (r — cr’ / 2)(T I) —(r — cr2 1 2)+ r — o’1 2 ° " cr 2 ! 2)( T — i) [ — — r[ I Hence the differential equation is satisfied. n

Problem 1d.12. Consider a derii ative fltat .pays off S at time T where S i.s the stock pri‹'e at that time. S When the stoe'k price %llows geome/ric Brownian motion, ii can be shot :n that its price at time i i T) has the form [ h(t, T).S ‘ r where S is the stock price at time t and h i.s n/iinc/ion or/y o/ / and T (a) By substituting into the — Black—Schole,s—Merton partial differential equation derive an ordinary differenliul equation .Watt sfied by h(t,T) . lb) Wharf is the boundary condition Jor the differential equation %r h(t,T) ? c (c) Show that r /$ / T) e'0. “ 7/ // ) ( u ’ 1)d{ J •t 1 here r is the risk-Jree interest 1 rnte and a is the stock price volutilitv. 2 ) ( T

97 â20 12 Pearson Lducation

lf G{S,t) = h(t,T) S" then éG I âi — — h,S' , â ! e3’ — — hns“°' , and c7'G I r°S ' — — hn{—n I)S"°’ where h, — — th I ñ/ . Substituting into the Black—Scholes—Merton diftcrcntial equation we obtain

The dcrivative is worth S“ when / = T . The boundary condition tor this differential cquation is therefore h{T, T) - 1 The equation h(t T) "*•-'I-'I"-'lit 'I

pt° ^

satisfies the boundary condition since it collapses to ñ = I 'hen / = T . It can also be shown that it satisfies the differential cquation in (a). Alternatively wc can solve the differential equation in (a) dircctly. The dif1’erential equation can be writtcn _

2g

2 The solution to this is In h — — [r(n - I ) + 2

n(n — 1)](T — /)

Problem 14.13. What i.s the price ofa European call option on a non-dividend-paving stock ii hen the slock pricc is $52, the striI:e price is $50, the risk-free interest rate iS l2‘Xo Jeer annum, the volalilit y is 3 H“Xo yer annum, and ltte time to maturity is three months'! In this case S, — — 52 , K — — sñ , d, — —

= 0.12 , cr - 0.30 and F — — 0.25 .

ln(52/ 50) * (0.12 + 0.3' / 2)0.25 - 0.5365 0.30 25

I, = I, — 0.30 0 25 = 0.3865 Thc price of the European call is 52 V(0.5365) — 50e '“ " Y(0.3865) = 52 x 0.7042 — 50e“ ” x 0.6504 = 5.06 or $5.06, Problem 14.14. What is the price ofa European put option on a non-dii'iJend-pp’ing stock when the stock price is $69, the strike price i.s $70, the risk-free interest i-ate is 5% per annum, Ihe volatilii) is 35 Xu per annum, anJ the time to maturit) i› six months.'* In this case I — 69, K — — 70, r = 0.05 , o = 0.35 and T — - 0.5 . 9fi tc:20 12 Pearson Education

J '

In(69 / 70) (0.05 + 0.35“ / 2) x 0.5 0.35

0 1666

6f, = d, — 0 . 3 5 a = —0.0809 The price of the European put is 70e “‘“"’ V(0.0809) — 69 Y(— 0.1666) — 70e '“ ' x 0.5323 — 69 x 0,433 b — 6.40 or $6.40. Problem 14.15. Con.wider an American call option on a stock. The stock price is $70, the time to inaturi5 i.s eight months, the risk-free r‹ife o/interest i.s 10"Xo per- annum, the exercise price is $65, and the olatili5 is 32%. A di»idend of $1 i.s e.xpe‹:ted after three month,s anJ again after six months. Short’ that it can never be optimal to r en fSC the option on ‹either of the two dividend dates. Use DerivaGem to calculate the price of the option. Using the notation of Section 14.12, D, — — D› — - I , K(—1 e "' ") = 65(1 — e ' "’ ’" ) — 1.07 , and K(—1 e°"' °"’) = 65(1 — e'' I" " ) = 1.TO . Since D < K(1 — e ’ " ° ' ’ )

and lt is never optimal to exercise the call tiption early. DcrivaGem shows that the value of the option is 10.94. Problem 14.16. A call option on a non-JividenJ-paying .s/ock fins a market price o/ y2.50 . The stock price is $ 15, the exert ise prim e i.s S 13, they time to maturi3' is thr‹•e months, and the risk-free interest rate i,s 5“Zo per annum. What i,s the implied volatility'." In the case c = 2. , S, = l5 , K — — 13, T — — 0.25 , r — — ñ.H5 . The implied volatility must be calculated using an iterative procedure. A volatility' o1 0.2 (or 20%« per annum) gives c = 2.20 . A volatility of 0.3 gives c = 2.32 . A volatility of 0.4 gives c = 2.507 . A vtilatility or 0.39 gives c = 2.487 . By interpolation the implied volatility is about 0.396 or 39.6fio per annum. The implied volatility can also be calculated using DerivaGem. Select equity as the Underlying Type in the first workshcet. Select Black-Scholcs European as the Option Type. Input stock price as 15, the risk-lrec rate as 5%, time to exercise as 0.25, and exercisc price as 13. Leave the dividend table blank because we are assuming no dividends. Selcct the button corresponding to call. Select thc implied volatility button. Input the Price as 2.5 in the second half o1 the option data table. Hit the Enter key and click on calculate. DerivaGem will show the volatility tif the option as 39.64%. Problem 14.17. With the ufO/fOf7 tts ed in thi.S Chapter

(b) Show that sñ'’(J, ) — — Ke '' "ñ"'(J_, ), v her-e S is the .stock price at time i

1n(S /

d, J_

ln(S I K)* (i‘ — m 1 2){T — t)

(c)Calculate id, IDSand ed,. IUS . (d)Show that w'hen c— — Sâ' (J, ) — Ke "'' ’ 'F (d, )

where ‹- i.y the price of a call option on a non-dividend-pg ing stock.

(e)Show that e°cIc°S—— .V(d, ) . Show that the c satisfies the Black—Scholes—Merton dijfere•ntial equation. Show that c satisfies the boundary condition for a European call option, i.e., that c = max(S — K,ñ) m /—+ T

(a)

Since .\'(x) is the cumulativc probability that a variable with a standardizcd normal distribution

wi!1 be less than , .V'( ) is the probability density function for a standardized ntirmal distribution, that is, (b) exp —

T — I—2 cr' (T

2

— i)

Because 2

1n(S / K) {r — o’/ 2)(T— t)

o IT — t

it follows that

As a result which is the required result. (c) d, — — In +

In S — In K (r

2012 Pearson Education

)(T — i)

Hence

Similarly

d,

In S —

and

Therefore:

(d)

c— — S.i(d, ) — Ke!"’ "L’(d ) — — SN'(d, )'’p' — rKe ' " "A ( I , ) — Xe ""*' ' N'(d2 ) Frtim (b): SN'(d, ) = Ke "’ "A”(I, )

Hence "

r“t

= —rKe '

.N(J ) + SL' (d ) "

Since

'

' ‹>/ fi/

'

Hence

(e) From differentiating the Black—Scholes—Merton formula for a call price we obtain =

(d,) + S.V’{d, )

dd

— Ke°’"!"N'(d ) dS

2

From the results in (b) and (c) it follows that " — — Nd ( ) ‹S '

(f) Differentiating the result in (e) and using the result in (c), we obtain ‹°D2012 Pearson Education

$

N d QS — - N'{d,)

From the results in d) and e) ’C

?S

— — —rKe°"’“'L’(J ) — Sh'{dv)2 T —

2

— — r[SN(d,) — Ke°"’ °"E(d,)] — rc

This shows that the Black-Scholes-Merton formuJa for a call option does indeed satisfy the Black-Schtiles-Merton differential equation (g)

Consider what happens in the formula for c in part (d) as apprtiaches T . If S > K d, and d,. tend to infinity and .Y(d,) and ,\'(d,) tcnd to 1. If S < K , d, and d tend to zero. It follow's that the formula for c tends to max(S — ñ, 0) .

Problem I 4.18. Shear that the Black—Scholes— Merton formulas for call and put options .Lori.(/i put—call parity. The Black-Scholes-Merton formula for a European call option is c = S,.\' (c/, ) — Ke• "’:v(d., )

so that

c [l — ñ'(d, )]

Ke "’ — — S„h’(d, ) Ke ''

or c Ke "' -- S N (J, ) + Ke "’ S' (— d, ) p — — d, —) putSoption is (— The Black—Scholes—Merton formula Ke for ’".N'(— a European d, ) so that * .St = Ke "’,U(— d.)— S S' (— J, ) .S p

S— — Ke “ L’(—d, ) S [1 — .'V(—c/, )] (d, )

p S — — Ke ” * (—>, ) S

This shows that the putwall parity result Problem 14.19. c A stock price i.s currently $5a und the ri.sk-See interest rate is 5"Xo. L'.se the DerivuGrin software to translate the Jollo›ving tahle o[European Ke "’ — — p „Sc'ull options on the ,stock into u table of 102 holds. i°/20 1 2 Pearson Education

implied volatilitie.s, a.seeming no di videnJs. Are the option f›rices consistent with the assumption.s underlying Black—Scholes -Xlerton? $tock Price

45 50 55

'faturiti’ — 3 nlonths | Maturin— 6 months

7.00 3.50 1.60 2.90

8.30 5.20 !

Moturiu — 12 months

10.50 7.50 5.10

Using DerivaGem we obtain thc follow ing table of implied volatilitics Stock Price 45 50 55

.Maturity — 3 months 37.75 32.12 31.98

Maturit—v 6 months 34.99 32. 75 30.77

MaturiN— 12 monids 34.02 32.03 30.45

To calculate first number, select equity as the Underlying Type in the first worksheet. Select Black-Scholes European as the Option Type. Inr ut stock price as 50, the risk-free rate as 5%, time to exercise as 0.25, and excrcise price as 45. leave the dix idend table blank bccause we are assuming no dividends. Select the button corresponding to call. Select the implied volatility button. Input the Price as 7.0 in the second half o1 thc option data table. Hit the Enter key and click on cdlCulate. DerivaGein will show the volatility of the option as 37.78%. Change the strike price and time to exercise and recompute to calculate the rest of the numbers in the tablc. The option prices are not exactly consistent u ith Black- Scholes—Merton. lf they were, the implied volatilities would be all the same, We usually find in practice that lo ' strike price options on a stock have significantly highcr implied volatilities than high strike pricc options on the same stock. This phenomenon is discussed in Chapter 19. Problem 14.20. E.xplain carefull y why Blac'k’s approach to evahiating an Amet‘fCGti c'uJJ Gption on a dividendI› aying stock may gi ve an approxiiilate an.swer e• v‹•n «hen only one Ji› idend i.s anticipated.

Doe.s the answer given by Black’.s aj›proach iinJerslale or a»erstate the true option value? Explain your answer. Black’s approach in ctfect assumes that the holder ot option must decide at time zero whether it is a European optitin maturing at time /, (the final ex-dividend date) or a Europcan option maturing at time Z . In fact the holder of the option has more flexibility than this. The holder can choose to exercise at time if the stock pricc at that time is abtive some level but not otheovise. Furthermore. i1 the option is not exercised at time / , it can still be exercised at time Z It appears that Black’s approach should undcrstate the true tiption value. This is becausc the holder of the option has more alternative strategies for deciding when to exercise the option than the two strategies implic ltly assumcd by thc approach. These alternative strategies add value to the tiption. However. this is not the whole story! The standard approach to valuing either an American or a European option on a stock paying a single div idend applies thc volatility to the stock price less the present value of the dividend. (The procedure for valuing an American option is exFlained in Chapter 20.) Black’s approach w’hen considering exercise just prior to the 103 2012 Pearson Education

dividend date applies the volatility to the stock price itself. Black’s approach therefore assumes more stock price variabiJity than the standard approach in some of its calculations. In some circumstances it can give a higher price than the standard approach. Problem 14.21. Consider an American call option on a stock. The stock price is $50, the time to maturity is 15 months, the ri.sk-free rate of interest is 8°Xo per annum, the exercise price is $55, and the vol‹itility is 25’Xo. Dividends o{$1.50 are eypycted in 4 monihs and 10 months. Show that it can never be optimal to exerci.be the option on either oj the two dividend dates. Calculate the price of the option. With the notation in the text D, — —D — — l .50, /, = 0.3333, i, = 0.8333, T — — 1.25,r = 0.08 and

K = 55

K 1 — e ' T°’ = 55(1 — e ” " ” ” ) = 1.80 Hence Also: ) = 2.16 Hence: lt follows from the conditions established in Section I 4.12 that the option should never be exercised early. The present value o1 the dividends is I.5e°' ”"”’” + I .5e ”’’”' " = 2.864 The option can be valued using the European pricing formula with: 50 = 5—0 2.564 = 47.136, K — — 55, cr = 0.25, r — — 0.08, T — — I .25 d,— -

In(47.136 / 55) *(0.08* 0.25' / 2)1.25 -0.0545 0.25

2 Y(c/, ) = 0.4783, d, — — d —0.3692 0.25 1 25 — —0.3340

F(d_,) =

and the call price is 47.136 x 0.4783 — 55e*' " ’ " x 0.3692 = 4.17 or $4.17. Problem 14.22. Show that the probability that a European call option will be exercised in a risk-neutral world is, with the notation introduced in this chapter, N(d›). What is an expression for the value ofa derivative that pays off $100 if the price o] a stock at time T is greater than K ? The probability that the call option will be exercised is the probability that S > K where S is the stock price at time T . In a risk neutral world In Sz - ‹p[ln S +(r —cr" 1 2)T, cr2 T] 104 02012 Pearson Education

The probability that S > K is the same as the probability that ln 6, > ln K . This is In x — In S—t (—r In(No / V) cr“ / 2)T

The expected value at time T in a risk nRutTal world of a dcrivative security which pays off $100 when S > K is therefore 100 \'(d_) From risk neutral valuation the value of the security at time / is 1 00e '.V(d2 ) Problem 14.23. Show that S°““ roulJ be the price ma trad‹•J deri vative set urity. If / = I

° then

= rS 2’ "“ — r This shows that the Black-Scholes-Merton cquation is satisfied. S ” °‘ could therefore be the price of a traded security. Problem 14.24. .4 company has an issue o[executi› e .stork options outstanJing. Should dilution be taken info account n'hei1 the options are valued'? Explain you answer. The answ er is no. 11 markets are efficient they have alrcady taken potential dilution into account in dctermining the stock price. This argument is explained in Business Snapshot 14.3. Problem 14.25. .1 company ’s .stock price is 3.f0 and 10 million .share,s are outs tanJing. The company i.s e'onsidering giving its employees three million at-the-money'five-year cv// options. Option exerci.se.s will he handled by thening ninre .shares. F/ie slork price volatility is 25“Zu, the five105 ‹ 201 2 Pearson tiducation

year ri.sk-free rate is 5% and the company doe.f not pay JiviJend '. Estimate the cost to the company of the emplovee stock option issue.

The Black-Scholes-Merton price of the option is given by setting I, = 50, — ñ0, r = 0.05 , o = 0.25 , and T — — 5 . It is 16.252. From an analysis similar to that in Section I 4.10 the cost to 10 1 6.2n2 — the company of the options is 10 + 3 12.5 or about $12.5 per option. The total cost is therefore 3 million times this or $37.5 million. It the market perceives no benefits fron the options the stock price will fall by $3.75.

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CHAPTER 15 Employee Stock Options Problem 15.1. Why was it attractive for companie.s to grant at-the-money .stock option.s prior to 2005? What changed in 2505? Prior to 2005 companies did not have to expense at-the-money options on the income statement. They merely had to report thc value of the options in notes to the accounts. FAS 123 and IAS 2 required the fair value of the options to be reported as a cost on the income statement starting in 2005. Problem 15.2. What are the main differences beta ecu a fypico/ employee stock option anJ an American call option iraJed on an exchange or in the over-the-counter market? The main differences are a) employee stock options last much longer than the typical exchange-traded or over-the-counter option, b) there is usually a vesting period during which thcy cannot be exerciscd, c) the options cannot be sold by the employee, d) if the employee leaves the company the options usually either expire wtirthless or have to be exercised immediately, and e) exercise of the tiptions usual ly leads to the company issuing more shares. Problem 15.3. Explain why employee ,sfoc/r options on a non-dividend-paying stock are frequently exercised before the end of their lives w'hereas an exchange-lraded call option on such a stock is never exercised early. It is always better for the option holder to sell a call option on a non-dividend -paying stock rather than exercise it. Employee stock options cannot be sold and so the only way an employee can monetize the option is to exercise the option and sell the stock. Problem 15.4. ”Stock option grant.s are good hecau.se they motivate executives to act in the best intere.sts of .shareholder.s.” Di.scuss tlii.s vien'J›oint. This is questionable. Executives benefit from share price increases but do not bear the costs of share price decreases. Employee stock options are liable to encourage executives to take decisions that boost the value of the stock in the short term at the expense of the long term health of the company. lt may even be the case that executives are encouraged to take high risks so as to maximize the value of their options. Problem 15.5. ’Granting stock options to executives i.s like allowing a j›rofe.s.sional footballer to het on the outcome of gases.” Discuss thi.s viewpoint. Professional footballers are not allowed to bet on the outcomes of games because they themselves influence the outcomes. Arguably, an executive should not be allowed to bet on 107 2012 Pearson liducation

the future stock price of her company because her actions influcncc that price. However, it could be argued that there is nothing wrong with a prtifessional footballer bctting that his team will »'in (but everything wrong with betti ng that it will lose). Similarly therc is nothing wrong with an cxecutive betting that her company * ill do well. Problem 15.6. Oy did some companies backdate .stock option grants in the 7’S prior to 2002? What changed in 2002.” Backdating alloy ed the company to issue employee stock options with a strike price equal to the price at somc previous date and claim that they were at the money. At the money options did not lead to an expense on the income statement until 2005. The amount recorded for the value of the tiptions in thc notes to the income was less than the actual cost on the true grant date. In 2002 the SEC required companies to report stock option grants within two business days of the grant date. This eliminated the possibility of backdating for companies that complied with this rule. Problem 15.7. /n what wav w'ould the beaef/s o[hacLdating be reJuceJ if a stock option grant had to be revalueJ at the end of each quarter? If a stock option grant had to be revalued each quarter the value of the option tif the grant date (however determined) would beconac less important. Stock price movements ftillowing the reportcd grant date would be incorporated in the next revaluation. The total cost of the options would bc independent of the stock pricc on the grant datc.

It ould be necessary to look at rctums on each stock in the sample (possibly adjusted for the returns on the market and the beta o1‘ thc stock) around the reported employee stock option grant date. One ctiuld designate Day 0 as the grant date and look at returns on each stock each day from Day —30 tti Day +30. The rcturns would then be a› eraged across the stocks. Problem 15.9. On hip› 31 a ‹Company ’s stock ¡mice i.s $7h. One million shares arc outstanding. An executive exercise.y 100,000 ›’iocl‹ options w'ith a .strike price of $50. What is the iwprfc/ of this on the stocL price? There should be no impact on the stock price because the stock price will already reflect the dilution expected from the executive’s exercise decision. Problem 15.10. The noir s ar.coin¡›an y’ing a con7,oani’ ’3 JiTuin ciul statements .say.’ “Our executive stock options lust 10 yrors and vr.st after fora years. We valued the options granted thi.s ear us rug the Black— Scholes— \ferton moclel i›’ith an expecteJ life of 5 years and a volatilit)’ o[20°Xn. "Whal does this mean.! Disrus.s the modc•linp a)›proach u›eJ by ltte company.

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The notes indicate that the Black—Scholes—Merton model was used to produce the valuation with T , the option life, being set equal to 5 years and the stock price volatility being set equal to 20%. Problem 15.11. In a Dutch uuclion of 10,000 options, hids are as follow's A biJs $30 for 3,000 B biJs $33 for 2,500 C biJs $29 for 5,ONO D bids $40 for 1,TOO E bids $22 %r 8,0ñ0 F bids $35 for 6,000 What is the re.cult of the auction? Who buys how many at what price.’* The price at which 10,000 options can be sold is $30. B, D, and F get their order completely filled at this price. A buys 500 options (out of its total bid for 3,000 options) at this price. Problem 15.12. A company has granteJ 500,0O6 options to its executives. The stock price and strike price are both $40. The options last for 12 year.s anJ vest ‹ijier four years. The company decides to value the options using an ex ›ected life of jive yeurs unJ a volatility of30‘Xo per annum. The company pays dividend.s the Scholes—Merton risk-]ree rate is 4°X. What companv report The options arenovalued usingand Black with S— —will 40,the K— — 40, T — — cras'=an 0.3 expense for the option.s on its income slutement? 5 , and r = 0.04 . The value of each option is $13.585. Thc total expense reported is 500, 000 x 13.585 or $6.792 million. Problem 15.13. A company’s CFO savs.- “The accounting treatment o] stock options is craz)›. We granted 10,000,000 at-the-money stock options to our employees last year lj hen the .stock price was $30. We estimated the value of each option on the grant Jate to be $5. At our year end the stock price had fallen to $4, hut we w'ere still ›'tuck »ith a $50 million char-ge to the P&L.” Discuss. The problem is that under the current rules the options are valued only once—on the grant date. Arguably it would wake sense to treat the options in the same way as other derivatives entered into by the company and revalue them on each reporting date. However, this does not happen under the current rules in the United States unless the options are scttled in cash.

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CHAPTER 16 Options on Stock Indices and Currencies Problem 16.1. A port%lio is cm i entlv worth ,$ IN million anJ has a beta of 1.0. An inJex is currently standing al 600. Explain he a put option on the inJex v ith a strike of 700 can be used to provide portfolio in.Durance. When the index goes down tti 700, the value of‘thc portfolio can be expected to be 10 x (700 / 800) - $8.75 million. (This assumes that the dividend yield on the portfolio equals the dividend yield on the index.) Buying put options on 10,000,000/800 — 12,500 times the index with a strike o1 700 therefore provides protection against a drop in the value of the portfolio below $8.75 million. If each contract is on 100 times the index a total of 125 contracts would be required. Problem 16.2. “Once we know how to value options on a stock pay ing u dividend yield, we know how to value options on .stock indice.s, currencies, anJ futures. ’Explain this statement. A sleek index is maJogu«s /u a sinck paying a cunlinuous dividend yield, the dividend y1eJd being the dividend yield on the index. A currency is analogous to a stock paying a continuous dividend yield, the dividend yield being the foreign risk-free interest rate. Problem J6.3. A .stock index is currently SO0, //te JiviJend yield on the iHdex iS 3“o per annum, and the riskfree interest rate is 8"Xo per annum. What i.s a lower hound%r the price oJ a six-month European call option on the inJex when the .y/r/keprice is 29a’? The lower bound is givcn by equation 16.1 as 300e° 0 "‘ 0— 29 D e ”“' ’ = 1 6. 90

Problem 4 6.4. A currenc is currently worth $0.80 and Isa.y a volatility of 12’No. Mo Jomestic and %reign riskfree inte•rest rales ure 6%‹ and 8°Xu, re.spectivel ’. Use a W o-step binomial tree to value a) a Ew-opean%ur-month call option u’ith a strike price of 30.79 anJ b j an American four- month call oplion with the same .y/i ikeyrice In this case « = 1 .0502 and = 0.4538 . The tree is shown in Figure 516.1. The value of the option if it is European is $0.0235: the value o1 the option if it is American is $0.0250.

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0.8 B24 0.0 924 0.0 924 0.BO0O 0.0100

0.8402 0.0469 0.ODW2 O.8OOD 0. 0235 0. 0230

A

fJ.G1fJfJ

OT253 00000

U.7610

0.0045

0.OO0O

UU045

Figure S16.1 Tree to evaluate European and American put option Problem 16.4, At each node, upper number is the stock price: next number is the European put price, final number is the American put price Problem 16.5. Explain how' corporations c'an use range [oru' arJ rontra‹ 1s to heJge their %reign exchange risk w'hen the are due to receive a certain amount oJ the foreign ‹’urrency in the future.

A range forward contract allows a corporation to ensure that the exchange rate applicable to a transaction will not be worse that one exchange rate and will not be better than another exchange rate. In this case, a corporation ould buy a put » ith the ltiwer exchange rate and sell a call with the higher exchange rate. Problem I 6.6. Calculate the value of ‹u three-month at-the-mourn' European cci/I option or n stock index n'hen the index is at 250, the risk-free intere.st rate is I H’Xo yet‘ atlrtHrn, the volatility of the itideX is I d’Xo per annutn, end the dividend yield on the inJex is 3’No per annum. In this case, S, — — 250, K — — 250, r — — 0.10 , cr = 0.1 S , T — — 0.25 , 9 = 0.03 and

J

1n(250/ 250) + (0.10 —0.03 * 0.1 8“ / 2)0.25 - 0 2394 '

0.1Sb0.25

d_, — — d, — ñ.1 S 0 25-

0.1494

and the call price is 250 V(0.2394)e " '" 250.6'(0. 1494)e "" "’ = 250

0.594(›e ’°' ' — 250 x 0.5594e "

”

or 11.15. Problem I 6.7. Calculate the value of an eight-month European put option oii a ‹’iirrenc'y with a strike Frice of 0.50. The ctii-rent exchange rate is 0.52, the volatility' of flue ex‹:hange rate i.s 12°X‹›, the domestic risk-jree interest rate i.s 4OX per annum, anal the]ot-ei for risk-f’ree infere.st rate i.s R°Xo per annum. 111 FU20 1 2 Pear.son Education

In this case I— 0.52 , K—— 0.50 , —r — 0.04 , r —— 0.08 . — 0.12 , T—— 0.6667 , and

d— -

In(0.52 / 0.50)

(0.04 — 0.05 0.122 / 2)0.fi66 0.12 0 6667

= 0.1 771

d, = ‹—/, 0.12 0 66 7- 0.0791 and the put price is 0.50:\' (—0.0791)e ”""‘—" 0.52.Y(—0. I 771)e "" “’ 0.50 x 0.4685e ' “ ' “‘" 0.52 x 0.4297e " ““ “” = 0.0162 Problem 16.8. Short that the formula in equation (I d.12) jor a taut option to .sell one ionit o[ctirrenc) A [or currency B at ylrike price n giv'es /hc sanic• value a.s equation (16.11) for a call option to hug a i/ ir of!cw-renc) B for t iirrency A at a strike price o[ i . K . A put option to sell one unit tif currency A for K units of currency B is worth Kx' — J —) s c " '(—a, ) wfierc

d — —

I K)

(. '“in and i; and ›; are the risk-free rates currencies A and B, rcspectivcly. The value or the option is measured in units of currency B. Defining S/ = 1 St and V“ = l ,' V d, —

—1n(S,* —1n(5,’ ,' K’ ) — (r, — i-z

o’ 1 2)T

The put price is therefore 1'here

This show s that put option is equip alent to K.S„ call options to buy 1 unit of‘ currency A for l .' K units or currency B. In this case the value of the option is measured in units of currency A. To obtain the call option value in units of’ciirrcncy B (thcsamc units as thc value of the put option was measured in) we must divide by S, . This proves the result.

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Problem 16.9. A fore•ign cm rencv is rurryntl) v orth $1.50. The dome.stic and foreiyit risk-]rer interest mtce are 5% and 9°X. respe •tively. Calculate a In er hound for the value of u si.x-month call option on thr c ur re m Frith a strike price of $1.40 if it i.s (a) F.uroJ›ean anJ (b) Anu•ri‹’an. Lower bound for European option is Nte — Ke°“ = 1.5e“ ” °— 1.4e '” ” = ().t)69 Low'er bound for American option is S, — K — — 0.10 Problem 16.10. Con.wider a stock index currentl y s I‹inJing at 250. The dividend yielJ on the rude.x is 4‘Xo per annum, and the risk-fre•e rate is 6% per annum. A three-month Eui-oj›ean rail opt iota on fhe index n'ith a strike price of 245 is currentlv ›s orth $1 H. What i.s the i'aluc• of a i/tree-month put option on the index iritli a ›'trike prim e of245" a

p

h case .S, = 250, y = 0.04. › = 0.06 , T — — 0.25 , ñ = 245 , and ‹ = 10 . Using pat call

Substituting: p —- 10 * 245e

“’ — 250c ”“’ “ — 3.84

The put price is .3.54. Problem 16.11. An rude.x currently efanele at 696 and he.s a volatility o/.10% peer annum. The iris h-Jrrr ratr of iniei-est is 7% per annum and the index I•Tovide.s a dii'idenJ i'ie•ld oJ 4‘Xo prr annum. C.all alote ihe value oJ a three-month European j›ut w'ith an exc•rcise f›rire• oJ 7T!LL In this case S — — 696, K — — 700, r — 0.07 . cr = 0.3 , T — — 0.25 and 9 = 0.04 . The tiption can be valued using equation (16.5). ln(696/ 700) (0.0—7 0.04 + 0.09 / 2) x 0.25 $$$$ 0.3 2 d

I, — 0.3 0 2- — 0.0632

and â(—cf, ) = 0.4654, .6’(—J„) = 0.5252 The value of the put, p , is given by: p = 700e“ " ” x 0.523—2 696e°' ‘“' ” x 0.4654 = 40.G i.e., it is $40.6. Problem 16.12.

where S, i.s the stock price, r i5’lho ri›k-free rate, and r > H . (Hint. To ol›tain the]irst half of the inequalit›’, consider possible value.s o/ Portfolio A, a European call option plu.s an amount K investeJ at the ri›h-free rate l°ortfolio B. an Amer-ican put option phi.v e "’ o/.slack iviiñ JiviJends being reinve,sted in the To obtain the second halfofthe in‹'‹fuality, consiJer possible• vn/t/e,i o Portfolio C. an American call option joins an amounl Ke°“ investcd at the ri.sk-free rate Porfolio D. a European put op/rort p/us one stock with dividend.f heing reinvested in the Following the hint, we first consider Portfolio A: A European call option plus an amount K invested at the risk-free rate Portfolio B: .4n American put tiption plus r " of stock with dividends being reinvested in the stock. Portfolio A is orth c K while portfolio B is worth P option is cxercised at time r(0 r < T) , portfolio B becomes: X — .Y, sK where S, is the stock price at time r . Portfolio A is worth c + yr’

S «

. If the prit S, e !' '

K Hence portfolio A is worth at least as much as portfölio B. If both portfolios are held to maturité (time /’ ), portfolio A is worth max(N, — K, 0)* Ke''

=inax(S„ K) K(e' —1) Portfolio h is worth max(S„ v) . Hence portfolio A is z'orth more than portfolio B. Becausc portfolio A is worth at least as much as portfolio B in all circumstances P+ S e “ c* K Bccause c

C:

.S,e “ '— K C—P This pro› es the first part of the incquality. For the second part consider. Portfolio C.' An American call option plus an amount Ke " invested at the risk-free rate Portfolio D. A European put option plus one stock with dividends bcing reinvested in the stock. Portfolio C is worth C + K ’ while portfolio D is worth p + S . lf the call option is exercised at time r(0 r < T) ponYolio C becomes: .Y, - K * Ke “’ " < S w'hile portfolio D is 'orth + S e"'’ “ > S Hence portfoli‹i D is worth more than portfolio C. lf both portftilios are held to maturity (time

Hencc portfolio 19 is worth at least as much as portfolio C. Since portfolio D is worth at lcast as much as portfolio C in all circumstances: U * Ke "' p S

Since pS P : This provcs the second part of the inequality. Hcncc: J„e •' K < C— P < S z — Ke ”

Problem 16.13. Shew that a European call option on a currencv has the same yrice a.s the corresponding European put oytion on the currency w'hen the fom»ard j›rice equals the strike price. This follows from put—call parity and the relationship beñveen the forward price, F , and the spot price, S, . ‹ j- Ke”’ — — p S e ""

and so that this reduces If K — —toNoc = , . The result that ‹ when K — — F, is true for options on all underlying assets, not just options on currencies. An at-the-money optitin is frequently defined as tone where K — — I, (or = y› ) rather than one w here K — — S, . Problem 16.14. Would 5 on expe‹'t the volatili of a stock rude.x to be greater or less than the volatility of a Iyyi‹a/ stock? E.xplain your an.swer. The ›'oIati1ity o1 a stock index can be expected to be less than the volatility of a typical stock. This is because some risk (i.e., return uncertainty) is diversified away when a portfolio of stocks is created. In capital asset pricing model terminology, there exists systematic and unsystematic risk in the returns from an individual stock. However, in a stock index, unsystcmatic risk has been diversilicd away and only the systematic risk contributes to volatility. Problem 16.15. Does the cost oJ portfolio insurance increase or- Jecrease as the beta ma portfolio increases? E. plain your answer. The cost of porttolio insurance increases as the beta o1 the portfolio increases. This is because portfolio insurance involves the purchase of a put option on the portftilio. As beta increases, the ›'o1atility of the portftilio increases causing thc cost of the put option to increase. When 115

index options are used to provide portfolio insurance, btith the number of options required and thc strike price increase as beta increases. Problem I 6.16. Tuppe.sr flint u portfolio is u orth $h0 million anJ the SFP 500 is at 120a. IQthe value of the portfolio mirrors the vo/tie o/ the index, what options should be purcha.sed to proviJe protection ognzn.st the › alue o) the portfolio [allin belch $54 million in one year’s time? II the value of the portfolio mirrors the value of the index, the index can be expected to have dropped by 10% when the value of the portfoliti drops by 10%. Hence when the walue of the portfolio drops to $54 million the value of the index can be expccted to be 1080. This indicates that put optitins w ith an exercise price tif 1050 should be purchased. The tiptions sh‹in1d be on: 60. 000. 50, 000 000 times the index. Each option contract is ftir $100 times the index. Hence 500 contracts should bc purchased. Problem 16.17. Consider again the .situation in Problem 16.16. Su¡›po.se that the portfolio has a heta of2.ñ, the rift-//’et' iHtel e.s I rate i.s 5% pet- umum, and fhe dividend yield on both the portfolio and thy inJe.x is 3’Xo per annum. tFliat options should be Jui-clan.red to proviJe protection again.st the alue of the portfolio [alling l›elow $54 million in one year’.s time? When the value of the portftilit› falls to $54 million the holder of the portfolio makes a capital loss of I 0% . After dividends are taken into account the loss is 7% during the ycar. This is 12% beltiw thc risk-free interest rate. According to the capital asset pricing model, the expected excess rcturn of the portfolita above thc risk-tree rate equals beta times the expected excess return of the markct above the risk-free rate. Therefore, when the portfolio provides a return 12%› below the risk-free interest rate, the market’s expectecl return is 6fi obelow the risk-free interest rate. As the index can be assumed to have a beta ‹›f

1.0, this is also the excess expected return (including dividends) front the index. The expected return trom the index is thereftire 10Z• per annum. Since the index providcs a 3*/o per annum dividend yicld, the expected movement in the indcx is —4* . Thus when thc portfolio s value is 554 million the expected value tif the index is 0.9f* x 1 200 = 1152 . Hence Uurtipean put options should be purchased with an exercise price ot 1152. Their maturity date should be in one year. The number of options rcquired is uvice the number required in Problem 16.16. This is because z'e wish to protect a portfolio which is to ice as sensitive to changes in market conditions as the portfolio› in Problem 16.16. Hence options tin $100,000 (or 1,000 contracts) should be purchased. TU check that thc answer is correct consider what happens when the value of’ thc portfolio declines by 20a to $45 million. The return including dividends is 17%›. This is 22* « less than the rlsk-free interest rate. The index cun be expectcd to provide a return (including dividends) which is 11% less than the risk-free interest rate, i.c. a return of — 6fio. The index can therefore be expected to drop by 9fio to 1092. The paytiff from thc put options is (I 152 — 1092) x 100. 000 = 86 million. This is exactly z'hat is required to restore the value o1 the portfolio to $54 million. 1 16 8.20 12 Pearson L ducatioii

The implied dividend yield is thc value of ‹¡ that satisfies thc put call parity equation. It is the value of y that solves 1.54 + l400e ”” — — 34.25 I 500e " This is 1.99%.

A total return index behaves like a stock paying no dividends. In a risk-neutral w orld it can be expected to grow on average at thc risk-free rate. Forward contracts and options on total retiini indices should bc i alued in the same way as form ard contractsaiid options on io i- dividendpaying stocks. Problem 16.20. Whiff fS the put—call j›ariti relutionsh ip [or Eiiroyecm currency opiioizs Thc put -call F• rity relationship for European currency options is To prove this result, the two portfolios to consider are: Portfolio A.’one call tiption plus one discount bond u hich will bc worth x at time T Portfi›lio B: one put option plus e ’ " tif foreign currency invested at the foreign risk-tree interest rate. Both portfolios are worth i ia x(S , K) at time y . They must therefore be worth the same today. The result follows. Problem 16.21. Can an option on the › en-euro exchange rate be created front tit o options, are on the dollare•uro e.« han,qe i-afe, and the other- on the Jollar-›’etc exchaitge rate.’* E.mphin ›’otii an.swear. There is no way of doing this. A natural idea is to create an option to cxchangc a euros for one yen mom an optitin to exchange › dollars f‹ir 1 yen and an option to exchange K euros for r dollars. The problem with this is that it assumes that cither both tiptions are exercised or that neither option is exerciscd. There are always some circumstance.s » here the first option is inthe-money at expiration w hile the second is not and vice versa.

In portfolio A, the cash, if it is invested at thc risk-free interest rate, will grow to s at time r . If S >K , the call option is exercised at time T and ponlolio A is w orth 5, . lt N, < K, the call tiption expires w orthless and the portfolio is u orth x . Hence, at time T. portfolio A is 117 2012 Pearson Education

Bccause of the reinvestment of dividends, portfolio B becomes one share at time T . It is, therefore, worth S at this time. It follows that portfolio A is always worth as much as, and is sometimes worth more than, portfolio B at tilTlC T . In thc absence or arbitrage opportunities, this must also be true today. Hence. c * Ke°"" ? She°” c Ke “

S e “—

This proves equation (16.1). In ptirtfolio C. thc rcinvcstment o1 dividends means that the portfolio is one put option plus one share at time T. If S < K , the put option is exercised at time T and portfolio C is worth a . If S > K, the put option cxpircs worthless and the portfolio is worth S . Hence, at time T , porttol to C is worth max (N , K) Portfolio D is worth at time r . It follo s that portfolio C is always worth as much as, and is sometimes ii orth more than, portfolio D at time T . In the absence of arbitrage oppormnities, this must also be true today. Hence, p S„e ” > Ke “ Ke " — S e " This proves cquation (16.2). Portfolios A and C are both worth max (6,, K) at time T. They must, therefore, be worth the same today, and the putwall parity result in cquation (16.3) follo s.

118 i8.›201? Pearson Education

CHAPTER 17 Futures Options Problem 17.1 Explain the difference between a call option on ; en and a call option on yen futures. A call option on yen gives the holder the right to buy yen in the spot market at an exchange rate equal to the strike price. A call option on yen futures gives the holder the right to receive the amount by which the futures price exceeds the strike price. If thc yen futures option is exercised, the holder also obtains a long position in the yen tiitures contract. Problem 17.2. Why are options on bond futures more actively traded than options on bonJs? The main reason is that a bond futures contract is a more liquid instrument than a bond. The price or a Treasury bond futures contract is known immediately from trading on the exchange. The price of a bond can be obtained only by contacting dealers. Problem 17.3. “A futures price is like a stock paying a dividend yield.” JJ'’hat is the dividend j'ield? A futures price behaves like a stock paying a dividend yield at the risk-free interest rate. Problem 17.4. A futures price is ‹’urrently 50. At the end o]›i. months it will he either 5d or 46. The riskfree interest rate is 6°/‹› jeer annum. What is the value ofa six-month F.uropean ‹’all option with a strike price of 58? In this case u = 1.12 and d — — 0.92 . The probability of an up movement in a risk-neutral world is 1 — 0.92 1.1 0.92 = 0.4 From risk-neutral valuation, the value of the —2call is e ' "” ‘ (0.4 x 6 • 0.6 x 0) = 2.33 Problem 17.5. Host' does the put—call J›arit)'formula]or u jiitures option di[fer (rom put—cull parity for an option on a non-dividend-paying slock? The put—call parity formula for futures options is the same as the put—call parity formula for stock options except that the stock price is replaced by F e '', where I, is the current futures price, r is the risk-free interest rate, and is the life of the option. Problem 17.6. ConsiJer an American futures call option where the futures contract anJ the option conti-act expire at the same time. UnJer what circum.stances is the futures option worth more than tht: 119 â. 2012 Pearson Education

corresponding American option on the unJerlying a,sset!! The American futures call option is worth more than the corresponding American option on the underlying asset whcn the futures price is greater than the spot price prior to the maturity of the matures contract. This is the case when the risk-free rate is greater than the income on the asset plus the convenience yield. Problem 17.7. Calculate the value ofa five-month European put fiitures option ›i'hen the futures price is $19, the strike price is $20, the risk-free interest rate is 12°Xu per annum, and the volatility of the futures price is 20"Xo per annum. In this case I, = 19, K — — 20, r — — 0.12 , o = 0.20 , and T — — 0.4167 . The value of the European put futures option is 20.V(—d,, )e ' '”' ’" — I ñ,V(— d, )e ’ '”’’'’ J ln(19/ 20) + (0.04/ 2)0.4167 0.2 0 4 This is

— 0.3327

7 d,_

d, ‘— 0.2 0 4 7 = —0.4618 e° ” "[20.V(0.46 18) —19 V(0.3327)] = e ° "”'"‘ (20 x 0.6778 —19 x 0.6303)

or $1.50.

= 1 .50

Problem 17.8. Suppo.se you buy a put option contract on October gold fiitures with a strike price of $1200 per ounces. Eaclt c oiitract is for- the delfV€f j’ of 1OH ounces. l47tat happen if you exercise v'hen the October futures price is $1,180? An amount (1,20—0 1,1S0)^100 = $2,000 is added to your margin account and you acquire a short futures position obligating you to sell 100 ounces of goJd in October. This position is marked to market in the usual way until you choose to close it out. Problem 17.9. .Suppose you scll a call option contract or April live ca//Je,/ii//irei ›i/ñ u strike prim e of 9F cents per pound. Each contract is]or ihe delivery of 40,000 pounds. What happens iJ the contract is exercised when the futures price is 95 cent.s? In this case an amount (0.95 — 0.90) x 40, 000 = $2, 000 is subtracted from your margin account and you acquire a short position in a live cattle futures contract to sell 40,000 pounds of cattle in April. This position is marked to market in the usual way until you choose to close it out. 120 2012 Pearson Education

Problem 17.10. Con.siJer a l›+'o-inonlh ccill futures option with a strike price oJ 40 when the ri.sk-free interest rate is I0°X per annum. The current futures price is 4.7. What is a loi+’er hound %r the value o[the future›' option if it is (a) Euroj›ean anJ Lb) Aineri‹’an? Lower bound if option is European is ( F), - K )e “ = (47

' = 6.8.8

40)c

Lower bound if option is American is Fz — K — — Problem 7.11. 7 a strike yr ic'e oJ 50 ConsiJer a Jotir-month put futures oj›tion with hen the ri.sk-fi-ee intere› t rate is I HXuper annum. The current fñtw‘es pri‹’e is 4.7. What i.y a lo›+’er hound for the value of the futures option i f it is t‹i) European and (b) Ameri‹’an? Loz’er bound if option is European is (K - I, )e ./= (o-0 — 47)c- i‹, : = 2.90 Lower bound it option is American is

Problem 17.12. A future ' price i› currently 60 and its volatilil y is 30°X. The risk-free interest r‹tte i,s 8’o yer annum. k'se a Evo-step binomial tree to calculate the value of a .ti.x-month F,uropean rall oplion on the fixtures with a .strike price of 60.! If the call ere American, would it ever be worth eie/-ci.sing it early? In this case n = e’"

= 1.16.18, d = 1 , u = 0.0. b607 ; and 1—

0.8607

= 0.4626

1.1(› IS — 0.8f›07 In the tree shown in Figure S17.1 the middlc number at each node is the price of the European option and the lower number is the price of the American option. The tree shows that the value of the European option is 4.3155 and the value of the American option is 4.4026. The American option should sometimes be exercised early. 80.991.5

9.5178 9.710 I

60.0 u00 0.0000

60.0000 4.4026

?164??

0.0000

Figure S17.1 Tree to evaluate European and American call options in Problem 17. 12 121 t°c2012 Pearson Education

Problem 17.13. In Problem 17.12 what doe.s the hinom ial tree give for a six-month European put option is the value of a six-month European put oj›tion on]uiures »ith a strike price of 60? If the put w'ere AwC
Problem 17.12 anJ the put prices calculated here satisfy put—call parity relationships. The parameters u , d and ¿› are the same as in Prtiblem 17.12. The tree in Figure S17.2 shows that the prices of the European and American put options are the same as those calculated for call optitins in Problem 17.12. This illustrates a symmctry that exists for at-themtiney futures options. The American option should sometimes be exerciscd early. Because K— — t oand , = y, , the European putwall parity result holds. c + Ke ” — — y Fz e "’

Also because C — — P , F e ”
0.0000

44.4491 0,0000 0.0000

tl.0000

Figure S17.2 Tree to evaluate Europcan and American put options in Problem 17.13 Problem 17.14. A futures ynice is currently 25, its volatility i.s 30% per annum, and thr risk-free interest rate is I 0"X per annum. What is the value ofa nine-month European call on the futures with a strike price of 26? In this case f, = 25, K — — 26, cr0.3 , i = 0.1 , T — — 0.75 '/ 2 d — In(N, /

— cr 2 F / 2

K)

e = e '"’[25 V(—0. 021 I) — 26'V( 0.2809)j 122 2012 Pearson Education

— 0.0211 — 0.2809

=e

’[25 x 0.491G— 26 x 0.3894] = 2.0 1

Problem 17.15. A [uture•s pric’e i.y ci1rrc•ntly 70, /fs olatilily is 20% Jeer cinnilm, and the risk-free intere.st i-atr i.s â"Xo per annwti. It hat is the valii e o/ a jii-e-month F-w-opean put on they future.s i+’itli a strihr y›ri‹ e of 65? In this case /{ = 70 . X = 65 , cr = 0.2 , - = 0.06 , F = 0.4167 T2 (L6386

d. — -

ln(I /

0.5095

2 y› = ‹•*’ " [65.V(—0.5095—) 70.V(—0.6386)j — — e ’”[65 x 0.3052 — 70 0.2G15] = 1.495

In this case

Put-cal I parity shows that we should buy one call, short one put and short a lustres contract. This costs nothing ur front. In one year, either we exercise the call or the put is exercised against us. In either case, u’c buy the asset for 34 rind close out the futures position. The gain on the short futures position is 35 — 34 = I

The put price is e ' [N.\’(—d, ) — F„ S‘(—d, )] Because .\ (—. ) = 1 — »''(i) for all .r the put price can also be written e "’ [K — Kñ' (d, ) — F,+ £{ \' (I, )]

Because £{ = K this is the same as the call price: e ’ O.F K * (‹*, )J

(‹*, ) —

This result also follows from put -call parity showing that it is not @idel dependent. Problem 17.18. Suppo› e that a future ' pri‹'e i.s currrntl y 30. The rig k-Jree intei-e.st rate is 5% j›‹•r umum. A three-month American call [utures option w ith a strike price o(28 is n orth 4. C"al‹u1l‹ite 123 F2H 12 Pearson Educ‹4 tion

bound.s [or the price cfa three-month American put funires oplion with a .emile price of ° 3

From equatitin (17.2), C— P must lie between 30c "

’ ' — 2 fi = 1 .63

a n d 3 — 0 2 S e * ‘ ' ” ‘ " = 2 . 3 5 Because C’= 4 is e iiiust hum e 1 .63 < —4 # < 2.35 or 1 .65 < P < 2.3.7

Prtiblem 17.19. Show thul if C. is the price of an Amerit an call option on a mimes c ontract when thr strike price is x anJ the niaturiH' i› T , iinJ p i.s the prire of an Aamerican put on the sawe fiitwe.. contract w'ith Use swine strike prim e unit exercise date, F„e '!" — K < C— P < F—z Ke “

when-e F i.y the]uture.s price and r i.s the risk-free rate. Assume that r > H und Ihat there i› no difference /ic6veen for w'ard anJ futures contracts. (Hint. Use an analogou.s approach to that indicated for Problem 16. 1-!.) ln this casc ne ctinsider Pot-;folio A: A tiuropean call option on fiitures plus an amount K invested at the risk-tree interest rate Pot lfolio B: M Americain ptit option on fiittires plus an amount F r " investcd at the rîskfree interest rate plus a long futures contract inaturing at time T Follti ing the arguments in C? hapter 5 we *’ill treat all matures contracts as forward contracts.

Por!jolio C. An American call futures option plus an amount Kc°" invested at the risk-free interest rate Portfolio D. A European put futures option plus an amount f„ invested at the risk-tree interest rate plus a long futures contract. Portfol iO C iS w’orth U + Ke "’ w’hile ptirtfolio D is « orth y +. If the call option is exercised at time T(ñ : T< T) portfolio C becomes:

while portfolio D is worth

Hencc portftilio D is worth more than portfolio C. If both portfolios are held to maturity (time T ), portfolio C is worth max(Nt, K) while portfolio D is worth

Hence portfolio D is u tirth more than portfolio C. Because portfolio D is worth more than portfolio C in all circumstances L!

Because

Ke "’ y L

P it follows that C + Ke " < P U — P < F — Ke "'

This proves the second part of the inequality. The result: h e "' K < C — P < F— Ke '" has thereforc been prtived. Problem 17.20. Culculate the ¡Price of a three-month European call oJ tion on the spot price o] silver. The three-month {u/ures price is S1_!, the .s/ffâe price i.s $ 13, the risk-free rate is 4’Xo, anJ the ro/a/i/ify of the price ‹j silver i.s 25“
three-month LIBOR is greater than 7fi at the option maturity, the Eurodollar futures quote at option maturity 'ill be less than 93 and there u’ill be no payoff from the option. lf the thrcemonth LIBOR is less than 7'Z«, one Eurodollar futures options provide a payotl of $25 per 0.01%. Each 0.01 o f of interest costs the corporation $125 (=5,000,000•0.0001•0.25). A total of I 25/25 = 5 contracts arc therefore required.

1 26a ^201 2 Pearson Education

CHAPTER 18 The Greek Letters Problem 18.1. 2xy/‹iia how a .stop-loss trading talr can he iniple•menteJ for the i›'ritet of an out-of-themone5' rall option. Why does it Frovide a relati› clv poor‘ hedge'! Suppose the strike price is 10.00. The optiun writer aims to be iiilly covered whenevcr the tiption is in the money and naked whenever it is out ot the money. Thc option writer attempts to achieve this by buying the assets underlying the option as soon as the asset pricc reaches 10.00 from below and selling as soon as the assct price reaches 10.00 from above. The trouble w’itli this scheme is that it assumes that when the asset price iiioves from 9.99 to 10.00, the next move will be to a price above 10.00. (In practice the next move might back to 9.99.) Similarly it assumes that when the asset price moves from 10.01 to 10.00. the next move is ill be to a price helow 10.00. (In practice the next move might be back to 10.0 .) The scheme can be implemented by buying at 10.01 and selling at 9.99. However, it is not a good hedge. The cost tif the trading strategy is zero if the assct price never reaches 10.00 and can be quite high if it reaches 10.00 many times. A good hedge has the property that its cost is always very close the value of the option. Problem 18.2. LThot doe.s it mean to assert tha I l!1e Jelio o(ci call option is 0.7.’" Heiv can a short position in 1,000 options he niaJe delta neutral when the delta o[each option i.y II.7!! A delta of 0.7 means that, when the price of thc stock increases by a small amount, the price of‘ the tiption increases by 70*o Oi this amount. Similarly, when the price o1 thc stock decrcases by a small ‹mount, the price of the option decreascs by 70% of this amount. A short position in 1.000 options has a delta of —700 and can be made delta neutral 1'ith the purchase of 700 shares. Problem 18.3. Calculate the Jelta o/ an of-he-moner 5f.I-mon/h European call oJ›tion on a non-dividendpaying stock when the risk-Ji’ee intei-est rate is I ñ% per‘ uiuium and the stock rice volatility i.s 25“Xo per- annum. In this case, S —- K , r — 0.1 , cr = 0.25 , and T - 0.5 . Also, ln{ S, I K)+ (0.1 + 0.25’ / 2)0.5 — 0.3712 0.25 The delta of the option is A'(6f,) or 0.64. d,

Prohlem 18.4. What does it mean to as›rrt that the thela of an oytion pcsition is —0. I is hen time is measureJ in venri.' I} a trader feels lliat neither ‹t stock price not- its iinf›lied volulility will change, what tvj›e o[option position is approJ›riate'! A theta of —0. i means that if ñ/ units of time pass with no change in either thc stock price or its vtilatility, the value of the optitrn declincs by 0.lA/ . A trader who feels that neither thc 127 2012 Pearson Education

stock price ntir its implied volatility will change shtiuld write an option with as high a negative theta as ptissible. Relativcly short-life at-the-money options have thc most negative thetas. Problem 18.5. the gamma ofa position is lorge anJ negative anJ the Jelta is Piero'? The gamma of an option position is the rate of changc ot’the delta of the posititin w ith rcspcct to the asset price. For example. a gamma of 0.1ould indicate that when the asset price increases by a certain small amount delta increases by 0.1 of this amount. When the gamma of an option lriter’s position is large and negative and the dclta is zero, the optitin writer w ill lose significant aintiunts of moncy i1 there is a large movement (eithcr an increase or a decrease) in the asset price. Problem 18.6. “The procedure for creating an oj›tion J›osition › ynllteticallv is the rever.be of the pi-ocrdure for hedging the option position.” kxj›lain this statement. To hedge an option position it is necessary to crcate the opposite option ptisition synthetically. For example, to hedge a lting position in a put it is necessary tti create a short position in a put synthetically. It folltiivs that the procedure for creating an option position synthetical ly is the revcrsc of the procedure for hedging thc option position.

Portfolio insurance involves crcating a put option synthetically. It assumes that as soon as a portfolio’s value declines by a small amount the portfoliti manager’s position is rebalanced by eithcr (a) selling part tif the portfolio, or (b) selling index futures. On October 19, 1987, the market declined str quickly that thc sort of rebalancing anticipated in portfolio insurance schemes could not be accomplished. Problem 18.8. The dlack-$ cljGleS- Hei‘loH /tri€.e o an oH/-G{-he-money ‹ all op I ion i»ith an c•xerci.ye price oj

$40 is $4. A traJer who has written the oJition j›lans to tise u stop-loss strutc•gv. The trader’s drill he hought or solJ. Thc strategy costs the trader o.1 0 cach time the stock is bought or sold. Thc total expected cost of the strategy, in present value terms, must be $4. This means that the expected number of times the stock will be bought or sold is approximatcly 40. The expected number of timcs it z ill be bought is approximately 20 and the expcctcd number of times it will be sold is also approximately 20. Thc buy and sell transactitins can takc place at any time during the life of the option. The above numbers are therefore only approximatcly correct because of the effects of disctiunting. Also the estimate is of the number of timcs the stock is bought ter sold in the risk-neutral wtir Id, not the ical w orld.

125 20l2re, onEjucGun

$25 is created s ’nthe•iicull y u.sing a continuullv changing J›osition in the stow k. Con.wider the following tiro scenarios.o) Stock prim e increases .steaJily front $20 to $35 during the liJe of the option. b) Stork yrice o6'c’i/la/es wildly, en6ting up at $35. Which sceiiorro nould makr the ›viilheticallv created option more expert.five'? E. pluin our The holding of the stock at any given time must be .i (‹/ ) . Hence the stock is bought just ner the price has risen and staid just aiier the price has fallen. (This is the buy high sell low strategy referred to in the text.) ln the first scenario the stock is continually bought. In second scenario the stock is bought, sold. bought again. sold again, etc. The final holding is the same in both scenarios. The buy, sell, buy, scll... situation clearly leads to higher costs than the buy, buy, buy... situation. This problem emphasizes one disadvantage tif creating options synthetically. Whereas the cost of an option that is purchased is know'n up front and depends on the forecasted x tilatility, the cost of an option that is created synthetically is not known up front and depends on the volatility actually encountercd. Problem 8.10. What i. the della O/ iy 3’IIOI“f flH.SiFiotl in 1,000 European call options on .silver/ii/urc•i i" The

option.s mature in eight month.s, and Ihe futures contract iinJerlving the option matures in nine month›. The cw+ent nine-month ft.itures pi’ice i.s â‹S User oun‹'e, the exerci.se price o] th‹• optiotls is $d, #ie risk-[ree intere› t rate is 12%^o per aiuium, and the volatilit v ofs il»er is 18°Xn The delta of‘a European futures call option is usually defincd as thc rate of change o1‘the option price with Tespect to the tutures price (not the spot price). It is e ’:S (d, ) r— — 0.12 , cr = 0.18, T — — 0.666 7 In this case F — — 8. K — — 8, / 8) + (0.18' / 2) 0.6607 = 0.18 0.6667 .v(d,) = 0.5293 and the delta of the option is 0.0735 e“ ”“ “" ' x 0.5293 = 0.488G The delta of a short ptisition in 1,000 futures options is therclore —48 b.6. Problem 18.11. In l°rohlem 18.10. what initial position in nine-month .Silver futures is nece.s.sara [or de•lta hedging‘.^ Ifsilver itsel{i.s u.sed, ii’liat is the initial ¡Position? IQ one-›’eyr .Silver futur es are useJ, ii’hat is tye initial position'! A5’5’ume no .Storage r.osls}or silver. In order to answer this problem it is important to distinguish between the rate o1 change o1 the option with respect to the futures price and the rate of change of its price w'ith respect to the spot price. The former will be referred to as the futures delta; thc latter will be referred to as the sptit delta. The futures delta of a nine-month futures contract to buy one ouncc of silver is by definition 1.0. Hence, from the answ’er to Problem 18.11, a long position in nine-month futures on 485.6 ounces is necessary to hedge thc option position. The spt›t delta of a nine-month futures contract is e’ " ' = 1.094 assuming no storage costs. l This is bccause silver can be treated in the same way as a non-dividend-paying stock when 129 rG›2012 Pearson Educaii on

there are no storage costs. F„ — — S„e'' sO that the spot delta is the futures delta times e ) Hence the spot delta of’ the option posititin is —4s 5.6 1.094 = —534.6 . Thus a long position in 534.6 ounces of sih er is necessary to hedge the tiption position. The spot delta ot a one-year silver futurcs contract to buy one ounce ot silver is e“ ' = .127a . Hence a long position in e ' ’° x 534.6 = 474.1 ounccs of one-year silver futures is necessary to hedge the option position. Problem 18.12. A company use ‹leltci heJging to hedgr a portfolio of long positions in put and call options

A long position in either a put or a call option has a positive gamma. From Figure 15.8, whcn gamma is positive the hedgcr gains from a largc change in the sttick price and loses from a small change in the stock price. Hence the hedpcr u ill fare better in case (b). Problem 18.13. /tcycor Problem J â.1s’fni a fincInc'i‹il insIilut¡on ›'vitit a purt]olio o] ,s//or/yo.si/io/is i//y‹// and ‹°‹/// oy›ricnâ on u ciIi“rency.

A short ptisition in either a put or a call option has a negative gamma. From Figrire 18.8, when gamma is negative the hedger gains trom a snail change in the stock price and loscs thorn a large change in the stock price. Hence the dredger » ill fare better in case (a). Problem 18.14. A finun‹ ial institution has jurt sold 1,000 .seven-mon/h European call options on the Japanese 'en. Suppose lhcit the sf›ot exe hanged rate i.s 0.80 cent Tier ’en, the exerci.se pric'e is 0.b1 cent per ) en, the rim k-Jr-ee interest rate in lhe United States is 8’No jO€F OrtiltiUi, the risk-free intere.f t

rafe in Japan i.s 5"Xo jeer annrim, anal tlie• volatili of IIIA YOit iE 15‘No fler annum. calculate the delta, gamma, vega. theta, ailJ rho of lhe fflancial institution’s position. Interpret each numl›er. II.08 , r— 0.05 , o = 0.1 5 , T — — 0.5833

In this case S — — 0. b0 , K — — 0.51, i 1n(0.8(1 ' 0.81) d, — d_, — — d—t 0.15 0

0.08 0.15—00.05 5 + fL15' . 2) 0.5833 ( 3 — —0.0130

\ (‹ri ) = 0.540a; (dz )'

.\' I).4995

Thc delta of tone call option is e ' ' \ (éf ) = e ' "’ x 0.5405 — so that 0.5250 . thc gamma of one call option is 0.3969 x 0.9713 = 4.206 ' .V (-S”(d,)e ,) = " — e 1).50 0.15 x 0.5833 S cr T “”" = 0.3969 130 â20 12 Pearson Edrtc a tion

The vega of one call option is .S IT.Y’ J )‹ ' ' — 0.80 0 5 0.2355 The theta of one call tiption is

3 x 0.39b9 x 0.9713 =

0.5 x 0.3969 x 0.15 x 0.9713 2 05 3 •0.05 x 0.5 x 0.5405 x 0.9713 — 0.08 x 0.81 x 0.9544 x 0.4948 = —0.0399 The rho of one call option is KTe!'"â'(J_,) — tJ.S1x 0.5833 x 0.9544 x 0.4948 = 0.2231 Delta can be interpreted as meaning that, when the spot price increases by a small amount (measured in cents). the value of an option to buy one ycn increases by 0.525 times that amount. Gamma can be interpreted as meaning that. when the sptit price increases by a small amount (measured in cents), the delta increases by 4.206 times that amount. Vega can be interpreted as meaning that, w hen the volatility (measured in decimal form) increases by a small amount. the option’s value increases by 0.2355 times that amount. When volatility increases by 1% (= 0.OJ ) the option price increases by 0.002355. Theta can be intcrprctcd as meaning that, when a small amount or time (measured in years) passes, the option’s value decreases by 0.0399 times that amount. In particular when one calendar day passes it decreases by 0.0399 / 365 = 0.000109 . Finally, rho can be interpretcd as meaning that, when the interest ratc (mcasured in decimal form) increases by a small amtiunt the option’s value increascs by 0.2231 times that amount. Whcn the interest rate increases by 1% (= 0.01), thc options value increases by 0.002231. Problem I 8.15. Under ii'ñn/ circtini,stOf?Cei fs it yossible to make a EnroJ›ean optien on ri stsck in‹1ex froth gamma neutral and vega rteu¿i ‹i/ by adding u j›o.sition in one other Eur opean option? Assume that S , K, r , c, T , q are the parameters for thc option held and £, , K‘ , r , cr, T’ , are thc parameters for another option. Suppose that 6/ has its usual meaning and is calculatcd on the basis tif the first set of parameters while d,’ is the ›'alue of di calculated on the basis of the second set ot parameters. S›uppose further that iv ot the second optitin are held for each o1’ the first optitin held. The gamma of’ thc portfolio is: where o is the number of the first option held. Since we require gamma to be zerti.

The vega of the portfolio is. 2012 Pearson Education

Since we reqiiirc vega to be zero: T X"'(d )e "" ’‘' "

T’

.X"'(d,“ )

Equating the tw ti expressions for ir T — —T

Hence the maturity of the tiption held must cqual the maturity of the option used for hedging. Problem 18.16. A flinJ manager has a w'ell-diver.sifted J›ortfolio that mirrors the per]ormance• o/tlie S'&P 5ñ H and i.y worth $3ñH million. The value of the SAP 50a is 1,200, anJ lhe portfolio manager w'ould like to buy in5’urunce against a reduction of more than 5’Xu in ltte i o/tie o) the portfolio overthe ne.xt six months. The risk-free in/era.s/ rule is 6°X‹› per annum. The dividend vield oti both the J›ortfolio and lhe S&P SOA is 3’
= 63.40 J32 The total cost of the insurance is therefore ‹ ’201 2 Pearson Education 300, 000 x 63.40 = .S19, 020, 000

0.4152

b) From put call parity

p — —c Ke

S„e “

This sho› s that a put optitin can be created by sclling (or sliorting) e ” of the index, buying a call optitin and investing thc remainder at the risk-free rate of intcrcst. Applying this to the situutitin under consideration, the fund manager should: 1. Sell 3h0e '”’”' ' = 53.54.64 million tif stock 2. Buy call options tin 300,000 times the S&P 500 with exercise price 1140 and maturity in six months. 3. Invest the remaining cash at the risk-free interest rate of G’/ per annum. This strategy gives the same result as buying put options di rectly. c) The Delta of true put option is e‘” [L fd, ) —1] ''(0.ñfi22 —1) —0.3327

e "’

This indicates that 33.270o Of the portfolio (i.c., $119.77 million) should be initially sold and invested in risk-free securities. The spot short position required is 1 99.808 . 1 times the index. Hence a short position 9 in , 7 1.023 x 250 7 futures contracts is rcquired. 0 , Problem 18.17. 0 firyerif Problem 18.16 on the o.s.siimption 0 Ihat the portfolio lia.s a heta of 1.5. As› ume Shut lhe dii-i‹1enJ vielJ on the f›ort(olio i.s 4“>oUser 0 uiiniioi. 1 When thc › alue of the portfolio goes down 2 S% in six months, the total return from the portfolio, including dii idends, in the six 0 months is —5 + 2 = —3 0 i.e.. —6 % per annum. This is 1 2"/» per annum less than the risk-free interest rate. Since the portftilio has a beta of 1.5 v c would expect the market to providc a return tif S% per annum less than the risk-free interest rate. i.e., iv c vould expect the market to prt» ide a return of —2*/» per annum. Since dividends on the markct index are 3% per annum, we would expcct l“3 'i'201 2 t'earson Education

the market index to have dropped at the rate of 5oñ per annum or 2. 5o/» per six months; i.e., we would expect the market to have dropped to l1 70. A total of 450, 000 = (1.5 300, 000) put options on the S&P 500 u ith exercise price 1170 and exercise date in six months are therefore required. a) Sz — 1200 , K — — 1170 , r — — 0.06 , n = 0.3. T — — o.5 and y dl

0.03 . Hcncc

1n(1200 / 1170) *(0.06 — 0.03 0.09 / 2) x 0.5 = 0.2961 0.3

if, - I, — 0.3

= 0.0540

S (d,) = 0 5335 Y 0.6164; .v(—d, ) = 0.3836; /Jz )' ñ.4665 The value of one put option is Ke ""L"(—d,) — She ”•v’(—d, )

A'(—I, ) =

= 1170c ° “ ' ‘ x 0.4665 — 1200c "" 5 x 0.3S3fi

= 76.28 The total cost of the insurance is thereforc 450, 000 x 76.25 = 534, 326, 000 Note that this is significantly greater than thc cost of the insurance in Problem 1 S.1 fi. b) As in Problem 15.1 6 the fund manager can 1) scll $354.64 million of stock, 2) buy call options on 450,00£i times the S&P 500 with exercise pricc 1170 and exercise date in six months and 3) invest the remaining cash at the risk-free interest rate. volatile than the S&P 500. When the insurance is considered as an option on c) The portfolio is 50fio more ln(360 / 342) 0.06 — 0.04 * 0.45 / 2) x

the portfolio thc parameters are as follows: S, — — 360, K — — 342, r — — 0.06 , 0 3517 ' (1.5 0.45 o - 0.45 , T — — 0.5 and g = 0.04 The delta of the option is = e "" ‘(0.6374 — 1)

—0.315 This indicates that 35.5% of the portfolio (i.e., 5127.5 million) should be sold and invested in riskless securities. d) We now rcturn to the situation considered in (a) where put options on the index are required. The delta of each put option is

134 \fi 201 2 Pearson Education

= e“ 0' ' (0.616—4 1) = —0.3779 The delta of the total position required in put options is —450, 000 x 0.3779 — = 170, 000 . Thc delta of a nine month index futures is (see Problem 18.16) 1.023. Hence a short position in 170, 000 = 665 1.023 x 250 index futures contracts. Problem 18.18. $how by substituting for the variou.s terms in equation (18.4 j that the equation i.s true for.a) A single European call option on a non-dividend-puying sto‹’k b) A single Etiropean put option on a non-diviJenJ-paying stock c) Anv port%lio of European yut anJ call options on a non-dividend-paying .stock a) For a call option on a non-dividend-paying stock A- .ñ'(dv

rKe°'’..N (d o Hence the left-hand side of— equation (18.4) is:

2

)

b) For a put option on a non-dividend-paying stock A= .N(d ) — 1 — —.V(—J, )

O= Hence the left-hand side of equatitin (1 8.4) is: 2

c) Ftir a portfolio of options, II, z , O and F are the sums of their values for the individual options in the portfolio. lt follows that equation (18.4) is true ftir any portfolio of European put and call options. 1 35 2012 Pe'eis vn Education

Problem 18.19 What is the equation corresponJing to equation (18.4) for (a) a portfolio o[derivatives on a currenc y and ) a portfolio of Jerivatives on a futures price.'! A currency is analogous to a stock paying a continuous dividend yield at rate r . The differential equation lor a portfolio of derivatives dependent on a currency is [sec equation 16.6)

Hence 1 2 Similarly, for a portfolio of derivatives dependent on a futures pricc (see equation 17.5) 2 Problem 18.20. Suppose that $7t1 billion of eqnit) o.s.set.y are the subject of portfolio insurance .schemes. A.s.sume that the schemes are designed to proviJe insurun‹ e again 'I they value of the a.s.sets declining by more than 5“fo within one year. Making whatever e•stimate•s you find neces.vary, use the Der/vaGeni ioJir'nre to calculatc ltte value of the sto‹'k or) / eel contracts that the udmitlistrators of the portfolio insurance schemes will attempt to sell iJ the mat-ket fcills hy 2.1‘No in a .single ‹my. We can re ard thc position of all ortfolio insurers taken together as a sin le ut o tion The three known parameters of the option, before the 23‘/ decline, are 5, = 70, K — — 66.5 , T — - i Other parameters can bc cstimated as r(0.0— — , = 0.25 and y = 0.03 . Thcn: 1n(70 / 66.5) —60.06 0.03 5 + 0.25' / 2) ().4t02 d= 2

The delta of thc option is

.Y(d, ) = 0.fi737 e ”[â"{d,) —1]

— — e°”'(0.6737 —1) = —0.3167 This show s that 31.b70o or $22.17 billion ot assets should have been sold beforc the decline. Thesc numbers can also be prtiducecl from DerivaGem by selecting Underlying Type and Index and Option Type as Black-Scholes European. After thc declinc, S — — 53.9 , ñ = 66.5 , T — — 1 , › = 0.06 , o = 0.25 and 9 = 0.03 . In(53 9 / ti6 5) + (0.06 — 0.03 * 0.25" / 2) d — ' — 0.25 0.5953 h"{d, ) = 0.2758 The delta of the option has droppcd to

136 ›20 12 Pearson Education

e '“' ’ (0.2758 — 1) = —().7025 This shou s that cuniulatively 70.2Sfi oof the assets originally held should be sold. An additiolaal 35.C 1 Noof

the original portfolio should be sold. The sales measured at pre-crash prices are about $27.t1 billion. At post-crash prices they are about $20.b billion. Problem 18.21. Dory a fomvurJ ‹ ontract on a .stock index have the same delta as the c'orre.sfloncling ft1tiire.f ‹ onlrut’t? Explain vour answer. With our usual notatitin the value of a forward contract on the asset is S,r "— K "' . When there is a small change, a.S , in S, the walue of the forward contract changes by c AS. The delta of the forward contract is therefore e " . Thc ifiturcs price is S„e' ”’ . When there is a small change, AS , in S, the futures price changes by A5"e" "’ . Given the ddily scttlcmcnt procedures in fixtures contracts, this is also the immediate change in the wealth of the holder of the matures contract. The delta of the futures contract is thcrelore c' "" . We ctinclude that the deltas of a futures and for ard contract are not the same. The delta of the tutures is greater than the delta of the corresponding forward by a factor of e"' . Problem 18.22. A bank ’.s yo.yi/ian in options on the dollar—euro exchange rate hn.y a delia of 3.0,000 and a gamma o/ —80, 000. E.mphin loo» these numbers can lie interpreted. The exchange rate (dollars per euro) is 0.9tt. What position would you take to make the position delia neutral’? After a whorl period o/free, the exchange rate move.s to 0.93. Estimate the ne›• Jelta. What additional trade is ncccs sury to keep the po.sition delta iie£fH‘o/'? AsSUIttiH@ lhr bank did set up a delta-netilrul pos ition oi‘iginallv, he.s it gaineJ or lost money from the c. change-rate movement? The delta indicates that when the walue of the euro exchange rate increascs by $0.01, the value of the bank’s pt sition increases by 0.0 1 30, 000 = $3OO . The gamma indicates that when the euro exchange rate increases by $0.01 thc dclta o1 the portfolio decreases by 1. x SO, 000 = 800 . For delta ncutrality 30,000 euros shtiuld be shorted. When the exchange rate moves up to 0.93, we expect the delta of the portfolio to decrease by (0.9—3 0.90) SO, 000 = 2. 400 so that it becomes 27,600. To maintain delta neutrality, it is therefore necessary for the bank to rim ind its shtirt ptisititin 2,4(l(l euros so that a net 27,600 have been shortcut. As shown in the text (see Figure 1 S.8). when a portfolio is dclta ncutral and has a negative gamma, a ltiss is experienced v. hen there is a largc movement in the undcrlying asset price. We can conclude that the bank is likely to ha›'c lost money. Problem 18.23. L'se the put all parity relationship to Jerive, for a non-dividend-j›aying stock, the relation.ship heni'een.(a) The delta of a Eur-opean c'all anJ the delta o(a European put. (b) The gamma ofa European ‹call and r/ie gamma o(a F-urope•an pul. (c) Th‹• veg‹i o/u European ‹ all anJ lhe ega of a European put. (d) Tlie• lhela of a Euro;›ean call and thcn the•ta of a Europeanput. 137 20.1.2 Pearson Education

(a4 For a non-dividend paying stock, put-call piiri ty gives at a general time : I* — — ‹ Ke°' '" ’ Differentiating with respcct to S : ”

—

ep _ c'r' US US This shows that thc delta tif a Furopean pia etluals the delta of the ctirresponding Eurtipean call less 1.0. (b) DifTerentiating with respect to S again US' Eyz Hence the gamma o1 a European put equals the gamma of a European call.

(c) Differentiating the put-call parity relationship 'ith respect to m ‘ TO TO show'ing that the vega of a Eurtipean put equals the vega of a European call. (d) Differentiatin s thc put-call parity relationship

ith respect to

0y

., ,.„ ‹‹

This is in agreement with the thetas of European calls and puts given in Scction 1S.5 since .V(d,) =—I x' —d_,) .

13fi

CHAPTER 19

Volatility Smiles Problem 19.J. What volatility smile is likely to he obj erveJ when (a) Both tvil ' oj the stock pt i‹:e distribution are less heavv than the.be of the lognormal distribution‘? (b) The right fail is heavier, anJ the left tail is /e.s.s heavy, than that ofa lognorinal distribution.’" (c) (d)

A smile similar to that in Figure 19.7 is observcd. An upward sloping volatility smile is tibserved

Problem 19.2. What volatility .smile i.s oh.sem'ed/or ‹equities.' A downward sloping vo!ati1ity smile is usually observed for equitics. Problem 19.3. What volatility smile is likely to be couseJ bi jumps in the underlying asset price? is the p‹illc.rn likelv to be more pronounceJ Jm a two-year option than/or a three-month option? Jumps tend to makc both tails of the stock price distribution heat ier than those of the lognormal distribution. This creates a volatility smile sirriilar to that in Figure 1.9.1. The volaiility smile is likely to be more pronounced for the three-month option because jumps tend to get “averaged out” o er longer perio‹ls. Problem 19.4. A Ltfropean call and put option ha»e the in/ne strike price and lime to inaturi 3. The call ha ' an implied volatility of 30“Xo oHd 111€ yttl how’ an implieJ volatility oJ 25%. Ref traJes w ould you do.'! The put has a price that is too low relative to thc call’s pricc. Thc correct trading strategy is to buy the put, buy the stock, and sell the call. Problem 19.5. Exf›lain carefully why’ a di.y/ribu/ion with a heavier le/t tail and less heavy right tail than the lognormal distribution give.s rise to a doivnA’ai-d.slo ring i'olatili3 simile. The heavier left tail should lead to high prices, and therefore high impl ied volatilities, for outof-the-money (low-strike-price) puts. Similarly the less heavy right tail should lead to low prices, and therefore low volatilities for out-of-the-money (high-strike-price) calls. A volatility smile where volatility is a decreasing function of strikc pricc rcsults. Problem 19.6. They market price ofa European call is 83.00 anal its price given hv Black-Schole.s-Merton model ›i'ith a volatili3 o] 30% is $3.50. The price given hy thi.s Black-Scholes-Merton model 139 20.12 Pearson Education

for a European yut option worth the same str ike prire anJ tin e to maturity is $1.00. Mat s houlJ the market price of the put option be.'* Exyluin the reasons f‹›r your an.s»er. With the notation in the text

It follows that ’6s

"md i

/’bs

hmm

In this case. c „, = 3.00 ; c„ = 3.50 ; and p„ — — 1.00 . lt follow’s that p„should be 0.50. Problem 19.7. Explain n’hat i.s meant hy cra.sliophohia. The crashophobia argument is an attempt to explain the prtintitinced x’olatility skew in equity markets since 1957. (This was the year equity markets shocked ex eryone by crashing more than 20% in one day). The argument is that traders are concerned about another crash and as a result incrcase the price of out-of-the-money puts. This creates the volatility skew. Problem 19.8. A stock price is currently 520. Tonlorro ', ne»s is expected to be announc'eJ that n ill either in‹+ease the price bv $5 or Jecrease the price hy $5. What are the problem.f in using Black— Scholes-Merton to value one-month options on the .stock? The probability distribution of the stock price in true month is not lognormal. Possibly it consists of two lognormal distributions superimposed upon each tether and is bimodal. Black— Scholes is clearly inappropriate, because it assumes that the stock price at any future time is lognormal. Problem I 9.9. What volatility sinile i› likel y to he observeJ for- si. -monlh options w'hen the volatility i.s uncertain and J›ositively correlateJ to the s toc'k pri‹’e’? When the asset price is positively correlated with volatility, thc volatility tends to increase as the asset price increases, producing less heavy left tails and hear icr right tails. Implied volatility then increases w’it1a the strike price. Problem 19.10. What problems ‹In ’on think w'oiilJ he encoiuitered in testinp a .stock option pricing model emf›iri‹’all ? There are a number or problems in testing an option pricing model empirically. These include the problem of‘ obtaining synchronous data on stock prices and option prices, the prtiblem tit estimating the dividcnds that will bc paid on the stock during the option’s life. the problem of distinguishing between situations w here the market is inefficient and situations z'here the option pricing model is incorrect. and the problcms ol cstimating stock price volatility. Problem 19.11. Suppose that a ce•ntral bank ’s yolicy i.s to allow an exchange rate to fluctuate betw een 0.97 140 ' 2012 Pcatson EJB catiun

and 1.03. What pattern o[in1plieJ vol‹itilities for options on the e.x‹'hange rate would vou expel I to see? In this case the probability distribution of the exchange ratc has a thin left tail and a thin right tail relative to the logntirmal distribution. We arc in thc opposite situatitin to that described for foreign currencies in Section 19.1. Both out-of-the-money and in-the-money calls and puts can be expected to have lower implied volatilities than at-the-money calls and puts. The pattern of impl ied volatilities is likcly to be similar tti Figure 19.7. Problem 19.12. Option traders sometimes i’efer to deep-out-of-the-money options a› being options on volatility. Why Jo you think tyey do thi.s? A deep-out-of-the-money optitin has a low value. Decreases in its volatility reduce its value. Howevcr, this reduction is srnah because the value can never go below zero. Increases in its volatility, on the other hand, can lead to significant percentage increases in the value of the option. The option does, therefore, have some of the samc attributes as an tiption on volatility. Problem 19.13. A European cull option on a c'ertain stock ha.s a .strike price oJ $3ñ, a time to Mofurifv oJ one year, anJ an implied volatilij of 30°X‹›. A European put option on the 'ame stock has a s tribe price of $50, a time to maturity ozone year, and an implie•J volatility o] 33°X‹›. What is the arbitrage opportunij open to a trader? Does the arbitrage work only when the log ormal assumption unJerlying Black—Schole.s—Merton holds." Explain carefully the reasons Jor vour answer. Put—call parity implies that European put and call options have thc same implied volatility. If a call option has an implied volatility of 30% and a put option has an implied volatility of 33% , the call is priced too low relative to the put. Thc correct trading strategy is tti buy the call, sell the put and short the stock. This does not depend on the lognonral assuinptitin underlying Black—Scholes—Merton. Putwall parity is true for any set of assumptions. Problem 19.14. Supposes that ltte rrsull oj! a major lav'snit affecting o company is duc to be announced tomorrow . The comp‹in y 's stock price i.s currently âd0. lfth • ruling is Javoruble to the company, the stock pric'e is expected to ;’uin¡› to $75. If it is unfa vorable, the stock is expe‹:teJ /oyumy to $50. What is he risk-neutral prohability ofa favorable ruling? Assume that the volatility of the company’s .stock n'ill he 25“Xo for si.x months ‹icier the riilinq i f the ruling i.s favorable and 4H°Xu if it i.s unfavorable. Usr Deriv'aGein to call ulate the relationship henveen implieJ volatili5 and.strike price for st. -month European options on the company today. The company doe.s not pay dividends. Assume lhat the six-month risk-free rate i.s 6%. Con.wider- call ovation.s n'ith .strike J›rices of $30, 840, $50, 560, $70, and 580. Suppose that p is the probability of a favorable ruling. The expected price of the company’s stock tomorrow is 75p 50(1 — /›) = 50 * 25a This must be the price of the stock today. (We ignore thc expected return tti an investor over one day.) Hence 50 * 25p — — 60 141 r0.›201 2 Pearson liducation

or p — — 0.4.

If the ruling is favorable, the volatility, o-, will be 25%. Other option parameters are Sz — — 75, r — — 0.06 , and T — — 0.5 . For a i-alue tif K equal to 50, DerivaGem gives the value of a European call option price as 26.502. If the ruling is unfavorable, the volatility, cr will be 40% Other option parameters are Sz — — 50, r — — 0.06 , and T — — 0.5 . For a value of K equal to 50, DerivaGem gives the value of a European call option price as 6.310. The value today of a European call option with a strikc price today is the weighted average of 26.502 and 6.310 or: 0.4 x 26.502 + 0.6 x 6.310 = 14.3 b7

DerivaGem can be used to calculate the implied volatility when the option has this price. The parameter values are N, = 60, K — — 50, T — — 0.5 , r — — 0.06 and c = 14.387 . The implied volatility is 47.76%. These calculations can be repeated for other strike prices. The results are shown in the table Strike fitThe i‹'e pattern Call Pri e: volatilities is Call Price. Weizhi‹c! empties V‹›iotiiity bclow'. of implied shown in Figure SU.1. 30

Favorable Outcome 45.fiS7

'

Lnlâi'orable Outcome 21.001

Price 30.955

(°Xu) 46.67

I 2.437 6.3 I O 2.S2G 1.161 0.451

21 .935 14.357 5.5f›4 4.430 1.934

47.75 47.76 46.05 43.22 40.36

36.152 2G.502 17.171 9.334 4.159

40

50

60 70 80

(%)¡o/\ p!aIdt

u j

50 48 46 44 42 40 38 20

40

60

80

Strike Price Figure S19.1

Implicd Volatilities in Problem 19.14

Problem I 9.15. An exchange rate i.s currently 0.R000. The volatility of the exchange rate is quoted ct.$ 12‘No and intere.st rate.s in the rim countrie.s are the .same. 7'.s ing the lognormal assumption, rstimute the F•^!••^ ility that the exchange rate in three months will he (a) les.s than 0.7000, (b) betw'een 0.7000 and 0.7500, (c j betw ecu ñ. 7500 and 0.b0t)0, (d1 hetw'een 0.8000 and 0.8500, (e) beti+'een iI.8500 and 0.9ñ00. and greater than 0. 9000. Base•d on the volatilitv 142 '2012 Pearson Education

smile usually observeJ in the market [or exchange rate›, which of these estimate.f would you expect to be too low and 'hich would you exye‹’t to be too high? As pointed out in Chapters 5 and 1 6 an exchange ratc behaves like a stock that provides a dividend yield equal to the foreign risk-free ratc. Whereas the grtiwtli rate in a non-dividendpaying stock in a risk-ncutral w orld is r , the growth rate in the exchange rate in a risk-neutral world iS r — r . Exchange rates havc low systematic risks and so z’e can reasonably assume that this is also the growth rate in the real world. In this case the foreign risk-free rate equals the domestic risk-free rate (r = r ). The expected growth rate in the exchange rate is therefore zero. If S is the exchange rate at tirllC T its probability distribution is given by equation ( 14.3) with =0: In S, ‹p(In TO — cr27 / 2, cr2zero 7 ) and o is the volatility of the exchange rate. In this where N, is the exchange rate at time case S, = 0.8000 and o = 0.12 , and T — — 0.25 so that In ST

e

(In 0— . 8 0.12' x 0.25 2, 0.12 x 0.25) In S, - ‹p(—0.2249. ñ.ñ6’ )

a) In 0.70 = W.3567. The probability that S, < 0.70 is the same as the probability that ln Sz < —0.35fi7 . It is —0.3J67 0.2249 0.06 This is 1 .41%. b) In 0.75 —0.2877. The probability that S,. < 0.75 is the same as the probability that ln N< —0.2877 . lt is —0.2877 + 0.2249 .\'(—1.0456) 0.06 This is 14.79%. The probability that the exchange rate is bctween 0.70 and 0.75 is therefore 14.79 — 1.41 = 13.38 . c) In 0.80 = —0.2231. The probability that S < 0.b0 is the same as the probability that In N, < —0.2231 . It is —0.2231 + 0.2249 0.06 This is 51.20%›. The probability that the exchangc rate is between 0.75 and 0.80 is therefore 51.2—0 14. 79 = 36.41fo . d) In 0.85 = —0.1 ti25. The probability that S < 0.85 is the same as the probability that ln N< —0. 1 625 . It is —0.1625 * 0.2249 0.06 This is 55.09%. The probability that the exchange mite is bemeen 0.b0 and 0.55 is therefore 85.0—95 1.20 33.89 % . 143 2012 Pearson Education

e) ln 0.90 = fi. 1054. The probability that N, < 0.90 is the same as the probability that In N< —0.1054. It is —0.1054 + 0.2249 : 0.06 V This is 97.69%. The probability that the exchange rate is between 0.55 and 0.90 is therefore 97.6—9S 5.09 = 1.2.fall* . D The probability that thc exchange rate is greater than 0.90 is 10—0 2.31 %

97.G9 =

The volatility smile encountered ftir ftireign cxchangc options is shown in Figure 1.9.1 of the text and implies the probabili5 distribution in Fisu e 19.2. Figurc 19.2 suggests that we z ould expect thc probabilities in (a), (c), (d), and (f) tti be too low anal the probabilities in (b) and (e) to be too high. Problem 1.9.16. A stock price i.s 540. A si.x-month European rall option on the .stock n›ith a .str ike price oJ S30 how’an implied volatility o(35°Xu. A six-month £nroyeuo c‹i// option oti the .stock ii’ith a strike pti‹ r of S.i0 has an imJ›lied volritili h’ of 28%. The siy-montlt i‘isk-free rate is 5’>n and no JividrnJy arch expected. E.xplain vi hi thy m.’o inip1ie‹l vohili1itie•s are Ji f[ercnt. Use Deriv'a Mem to c.alculote the j›rices of the Evo options. k's‹• piil—‹Neill purit5 to calculate the The difference beta een the two implied i olatilities is consistent with Figure 1.9.3 in the text. For cquities the volatili smile is dtiw nw ard sloping. A high strike price option has a lower implicd volatility than a lo1' strikc price tiption. The reason is that traders consider that the probability ot a large downz'ard movement in the stock pi-ice is higher than that predicted by the lognomial probability distr ibution. The implied distribution assumed by tradcrs is shown in Figure 19.4. T‹› use DerivaGein to calculate the price of the first tiption, proceed as follow s. Sclect Equity as the Underlying Type in the first worksheet. Select B lack-Scholes European as thc Option Type. Input the stock pricc as 40, volatility as 35%, risk-free rateaS 5fi o, time tti exercise as

0.5 year, and exercise price as 30. Leave the dividend table blank because we are assuming no div idencls. Select thc button corresponding to call. Cio not select the implied i olatility button. Hit the Enter ka y and click on calculate. DerivaG em 'ill sliov the price of the tiption as 11. . 1.55. Change the volatility to 28k o and the strike price to 5tL Hit the F-nter ke yand click on calculate. DerivaGem w ill show

the pricc o1 the option as 0.725. Put—call parity is c + Kc ’ = * St

so that For the first tiption, ‹ = 11 .1 55 , J„ = 40, r = 0.054 , K — — 30 . and T — — 0.5 so that p— — 11. 155 30e ' '”' 40 0.414 For the second option, r = 0.725 , S„ — — 4ñ, r— — ñ.n6 . A— 50 . and T — — 0.5 so that /› = 0.725 5(Ie " "" — 40 = 9.490 To iisc DerivaGeir to calculate the implied volatility o1’ thc tirst put option. input

price as 40, the risk-free ratc as 5%. time to exercise as 0.5 year . and the exercise price as 30. Input thc pricc as 0.414 in the second half of the fJption data table. Select the buttons for a put option and implicd volatility. Hit the Lnter key and click on calculate. Dcri›'aGem will show the implicd ›'olatility as 34.99%. Similarly, to use DerivaGem tti calculate the implied volatility o1 the tirst put option, input the stock price as 40, the risk-free rate as 5 . time to exercise as 0. 5 year. and thc cxcrcisc pricc as 50. Input the price as 9.490 in the second half ot the Option Data table. Select the buttons for a put optitin and implied volatility. Hit the Enter key and click tin calculate.

DerivaGena will sht»v the implied volatili9 aS 27.99k o.

These results are what we would expect. DcrivaGem gives the implied volatility of a put vitli strike pricc 30 tti be almost exactly the same as the implied › olatility of a call u’itli a strike price of 30. Similarly, it gives the implied › olatility o1’ a put with strike price 50 to be almost exactly the same as the implied volatility of a call moth a strike price of 5fL Problem 19.17. “The Black—Scholes—over ton model r.s u.red log' trader.s as an interpolation tool. ” Discuss thiy When plain vanilla call and put optitins are being priced, traders do use thc Black-ScholcsMerton model as an interpolatitin tool. They calculate implied volatilitics for the options whose prices they can observe in the market. By interpolating beta een strike prices and betwccn times to maturity, thcy estimate implied volatilities for other options. Thesc implied volatilities are then substituted into Black-Scholes-Merton to calculate prices for these options. In practice much of the work in producing a tablc such as Table 19.2 in the over-the- counter market is done by brokers. Brokers olicn act as intermediaries between p‹irticipants in the over-thecounter market and usually ha›'c more information on the trades taking place than any individual financial institution. The brokcrs provide a table such as Table 19.2 to their clients as a service. Problem 19.18 k!sing Toble 19.2 calculate the implied 'olatility a ti ader ii -oald ii.se for an d-month option with K/S — 1.04. The implicd ›'o1atility is 13.45%›. We get the same ans* er by (a) interpolatiiig between strike priccs ot‘1.00 and 1.05 and then between maturities six months and one ycar and (b) intcrpolating between maturities tif six months and one year and then betwccn strike prices of 1.00 and 1.05.

145 â'2012 Pearson Editca tion

A binomial tree cannot be used in the w ay described in this chapter. This is an example of what is known as a history-dependent option. The payoff depends on the path followed by the stock price as veil as its final value. The option cannot be valued by starting at the end of the tree and v orking bucks ard since the payo1‘1 at the final branchcs is not known unambiguously. € hapter 26 describes an extension o1 the bintimial tree approach that can be used to handle options where thc payoff depends on the average value ot the stock price. Problem 20.6. “For a dividen‹l-paying .stock, the tree Jor the .stocL price Joes not recombine, hut tl1‹• tree for the s lock yrice less lhe pre.senf 'alti0 of future dividen‹ls does reconihine. " E. ph in thi.s ,s/u/e/neiif.

Suppose a dividend equal tO D iS paid during a certain time interval. If S is the stock price at the beginning of thc time intern at, it will be either .S« — D or S‹—/ 6 at thc end of the time interval. At the end of the next time interval. it trill be one tit (N« — D)u , (Si—i N)af , tT .e—l D)u and (id— D)d . Sincc (.S— D)d does not equal ‹id — D)u the tree does not recombine. 1 f 6 is cqual to the stock price less the prescnt value of future dividends, this problem is ay oided. Problem 20.7. Shot» that /he pro6‹/bi/ivies iii u Cox. Ross. and Rubinstein binomial tree are neg‹itive when hue c oitclition in footnote 9 fioJr/s. With the usual notation

II r < I or o > ii , one ‹if the two probabilities is ncgative. This happens u hen e"” ' < e ° OF

This in turn happens when (y —rJ ticcur u hen

> o OF { r —g)> cr Hence negative probabilities

This is the c‹indition in footnote 9.

In Table 20. cells A1, A2, A3,..., A100 are random numbcrs bemeen 0 and 1 defining how far to the right in the square the dart lands. Cells B1, B2, B3,...,B100 are random numbers between 0 and 1 defining how high tip in the square the dart lands. For stratificd sampling we could choose equally spaced values for thc A’s and the B’s and consider every possible conilsination. To generate 100 samples w e nced ten equally spaced values ftir the A’s and the 145 *.:2012 Pearson Education

B’s so that there are 11) i 0 = 1.00 combinations. The equally spaced i alues should be 0.05. 0.15, 0.25,..., 0.9.5. We could thcrcfore set the A s and S’s as follow s: A1 = .42 — A3 = ... = A10 = 0.05 A1 I = A1.2 = A13 = ... = A 20 = 0.15

and

A 91 = A 92 = A 93 — ... = A100 = 0.95 BI = B1.1 = H 2.1 = .... = B9.1 = 0.05 B2 = BI 2 — B2.2 = ... = H 9.2 = 0.15

Bl 0 = B 20 — h3.11 = . . . = B1 t10 = 0.95

We get a valuse to cqual to 3.2, which is closer to the truc value than the i-alue of 3.04 obtained with randtiin sainpl ing in Table 20.1. Because samples are not random we cannot easily calculate a standard error of the cstimate.

In Monte Carlo simulation sample values for the derivative sccurity in a risk-neutr‹i1 ii ‹orld are obtained by simulating paths ftir the underlying variables. On each simulatitin run, ›-alues for thc underlying variables are first determincd at time A/ , then at time 21/ , then at time 3a/ , etc. At time it/(i = 0,1, 2 ...) it is not Possible to determinc u hcther early exercise is optimal sincc the range of paths u’hich inig•ht occur atter time i.A/ ha› e ntit been investigated. ln short, klonte Carlo simulation w orks by mov ing forward from time i to time /. Other numerical proceclures is hich accommodate early exercisc work by rr1O\‘ing backwards from time r to time i . Problem 20.10 A nine-month Ainericcin put motion on u non-‹li Us nJ-pan in rtock he.s a .strike Eric.e o[S49. The slow k prim e i› $50, the riwk-fr‹•c• ratt• i8 "No y£'t tlnntiin, and the volalili n’ i s 30’>o per rinai/m. k‘›'e a three-.stej› l›inomial tree. to cah ulute the option price. In this case, S, = 50 , ñ = 49, - = 0.05 .

0.3f) , T — — 11.75 , and ñ/ = 0.25 . Also

149 20.1.2 Pearson Lducation

0

‘0

= l . i 61 8

d— — = 0.8607 a = e"“' = e’ ’°” = 1.0126 = 0.5043 1 — y› = 0.4957 The output trom DerivaGein for this example is shown in the Figure 520.3. The calculated price of the option is $4.29. Using 100 steps thc price obtained is 53.91

Bolded val ues are a result of e xe rcise irowth factor pe r ste p, a = 1.0126 Probability of up move, p = 0.5043 Up ate p size, u = 1.161B Down ste p size, d = 0.8607

78.41561 0 67.49294 0

5B.0917T

58.09171 0

L.429187

50 2,91968

50 4.289225

43.0354

43.03S4 7.308214

5.964601

37.04091

31.88141 17.1 1859 Node T! me:

o,aoo

0.5000

0.2500

Figure 520.3

0.7500

Tree for Pi oblem 20.10

Problem 20.11. Use a three-time-step tree to volue a nine-month Amer ican call option on ›il1eat futures. The current futures price is 40a ce•nts, the .strike price is 420 cents, the risk-free rate is 6’u, and the volatility is 35% per annum. E,stimate the delta of the option]rom your tree. In this case I, — — 400, K — — 420, r = 0.06 ,q = 0.35 , T — — 0.75, and Al = 0.25 . Also

150 2012 Pearson Education

a p1

a —d 0.4564

—1 p— — 0.5436 The tiutput from Derii aGem for this example is shown in the Figure S20.4. The calculated price of the option is 42,07 cents. Using 100 time steps the price obtained is 3S.ti4. The optitrn’s delta is calculated from the tree is (79.97—1 11.419) / (476.49—8 33.5.753) = 0.487 When 100 steps are used the estimatc of the option’s delta is 0.4b3.

At each node: Upper value = Underlying Asset Price Lower value = Option Price Bolded values are a result of exercise

Growth factor per step, a - 1.0£00 Probability of up move, p = 0.4564 Up step size, u = 1.1922 Down Step size, d = 0.8395

676.1835 256.1835 567.627 147.627

476.4985 79.971

476.4985 56.49849

335.7828 11.41894

335.7828 0

400 42.06767

281.8752 0

236.6221 0 Node Time: 0.00£0

0.2500

Figure S20.4

0.5X 0

0.7500

Tree for Problem 20.11

Problem 20.12 .4 three-month American call opiion on u s tock 11a.s a .strike price of $2a. The sto‹’k pri‹’e is .$20, the risk-fi-ee rale is 3"X per annum, and the volatility i5 2§ o per annum. A dividend of 52 i.s expected in 1.I months. Use a three-.step binoii›tal trees In ‹’al‹’ulate the option price.

In this case the present valuc of the di 'idend is 2e "”’ '” — — 1.9925 . We first build a tree for S, — — 20 —1.9925 = 1 8.0075 , K - 20,r = 0.03 , n = 0.25 , and T — — 0.25 u’ith Al = 0.0b333 . This gives Figurc S20.S. For nodes between times0 and 1.5 months w e then add the present 151 ?"^20l2 Pearson Educati on

value of the dividend to the stock price. The result is the tree in Figure S20.6. The price of the option calculated from the tree is 0.674. When 100 steps are used the price obtained is 0.690. Tree shows stock prices less PV of dividend at 0.125 years Growth factor per step, a = 1.0025

Node Time: 0.0000

0.0833

Figure 520.5

0.1667

0.2MD

First tree for Problem 20.12

At each node: Uppe r value = Underlying Asset Price Lowe r value = Option Price Bolded values are a result of exercise

Node Ti me: 0.0000

0.0833

0.1667

0.2500

Figure S20.6 Final Tree for Problem 20.12 152 ’fi›201 2 Pearson Education

Problem 20.13 A one-y’c›ar Ame•rirun pul option on a non-dii'idend-pa i ing s low k has an exerci.se firir.e of $18. The cut-rcnt stock price is $20, the ri.@-/rec interest ralr is IS°X• per annum, cmd the volatiliN of the .stoc.L is 40’X p‹•r annum. K.be the f9erivuGem softirurc with [car three-month time steps to e.stimate ltte vulue o]! the oytion. Display’ the tree anal verts that the option pi‘ices of the final and prniilliinute noJes are correct. L’se Det-ii uGem to value the European vrrs ion of the option. k'“se i/ie ‹’ontrol vm-inc tecltniqur In impro› e your e.stimate of tlte. pri‹-e o] the Amer ic'‹in option. In this casc S, — — 20 . K — — 18, r = 0.15 , cr = 0.40 , T — — 1 , and Al = 0.25 . The parameters for the tree are ii = e°

= e“

= I .2214

d— —1/u— — 0.S1.87

P— °

a —‹/

—

1.038-2 0.8187 =

u d 1.2214 — 0.545 0.8187 The tree produccd by DerivaCiem for the American option is shown in Figure S20.7. The estimated value of‘ the American option is $1.29.

Uppe r Tal uc = Unde rlyi ng Ass I P n ce Lowe r Tal ue = 0 ption Prt¢e Bal d ed yalues as a resul I of eve ro se Growth fartot per she p, a = I 03B2

P ro babil'h of up movt, p = 0 SdS1 Up ate p stzc, u • 1.Z214 Doc n s te p si ze, d = 0.8T87

16.37462 2.475954

2 9 . 8 3 6 4 9

16.37't62 2 clz886

24.4

0.882 034

i' igure S20.7

Tree to cvaluate American option for Problcm 20.13

153 ' :20l 2 Pearson Education

The tree produced by DerivaGem is shown in the I igure S20.9. The estimated price of the option is $14.93.

Uppe r value = Underlying As3et Pri¢e Lower val ue = Option Price eol ded values a re a re s ult o I exercise Probabil ity of up move, p = 0 sJS9 Up ste p size, u = T.0524 Down ste p size, ds 0.9'202

0 564.0698

836.0069 0

536.0069 0 509.3401 4.8148z1

509.340J 0

9.986432 459 9206 26.84302

2o.71293 d37.0392

437.0392 32.9608fi

42.8606S

415.296I

0.0833

0.1260

N ad e ii me

Figure S20.9

0.1667

Tree tti evaluatc optitin in Problem 20.14

Problem 20.15 How can the ‹’ontrol var/‹ifc approach to improve the estimate o[thr delta man option when the binomial free aypi’oach is used?

Ameri‹an

First the delta of the American option is estimated in the usual way from the tree. Denote this by a’, . Then the delta o1 a European option z hich has the same parameters as the American option is calculated in tlxe same w ay using the same trec. Denote this by ñ} . Finally the true European delta, A, , is calculated using the formulas in Chapter IS. The control variate estimate of delta is then:

Problem 20.16. Suppo.se that Monte Carlo siinul‹ition is hring ii.sed to evaluate a European call option on a non-dividend-paying .stock when the volatilij i.s .stochastic. How coulJ the control variate and aiItitheli‹ varial›le technique he useJ to improv'e numerical e/]i‹ ieney? Explain why it i.s nece.s.vary fc' calculate six values of thr option in eric/t sim elation lt-ial 'hen both the control vat-rate anJ the antithetic variable technique are used.

In this casc a simulation requires two sets tif samplcs frum standardized normal distributions. The first is to generate the volatility movements. The second is to generate thc stock price movements olice the volatility movcments are know n. The control variate technique involves carrying out a second simulation on the assumptitin that the volatility is constant. The same 155 fi20 12 Pearson Education

random number stream is used to generate stock price movements as in the first simulation. An improved estimate of the option price is where /is the option value from the first simulation (w’hen the volatility is stochastic), / is the option v-alue from the second simulation (when the volatility is constant) and /, is the true Black-Scholes-Merton value when the volatility is constant. To use the antithetic variable technique, two sets of samples from standardized nonnal distributions must be used for each o1’ volatility and stock price. Denote the volatility samples by (V, * and I? J and the stock price samples by (S, / and , S, ,'. U, is antithetic to ( Hz / and /S, J is antithetic to J 2 /. Thus it (V, ) — — 0.53, +0.41, —0.21... then {V› z' — — —0.b3, 0.41, +0.2 I ... Similarly for 'S, ,! and I, . An efficicnt way of proceeding is to carry out six simulations in parallel: Simulation 1: Use t'S, ,' z ith volatility constant with volatility constant Simulation 2. Use ,I Simulation 3: Use (S, and U, Simulation 4. U.se ,'S, and ( I<, and JIC, , Simulation 5: Use N Simulation 6: Use /J z and U, ,' i , simulations 3 and 4 provide an estimate It f is the option pricc from simulation 0.5( /} + /, ) for the option price. When thc control v'ariate technique is used u e combine this estimate wrth thc result of simulation 1 to obtain 0.5( /, + 5—), / + /\ as an estimate tif the price where {, is, as above, the Black-Scholes-Merton option price. Similarly simulations 2, 5 and 6 provide an estimate 0.5(f, + J,— j,_ fB . Overall the best estimatc is: 0.5[0.5( * f,—) f, + f!z + H.5(f, J,) — /) + J D ) Problem 20.17. 6. ya/niri finn equations (20.27) to (2ñ.3i1 j changr when the implicit Jinile difference method is heing u.seJ to evu/tia/e an American rall oytion on a currenc . For an American call tiption on a currency

With the mutation in the text this bccomes 2/ 24S for y = 1. 2. . ti — 1 and i 0, 1 ... Y — 1 . Rearranging terms we obtain , ’-,

+ b ,.

‘'r „

,

where 156 â20l2 l'earscn Education

-/'

,

2

1 1 Equations (20.28), (20.29) and (20.30) become /,; — max[yAS — K, 0]

c, = —— r - ry ) /A/ —— cr

/ = ñ,i ...M

=0 i = 0,1...,V /„ = M US — K / = 0,1... N Problem 20.18. An American put option on a non-dividend-paying stock has%ur months to maturit) . The exercise price is $21, the. stock price is 520, the ri.wk-free rufe o/ intere.at iS 1K’Xo per annum, anJ the volatility is 30’Xo yer annum. Use the e.xylicit version a} the finite dijferen‹ e approach to value the option. Use stock price intervals of $4 and time interval.s of one month. We consider stock prices of $0, $4, $S, $12, $1 fi, S20, $24, $28, $32, $36 and 540. Using equation (20.34) with r = 0.10, Al — 0.0833 , US — — 4, rr 0.30, K — — 21 , T — — 0.3333 we obtain thc grid shown below. The option price is $1.56. Stock Price Time to Maturity (Months) ($) 4 3 40 0.00 0.00 0.00 0.00 0.00 36 0.00 0.00 0.00 0.00 0.00 32 0.01 0.00 0.0tJ 0.00 0.00 28 0.07 | 0.04 0.00 0.00 0.02 24 0.1 I 0.00 0.38 0.30 0.21 1.56 1.44 1.31 1.00 20 1.17 16 5.00 5.00 5.00 5.00 5.00 12 9.00 9.00 9.00 9.00 9.00 8 13.00 13.00 13.00 13.00 13.00 4 I 7.00 | 17.00 17.00 17.00 17.00 0 21.00 21.00 21.00 21.00 21.00 Problem 20.19. The spot price of copper is $0.60 per pound. Suppose that the futures prices (dollars per pounJ) are as follow.s.' 3 months 0.59 6 months ñ.57 9 months 12 months The volatility0.54 o/ the price oj copper’ is 40’Xn per annum and thr risk-See rate is 6’X per annual. U.se a0.50 binomial tree to value an American call option on coJ›per with an e.xercise price of $0.60 and a time to maturi5' o] one year. Divide the life of the oplion into four 3month Feriods for the puryo.se.s of constructing the tree. (Hint. A.s explained in Chapter 17, the futures price o) a variable is its expected Juture price in a ri.sk-neutral w'orld.) In this case At — 0.25 and o- = 0.4 so that 157 â›2012 Pearson Education

e’2 ‘

— ii

1

— 1.2214

1 = 0.81 87

The futures prices provide estimates of the growth rate in copper in a risk-neutral world. During the first three months this growth rate (with continuous compounding) is 41n $ = —6.72% per 6 annum The parameter for the first three mtinths is therefore e“ °‘"“" — 0.8187 0 4088 1.2214 0.8187 The growth rate in copper is cqual to 13.79%â, — 21.63% and 30.75% in the following three quarters. Therefore, the paraincter p for the second three months is e ' ” “’ 2‘ 0 3660 0.8187 1 .2214 — 0.81 For the third quarter it is b7 e "*”"— 0.8187 0.3195 1.2214 — 0.8187 For the final quarter, it is — IN.8187 = 0.2f›f›3 1.2214 0.8187 The tree ftir thc movements in copper prices in a risk-neutral w orld is shown in Figure 520.10. The value of the option is $0.0ti2. 0.895 0 295 0.733

0.133

0.60

0 006 £

1.093 .733 0.493 0.133

0.600

0.600 0.042

U.401

1.335 0.895 0735 0 295

0.000 0.491

0.015

0.000

0402

0.000

0.329 0.000

0.40 2 0.000

0.270 ODOO

Figure S20.10 Trec to value option in Problem 20.19: At each node, upper number is price o1‘ copper and lower number is option price Problem 20.20. Use the binomial tree in Problem 2ñ.19 to value o security that pays oJJ x2 in one year where x i.‹ the price o/ copper. In this problem wc use exactly the same tree for copper prices as in Problem 20.19. How ever, the values of the derivative are differcnt. On thc final nodes the values or the derivative cqual 15S â2012 Pearson Education

the square of the price of coFper. On othcr nodes they are calculated in the usual way. The current value of the security is $0.275 (see Figure 520.11).

Figure S20.11 Tree to value derivative in Problem 20.20. At each node, upper number is price of copper and lower number is derivative security price. Problem 20.21. Men do the boundary conditions %r S — — 0 and S prices in the explicit finite Jflerence method?

z affect the estimate.s of derivative

Define S as the current asset price, S,p as the highest asset price considered and S lowest asset price considered. (In the example in the text 5„, = 0). Let

as the

,— and Q, S , and let V be the number of time inten'als considered. From the triangular structure of the calculations in the explicit version of the finite difference mcthod, w e can see that the values assumed for the derivative security at S — $„, and S - S„, affect the derivative’s value if Q

S , —'’

Problem 20.22. How would you u,se the antitheiic voriahle• method to imf›rove ltte estimate of the European option in Bit.Hne.s.s Snapshot 20.2 and Tahle 20.2? The following changes could be made. Set LI as = NORMSINV(RAND()) A1 as =$C$*EXP(($ E$2-$FS2*$F$2/2)*$G$2*$FS2*L2*SQRT(SG$2)) Hl as =$C$*EXP((SE$2-$F$2*$F$2/2)*$G$2-$F$2*L2*SQRT($G$2)). 11 as = EXP(-$E$2*$G$2)*MAX(HI-$D$2,0) and /1 as

159 v2012 £earSc£t Education

=0.5*(B I*.11) Other entries in columns L, A, II, and I are defined siirlilarly. The estimate of the value o1 the option is the average of the values in the J column Problem 20.23. A company’ has issued a three-year con vertible 6ond that has a face value of $25 unJ con be exchanged for Evo of the c.oinJ1any’s shares at any time. The e'ompuny can call the issues, forcing conver.sion, when thc share pri‹'e is greater than or equal to $16. ns.string that they company will force conversion at the eat liesl opportunity, what vre the boundot9 ‹!oné/i1fOfi5 for the price of the convertibles.'" De.scrihe how' you w'ould u.se finite di/fer‹•ncc• metlioJs to company defaulting. The basic approach is similar to that described in Section 20.5. The only difference is the boundary conditions. For a sufficiently small value of the stock price, S ., , it can be assumed that conversion will ncver take place and the convertible can be v alued as a .straight bond. The highest stock price which needs to be considcrcd, N, „ , is $1 S. When this is reached the value o1 the convertible bond is $36. At maturity the convertible is worth the grcater ot 2N and $25 where S is the stock price. The convertible can bc valued by working backwards through the grid using either the explicit or the implicit finite difference method in conjunction with the boundary conditions. In formulas (20.25) and (20.32) the present value of the inctime on the convertible betwcen time I i at and t + (i +1) at discounted to time r + i At must be addcd to the right-hand side. Chapter 26 considers the pricing of convertibles in more detaiL Problem 20.24. Provide formulas thai can be u.sed for obtaining threc ranJom s‹imples from .stanJard normal disti-ibutions when the correlation helm een sample i and.sample ; i5' p . Suppose x, , .y, and y are random samplcs from three independent normal distributions. Random samples with the required correlation structure are ci , „ r, w'here

and where

and This means that

â^201 2 Pearson Lducation

161 â2012 Pcarson Education

CHAPTER 21 Value at Risk Problem 21.1. CoHSfdet dyos’/ffo/i COnsisting o[a 5100,000 investment in asset A anJ o $100,000 investment in n.sse/ B. Assnine that the daily volatilities of hoth assets are 1’Xo turd that the coe]jicient of correlation between their returns is 0.3. What i. the 5-day 99°Xo VaR for the portfolio'? The standard dev iation of the daily change in the investment in each asset is $1,000. The variance of the portfolio’s daily change is 1, 000" + 1, 000" * 2 x 0.3 x 1, 000 x 1, 000 = 2, G00, 000 The standard deviation of the portfolio’s daily change is the square root of this or $1,612.45. The standard deviation of the 5-day change is 1, 6 I 2.45 x 5 = $3, 505.35 From the tables of ,\'(i) we see that N(—2.33) = 0.01 . This means that 1% of a normal distribution lies more than 2.33 standard deviations bclow the mean. The 5-day 99 percent i alue at risk is therefore °.33. 3, 605.55 $S, 401 . Problem 21.2. Dese'ribe three way’s of hanJling interest-rate-depenJent instrument.s when the moJel huilJing approach is used to calculate VaR. How ›voulf you handle intere.st-rate-Jependent in.s/rumen/s ii’hen historical siinulation is u.sed to calculate VAR,’! The thrcc alternative procedures mentioned in the chapter for handling interest rates whcn the mtidel building approach is used to calculate VaR involve (a) the use of the duration modcl, (b) the use of cash low mapping, and (c) the use of principal components analysis. When historical simulation is used we need to assume that the change in the zero-coupon yield curve between Day m and Day m + 1 is thc same as that bebveen Day i and Day i 1 for different values of i . In the casc of a LIBOR, the zero curvc is usually calculated from deposit rates, Eurodollar futures quotes, and swap rates. We can assume that the percentage change in each of these between Day m and Day m I is the same as that bctween Day i and Day i + 1 . In the case of a Treasury curve it is usually calculated from thc yields on Treasury instruments. Again we can assume that the percentagc change in each of these between Day m and Day w + 1 is the samc as that between Day i and Day i + 1 . Problem 21.3. A financial institution ov n› a portfolio of options on the U.S. do/law—,sterling exchange mite. The Jclta of the portfolio is 56.0. The current e.re/iange ratr is I.5000. Henive an appro.ximute• linear relationship bel seen the change in the portfolio value and the percentage change in the excñnge rate. If!the duilv volatili5' of the e*c/tange rate is 0. 7°/o, estimate the 10-day 99°X» VaR. The approximate relationship between the daily change in the portfolio value, 6s , and the daily change in the exchange ratc, AS . is IP— 56AS

The pcrcentage daily change in the exchange rate, At , equals aS / 1.5 . lt follow's that 1 62 "°. 2012 Pearson Education

4 P = 84A

The standard deviation of‘ At equals the daily volatility of the exchange ratc, or 0.7 percent. The standard deviation of 6s is therefore 84 x 0.007 0.58S. It follows that the l 0-day 99 percent VaR for the portfolio is 0.588 x 2.33 x = 4.33 Problem 21.4. Suppose yoti know that the gamma of the poi-t%lio in the firevious question is 16.2. How does this change your estimate of the relation.ship hetw'een the change in the portfolio value and the percentage chunge in the exchange rate'? The relationship is W = 56 x I .5Ar -F1 1 .52 x 16.2 x A ' k P— — 84Ax + 18,225ñi’ Problem 21.5. Suppose that the daily change in the value oJ a poi‘tfolio i.y, fo a good appro.ximation, linearly dependent on tw›o factors, calculated]i-om a principal components analpsis. The delta o] a portfolio with respect to lhe first]i« ter is 6 and the delta ith re,spect to the second]acior is 4. The standard de•viations of the}a‹ ter are 2ñ and R, re.spectivelv. What is the 5-day 9K‘Xo VaR'? The factors calculated thorn a principal ctimponents analysis are uncorrelated. The daily variance of the portfolio is 6’ x 202 * 42 x 8’ = 1 5, 424 and the daily standard deviation is 5 424 = $1 24.19 . Sincc .V(— l .282) = 0.9, the 5day 90% value at risk is 124.19 x x 1.2b2 = $35a›.0 1 Problem 21.6. Suppose a company has a portfolio consisting of positions in stoc'ks, bonds, foreign e.xchange, and commodities. As.sumy there are no dci-ivaiives. Explain the assumptions underlying (a) the linear model and C) the historical simulation moJel for ‹’alculating VaR. The linear model assumes that the percentage daily change in each market variable has a normal probability distribution. The historical simulation model assumes that the probability distribution observed for the percentage daily changes in the market variables in the past is the probability distribution that will apply over the next day. Problem 21.7. Expl‹iin how' an interest rate .yway i.s maF ped into a portfolio oj zero-‹’oupon bonds with 'tanJarJ maturities for the purpo.se.s ma VaR colculal ion. When a final exchange of principal is added in, the floating side is equivalent a zero coupon

bearing bond. which is equivalent to a portfolio of zero-coupon bonds. The swap can therefore be mapped into a portfolio of zero-coupon bonds with maturity dates corresponding to the payment dates. Each of the zero-coupon bonds can then be mapped into positions in the adjacent standard-maturity zero-coupon bonds. Problem 21.8. Explain the Jijferenr.e betw'een Value at Risk and Expected Shortfâll. Value at risk is the loss that is expected to be exceeded (10—0 J)% of the time in N days for specifled parameter values, x and .V . Expected shortfall is the expected loss conditional that the loss is greater than the Value at Risk. Problem 21.9. Explain why the linear model can j›rovide only uppro.ximate e.stimates of VaR for a portfolio containing options. The change in the value of an option is not linearly related to the change in the value of the underlying variablcs. When the change in the values of underlying variables is normal, the change in the value o1 the option is non-normal. The linear model assumes that it is normal and is, therefore, only an approximation. Problem 21.10. Some time ago a company has entered into a forward contract to buy £1 million [or 5/.5 million. The contract now has six months to maturity. The daily volutilii y’ ofa six-month zero- couj›on sterling hond (when its pri‹’e is truns lated to dollars) is 0.06‘Xo OHd thE daily volatility of a .six-month zero-coupon dollar bonJ is H.HS‘No. The correlation between returns fiom the two honds is 0.d. The current exchange rate is I. 53. Calculate the standard deviation of the change in the Jollor value of the [orwarJ contract in one day. What is the 10-dO}› 99’Zo FOR? Assume that the six-month interest rate in both sterling and dollars is 5°X‹› per annum with continuou.s compounding. The contract is a long position in a sterling bond combined with a short position in a dollar bond. The value of thc sterling bond is 1.53c*’”“' or $1.492 million. The value of the dollar bond is l .5c '”" ’ or $1 .463 million. The variance of the change in the value of the contract in one day is 1.492' x 0.0006’ + 1.463’ 0.0005' — 2 x 0.S 1 . 492 x 0.0006 x 1.463 0.0005 0.000000288 The standard deviation is therefore $0.000537 million. The 10-day 99% VaR is 0.000537 x x 2.33 = 30.00396 million. Problem 2L11. The text calculate.s a VaR estimate jor the examjlle in Table 21.9 assuming m'o factors. How does the estimate change if you assume (a) one factor and (b) three factors. II we assume only one factor, the model is IP — — —0.08/ The standard deviation of f is 17.49. The standard deviation of a/• is therefore 0.08 x 17.49 = 1.40 and the 1-day 99 percent value at risk is 1.40 2.33 = 3.26 . If we assume three factors, our exposure to the third factor is 164 2012 Pearson Education

10 x (—0.3 7) + 4 x (—0.3 8) — 8 x (—0.30—) 7 x (—0. 12) * 2 x (—0.04) = —2. 06 The model is thcrclorc AP = —0.08 — 4.40 {—2 2.Of› /\ The variance of s P is 0.08' x 17.49' * 4.40' x 6.052 + 2.06' x 3.10' = 751.3G The standard deviation of 6s is 51 3 = 27.41 and the 1-day 99ko Value at risk is 27.4 I x 2.33

$63.S7 .

The exmple i11usi:ates that the relative importance ot diffcrcnt factors tlepends on the portfolio being considered. Normally the second tâctor is less important than the first, but in this case it is much more important. Problem 21. 12. A bank Isa.s a portfolio o[options on an asset. The delta of the options i ' —30 anJ the gamma is —5. E-xplain how these numbers ‹ an be /nf6ryire/ed. The asset prim e is 20 anJ its volatility per day i.s I“to. Adcipt !Sample Application E in the DerivaGem Ap ›lication Builder software The delta of the options is the rale if change of the value of the options with respect to the price of the asset. When lhe asset price increases by a small amount the value of the options decrease by 30 times this amount. The gamma of the options is thc rate of change of their delta with rcspect to the price of the asset. When the assct price increases by a small amount, the delta o1‘ the portfolio decreases by five times this amount. By entering 20 ftir S , 1 % for the volatility per day, -30 for delta, -5 for gamma. and recomputing we see that Eik P)= —n. Iñ , E(L P“) = 36.03 , and £(AP’) = —32.415 . The 1day, 99% VaR given by the softwarc tor the quadratic approximation is 14.5. This is a 99% 1day VaR. The VaR is calculated using lhe formulas in footnote 9 and the results in Technical Note 10. Problem 21.13. Suppose lhat in Problem °1.12 the vega o[the• portfolio is -2 per I’X• change in the annual volutilitv. Derive a model relating the change in lhe portfolio value in one day to delta, gammu, anJ vega. E.xplain i+'ithout doing Jelailed call ulutions how you would u.se the model to e’alrulate a VaR estimate. Define as the volatility per year, A o as thc change in in one day, and An and the proportional change in n in one day. We measure in o- as a multiple o1 1% so that the current value of cr is I x 252 = 15.87 . The delta-gamma-vega model is ñP = —30iSS — .5 x 5 x (AS)' — 2Acr AP = —30 20s (Ar)' — 2 x 15.87Aw

— 0.5 x 5 x 202

w’hich simplifies to W = —600A—v 1, 000(Ar) — 31 .74Aw Thc change in the portftilio value now depends on two market variables. Once the daily volatility ‹if o- and the correlation be l ce n o- and S have been estimated we can estimate moments of nr anö usc a Cornish—Fisher expansion. Problem 21.14. The one-Jan 99’X VOR i.s CO/C1f/6fred/or the Jour- inJe.x examj›le in Section 21.2 as $253, 3d5.

Lent ml he underlying s]›read.sheet.s on the author 's website anJ cal‹’ulate the o1 the 95’No one-Jan VuR and b) the 97"Xo one-day VaR. The 95% one-day VaR is the 25th worst loss. This is $156,511. The 97% one-day VaR is the 15th worst loss. This is $172,224. Problem 21.15. Use Ihe spread.sheets on the author’s web site to calculate ihe one-Jay 99“Xo VctR, using the hasie methodology in Section 21.2 if the %ur-inJex portfolio considered in Section 21.2 is equally Jivided between the font- indice.s. In the “Scenarios” worksheet the pvrtto! in investmcnts are changed to 250tl in cells L2:02. The losses are then sorted from the largest tti the smallest. Thc lifth worst loss is $238,526. Tims is the one-day 99 o VaR.

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CHAPTER 22 Estimating Volatilities and Correlations Problem 22.1. Explain fire exponentiallv weighted moving average (EW:M) model/or e.stimatiiig volatility fi o i hi›'torical Jata. Dcfine u as (S — N ) / , , where is value of a iiiarket variable on day i . In the EWMA model, the variance rate o1 the market variable (i.e., the square o1 its volatility) calculated for day n is a w'eighted average of the u{ ’s ( i — 1, 2, 3. ... ). For stime constant ? ( 0 < ? < 1 ) the weight given to «{ , is 2 times the weight given to «{ . Thc volatility estiiiiated ftir day ii , rr, , is related to the volatility estimated for day — n 1, o- , , by n — âo* , (l — J.)ii, , This formula shti»'s that the EWMA model has one very attractive property. To calculatc the volatility estimate for day n , it is sufficient to know' the volatility estimate for day n — 1 and ". i Problem 22.2. What is //ie Jifjerence between the exponen!iullv and the MARCH(1, 1) mo‹Iel for updating volatilities?

eighteJ moving average model

The EWMA mtidel produces a forecast o1 the daily variance rate for day n which is a weighted average of(i) the forecast for day 1, and (ii) the square of the prtiportional change on day n — 1 . The GARCH (1.1) model produces a forecast of the daily variance for day u which is a w’eighted average of(i) the lorccast for day ii — 1, (ii) the square of the proportional change on day n — 1 . and (iii) a long run average variancc rate. GARCH (1,1) adapts the EWMA model by giving some weight to a long run average variance rate. Whereas the EWMA has no mean reversion, GARCH (1,1) is consistent with a meanreverting variance rate model. Problem 22.3. The mo.st recent estimate of the Jailv volatilih of an a.s.set is 1..f*o ooJ f/7e pric e ol the asset at the close of traJing ye› terHai was $30.00. The parameter ñ in the £ Web model is 0.94. Suppo.se that the price of the oss'ef a/ the e’lo.se o[trading today is $30. 50. How will this cuuse the volatility to bv• updated b the EWMA model? In this case cr, i — 0.01 5 and u,

0.5 / 30 = 0.01667 , so that cquation

(22.7) gives rrj = 0.94 x 0.015 + 0.06 x 0.01667’ = 0.00022 S1 The volatility estimate on day n is therefore

0 00228 = 0.015103 or 1.5103%.

Problem 22.4. A comJ›an user an EWS moJel for foreca,sting volatility. It JeciJes to change the parameter 2 from 0.95 to 0.85. Exj›lain the likely impact on the forecast.s. 167

Reducing d from 0.95 to 0.85 means that mtire weight is put on recent observations of «and less weight is given to older observations. Volatilities calculated with I — 0.85 will react more quickly to new information and will “bounce artiund” much more than volatilities calculated withâ = 0.95 Problem 22.5. The volatility ofa certain market variable i.s 30% jeer annum. Calculate a 99% confiJence internal %r the size o] the percentages daily change in the variable. The volatility per day is 30/ 2 2 = 1.89% . There is a 99% chance that a normally distributed variable will be within 2.57 standard deviations. We are therefore 99% confident that the daily change will be less than 2.57 x 1. b9 = 4.86% . Problem 22.6. A ‹’ompany uses the GA RCH(1,1) model/or upJating volatilit). The three parameters are ai, a, and p. the.ycrif›e the impact ol making a small increase in each of the parameters while keeping the others fixed. The weight given to the long-run average variance rate is 1— a — Q and the long-run average variance rate is m / (1 — o — 9). Increasing o increases the long-run average variance rate; increasing a increascs thc weight given to the most recent data item, reduces the weight given to the long-run average variance rate, and incrcases the level o1 the long-run average variance rate. Increasing 9 increases the weight given to the presions variance estimatc, reduces the wcight given to the long-run average variance rate, and increases thc level of the long-run average variance rate. Problem 22.7. The most recent estimate of the Jaily volutilil y of they US dollar—sterling exchange rate i.s 0.6’Xo Ond the exchange rate at 4 p.m. yesterJuy n'us 1.5000. The parameter L in the EWMA model is 0.9. Suppose thal the exchange rate at 4 p.m. today prove.f to he 1.4950. How ›‹ould the e.stimate of the Jaily vol‹iiility be upJated? The proportional daily change is —0.005 1 .5000 = —0.003333 . The current daily variance estimate is 0.006’ = 0.000036. The nez' daily i ariance estimate is 0.9 x 0.000036 * 0.1s 0.0033332 = 0.000033511 The new volatility is the square root of this. It is 0.00579 or 0.579%. Problem 22.8. Assume that S&P 500 at close of troJing yesterJuy ii'‹/s I, 04s and the daily volatility o/the index wa.s e.siima/eif as 1°Xu per Jay at Ihat lime. Th€ p6fKometers in a GARCH(1, 1) model are m — — 0.000002 , a = 0.06, and Q 0.92 . IJ the level of ihe inJex at close of trading today i.s 1,060, what is the new volatilit)' e.yfimaie? With the usual notation u, , = 20 / 1040 = 0.01923 so that cr = 0.000002 * 0.06 x 0.019232 + 0.92 x 0.01' = 0.0001 1 62 so that cr = 0.01078 . The new volatility estimate is therefore 1.078% per day.

165 0c?20I 2 Pearson Education

Problem 22.9. >!PF! e that the daily volatilities of asset A and asset B calculated at the close of trading

ye,sterday’ are 1.6“Xu add 2.5’Xo, respectively. The prices of the assets at close of trading yesterday v ere 820 and 540 and the rstim‹tte of the coe icieni of correlation between the returns on the two assets was 0.25. The pai-ameter ? useJ in the EWMA moJel is 0.95. fa) Calctilate the current e.stimate of the covariance l›etw'een the assets. (h) in the a.s.sumption that the price.s of the assets at close of trading today area $20.5 and $40.5, u¡›date the correlation estimate. (a) The volatilities and correlation imply that the currcnt estimate of the covariance is 0.25 x 0.016 x 0.025 = 0.0001 . (b) If the prices of the assets at close of trading are $20.5 and $40.5, the proportional changes are 0.5 20 = 0.025 and 0.5 / 40 = 0.0125 . Thc new covariance estimate is 0.95 x 0.0001 * 0.05 x 0.025 x 0.0125 = 0.0001106 The new variance estimate for assct A is 0.95 x 0.016’ * 0.05 x 0.025’ = 0.00027445 so that the new volatility is 0.01 fi6. The new variance estimate for asset B is 0.95 x 0.0252 + 0.05 x 0.0125" = 0.000601 562 so that the new volatility is 0.0245. The new correlation estimate is 0.00t)1 I Of

—

0.272

0.0166 x 0.0245 Problem 22.10. The parameter.s o/a 6iA RCH(1, 1) modc•l are estimated as n› — — 0 000004, ri 0 05, oud Q— — 0.92 . What is the long-run a erage volatility and ›rhat is the equation de•scribing the u’ay that the variance rate revert.s to it.s long-run average? If the current volatilil y is 2O’Xo per year, what i.s the exj›ected volatilitv in 20 days.'! The long-run average variance rate is / (1 — a — ) or 0.000004 / 0.03 = 0.0(i01 333 . Thc long-run average ›'olatility is 0 0001333 or 1.155%. The equation describing the way the variance rate reverts to its long-run average is equation (22. 13) F. a'"

—V + u a’ — V

In this case E(,c 0.0001 333)

] = 0.0001 333 * 0.97‘ (rr° —

If the current volatility is 20% per ycar, cr, = 0.2,* 252 = I).(J126 . The expected variance rate in 20 days is 0.0001333 11.97 2’ (0.01 26 — 0.000 1333) = 0.000 1471 The expected volatility in 20 days is therefore 47 = 0.0 1 2 I or 1.21% per day. Problem 22.11. Suppose that the curre•nt daily vo/a/i/rfie* of asset X anJ asset Y are 1.0’X‹› and 1.2°X‹›, respecli vely. The price›' of the assets at close o[trading ve.sterday were $30 and $50 and the estimate oJ the ‹’oe fit rent of correlation between the return.s on the type assets made at thi,s lime was 0.50. Correlations and volatilitie.s are updated itsing a GA RC H(I, I) moJel. The estimates of the model’s parameter.f are a — — 0.04 and Q — — 0.94. for the correlation

Using the notatitin in the text o-, , — 0.0 l and

, - 0.012 and the most recent cstimatc of

o, the covariance between the asset returns is »„ , = 0.01 x 0.01 2 x 0.50 0.00006 . The variable i‹ = 1/ 30 = 0.03333 and the variable v, , = 1/ 50 = 0.02. The new estimate of the covari‹ince, cov , is 0.000001 + 0.04 x 0.03333 x 0.02 tl.94 x 0.00006 = (L000(184 1 The nez estimate of the0.000003 variance*of thexfirst asset, cr'0.94 is x 0.0 I’ = 0.0001 414 0.04 0.03333“ so that o-, = second 14 4 = 0.011 s9 or I . I 89%. The new estimate of the variance of the asset, o-}, is 0.000003 + 0.04 x 0.02" -F 0.94 x 0.012" = 0.0001544 so that u , = 0 000 544 — 0.01242 or 1.242%. The ne estimatc ot‘thc correlation between the assets is therefore 0.0000841 / (0.011 59 x 0.01242) = 0.569 . Problem 22.12. Suppose that the daily volatility o / / h c FTSE ION stock index fn7eastir’ed in poULlJs stei‘ling) iS

7.8% and the Jaily volatilin of the dollar/sterling e.xchange rate is 0.9%. Suppose further that the correlation between the L TSL 100 anJ the dollar?sterling exchange rate is 0.4. What is the volatilil y of the FTSE I fJ0 when it is tran.Plated to L’.S. dollar.s? A.Lennie that the dollar/sterling exchange rate is expressed a.s the numher o[k’.S. dollar.s per pound sterling. (Hint.’ When Z.— X 1 , lhe percentage daily change in Z is aJ›pro.xiniatel equal to the J›ercentage Jails' e’hunge in X plus the perce•ntage daily change in i .) The FTSE exprcsscd in dollars is xr where z is the FTSE expressed in sterling and r is the exchange rate (value of one pound in dollars). Define i as thc proportional change in x on day i and ¿ as thc proportional change in r on day i . The proportional change in ,Ye is approximately x. y. . The standard de› iatitin tif x is 0.01S and the standard deviation of t, is 0.009. The correlation between the two is 0.4. The variance o1 x. * y. is therefore 0.0 IS' -F 0.009’ -P 2 x 0.018 x 0.009 x 0.4 0.0005346 so that the volatility tif .r +/ is 0.023 I or 2.31%. This is the volatility o1 the FTSE expressed in dollars. Note that it is greater than the volatility of the FTSE expressed in sterling. This is the impact of the positive correlation. When the FTSE increases the value of stcrling measured in dollars also tends to increase. This creates an even bigger increase in the value of FTSE measured in dollars. Similarly for a decrease in the FTSF. Problem 22.13. Suyyose that in Problem 22.12 the ‹ orrelation betw ecu the S&P 500 InJex (nieusureJ in dollars) and the FT-SE 1OH InJex (measureJ in sterling) is 0.7, the c'orrelation hefween the SAP 500 index (mea.cured in Jollars) and the Jollar-sterling exe hunz«e rate is 0.3, anJ ltte Jaily volatility of the S& I° 500 Index i.s 1.6%. What i.s the correlation heRveen the S&P 500 Index (measured in dohers) and the FT-SE 100 Index n›hen it i.s tran.slated to dollars? (Hint.’ h’or three variable› X, i anJ z , the cov‹triance l›etw een X Y and z equal.‹ the covariance hetv’een Xand Z plus the envariance bet ‹en z and z .1

170 2012 Pearson Education

Continuing with the notation in Problem 22.12, define r as the proportional change in the value o1 the S&P 500 on day i . The covariance between x and z. is 0.7 x 0.018 x 0.016 = 0.000201 6 . The covariance betw'een y and z is 0.3 x (L 0149 x 0.tl1 6 = 0.0000432 . The covariance between . + j and z equals the covariance between i and c plus thc covariance betw een y and z . It is 0.000201 6 + 0.0000432 = 0.0002445 The correlation between i + J and z is 0.000244b = 0.662 0.0 1 6 x 0.0231 Note that the volatility of the S&P 500 drtips out in this calculation. Problem 22.14. Show that the GARCI-I (1, I) model n — m + an' ,+ o, ,

in equation (22.9) is equivalent to the .stochaslf€ YOlOtility fOdFl = a2 dV tic — —volatilitv aV V) df hen V dWhat is the stocha› model time i.s urea.sure in years? where The timevariable is measured Jays anJ on V i.sthe theasset .square o[the (Hint. ti, , i.yinthe return price in time Al. It can be assumed to be i'olatiliti of they asset price and normolly Jistribule.J with mean z-ero and.yfandnrd devio/ion cr, , . It follow's that the mean of u"

Ould u , are a , and 3

, , respecli v'ely.)

rr,'

ft + as , , + 9o-/ ,

so that '

' — a, ,

(,0

1) crj

The variable u , has a mean of u- and a variancc o1‘ tu,— [*(•/ , )]' — 2o‘/., 2rr , . Assuming the i are generated by a Wiener process, The standard deviation ot «{ , is , we can therefore w'rite wheres is a random sample front a standard normal distribution. Substituting this into the equation for o-} o-/ we get

171 2012 Pearson Educe tion

,

Because time is measured in days, Al = I and The result follows. When time is measured in years Al = 1/ 252 so that and the prticess for I< is

Problem 22.15 At the end o/Section 22.R, the VaR %r the four-inJex example eras calculated using the model-building approach. How doe.s the VaR calculateJ chunge ifthe investment is !2.5 mil/ion /0 each index!! Cary out calculations when a) volatilities anJ correlations ure estimuteJ using the equa//v weighted model and h) when fhey aide estimated using the EWMA moJel with ? — — 0.94 . Use the ,spreadsheets on the author ’.s web site. The alphas (row 21 for equal weights and row 7 for EWMA) should be changed to 2,500. This changes the tone-day 99%» VaR to $226,836 z'hen volatilities and correlations are estinuted using the equally z eighted mtidel and to $487,737 n'hen EWMA with = 0.94 is used. Problem 22.16. What is the effect of changing 2 from 0.94 to 0.97 in the EWMA co/cuJafioas in the%urit1de.x example at the enJ oJ!Section 22.$.! 7’se the spreadsheets on the author’.s wehs ite. The parameter 7. is in cell N3 of the EWMA worksheet. Changing it to 0.97 reduces the oneday 99% VaR from $471,025 to $389,290. This is because less weight is given to recent observations.

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CHAPTER 23 Credit Risk Problem 23.1. The spreuJ betw een the yield on a three-year corporate bond and the ) ield on a similar risk-free bonJ is 50 hasi.s points. The recove ry rate is 30%u. Estimate the average huzurd rate per vear over the three-year period. From equation (23.2) the average hazard rate over thc three years is 0.0050 / (1 — 0.3) = 0.0071 or 0.71'i› per ycar. Problem 23.2. Suppose that in Problem 23.1 the .spread between the yield on a five-year bonJ issued hy the same company anJ the yield on a .similar risk-free bond is 60 basis points. Assume the same recovery rate of30°Xu. Estimate the average hazard rnte per yrar over the five-year period. What Jo your result.s indicate about the average huzard rate in year.s 4 and 5 From equation (23.2) the average hazard rate over the five years is 0.0060 / (1— 0.3) = 0.008s or 0.b6o ñ per ycar. Using the results in the previous question, the hazard rate is 0.71% per year for the first three years and 0.0086 x 5 —0.0071 x 3

- 0.0107

2 or 1.07% per ycar in ycars 4 and 5. Problem 23.3. Should re.searchers use real-»orlJ or risk-neutral default prohal›ilities for a) calculating credit value at risk anJ b) a usting the price ofa derivative %r defaults'? Real-world probabilities of default should be used for calculating credit value at risk. Risk-neutral probabilities of default should be used for adjusting the price of a derivative for default. Problem 23.4. How are recover5 rates u.stia//y defined? The recovery rate for a bond is the value of the bond immediately after the issuer defaults as a percent of its face value. Problem 23.5. Explain the difference betw'ern an unc'onJitional default prohal›ility den,sity and a hazard rate. The hazard rate, ñ(/) at time i is defined so that h(t)bi is the probability of default between times / and / Al conditional on no default prior to time / . Thc unconditional default probability density g(/) iS dcfined so that y(/)A/ is the probability o1 default between times / and / At as seen at time xero. 173 ?c 2012 Pearson Education

Problem 23.6. Veri/j' a) lhat the numbers in the second column of Tahle 23.4 are consistent with the numbers in Table 23.1 anJ b) tltat the iuimloer.s in the fourth column o[Tahle 23.5 are can.st.Alex/ n i/ñ the niimbet-y in Tahle 23.4 and a recovery rate o[40%. The first number in the second column of Table 23.4 is calculated as —7 1n(1 — 0.00245) = (I.()(lfJ3504 or 0.04% per year when rounded. Other numbers in thc column are calculated simi larly. The numbers in the fourth column of Table 23.5 are the numbers in thc sccond column or Table 23.4 multiplicd by one minus the expected rectivery rate. In this case the expccted recovery rate is 0.4. Problem 23.7. De.scrihe how netting works. A bank already has one tran.faction n›ith a counterpart on its books. Explain who a ne»’ transaction by a bank w'ith a counterpart can have the e]fect of increa.sing or reducing the hank’s credit exposure lo the cor.interpart y. Suppose ctinipany A goes bankrupt when it has a number of outstanding contracts with company H. Netting means that the contracts with a positive value to A are netted against those 'ith a negative value in order to determine how much, i1 anything. company A owes company B. C ompsny A is not alltiwed to “cherry pick” by keeping the positive-value contracts and defaulting on the negative-›'alue contracts. The ncw transaction will increase the bank’s exposure to the connlerparty if the ctintract tends to have a positive value whenever the existing contract has a positive value and a negative value whcnever the existing contract has a negatix e value. However, if the ne transaction tends to offset the existing transaction, it is likely to leave the incremental effect or reducing credit risk. Problem 23.8. What is meant hy a haircut in a collaterali ation az«reernent? A comp‹fnv of/ers to poor its own eqnity as collateral. How’ ould you responJ? When securities are pledged as coI1atera1 the haircut is thc discount applied to their market value for margin calciilatitins. A company’s own equity would not bc good collateral. When thc company defaults on its contracts its ••iuity is likely to be w'orth very little. Problem 23.9. Explain the Ji})erence bel›reen the Gaussian copula model%r the time to de[aull and CreditMetrics as Jar as the jollo ing are concerned.’ u) the definition o[a credit /o.s.s and h) the way' in n’hich default correlation is mo‹fefcé/. (a) (b)

I n the Gaussian copula model for time to default a credit loss is recognized only when a default occurs. In CreditMetrics it is rccognizcd when there is a credit downgrade as well as when therc is a default. In the Gaussian copula infidel of time to default, the dclault correlation arises because the value of the factor .u . This defines the default environment or average default 174 2012 Pearson Education

rate in the economy. In CreditMetrics a copula model is applied to credit ratings migration and this determines the joint probability of particular changes in the credit ratings of two companies. Problem 23.10. Suppor e that the LIBOR/.swap curve is fiat Ot 6’No with continuous compounding anJ a hve-year bonJ with a coupon of5’No (yaiJ semiannually) sell.s for 9ñ.00. How ii ould an asset .swap on the bond be structured? What is the asset swap .spread that ould he calculated in this situation? Suppose that the principal is $100. The asset swap is structured so that the 510 is paid initially. After that $2.50 is paid every six months. In return LIBOR plus a spread is received on the principal of $100. The prescnt value ot the fixed payments is 10 -+ 2.5e°’*”' * 2.5e°’"" * ... + 2.5e ' " ’ + 100e ’“’ = 105.3579 The spread over LIBOR must therefore have a present value of 5.3579. The present value of $1 received every six months for five years is b.5105. The spread received every six months must therefore be 5.3579 / 8.5105 = i0.5296 . The asset swap spread is therefore 2 x 0.6296 = 1 .2592 % per annum. Problem 23.11. Show that the value of a coupon-bearing corporate hond is the sum of the values of its constituent zero-coupon b e n t when the amount claimeJ in the event of Je[ault is the nodefault value of the bond, but that this is not .so w'hen the claim amount is the face value of the bond plus occrued interest. When the claim amount is the no-default value, the loss for a corporate bond arising from a default at time r is v(/)(1 — jB’

where v(/) is the discount factor for time / and B“ is the no-default value of the bond at time i . Suppose that the zero-coupon bonds comprising the corporate bond have no-default values at time i of Z, , Zz , ... , Z,, respectively. The loss from the i th zero-coupon bond arising from a default at time / is

The total loss from all the zero-coupon bonds is ( )(1 — )

— — (/)(1 — ›B’

This shows that the loss arising from a default at time is the same for the corporate bond as for the portfolio of its constituent zero-coupon bonds. It follows that the value of the corporate bond is the same as the value of its constituent zero-coupon bonds. When the claim amount is the face value plus accrued interest, the loss for a corporate bond arising from a default at time i is

where r is the face value and a(/) is the accrued intcrcst at time i . In general this is ntit the samc as the loss from the sum tif the losses on thc constituent zero-coupon bonds. Problem 23.12. A/our-year corporate bonJ provides a coupon o[4% per ›’ear payable semiannuall y and has a yielJ o] 5’X» expresseJ with continuoti.s compounding. The risk-free yielJ e'urve is flat at 3°X‹› with continuous compounding. A.ssume that de/rim/ry can take ylace at the ‹ nil of each y’ear (immeHiatel» before o coupon ar principal Payment and the recover) rate is 30%. E.stimate the risk-neutral Je]ault probability on the a.s.suniption that it is the same each vear. Define

as the risk-free rate. The calculations are as follows

1 .0

30

1.04.75

74.75

0.9704

72,57

2.0

30

1.03.55

73.88

0.941 8

69.5 80

3,0

30

102.96

72.96

0.9139

40

30

1.02.00

72.00

0.8569

63.860 272.69)

Total

The bond pays a coupon of 2 every six mtinths and has a continuously compoundcd yield of 5% per year. Its market price is 96.19. The risk-free value of the bond is obtaincd by discounting the promised cash flows at 3%. lt is 103.t›fi. The ttital loss from defaults should therefore be equated to 103.66 — 96.1 9 = 7.46. The value of Q implied by the bond price is thcrefore given by 2.72.G9 = 7.4G. or @ = 0.0274. The implied probability ot default is 2.74 per year. Problem 23.13. A companv has issued 3- and 5-y’ear hond.s v'itli a coupon o[4% per annym payable anittlallv. The vields on the' bonds (expressed with contintiou.s comJ›ounding) are 4.5% anJ maturities. The rec'ovei y rule is 40°>». Defaults can take place half ii’ay through each year. The ri.sk-neutral defâult rates per year arr Q, %r years 1 to 3 and Q,%r 5’ears 4 and 5. E.stiinate Q, and (_. The table for the first bond is

176 20.12 Pearson Education

Time

Dej. Proh

Rrc‹ ver y Amount ($)

Risk-free Volue ($)

Loss Given Default ($)

Di.s ount Factor

PV of L. p‹•cic•J Loss ($)

0.5

Q,

40

103.01

63.01

0.9527

61.92}9,

1.5

Q,

40

102.61

62.61

0.9489

59.41a,

40

102.20

62.20

0.9162

56.95),

2.5 Total

178.31f9,

The market price of the bond is '7S.35 and the risk-free value is 101.23. It follows that is given by 178.31Q, — — 101.23 — 95.35 so that Q, — — 0.0161 . The table for the second bond is

Time (yrs)

Def. Proh.

Reco very Amount ($)

Ri.sk-{re.e Value (S)

Loss Gii'cn Default

Dis cotlnl Factor

PV of Expec'ted Loss ($)

0.5

Q,

40

103.77

63.77

0.9827

62.67)

1.5

@,

40

I 03.40

63.40

0.9489

60.1 6Q,

2.5

40

103 01

ti3.01

0,9162

57.730,

3.5

40

102.61

62.6 l

0.8847

55.390,

40

102.20

62.20

0.5543

53.13Q

4.5

2

Ttital

180.56Q, +108.53Q

The market price of the bond is 96.24 is and the risk-free value is 10 1.97. It tollo s that 180.56 ,

105.53(j, = 101 .97 — 96.24

From which we get 9 z = 0.0260 The bond prices therefore imply a probability of default tit 1.61% per year for the first three years and 2.60% for the next twti years. 177 ñ*20 12 Pearson Education

Problem 23.14. Suppose that a financ'ial institution has entered into a swap depenJenl on the sterling interest rate with counterpart› X and an exactly offsetting .swap with counterparty Y. Which of the j'ollowing statement.s are true unJ w'liicli area fat.‹e. (a) The lotul pre.sent value of the cost o] JeJoults is the sum o[the J›resent value of the cost of Jefault.s on the contract with X plus the present value of the cost of deJuults on the corifrucf w'ilh Y. (b) The expycteJ expos urea in one year on hoth ‹ ontracts is the sum of the expected expo.sure on the contract with X and flue expected expo›'iire on the contract with Y. (c) The 9J°‹› t/pper conJ‹fence limit [or the exposure in one year- on both contracts is the ,sent o[tl1e 95% upper confide•nce limit %r the expos ure in one ycar oy the contras t with X and the 95°Xu upper confidence limit for the exposure in one vear on the contract with Y. E.xplain your an,sn’cr.s. The statements in (a) and (b) are true. The statement in (c) is not. Supposc that v, and v, are the exposures to X and Y. The expected value of v v, is thc expected value of v plus the expected valuc of v, . The sane is not true of 95% confidence limits. Problem 23.15. A cocipun enters into a one-year forir urd contract to .sell $100 %r AUD150. The contract is initially at the money’. In other words, the%rw'ard exchange rate is 1.50. The oneyear dollar risk-Jree rate of intere.at i.s 5% per- annum. The one-year dollar rate of interest at which the counterpart)' can borrow i.s 6% per annum. The exchange rate volatility is 12’No per annum. E.stimate the pres ent pa/ue of the co.s/ of Hefuults on the contract As.crime that JeJuults' are recognized only at the end of the life o}’the contra‹’t. The cost of defaults is w where o is percentage loss trom defaults during the life of the contract and v is the value of an option that pays of1 max(1505, 100, 0) in one year and S is the value in dollars of one AUD. The alue of u is u =—1 e '° ”'' ' ’ ' = 0.009950 The variable v is 150 times a call option to buy one AUD for 0.66fi7. The formula for the call option in tennis o1 forward prices is d d_, —- d, — cr T In this case, = 0.6667 , K — 0.6667 , o = 0.12 , T — — i , and r = 0.05 so that d, — — 0.06 , d 2 -—^.06 and the ›'alue tif the call option is 0.0303. It follows that › = 150 x 0.030? — — 4.545 so that the cost of default.s is 175 201 2 Pearson Education

4.345 x 0.009950 = 0.04522 Problem 23.16. Suppose that in Problem 23.15, the .six-month forward rate is also 1.50 anJ the sixmonth dollur risk-jree interest rate i.s 5°Xu jeer annum. Suppose further that the six-month do Her rnte oj interest at which the counterj›arty can borrow is 5.5“Xo yer annum. Estimate the present value oJ the cost of de/ault.s os.suming that defaults can occut- either at the six-month point or at the one-year point? (Ifa default occurs at ltte six-month point, the comJ›an) ’s potential lo.ss i.s the market value of the conta uct.) In this case the costs of defaults is ii, r, + uz v, w here u, = 1 — e '""” ' ’“’ ’ = 0.002497

v, is the value or an option that pays off max(1505, —100, 0) in six months and v is the value of an option that pays off max(150a— 100, 0) in one year. The calculations in Problem 23.17 show that v is 4.545. Similarly i', = 3.300 so that the cost o1 dcfaults is 0.002497 x 3.300 + 0.007453 x 4.545 = 0.04211 Problem 23.4 7. “A long form ard contra‹’t subject to credit risk i.s a combination ofa short position in a no-default put and a long position in a call stlhject to credit ri.wk. ” Exj›laii1 this statement. Assume that defaults happen only at the end of the life of the forward contract. In a default-free world the forward contract is the combination of a long European call and a short European put where thc strike price of the options equals the delivery price and the maturity o1 the options equals the maturity of the ftirw’ard contract. lf the no-default v'a1ue of the contract is positivc at maturity, the call has a positive value and the put is worth zero. The impact of defaults on the ftirward contract is the same as that on the call. II the no-dclault value of the contract is negative at maturity, the call has a zero value and the put has a positive value. In this case defaults have no etfect. Again the impact of defaults on the forward contract is the same as that on the call. It follows that the contract has a value equal to a long ptisititin in a cal1 that is subject to default risk and short position in a default-free P"' Problem 23.18. Explain w'hy the creJit evjao.yore on a matched pair of forward contracts res embles a sirudJle. Suppose that the forward contract provides a payoff at time T . With our usual notation, the value of a long forward contract is â—., Ke ' . The credit exposure on a long forward contract is therefore mai(S—, K‹• ’, 0) ; that is, it is a call on the asset price with strike price Ke "’ . Siinilarly the credit exposure on a short forward contract is T maz(Kc '’ — .9 ,n) ; that is, itaisstraddle a put onwith the strike asset price price Ke ’ . The total credit exposure is, therefore, pricewith Ke strike 179 ‹â>20 12 Pearson Lducation

Problem 23.19. Explain why the impact of ‹’refit risk on a matched yair of interest rate s›i'aps tends to he less than that on a matched paM of ‹’urrcr«‘} S Pt’Ofi.S. The credit risk on a matched pair of interest rate swaps is | B,., B„„„ . As maturity is approached all bond prices tend to par and this tends to zerti. The credit risk on a matched pair of currency swaps is | SB„„ „— B„, | w hcre S is the exchange rate. The expccted value of this tends to increase as thc swap maturity is approached because of the uncertainty in I . Problem 23.20. ’When a bank i.s negotiating currency .sung.s, it .yhnu/d /ry /o ensiire that it i.s receiving the lo ver intei-e.st rate currency jrom a company with a low creJit risk.” Explain. As time passes there is a tendcncy for the currency which has the low er interest rate tti strcngthen. This means that a swap z hcre we are receiving this currency will tend to move in the money (i.e., have a positive value). Similarly a swap where we are paying the currency will tend to move out of the money (i.e., havc a negative value). From this it follows that our expected exposure on the swap where we are receiving the low-interest currency is much greater tJian our expected exposure on the swap whcrc we are receiving the high-interest currency. We should therefore look for counterpartics v ith a low credit risk on the side of the swap where we are rcceiving the low- interest currency. On the other side of the swap we are far less concerned about the creditworthiness of the counterparty. Problem 23.21. Does put—call parity hold when there is default risk." ExJ›lain your ans›rer. to, putwall parity docs not hold when there is default risk. Suppose c‘ and p‘ are the nodefault prices of a European call and put with strike price and maturity r on a nondividend-paying stock whose price is .S , and that c and p are the corresponding values z lien there is default risk. The test shows that when u e make the independencc assumption (that is, we assume that the variables determining the no-default value of the option are independent of the variables determining default probabilities and recovery rates), c = c’e°""’'° ’ '" '1" and = -[ “ . The relationship

hich holds in a no-default world therefore becomes

when there is default risk. This is not the same as regular put—call parity. What is more, the relationship depends on the independence assumption and cannot be deduced frtxn the same sort of simple no-arbitrage arguments that we used in Chapter 10 for the put—call parity relationship in a no-default z'or1d. Problem 23.22. Suppose that in an a.s.set .swap p is the market price of the hond per Jollar of ‹ .2012 Pcs

on L ducation

f›rincipal, ñ“ is the default-[ree value of the bonJ per Jollar oJ print ipal, and V is lltr. Jure.sent value o[tlie n.s.yet.away .spread per doll‹ir of principal. .Who that Y — — B—‘ B. We can assume that the principal is paid and rcccivcd at the end of the life of‘ the swap without changing the sw ap’s ›'aIue. It thc spread crc zero the present › alue of the floating payments per dollar of principal would be 1. The payment of LIBOR plus the spread therefore has a present value of 1 + I . The payment of the bond cash tows has a present value per dollar of principal of B’ . The initial payment rcquired from the payer of the bond cash flows per dollar of principal is —i r . (This may be negative; an initial amount of s— i is then paid by the payer ot‘the floating rate). Because the asset swap is initially ivtirth zero we have 1 -F

= fi" +—1 fi

so that L — — D — D

Problem 23.23. Show' that iindc•r .Merton’s model in Section 23.6 ltte creJii spreciJ on a T -year emcoupon l›onJ i ' In[ Y(d_, ) X'(—d, ) ! L] ' T there L — — De ' I j; . The value debt—inr„.› Merton’s De of ' the v ‹d.› r,› r„model = De is .›£ — (c/.E) or + r, (— ) If the credit spread is this should equal De ' ' so that Substituting De ' — —L V„

De ' " ’"

— De ' â'(cl ) + U .\ (—‹/, )

so that .i — — I n[.\'(éf 2 ) * \ (—c/ ) / I] /' T

V'hen the dcfault risk of the seller of the option is taken into acctiunt the tiption value is the Black-Scholes-Merton price multiplied by e ”“’ = 0.9704 . h lack-Scholes-Merton overprices the option by about 3'?o. Problem 23.25. Gi e an example o(a) riqht-way ri.sk and /ij wrong- 'a;' ri.sk. (a) Right way risk describes the situation when a default by the counterparty is mtist likely to occur when the contract has a ptisitix e value to the counterparty. An example of right way risk wtiuld be w hen a counterparty’s future depends on the price of a ctiinmodity and it enters into a contract to partially hedging that exposure. 181 â20 12 Pearson Education

(b) Wrong way risk describes the situation when a default by the counterparty is most likely to occur when the contract has a negative value to the counterparty. An example of right way risk ould be when a counterparty is a speculator and the contract has thc same exposure as the rest of the counterparty’s portfolio.

182 fi2012 Pearson Educati on

Cl—fAPTER 24 Credit Derivatives Problem 24. I. E.xplain flue dijference betw een a regul‹ir credit default sii’‹ip and a hinar5' ‹•rec!it Jefault Both provide insurance against a particular company defaulting during a period o1 time. In a credit default swap the paytiff is the notional principal amount multiplied by one ithin us the recovery rate. In a binary swap the payoff is the notional principal. Problem 24.2. A rredit JeJâult .swaf› requires a seinianniicil payment at the ralr o] 60 ba.si.s points per year. Th‹• print ipal i.s 5300 million and the credit Jefault .s»'ny is s ellled in ca.sh. A dejuull o‹ ‹’urs afler few- ear.s anJ tw'o mould.y, arid the call ulation ugent estimates that the price oj the cheapest deliveral›ly bond is 4H’Xo of its face valtie .sliortly ajier the default. Lis i the cash flows and their timing for the .seller o] the ‹’redit default swup. The seller receives 300, 000, 000 x 0.0060 x 0.5 = 5900. 000 at times 0.5, 1.0, 1.5, 2.0. 2.5, 3.0, 3.5, and 4.0 years. The seller also receives a fi nal accrual payment of about $300,000 ( = 5300, 000, 000 x 0.060 2 / 1 2 ) at the time of the default (4 years and two months). The seller pays 300, 000, 000 0.6 S150, 000, 000 at the time of the default. Problem 24.3. E.xplain the tw'o ways a c'redit default swcip can he settled. Sometimes there is physical settlement and stinaetimes there is cash settlement. In the event of a default whcn there is physical scttlement the buyer ot protection sells bonds issued by the reference entity for their face v’alue. Bonds cvi th a total face value equal to thc notional principal can be sold. Cash settlement is non' more usual. A calculation agent, often using an auction process, cstimates the value of the cheapest-to-deliver btinds issued by the reference entity a specified number of days after the default event. The cash payoft is then baseil on the excess o1 the force value o1 these bonds over thc cstimated value. Problem 24.4. Explain how a cash CDO and a .s5 nthelic CDO are create•J. A cash CDO is created by forming a portfolio tif credit sensitive instruments such as bonds. The returns from the bond portfolio tion to a number or tranches (i.e., different categories of investors). A waterfall defines the way interest and principal payments flow to tranches. The more senior a tranche the more likely it is receive promised payments. In a synthetic CDO n portfolio is creatcd. lnstead a poitfolio of credit default sw ape is sold and the resulting credit risks are allocated tti tranches. 2012 Pearson Education

Problem 24.5. Explain what u first-to-default credit default sn›ap is. does its value increa.se or decrease a.‹ fhe default correlation hetw ren thc companies in the ba.sket increases'* Explain. In a first-to-default basket CDS there are a number ot referencc entities. When the tirst one defaults there is a payoff(calculated in the usual way for a CDS) and basket CDS terminates. The value of a first-to-default baskct CDS decreases as the correlation between the reference entities in the basket increases. This is because the probability of a default is high w'hcn the correlation is zero and decreases as the correlation increases. In the limit when the corrclation is one there is in effect only one company and the probability of a default is quite low. Problem 24.6. Explain the J ference henveen risk-neutral and real-world default prohahilifres. Risk-neutral default probabilities are backed out from creclit dcfault swaps or bond prices. Real-w orld default probabilities are calculated Lom historical data.

Suppose a company wants to buy some assets. If a total return swap is used, a financial institution buys the assets and enters into a swap with the company where it pa}s the company the return on the assets and reccives from the company LIBOR plus a spread. The tlnancial institution has lcss risk than it w’ould have if it lent the company money and used the assets as collateral This is because, in the cvent of a default by the company, it owns the assets. Problem 24.8. Suppose that the risk-free zero curve is flat at 7’X• per annum with continuou.s compounding anJ lhat defaults ran occur half wuy through each year in a new five-yeur credit deJuult swap. Suppo.be thal the recovery rate is 3HXo anJ the default prohab ilities each year conditional on no earlier default are 3°Xo. Estimate the credit default swup spread. Assume pa) ments are made annually. The table corresponding to Tables 24.1, giving unconditional default probabilities, is Time (years) 1 2 3 4 5

Default Probabilit y 0.0300 0.0291 0.0282 0,0274 0.0266

Survival Probahiliiy 0.9700 0.9409 0.9127 0.8853 0.SJ87

The table corrcspondiny to T’ablc 24.2, giñiig the i n s e n t value of the expected regular payments payment ratc is s per year), is

184 2012 Pearson Education

Time Lyrs) 2 4 5 Total

I’i-obabiliz of survival 0.9700 0.9409 0.9127

Expected Pen ment 0.9700s 0.9409.i 0.9127s

Discount Factor

0.Sb53 0.5587

0.5553 0.fi587s

0.7555 0.7047

PV of Expected Pai’ment 0.90J4i 0.8180i 0.7398a

0.9324 0.8694 0.8106

0.6691s ú.6ü51s 3.7364.i

The table corresponding to Table 24.3, giving the present value of the expected payotl‘s (notional principal =$l), is Time (yr.s)

Prob‹tbili3 of deJâult

0.5 1 .5 2.5 3.5 J.5 Total

300 0.0291 0.0282 0.0274 0.0266

' | |

Rec.overy Rate

ExpecteJ Payof

Di.scount Factor

0.3 0.3 0.3 0.3

0.0210 0.0204 0.0198 0.0192 01

0.9656 0.9003 0.8395 0.7527 0.7295

PV of Expecte d Payment 0.0203 0.01 83 0.0166 0.0150 0.0136 0.083S

The table corresponding to Table 24.4, giving the present value of accrual payments, is Time (yr› I 0.5 1.5 2,5 3.5 4.5 Total

Prohab ility of default 0.0300 0.0291 0.0252 0.0274 0.0266

Expected Accrual QP ment 0.0150s | 0.0146s 0.0141s 0.0137s 0.0133s

Discount Foctor 0.9656 0.9003 0.5395 0.7827 0.7298

PV oJ E.T y ected Accrual Payment 0,0141s 0.0131s 0.0l1Ss 0.0107s 0.0097s 0.0595s

The crcdit default swap spread s is given by: 3.7364s + 0.0598.i = f1.053 d It is 0.0221 or 22 l basis points. Problem 24.9. What is the value ol the .s»ap in Problem 24.‹3 per dollar of notional principal to the protection buyer iJ the credit de•]ault swap .spreaJ i ' 150 basis point.s.' If the credit default swap spread is 150 basis points, the value of the swap to the buyer of protection is: 0.0838 — (3.7364 0.0598) x 0.0150 0.02(›9 per dollar of notional principal. Problem 24.10. What is the c'redit default swap spread in I°rohlein 24.8 i[it i.s a hinary CDS? 185 .201 2 Pearson Education

If the swap is a binary CDS, the present value of‘expected payoffs is calculated as follows ExpecteJ Payojj

Discount Factor

0.5

Probability of Default 0.0300

0.0300

0.9656

1.5 2.5 3.5 4.5

0 0291 00282 0.0274 0.0266

0.0291 0.0252 0.0274 0.0266

0.9003 0.8395 0.7827 0.7298

Time (years)

PV of expecteJ Payoff 0.0290 0 02ti2 0 0237 0.0214 0.0194 0.1197

Thccuditdelaultswapsqrcad .‹ isgivcnby: 3.7364s+0.050hs= 0.1 197 It is 0.0315 or 315 basis points. Problem 24.11. Ho w doe.s a five-year n ih-to-de/ault credit default sw'uj:i w'ork'? Consider a tea.yker oJ 1tltl reference entities where each reference emity has a pi-ol ahiliU of defaulting in each year of “No. AS the Je}uult corrclation l›etw'een the reference entitie.s increas es what would you expect to happen to the value of the .fv'ap ›rhen a) n — — I and h) n — — °5 . Explain your answ'er. A five-year nii to default credit default swap works in the same way as a regular credit default swap except that there is a basket of companies. Thc payoff occurs when the n th default from the companies in the basket occurs. After the nth default has occurred the swap ceases to exist. When u = I (so that the swap is a “first to default”) an increase in the default correlation lowers the value of the sw ap. When the default correlation is zero thcre are 100 independent evcnts that can lead to a payoff. As the correlation increases the probability of a payoff decreases. In the limit when the correlation is perfect there is in effect only onc company and therefore only one event that can lead to a payoff. When n — — 25 (so that the swap is a 25th to default) an increasc in the default correlation increases the value of the swap. When thc detault correlation is zero there is virtually no chance that there will be 25 defaults and the value of the swap is very closc to zero. As the correlation incrcases the probability of multiplc defaults increases. I n the limit when the correlation is perfect there is in effect only one company and the value of a 25th-to-default credit default swap is the same as the value of a first-to-dcfault swap. Problem 24.12. What is the formula relating the pc9 o]fon a CDS to the notional print ipal anJ the recover’ rate? The payoff is L 1 — R) where Lis the notional principal and p is the recovery rate. Problem 24.13. Show that the .sj›read %r a new plain v'anilla CDS should be (1 — R) times the spread for a similar new binar) CDS where R is fhe recovery rate. The payoff from a plain vanilla CDS is 1 — R times the payoff ñom a binary CDS with the same principal. The paytiff always occurs at the same time on the two instruments. It folltiws 1S6 201? PearsonEducation

that the regular payments on a new plain vanilla CDS must be —i x times the payments on a new binary CDS. Otherwise there would bc an arbitrage tipportunity. Problem 24.14. Veriff that i f the CDS spreciJ for the example in Tahles 24.1 to 24.4 is 100 ha.st.s points and the yrohahilitv o[de[ault in a vear (e'onditional or no earlier default j intist he 1.61’No. How Joes the J›rohahilik' of Jefault chan,qe when the rero› er) i‘ute is 2H°Xu in.stead o/ 4H‘Xo'? Verify that your an,sii’er is ‹ on5'istent with the implieJ prohabiliH oJ default being approximately j›royortionul to I I (1 — R) where R i.s the re•co ver rate. The 1 .61'?o implied dcfault probability can be calculated by setting up a orkshcct in Excel and using Solver. To verify that 1.61*o is correct we note that. with a conditional default probability o1 1.61%, the unconditional probabilities are: Demin// ProbabiliH 0.01 I›1 0.0155 0.0156 0.0153 0.0151

Survi val Probability 0.9539 0.9681 0.9525 0.9371 0.9221

The present value of the regular payments becomes 4 11 70 s , the present value of the expected payoffs becomes 0.0415, and thc present value of the expected accrual payments becomes tl.0346 i .When i = 0.01 the present value of the expectcd payments equals the present value of the expected paytiff.s. When the recovery ratc is 20%i the imps ied default probability (calciilated using Solver) is 1.21% pcr year. Note that 1.21/1.61 is approximately equal to — ( 1 0.4) / (1 — 0.2) showing that the implied default probability is approximately proportional to l / (l — ñ). In passing we ntite that i1 the CDS spread is used to imply an unconditional default probability (assumed to be the samc each year) then this implied unconditional default prtibability is exactly proportional to 1 / (1 — fi) . When wc use the CDS spread to imply a conditional default probability (assumed to be the samc each year) it is only approximately proportional to I / (1 — ñ) . Problem 24.15. A ‹Company enter.f into a total relurn swap where i1 receives the i-eIurn on a corporate Idend paying a common of 5°Xo and pay› LIBOR. Ex Flain the difference bend een this and a regular sv›ay whei-e 5 Xois exchange J for LIBOR. In the case of a total return swap a cc°nspany receives tpays) the increase (decrease) in the value of the bond. In the rcgular swap this does not happcn. Problem 24.16. Explain hour forward contracts and options on credit default swaps are structured. When a company enters into a long (short) forward cuntract it is obligated to buy (sell) the protection given by a specified credit default swap w ith a specified sprcad at a specified futurc time. When a company buys a call (put) option contract it has the option to buy (sell) thc protection given by a specified credit default sla p with a specified spread at a specified 157 â20 12 Pearson Education

future time. Both contracts are normally structured so that they cease to exist if a default ticcurs during the life of the ctintract. Problem 24.17. '’The position of a buyer of a credit deJaull s may i.s .f inlilar to the yo.ritten of someone who is long a risk-free hond anJ short a corporate bond.” Explain this statement. A credit default svap insures a eoipornte bond issued by the reference entity against default. Its apprtiximate effect is to convert the corporate bond into a risk-free bond. The buyer of a credit default swap has therefore chosen tu exchange a corporatc bond for a risk-free bond. This means lhat the buyer is long a risk-free bond and short a similar corporate bond.

Payoffs trom credit dcfault swaps depcnd on whether a particular company defaults. Arguably some market participants hav'c more information about this than other market participants. (See Business Snapshot 24.3.) Problem 24.19. Does valuing a CDS u.sing real-world default probabilitie.‹ rather than ri,sk-neutral Jefault probabilities overstate or understate its value? E. plain vour answer. Real world default probabilities are less than risk-neutral default probabilities. It follows that the use of real world (historical) default probabilitics will tend to understate the value of a CDS. Problem 24.20. Mat is the difference benrecn a total return s nap and an asset snap? In an asset swap the bond s promised payments are swappcd for Ll BOR plus a spread. In a total return swap the bond’s actual payments are sw’apped for LIBOR plus a spread. Problem 24.21. Suppo.se r/inf in u one-factor Gaussian coyula nloJcl the five-year probah ilit)' o/defaull for each of 125 navies is 3"Xo and the pair ivi,se copula correlation i.s 0.2. Calculate, for fâctor volues o/—2, —/, 0, I, mud 2. a) the Je]ault f›rohabilitv conditional on the]uctor value anal b) the probab ilit) of more than 10 defaults conditional on the factor value. Using equation (24.2) the probability o1 default conditional on a factor valuc of .o is —02 For .u equal to —2,—1, 0, 1, and 2 the probabilities of default are 0.135, 0.054, 0.0 lb, 0.005, and 0.001 respectively. To six decimal places the probability of more than 10 defaults for these vat ues of w can be calculated using the BINOM AlAST function in Excel. Thcy are 0.959284, 0.79851, 0.000016, 0, and 0, respectivcly. Problem 24.22.

E*F!°!• thy difference between ha.se correlation and compound correlation

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Compound correlation for a tranche is the correlation which when substituted into the onefactor Gaussian copula model produces the market quote tor the tranche. Base correlation is the correlation which is consistent with the one-factor Gaussian copula and market quotes for the 0 to X% tranche where X% is a detachment point. It ensures that the expected loss on the 0 to X % tranche equals the sum of the expected losses on the underlying traded tranches. Problem 24.23. In Example 24.2, what is the tranche spread for the 9X» to I2°X‹ tranche?

In this case a, — 0.09 and ‹ip = 0.12 . Proceeding similarly to Example 24.2 the tranche spread is calculated as 30 basis points.

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CHAPTER 25 Exotic Options Problem 25.1. Explain the di]fet-ence betv’een a forw'cir‹I start option and a chooser option. A forward start option is an option that is paid for non' but * ill start at some time in the future. The strike price is usually equal to the price o1‘the asset at the time the optitin starts. A chooscr option is an option whcre, at some ti me in the ñiture, the holder chooses whether the option is a call or a put. Problem 25.2. Descril›e the payo/f from a portfolio con.si.sting of a floating lookb‹1cL call and a floating lookl›ark put › 'iih the .same maturity.. A floating lotikback call provides a payoff of S — S . A floating lookback put provides a payoff of 6„, — S, . A combination of a floating lookback call and a floating lookback put therefore provides a payotl of S „, — .S„, .

optimal to make the ‹ hoice hefore the rnd of the type-year period!* Explain your answer. No, it is never optimal to choose early. The resulting cash flow s are thc samc regardless of when the choice is made. There is no point in the htilder making a commitment earlier than necessary. This argumcnt applies whcn the holder chooses between mo American options pr‹ii iding the options cannot be exercisecl before the 2-year point. lf the early exercise period starts as soon as the choice is made, the argument does not hold. For example, if the stock price fell to almost nothing in the first six months, the holder is ould choose a put option at this time and exercise it immediately. Problem 25.4. Suppose thot ‹, anJ p are the prices oJ a Ew oj›ean average price call and a European

The payoffs from ci , c , c,.yi, y°, /›s are, respectively, as follows: max(.Y — K. 0)

max(S— S, 0) max(N—,

K,0) 190 ?2012 Pearson Education

max(ñ - S, O)

max( — N„ 0) rnux(K-

S,., 0)

The payott from c, — y, is always S — K , The payoff from c — p2 is always S, — S ; The payoff from c, — p,

is always .St — K ; It follows that c — — — c. —

+ c.

p,

at time T› and a put maturing at time I I . Derive an alternati v'c Jecomyo.sition info a ‹call maturing at time T and a put maturing at time T_.

Substituting for c , put-call parity gives max(r, p) — max

›, p+ S,e "’" ‘ ' — Ac "" '’

= ya+max 0, S,e ’!'’" "'' — Ke "'"-’ "''

This shows that the chooscr optit›n can be dccomposed into: A put option with strike price K and maturity T, , an‹J e *'"-*" call options with strike price Ke ' "'’- ' and maturity T. Problem 25.6. .Section 25.8 gives tv’o foriiiiil‹is for a don'n-anJ-out c ull. The Fu st applies to the .yi/iin/ion where the barrier. II i.s less f/ion or equal to lh‹• slrike price, K . Th • srcond apf›lie.s to thcn .f ituation where It > F . Show thot the H+ o formulas ure the .same wlieti H — —r .

+Ke”!’ (H I S )‘"" Substituting H — — c and noting that r — q * w" / 2 * 2

we obtain .r, = ‹I, so that c„ — — c — She”” (H ! Sz )“’.N (v, ) + Ke°"’ (ñ / N, ) ’ ’ Y( y, — crT) The formula for c„ when o x is c — S e "’ (H ,* St)’’ N(5›—) Ke°'' [H / .S, )" " .\'(— c r T ) Since c„ — c — c d,

c„

iS' y, — c r T )

Problem 25.7. Explain why a do n-and-out put is worth zero when the barrier i.s greater than the strike The option is in the money only when the asset price is less than the strikc price. However, in these circumstances the barrier has been hit and the option has ceased to exist. Problem 25.8. Suppose that the strike price o/ an American call option on a non-di vidend-paying stock gi-ow at rate z . Show that if z i.s Ie.‹s thOH thy ff I k - s e e rDte, i‘ , it is nl?vHr Optical to

exercise the call early. The argument is similar to that given in Chapter 10 for a regular option on a non-dividendpaying stock. Consider a portfolio consisting of the option and cash equal to the present value of the terminal strike price. The initial cash position is Ke”°"" By time r ( 0 r s T ), the cash grow s tti Kp -’'’ —‹) ' x*

ff q x' ,- i —x j—!’ ‹I

Since r > g , this is less than Ke’’ and therefore is less than the amount required to exercise the option. It follows that, if the option is exercised early, the terminal value of the portfolio is less than S . At time T the cash balance is Ke . This is exactly what is required to exercise the option. 11 the early exercise decision is delayed until time T , the terminal value of the portfolio is therefore This is at least as great as S . It follows that early exercise cannot be optillnal. Problem 25.9. How can the value of a forwarJ s tart put option on a now-dividend-{ip ing stock be calculated if it is agreed that the strike price will be 10% greater than the stock price at the time the option starts'? When the strike price of an option on a non-dividend-paying stock is defined as 10% greater that the stock price, the vaJue of the option is proportional to the stock price. The same argument as that given in the text for forward start options shows that if i, is the time z'hcn the option starts and i is the time when it finishes, the option has the same value as an option starting today with a life of /, —/, and a strike price of 1.1 times the current stock price. Problem 25.10. Ifa stock price %llows geometric Brownian motion, what jlrocess does A(t) follow where A(i) is the arithmetic average stock price between time zero and time t ? Assume that we start calculating averages from time zero. The relationship betwcen A(i • fit) and A(i) is A(I

a t) • (i + a t) — A (t ) t + s ( i ) a t

where S(i) is the stock price at time i and tems of higher order than fi are ignored. If we continue to ignore terms of higher order than fi, it follows that 192 02012 Pearson Education

Af

Taking limits as o tends to zero

The process for A(i) has a stochastic drift and nt› fi tenn. The process makes sense intuitively. Once some time has passed, the change in S in the next small portion of time has only a second order effect on the average. It S equals the average has no drift; if S > H the avcrage is drifting up; if S < A the average is drilling down. Problem 25.11. Explain why Jelta hedging is easier jor Asian options than jor regular options. In an Asian option the payoff becomes more certain as time passes and the delta always approaches zero as the maturity date is approached. This makes delta hedging easy. Barrier options cause problems for delta hedgers when the asset price is close to the barrier bccause delta is discontinuous. Problem 25.12. Calculate the price ofa one-year F,uropean option to give ufi 100 ounces of.silver in exchange for one ount e of gold. The coizerif price.s of gold and silver are $380 and $4, re.spectively, the risk-free intere.st rate is 10% per annum,' the volatility o[eacl1 commodity price is 20"Xu, anJ the correlation brm een the iv'o price.s is 0.7. Ignore storage costs. The value of the option is given by the formula in the text i'te.X [d z ) w'here

V(‹I, ) — b c '

and cr = -+- w/

— 2 U›cr, w

In this case, I', = 380 . L = 400, q, = 0, q2 = 0, T — — i , 2 = 0.1549 and or $15.38.

7) — 400 A'(—0.4086) = cr380i\'(-0.253 = 15.38 2 * 0 2'

Problem 25.13. I.s a Luropeun Jo›i'n-and-out 2 0option 7 0 2 on an asset n orth the sumc as a European doc n-and-out option on the asset’.s futures price]or a future.s maturing at the .same time as the Because d, — — —0.2537 and d. —0.4086 , thecontract option price option."! is No. If thc future’s price is above the spot price during the lif’e of the option, it is possible that the spot price will hit the barrier when the futures price does not.

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Problem 25.14. Answer the following questions ahout compound option.s (a) What put—‹ all parity relation.ship exisls between the price of a European call on a call and a European put on a call? Show that the formulas given in the text satisp the relation.ship. (b)What put—call party relationship exists between the price of a European call on a put anJ a European put on a put.’* Show› that the formulas given in the text satisfy the relationship.

(c)The putwall relatitinship is cc * K e "" — pc

+ c

where cc is the price o1 thc call on the call, y,‹ is the price of the put on the call, c is the price today of the call into which the options can be exercised at time T , and K, is the exercise price for rc and p . The proof is similar to that in Chapter 10 ftir the usual put—call parity relationship. Both sides o1 the equation represent the values of portfolios that will be worth wax(c, A,) at time T . Because and we obtain Siiice put—call parity is consistent with the formulas

(b)The put— call relationship is cp K, e ”' — — p f› p

where cp i$ the price of the call on the put, p , is thc price of the put on the pute p Îs the price today oï the put into which the options can be excrcised at time T , and *, is the exercise price for ry, and pp . The proof is similar to that in Chapter 10 for the usual put—call parity relationship. Both sides o1 the equation represent the values of portfolios that will be worth max(y, K,) at time T, . Because U(a, è; p) = iY(‹/) — M(a,— b; — p) — N(h) — M(— a,b,; —p) and A (x) — l — .\'(—.r) it follows that c f—

pp — — Se°”’ N (— b, ) K,,e. ”’ -N (—h ) — K, r ""'

Because —.Yc "'

N’ (— b, j

K

value of a floating lookback call. Problem 25.16. Does a down-and-out call become more valuable or less valuable as we increase the frequency with which we observe the asset pri‹’e in Jelerni ining whether the harrier has been crossed'! What is the answer to the same question for a down-and-in call!" As we increase the frequency with which the asset pricc is observed, the asset price becomes more likely to hit the barrier and the value tif a down-and-out call goes down. For a simi lar reason thc value of a down-and-in call goes up. Thc adjustment mentioncd in the text, suggested by Broadie, Glasserinan, and Kou, intives the barrier liirther out as the asset price is observcd less frequently. This increases the price of a down-and-out option and reduces the price of a down-and-in option. Problem 25.17. Explain why a regular European call option is the sum oj a Jown-and-out EuroJ›ean call and a dow'n-and-in European call. Is lhe same true for American call option,s.'* If the barrier is reached thc dtiwn-and-out option is worth nothing while the dozen-and-in option has the same value as a regular option. If the barrier is not reached the down-and-in option is wonh nothing while the down-and-out option has thc same value as a regular option. This is why a down-and-out call option plus a down-and-in call option is worth the same as a regular option. A similar argument cannot be used for American options. Problem 25.18. What is the value of a derivative that puys of] $100 in six months if the S&P 5 HO inJex is greater than 1,000 end zero otherw tee." Assume that the current level of the inJex is 9ñiH, the risk-free rate is 8’"u yer annum, the HiviJenJ yielJ on the rude.x is 3’k per annum, anal the volatility of the index i.y 20%. This is a cash-or-nothing call. The value is 100 W' (d. )e ”"” where Jg In(960 / 1000) (0.05 =— — 0.03 — 0.2 z / 2) 0.5 0.1826 0.2 x Since N(d_) = 0.4276 the value of the derivative is $41.08. Problem 25.19. Jn a three-monlh Jon n-unJ-out call option on silver fumre.s the .strike prire is $20 per ounce and the harrier i.s I R. The current futures price is SI 9, the risk-free intere.st rate is 5’X, and the volatility ofsilver/utures is 40% per anni.mm. E. plain ho w' the option z orks anal ‹ al‹’ulote its valtie. What is the value o[a regular call option on silver futures w'ith the .same terms? What is the value ofa doirn-and-in call option on silver futures w'ith the .same terms? This is a regular call with a strike price of $20 that ceases to exist if the futures price hits $18. With the notation in the text H — — 18, K — — 20, S — — 19 , r — — 0.05 , cr = 0.4 , y = 0.05 , T — — 0.25 From this 2. = 0.5 and In[182 / (19 x 20)] 0.5 x 0.4 25 = — ' 0.4 0 25 + 0.69714 The value or a down-and-out call plus a down-and-in call equals the value of a regular call. 195 ^U2012 Pearson Education

Substituting into the formula given when n < K we gct c = 0.4638. The regular BlackScholes-Merton formula gives c = 1.0902 . Hcnce c„ = 0.6264 . (These answers can be chccked with DerivaGem.) Problem 25.20. A nrw European-style fioating lookh‹ick ca11 option on a stock index has a maturi5 of nine month.s. The current level of flue index is 400, the risk-[ree rate is 6% per annum, the dividend yield on the index is 4% per annum, anJ the volatility of the index is 20°Xo. Use DerivaGem to value the option. DerivaGem shows that the value is 53.38. Note that the Minimum to date and Maximum to date should be set equal to the current alue of the index for a new deal. (Sce material on DerivaGem at the end of the book.) Problem 25.21. E.stimate the value ofa new' .six-month European-s}’le uve•rage yrice call oj›tion on a nondividend-paying stock. The initial stock price is 830, the strike price is $30, the risk-Fee interest rate iS 5’Xo, iHd the stock price volatility is 30%. We can use the analytic approximation given in the tcxt. ' —1)x 30 M, — (e’"" 0.05 x

30.378

Also II, — — 936.9 so that cr = 1 7.41 % . The 0.5 option can be v alued as a futures option with Up = 30.375 , K — 30, r - 5 y fi - I 7.41’No , and — 0.5 . The price is 1.637. Problem 25.22. Use DerivaGem to calculate the value of.

(a)

A regular European call option on u non-dividend-paying .stock where the stock price is $50, the strike price i.s S5h, the risk-jiee rale is 5’No per annum, the volatility is 30°Xn, and the time to maturity i.s one vear. (b) A Jo›•'n-and-out European call which is as in (u) n’ith the barricr at 545. (c) A Jo n-and-in European call n'hich is as in (u) ›•'ith the barrier at $45. Show that the option in (a) is worth the .sum o[tlie values of the options in (b) and (c). The price o1 a regular European call tiption is 7.116. The price of the down-and-out call option is 4.696. The price of the down-and-in call option is 2.419. The price of a regular European call is the sum of the prices or down-and-out and downandin options. Problem 25.23. Explain adjustments that hate to be inaJe when r — - q %r a) the aluation formulas for lookback call options in Section 25. 10 ond b) the jorinulas]or M, anJ ,H,, in Section 25.12. When r - y in the expression for a floating lookback call in Section 25.10 a, = o. and I I = In(S, / S,„, ) so that the cxpression for a floating lookback call becomes S, e " ñ' (a,—) S „e "'A'(‹i, )

As y approaches r in Section 25.12

e get 196

o 4T'

T"

cr‘

Problem 25.24. Value the variance swap in Example 25.4 oj Section 25./5 assuming that the implied volatilities for option.s with ,strike prices 800, 850, 900, 950, 1,000, I,050, 1,100, 1,150, 1,200 are 20’X•, 20.5°X‹:, 2 ’No, 21.5‘Zo, 22’No, 22.5 Xo, 23’X‹›, 23.5%, 24"Xo, respectively.

In this case, DerivaGem shows that Q(K,) — 0.1772 , Q(H z ) = 1.1857, Q(K ) = 4.9123, Q(K ) = I 4.2374, Q(K ) = 45.3738, Q(K ) — — 35.9243 , Q(K-) = 20.65b3, Q(K ) = 11.4135 , Q(K ) — — €.1043 . E(U) = 0.0502 . The value of the variance swap is $0.51 million.

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CHAPTER 26 More on Models and Numerical Procedures Pt obleca 26.1.

Confirm that the CEV model (cannulas satisp foul—‹ all parity. It follows immediately from the ec|nations in Section 26.1 that

in all cases. Problem 26.2. U.se Monte Carlo s Emulation to .‹how that Verton’s vn/i e/or a Euro mean option is In this case K’= 0.3• 1.5=0.45. The variable/, is the Black-Scholes-Merton price when the v‹ilatility is 0.25'*n.i"?T and the risk-free rate is -0.1+n^ In(I .5)/T. A spreadsheet can be constructed to a) Valuc the option using the first (say) 20 terms in the Merton expansion.

lt is convenient to usc the DerivaGeir functions to do this.: b) Value the option using Monte Carlo simulation. There are a number o1 alternative approaches. One is as follows. Sample to determine the number ofjumps in time T as described in the text. The probability of \’ jumps is e°" ( T) '

The proportional increase in thc stock price arising from jumps is the product off random samples from lognormal distributions. Each random sample is exp(N where A is a random sample from a normal distribution z'ith mean ln(1.5)—s’/2 and standard deviation s. (Note that it is one plus the percentage jump that is lognormal.) The proportional increase in the stock pricc arising from the diffusion component o1 the process is exp((0.05 — 0.3 x 0.5 — 0.25' / 2)T + 0.25tT) where e is a random sample from a standard normal distribution. The final stock price is 30 times the product of the increase arising from jumps and the increase from the diffusion component. The payoff from the option can be calculated from this and the present value of the average payoff over many simulations is the estimate of the v alue of the option. I mud that the two approaches give similar answ'ers. For examplc, for a call option with T — 1, =0.5, and K—3 ñ, the option value given by both methods is about 5.47. 'i°.201 2 Pearson Education

Problem 26.3. Confirm that Merton’.s jump differs ion moJel satin fies yutmall parin ii’lien the jump size is lognormal. With the notation in the text the value of a call option, c is

where c, is the Black-Scholcs-Merton price of a call option where the variance rate is

and the risk-free rate is T where y = ln(1 £) . Similarly the v alue of a put option p is {Z'T)" where y, is the Black-Scholes-Merton price of a put option 1'ith this variance rate and riskfree rate. It follow's that ’ T (?’T)"

From putoall parity

Because it follows that

T)" s,‹ n!

Using 2’ — ?(1 â) this bcctimes 1 n! e z From the expansion of the exponential fiinction we get (IT j“ n! (?’T)“ n! H e n c e T

n— 0

„ 0

— p c -

The average variance rate is 6 x 0.22 + 6 x 0.22’ + 12 x 0.242 24

= 0.0509

The v'o1atility used should be 0 0509 = 0.2256 or 22.56%. Problem 26.5. Consider the case of Merton /f jump diffus ion model where jumps uln'uys reduce the asset price to zero. Assume that the average number of jumps per year is L . Show that the price ofa European call option is the same as in a world with no jumps except that the ris/rfree rate is r ñ rather- than r . Does the possibility oj jumps increase or reduce the value of Uf? YOU GflliMlt fK /hf3 C'ffJ’6.° (Hint. Uo/t/e the option ass:uming no j:umps and assuming one oy

more jump.f. The probability of no jumps in time T is e°’'). d (r —q — Zk) dt + cr In a risk-neutral world the process for the asset price exclusive of jumps is Jz In this case k = —1 so that the process is dS S The asset behaves like a stock paying a dividend yield of —9 I . This shows that, conditional on no jumps, call price S e ' ' ‘ ’N(d,—) Ke "" d

1n(S, / K)+(r

d_, — — d, — mT There is a probability of e " that there will be no jumps and a probability of 1 — e “ that there will be one or more jumps so that the tinal asset price is zero. It follows that there is a probability of e°” that the value of the call is given by the above equation and —1 e “' that it will be zero. Because jumps have no systematic risk it follows that the value of the call option is e ” [S „e '* "' .V(d, ) — K e ' ]

This is the required result. The value of a call option is an increasing liinction of the risk-free interest rate (see Chapter 10). lt follows that the possibility ofjumps increases the value of the call option in this case. Problem 26.6. At time zero the price of a non-dividend-paying stock is So . Suppose that the time interval between 0 and T is divided into two .fuhintervals of length 1, anJ i2 . Dui-ing the ffst subinterval, the risk-free interest rate and volatilitv are r, and m,, respectively. During the 200 ?°^2012 Pearson liducation

.set onJ subinterval, thcy are r, anJ m,, r‹spectivcly. Asswne that the w'orlJ is risk neutral. (a) L' e the re.cults in Chapter 14 to Jeterinine the stock price distribution at time T in terms of r, . r,_ , a, , cr,, t , t, . and Sz . (ñ) Suppose that r is the average interest rate between time zero and T and that I is the average variance rate between time.s zero and T . ft hat is the stock price (c) What are the res ults ‹'orresponJing to i'a) and (b) ›i lien there are three subinterval.s with Jifferent interest rate.s and volutrlities? (d) Show that if the risk-free role, r , and the volatility, rr, ai-e knelt n function.s o[ time, the stock J›rice Jistribution at f/m‹' T in a ri.‹k-neutral w‹ i Id is

(a) Suppose that S is the stock price at time / and S, is the stock price at time r . From cquation (14.3), it fol1t»vs that in a risk- neutral world:

Since thc sum of two independent normal distributions is normal with mean equal to the sum of the means and ›'ariance equal to the sum of the variances

'

2/

(b) Because and it follows that: 2 (c) If o, and r are the volatility and risk-free interest rate during the i th subinterval (i — 1, 2, 3) , an argument similar to that in (a) shows that: 20 I ?"^2012 Pearson Education

where /, , t,_ and t are the lengths of the three subintervals. It follows that the rcsult in (b) is still true. (d) The result in (b) remains true as the timc bcnvccn time zero and time T is divided into more subinteo-als, each hay ing its o n risk-ties interest rate and volatility. In the limit. it follows that, if r and cr are known functions of time, thc stock price distribution at time T is the same as that for a stock with a constant interest rate and variance rate w'ith the constant interest rate equal to the average interest rate and the constant variance rate equal to the avcrage variance rate. Problem 26.7. Write doour the equations/or simulating flue path followed by the o›sef price in the stochastic volalilit y mode•l in equation (26.2) and t26.3). Thc cquations are+g hn „

S(i) exp[(r — 9 — I"(I) / 2)A/ s, U(t)Ai]

Problem 26.8. “The IVF moJel Joes not necessarilv get the ct olution o[the volatiliH surface correct. ’ ExJ›lain this statement. The IVF model is dcsigned to match the volatility surface today. There is no guarantee that the volatility surface givcn by the model at future times will reflect the true evolution of the volatility surface. Problem 26.9. “When interest rate.s are con.stant the IVF model corrr‹ tly values any Rei-ivati ve whose payofJ depends on the value of the underl) ink a.f.set at only one time.” Explain why. The 1VF model ensures that the risk-neutral probability distribution o1 the asset price at any future time conditional tin its value today is correct (or at lcast consistent with the market prices of options). When a derivative‘s payoff depends on thc valuc ot the asset at only one time the lVF model therefore calculates the expected payoff from the assct correctly. The value ot the derivative is the present i aluc of the expected payoff. Whcn interest rates are constant the I VF model calculates this present walrie correctly. Problem 26.10. U.se a three-time-step tree to value an Amrrican fioaling lookback call option on a cw-rency iv hen the initial e.xchange rate is 1.6, the Jonestic risk-fre•e• rate iS 5“Noher atiiiititi, the foreign ri.sk-free intere.at rate is 8% per annum, flue exrhatige rate olutilit is /5*, and the timc to maturity /.s 1‹S months. Use the approa‹’h in Secfion 2h.5. In this case S — — 1.6 , - = 0.05 , r, = 0.05 o — 0.1 5, T — — i .5 , At = 0.5 . This means that 202 ’ 201 2 Pearson Education

ti — — e’ ' J

1.1119

= 0.8994

o = e"” ’” '“''’ = 0.9551 p d

—a

0.4033

1 — = 0.5967

The option pays off

2.1995 1 6000

0 6995 4. 6000

4 6000 4. 6000 04 307

0.0781

4 7790 1 i3000 4. 8970

1 6000 1 4390 1.6000 0 1610 0.0704

1 4390 1.4390 0.0964

1.7790 1 6000 1 439 0 1790 0 340 1.4390 1. 4390 1 294 0. 0000 0 144

1.2943 0.0569 1.1641

0.0000

Figure 526.1

Binomial tree for Problcm 26.10.

The tree is shown in Figure S26.1. At cach node, the upper number is the exchange rate, the middle number(s) are the minimum exchange rate(s) so far, and the lower numberts) are the value(s) of the option. The trcc shows that the xalue of the option today is 0.131. Problem 26.11. th at happens to the variance-gamma model a.s the ›aranietet- v ren#i to zero.'" As v tcnds to zero the value of g becomes T with certainty. This can be demonstratcd using thc GAMMADIST functitin in Excel. By using a series expansion for the In function w'e see that z becomes —dT . In the limit the distribution ofln Sz therefore has a

mean of In Sz + {r — q)T and a standard deviation of crT becomes geometric Brownian nitititin.

str that the model

Problem 26.12. Use a tlire•e-time-s lep free to alue an Ameri‹ an j ut option on the geometric average 203 201 2 Pearson Educatio n

of the price ofa tion-dividend-pay’ing .stock v:hen the sto‹’k price is $40, the strike price is S4i), the risL-]ree interest rate is ll1% pe awnum, the »olatility i.s 35.°<‹› per annum, anal the time to maturity is lhree• months. The geometric average i.s mea.sured from fodoy until the oJ›tion matures. In this case N, = 40, K = 40, r — — 0.1 , cr = 0.35 , T — — 0.25 , fit = 0.08333 . This means that ii — — e'

= 1.1063

'5

n = e' ‘“ ” 368 a

The option pays off

= 1 . OOS d

u — = 0.5161 d I —— 0.4539

where S denotes the geometric average. The trec is shown in Figure 526.2. At each node, the upper number is the stock price, the middle number(s) are the geometric average(s), and the lower number(s) are the value(s) of the option. The $eornetiir av'eras es are calculatcd using the first, the last and all intermediate stock prices on the path. The tree shows that the value of’ the option today is $1.40.

204 20i 2 Pearson Education

Figure S26.2

Binomial tree for Problem 26.12.

Problem 26.13. Can the approach for valuing path depenJent options in Section 26.5 be used%r a tw o-year American-.style option that provides a payoff equal to max(Ca„— K,ñ) where S„., is the average asset price over the three months preceJing exercise? Exj›lain your- answer. As mentioned in Section 26.5, for the procedure to z ork it moist be possible to calculate the value of the path function at time r Al from the value of the path function at time r and the value of the underlying asset at time r + ml . When S„,e is calculated from time zero until the end of the life of the option (as in the example considered in Section 26.5) this condition is satisfied. When it is calculated over the last three months it is not satisfied. This is because, in order to update the average with a new observation on .S , it is necessary to know the observation on S from three months ago that is now no longer part Of the average calculation. Problem 26.14. Verify that the 6.492 numher in Figure 26.4 is ‹ erred:t. We consider the situation where the average at node X is 53.83. It therc is an up movement to node Y the new average becomes: 53.83 x 5 + 53.97 54.68 Interpolating, the value of the option at node Y when the average is 53.97 is (53.97 — 51.12) x 8.635 + (54.26 — 53.97) x 8.101 54.26 — 51.12 Similarly if there is a down movement the new avcrage will be

‹C20l2 Pearson Education

8.5S6

53.83 x 5 45.72

52.45

In this case the option price is 4.416. The option price at node X when the average is 53.83 is therefore: 8.586 x 0.5056 * 4.41 6 x 0.4944)e ' ' " = 6.492 Problem 26.15. E. amine the early exercise policy for the eight paths cons idereJ in the example in Section 26.8. What is the difference between the early exercise policy given by the least squares apj›roacli anJ the e.xercise bounJary parameteriz•ation approach? Which gives a higher option price for the paths sampleJ? Under the least squares approach we exercise at time i = 1 in paths 4, 6, 7, and 8. We exercise at time t — — 2 for none of the paths. We exercise at time i = 3 for path 3. Under the exercise boundary parameterization approach we exercisc at time i = 1 for paths 6 and S. We exercise at time r = 2 for path 7. We exercise at time i = 3 for paths 3 and 4. For the paths sampled the exercise boundary pararneterization approach gives a higher value for the option. However, it may be biased upward. As mentioned in the text, once the early exercise boundary has been determined in the exercise boundary parameterization approach a new Monte Carlo simulation should be carried out. Problem 26.16. Consider a European put option or a non-dividend paving .s/ock v’l1en the stock price is $1 Hil, the strike price i.s ,$110, the ri›'k-free rate is 5"Xo per annum, and the time to maturity is one year. Suypo.se that the average variance rate dtiring the I'Cfc ofan option has a 0.20 probabili5 ofheing 0.06, a 0.5 probability oJ being 0.09, and u 0.3 probabilij' of heing 8.12. The volutilily is uncorrelated 'ith the .stock price. Estimate the value of lhe option. Use DerivaGem. If the average variance rate is 0.06, the value of the option is given by Black-Scholes with a volatility of 0 06 = 24.495 % ; it is 12.460. If the average variance rate is 0.09, the value of the option is given by Black-Scholes with a volatility of 0 09 = 30.000% ; it is l 4.655. If the avcrage variance rate is 0.12, the value of the option is given by BlackScholesMerton u ith a volatility of 0 2 = 34.641% ; it is 16.506. The value of the option is the Black-Scholes-Merton price integrated over the probability distribution of’ the average variance rate. It is 0.2 x 12.460 + 0.5 x 14.655 * 0.3 x 16.506 = 14.77 Problem 26.17. When there are U'o barriers how can a tree he de.signed.so that nodes lie on both barriers? Suppose that there are try o horizontal barriers.ñ and ñ, , with /-I, < H, and that the underlying stock price follows geometric Brownian motion. In a trinomial tree, there are three possible movcments in the asset’s price at each nodc: up by a proportional amount « ; 206

stay the same; and down by a proportional amount d where d — — i u . We can always choose o so that nodes lie on both barriers. The condition that must be satisfied by u is H _,— — TI,u’

or In // ln i/

= In H, .\

for some integer Y . When discussing trinomial trees in Section 20.4, the value suggested for n was e" so that In o = cr 3A/ . In the situutir›n considered hcre, a good rule is to choose lr « as close as possible to this value. consistent ith the condition givcn abov e. This means that we ln N, — set In ln //, where .V

int In li, — In H,

and int(i) is the integral part of . This means that nodes are at values of the stock pricc equal to H, , P,« . H,u' , . , H,ii U, Normally the trinomial stock price tree is constructed so that the central node is the initial stock price. In this case, it is unlikely that the current sttick price happens to be o, for some / . To deal with this, the first trinoinial movement should be from the initial stock price to o u’ , N u‘ and H,ii"' 1'here / is chosen str that H,u' is closest to the current stock price. The probabilities on all branches of the tree are chosen, as usual, to match the fii‘st two moments of the stochastic proccss follow en by the asset price. The approach works well except when the initial asset price is close to a barrier. Problem 26.18. ConsiJer an 18-month zero-couyon bond with a face › alue of $100 lhat ‹.’an be converted into five shares o(the company ’s stock at any time dtiring ifs life. Stippo› e hiat the current share price is .$20, no divicfe0éf6 ore paid on rite .stock, the risk-free rate for all maturities is 6Xu per annum worth continuous comj›ounding, and the share price volatiliu i.s 25o> per annum. fi.s.stime that the JeJault inten.City /.y 3’No per ve•ar anJ the recovery rate i.y 35%. The bond i.s callal›le at $110. L’se a three-tinier-.st‹•p tree lo ‹all ulate the value of the bonJ. What is the value of the ‹ oni ersion option (net o/the issuer ’s ‹’ull option)? In this case ai = 0.5 , 2. = 0.03 , o = 0.25 , r = 0.06 and 9 = 0 so that u = 1.1360 . d— — 0.8803, a = 1.0305 , p„ — — 0.6386, , = 0.3465, and the probability on dcfault branches is 0.0149. This leads to the tree shown in Figure S2G.3. The bond is called at nodes B and D and this forces exercise. Without the call the value at node D would bc 129.55, the value at node B would be l 15.94, and the value at node A would be 105.18. The v-alue of the call tiption to the bond issuer is therefore 105.18 — 1 03.72 - l .46 .

207 .â2012 Pearson Education

43.44 36.14 131.47 30.06 18 74

fi 05.00 30.06 10500

25.00

25.00 105.21 20.79 100.90

102.8 7

20.79 105.00

17.30 102.87 14.39 105.00

Default 0.00 30.00

Figure S26.3

Default

0.00 30.00

Tree for Problem 26.18

205 G20l2 Pearson Education

Default 0.0 0 30.0 0

CHAPTER 27 Martingales and Measures Protilem 27.1. How is the market price of risk defined[or a variable that is nof the price of an investment asset'? The market pricc of‘risk for a variable that is not the price or’ an investment assct is the market price of risk of an investment asset whose price is instantaneously perfectly positively correlated with the variable. Problem 27.2. Suppose that the mai-ket pricr of risk for golJ is zero. IQthe .storage costs are 1“Xupe•r annum and the ri.sk-free rate of interest is 6’X• per annum, what is the expected gi-o›vlh rate in the price of gold? s.s.suD7e hat golJ proviJes no income. lf its market price of risk is zero, gold must, after storage costs have becn paid, provide an expected return equal to the risk-free rate of interest. In this case, the cxpcctcd retuni after storage costs must be 6% per annum. lt follows that the expectcd grow th rate in the price of gold must be 7% per annum. Problem 27.3. ConsiJer Evo securitie.s hoth of which urc depenJent on the same market varial›lc. The expected returns from the securities at E 8’Zo tlRtl 12%. The ve latiliiy of the (tt-st secw it5’is 15%. The instnntaneoti.s ri.s/r-{see rate is 4°/o. Whnt is the volatili3 o(the .second security:! The market price of risk is

This is the same for both securitics. From the first security z'e know' it must be 0.0b — 0.04 - 0.266G7 0.15 The volatility, cr for the second security is givcn by 0.12 ().04 - 0.26667 The volatility is 30%. Problem 27.4. An oil company i.s set up solely jor the purpose o[e.xploring for oil in u ‹ ertain 'mall area o(Texa.s. Its value depenJs primarily on two stocha.stic variahlc•.s. the f›ri‹•e of oil anJ the quanti5' o/ J›roven oil reserves. Discuss whether the market price o/ risk Jor the second of these tw’o varial›les is likely In be positive, negative, or zero. It can be argued that the market price of risk for the second variable is zero. This is because the risk is unsystematic, i.e., it is totally unrelated to other risks in the economy. To put this another way, there is no reason why inv estors should demand a higher return for bearing the risk since the risk can be totally diversified aw'ay. 209 ' 2012 Pearson Education

Problem 27.5. Deduce the diferential equation for a derfvativ'c dep€Hdenl on the yrice.s of two nondividenJ-paying ti-udeJ securities by forming a riskless portfolio consi.sting of the derivative and the two tradeJ securities. Suppose that the price, /, of the derivative depends on the prices, and S, , of two traded securities. Suppose further that: dS, — - y,S d/ + a,S,d-,

where d , and drz are Wiener processes with correlation

. From Ito’s lemma

df — — p S,

To eliminate the d-, and dc u’e chtiosc a portfolio, fJ, consisting of — derivative 1: first traded security + " : second traded security °S

s

3

Since the portfolio is instantaneously risk-free it must instantaneously earn the risk-free rate of interest. Hence Combining the above equations

bt 2

2 ’

+’

— — r—

so that:

f "

bs

S

210 2012 Pearson Education

“

ñ s

S dt

rS,

2

Problem 27.6. Suppose that an interest rate, x, follow.s the process

where a, q, anJ c °r* F^ •!!!'N€ cOTtstanly. Suppose further that the market price oj risk for x is 2 . What i.s the process for x in the traditional ri.sk-neutral world The process for

can be written

Hcnce the expectcd growth rate in x is: a r —.

and the volatility of x is

In a risk neutral world the expected growth rate should be changed to

so that the process is

i.e . Hence the ‹lrift rate should be reduced by Act . Problem 27.7. Prove that v'hen the seruri3’ f proviJes income at rate eqtiation (27.9) becomes y + ; — r — — al . (Hint.' Form a new security, [’ that rovide.s no income by assuming that all the income from f is reinvested in f.) As suggcstcd in the hint we form a new security ' which is the same as f exeepi that all income produced by J is reinvestcd in / . Assuming we start doing this at time zero, the relationship betwcen and J" is 211 E2012 Pearson Education

If y’ and cr are the expected return and volatility of /’ , 1to’s lemma shows that

b

From equation (27.9) It follows that

Problem 27.8. She equation

thal ›i hen f and z yrcvide income of rGlt?S q . Otld (g re.syectivelv,

(Hint. Form nc•iv ,securities ’fanJ g’ that proviJe no income h)’ a.ssuming that all the income from l is reinve,sted in f and all the ink ome in g is reit1ve.ited in g .) As suggested in the hint, we form two new securities /“ and p’ which are the same as / and g at time zero, but are such that income from f is reinvested in f and income trom g is reinvested in . By construction J" and g’ are non-income producing and their values at time / are related to / and g by

From lto’s lemma, the securities g and g* have the same volatility. We can apply the analysis given in Section 27.3 to /' and g’ so that room equation (27.15)

or

Problem 27.9. ”The expecte d juture value of an

i.y in the real i• orld.” What Joes this statement implv ahout the market price of risk for (a) an intere.st rate and (h) a bonJ price. Do you think the statement is likely to be true? Give reasons. This statement implies that the interest rate has a negative market price of risk. Since bond prices and interest rates are negatively correlated, the statement implies that the market price of risk for a bond price is positive. The statement is rcasonable. When interest rates increase, there is a tendency for the stock markct to decrease. This implies that interest rates have negative systematic risk, or equivalently that bond prices have positive systematic risk. Problem 27.10. The variable S /z' an investment as.ye/yrovid/rig income th YffU q men.sured in currency A. It%llo ws the process

in the real w orld. Defining nay variables as neces vary, give the pt-ocess followed hy S , and ihe corresponding market price of risk, in (a) A world that i.s the traditional risk-neutral orld[or currenc A. (b) A world that i.s the traJilional risk-neutral 'orld for ciirrencJ’ D. (c) A world that i,s forward risk neutral with respect to a zero-coupon currency A hond mat yring at time T .

(d) A world that is]orw arJ risk neutral with respect to a zero coupon currency B bond mci/wring of time T (e)

In the traditional risk-neutral world the process followed by .8 is JS

(r — q)S dl + cr yS dz

where r is the instantaneous risk-free rate. The market price of dr -risk is zero. (b)

In the traditional risk-neutral world for currency B the proccss is

where (j is the exchange rate (units of A per unit of B), p„ is the volatility of Q and , is the coefficient of correlation between 9 and S . The markct price of -risk is

(c)

In a world thai is forward risk neutral with respect to a zero-coupon bond in currency A maturing at time Z

where cr, is the bond price volatility. The market price o1 ñ -risk is cr, (d)

In a world that is forward risk neutral with rcspect to a zero-coupon bond in currency B maturing at time T 213 201 2 Pearson fiducation

^ * 7 > s ^ r # » i vhGrC F is the forward exchangc rate. cr, is the volatility of r’ (units of A per unit of B, and p is the correlation betweenr and S . The market price of A -risk is

Problem 27.11. Explain the difference l›eki ecu the. »ay a forward interest rate is He]incd and the way the forw arJ values of other variables su‹:h as stock price.s, comnioJitf pricrs, and exchange rate.s are JefineJ. The forw'ard value of a stock price, commodity price, or exchange rate is the delivery price in a forward contract that causes the valuc ot the forw'ard contract to be zcro. A forward bond price is calculatcd in this ay. However, a tñrward interest rate is the intercst rate implied by the form ard bond price. Problem 27.12. Pro 'e the result in Section 2.7.5 that • hen dz and .i f g i5’ a inartingale [or ¿— a (13A. I I) to gct the proc ess e› for ln fand fn g.) with the dz uncorrel‹iteU,

. (Hint. Start bv ii sing equation

so that

Applying Ito’s lemma again

When â = cr, the coefficient of J/ is zero and //g is a martingale. Problem 27.13. 3how that w'he•n » — h g and h andf g ore each deJ›endent on n Wienr.r processes, //ie i th comjlonenl oj ihr v'olatili3’ of w is the i lh component of the volatili3 o} h minus the i th component of the olatili3 of g . 2'se Ihis to pro+'c› the result that iJ cz r is the volatility of L and a,. i.s the volulility of t' then the volatililv o] G'' i V is 214 'â'2012 Pearson Education

— 2pcr, a„ . (Hint.'

j|nsaa age saxozd S!h1

— ’!’ D

b'

’Zp( ' ’D

)

-}-

UI

CHAPTER 28 Interest Rate Derivatives: The Standard Market Models Problem 28.1. A company caps thrce-month LIBOR at 10°Xo per annum. The principal amount is $20 i’tlfllion. On a reset date. three-month LIBOR i.s 12oX‹›per annum. What payment would this lead to under the cap? When Iron/d fhe payment be made? An amount $20, 000, 000 x 0.02 x 0.25 = $100, 000 would be paid out 3 months later. Problem 28.2. Explain why a sv ap option can he regardeJ as a type of bond option. A swap option (or swaption) is an option to enter into an interest rate swap at a certain time in the future with a ceriain fixed rate being used. An interest rate swap can be regarded as the exchange of a fixed-rate bond for a floating-rate bond. A so aption is therefore the option to exchange a fixed-rate bond for a floating-rate bond. The floating-rate bond will be worth its face value at the beginning of the life of the swap. The swaption is therefore an option on a fixed-rate bond with the strike price equal to the face value of the bond. Problem 28.3. Use ltte Black ’s model to value a one-year European put option on a 10-year bond. Assume t,hat the current value of the hond is $125, the strike price is $ 11ñ, the one-year interest rate is 10’X per annum, the bond’.s/orward price vol‹itility is d“
— (125 —10)c'“" = 127.09 , K — — 1 10, P(0,T) = c " , crB— — 0.0b , and T — — In(127.09/ I 10) (0.08' / 2)

— 1.8456

0.08 d, — ‹I— N 0.08 = 1.7656 firom equation (25.2) the value of the put option is 11 0e “' N(— 1.7656—) 127.09e*’ ’“’.Y(— I. 5456) = 0.12 or $0.12. Problem 28.4. Explain carefully hon› you would use (a) spof volatilities and (b) fiat volatilities to value a 21 6 five-year cap. 2012 Pcarson Education

When spot x olatilities are used to value a cap, a diflércnt volatility is used to value each

Problem 28.5. Calculate the Price of an oj›tion that r.aps the thre•e-month rate, starting in 15 month’ time, at 73“ (quoted with qiiarterly c.ompounding) on a principal amount of $1,000. The forw'ard inte•rest rate for the period in que.stion i.s 12% per annum (quoted with quarlerl v compounding), the I 8-month ri.sk-/ree interc•st rate (continuously compounded) i.S 11.5‘Xo yer annum, ond the volatility of the form ard rate iS 12‘Zo fiñt‘ NTltttlttt.

In this case I = 1000, b, = 0.25. P(0,tz, ) = 0.8416 .

= 0.1 2, ñ, = 0.1 3, r = 0.11 5 , rr, = 0.1 2,

=I . 25,

L6, = 250

d, —

In(0.12 / 0.13) + 0.12 x 1.25/ 2 0.12 1 25

— 0,5295

— —0.5295 — 0.12 25 = — The value of the option is d — 250 0.bti37 x 0.541 6 x [0.1 2 A"(—0.5 295) — 0.1 3.V(—0. 6637)] = 0,59

or $0.59. Problem 28.6. A hank uses Black ’s model to price European hond option.s. Suppose that an implicd prim.e volatility for a 5-year option on a hond inoniring in 10 years is used to price a 9-year option on the bond. WoulJ vou e.xpect the re.sultant yrice to he too high or too low? Explain. The implied volatility measures the standard deviation of the logarithm of the bond price at the maturity of the option divided by the square root of the time to maturity. In the case of a five year tiptitin on a ten year bond. the bond has five years left at option maturity. In thc case of u nine y••• '•Ftion on a ten year bond it has one year led. The standard deviation ot‘a one year bond price

obsein ed in nine years can bc normally bc cxpccted to be considerably less than that of a five year bond price observcd in fivc ycars. (See Figure 28.1.) We would therefore expect the price to be too high. Problem 28.7. Calculate the vulue of a four-year European call option on Idend that will mature five years /i cm todg using Black’s model. The five-yecit- cash hond price is $ 1 HS, the cas h prices of a four-year hond 'ith the .same coupon i,s S10z*, the strikes price is $100, the four-i ear risk-free intere.f I rme i.s 10’No yer annum rith continuous compounding, am the volatility for the bonJ price in four year.s i.s 2”Noper annum. The present value of the principal in the tour ycar bond is 100c "’ ‘ — 67.032 . The present value of the coupons is, thereftire, 102 67.032 = 34.968 . This means that the forward price of the five-year bond is (10—5 34.9(›8)e”’ ' = 104.475 The parameters in i3lack"s modcl arc therefore F = 104.475, K — — 100, i = 0.1, T — — 4, and — 0.02 .

In 1.04475 + 0.5 x 0.02' x 1.11 4 '' 44 d, — — d, — 0 . 002. 04 2=4 1 . 0744 The price of the Europeane call is104.475N(1. 1144) — 100 Y(1.0744)) = 0 3.19 or $3.19. Problem 28.8. I] Ihe yield volatility for a five-year put G /fO O1't a hond maturing in 10 years time is specified as 22“Xo, how should the option he valued." Assume /ñ«/, baseJ on toJa ’s interest rates the moJiJied duration o] ihe bond at the nialurij oJ the option v’ill he 4.2 years anJ the fomvard yield on the bonJ is 7%›. The option should bc valued using Black’s model in equation (28.2) with the bond price volatility being 4.2 x 0.07 x 0.22 = 0.0647 or 6.47%. Problem 28.9.

that other instrument is the same as a five-year zero-co.at collar n'here the .strike jar-ice o[the cap equafs the strike price of the i!oori What does the common .strike Price equal? A 5-year zero-cost collar where the strike price of the cap equals the strike price of the tloor is the same as an interest rate swap agreement to receive floating and pay a fixed rate equal to the strike price. The common strike price is the swap rate. Note that the swap is actually a forward swap that excludes the fi rst exchange. (See Business Snapshot 28.1) Problem 28.10. Derive a pu/—call pariH relations hip for European bonJ options. Thcrc arc two way of exprcssing the put call parity relationship for bond tiptions. The first is in terms of’bond priccs: c + I + Ke• — — pD

wherc c is thc price of a Europcan call option, › is the price of the ctirresponding European put option, / is thc prcsent value of the bond coupon payments during the life of the option, K is the strike price, T is the time to maturity, B is the bond price, and R is the risk-free interest rate for a maturity equal to the life of the options. To prove this we can consider to o portfolios. The first consists o1 a Europcan put option plus the bond; the second consists of the European call option, and an amount of’ cash cqual to the present value of the coupons plus the present value of the strike pricc. Both can be seen to be worth the same at the maturity o1 the options. The second way of expressing the putwall parity relationship is c * be ” = yr + zF e ”

where F is the forward bond price. This can also be proved by considering two portfolios. The first consists of a European put option plus a forward contract on the bond plus the present value of the forward price; the second consists of a European caI1 option plus thc present value of the strike price. Both can be seen to be worth the same at the maturity of the options. 21 8 2012 Pearson Education

Problem 28.11. Derive a put—call parity relationship for European sw'ap options. The putwall parity relationship for European swap options is wherc c is the value of a call option to pay a fixed rate o1 .t, and receive floating, p is the value of a put option to receive a fixed rate of ,. and pay floating, and C is the value of the forward swap underlying the swap option wherc .s, is receix ed and floating is paid. This can be proved by considcring tw'o portfolios. The first consists of the put option; the second consists o1 thc call option and the swap. Suppose that the actual sz'ap rate at the maturity of the options is greater than s . The call will be exercised and the put will not be exercised. Both portfolios are then worth zero. Suppose next that the actual swap ratc at the maturity of the options is less than s, . The put option is exercised and the call option is not exercised. Both portfolios are equivalent to a swap where .‹ is received and floating is paid. In all states of the world the two portfolios are worth the same at time T . They must therefore be worth the same today. This proves the rcsult. Problem 28.12. Explain why there is vn arbitrage op ›ortunitv ij the implied Black (ilat) volatility of a cap is different from that off jloor. Do the hroker quotes in Tahle 28. 7 yresenf an arbitrage •PF

Ttit it

ity ’?

Suppose that the cap and floor have the same strike price and the same time to maturity. The following put—call parity relationship must hold: cap swap = floor where the swap is an as•eement to receive the cap ratc and pay floating over the wholc life of the cap/floor. If the implied Black volatilities for the cap equal those for the floor, the Black formulas show that this relationship holds. In other circumstances it does not hold and thcre is an arbitrage opportunity. The broker quotes in Table 28.1 do not present an arbitrage opportunity because the cap offer is always higher than the floor bid and the floor offer is always higher than the cap bid. Problem 28.13. When a bond’.s price i.s lognormal c'an the hem’.s yield be negative? Explain your on.swer. Yes. If a zero-coupon bond price at some future time is lognormal, there is some chance that the price will be above par. This in turn implies that the yield to maturity on the bond is negative. Problem 28.14. What is the valtie ofu EuroFean .swap option that gives the holder the right to enter into a 3year annual-pay swap in %ur vears where a fixed rate of 5‘o is paiJ anJ LIBOR i.s received'? The smap principal i.s $10 million. Assume that the yield curve is}lOt Ot S olo Jeer annum with annual compounding and the volatility of the .swap rnte is 20%^•. Compare your answer to that given by DerivaGem. In equatitin (28.10), L = 10, 000, 000 , s, = 0.05 , sz — — 0.05 , d — 0 . 2 4 / 2 = 0.2 , d 2 = —.2 , and 219 â201 2 Pearson £iducation

l

= 2,2404 1.05 “ 1 .05‘ The value tif thc swap option (in millions 1.05’ of dollars) is 10 2.2404[0.05 A'(0.2) — 0.05A'(—0.2)] = 0.178 This is the same as the answer given by DerivaGem. (F-or the purposes of using the DerivaGein sottw'are, mite that the interest rate is 4.579% with continuous compounding for all maturities.) Problem 28.15. Suppo c that the yiefd, p , on a zero-‹'ovpon hoiid %11c» s the proce.s.s dR — — Et dt + a dz

The price of the bond at time is e*"’ " where T is thc time when the bond matures. Using Itô’s lemma the volatility of the bond price is w' t)e°’"” '

e°"’”' = -cr[T—

This tends to zero as i approaches / . Problem 28.16. Carry out a manual calculation to verify the option prices in Example 28.2. The cash price of the bond is 4e ”’ " * 4e*’ ” ’ " * ... + 4e ”"’ * 100c ' "' = I 22.82 As there is no accrued intercst this is also the quoted price of the bond. The interest paid during the life of the option has a present value of 4e = I 5.04

” + 4e ' '”' + 4 e ' ' " “ ’ + 4e ' ' ‘ 2

The for ard price of the bond is therefore (1 22.5—2 15.04)e'’““ " = 120.61 The yield with semiannual compounding is 5.0630%«. The duration of thc bond at option maturity is 0.25 x 4e°’'”’ "‘ ... 7.75 x 4e‘’"” + 7.75 x l00e ’” 4 -’ + 4. ‘ (c) The 5-› ccii‘flut volatili for coy,s and floors (d) The floor rale in a..-zero-cost 4 " 5-year collar when the cay rate is 8 Xa 100 “ "

220 'i€“201 2 Pcarson fiducati on

Wc choose the Caps and Sv ap Options workshcet of DcrivaGem and choose Cap/F1t›or as the Underlying Type. We enter the 1 -, 2-, 3-, 4-, 5-year zero rates as 6 o, 6.4%t 6.7%t 6.9@ », and 7.0%› in the Term Structure table. We entcr Semiannual for the Settlement Frequency, 100 for the Principal, 0 for the Start (Years), 5 for thc End (Years), 8% for the Cap/filoor Rate, and $3 for the Price. We select Black-European as the Pricing Model and choose the Cap button. We check the Imply Volatility box and Calculate. The implied volatllity is 24.90%. We then uncheck Implied Volatility, select Floor, check Imply Breakeven Rate. The fltitir rate that is calculated is 6.71%. This is the moor rate tor which the floor is worth $3.A collar when the floor rate is 6.71% and the cap rate is 8% has zero cost. Problem 28.18. Shear that V, — — V, where

i.s the value ofa .s›iaJ›tion to puy a jixed rate of s and

We prove this result by considering two ptirtfolios. The first consists of the swap option to rcceive s, ; the second consists cf the swap option to pay ›, and the forward swap. Suppose that the actual swap rate at the maturity of the options is greater than s . The swap option to pay s will be excrcised and the swap option to receive s, u ill nol be exercised. Both portfolios are then worth zero since the sw ap option to pay sq is neutralized by the foovard swap. Suppose next that the actual swap rate at the maturity of the tiptions is less than .‹, . The swap option to receive s, is exercised and the swap option to pay .i, is not exercised. Both portfolitis are then equivalent t‹i a swap whcre .i, is received and floating is paid. In all states of thc world the two portfolios are wtirth the same at time T . They must therefore be u orth the same today. This proves the result. When ›g equals the current forw'ard swap rate — 0 and U,— U, . A swap option to pay fixed is thcrcfore ivtirth the same as a similar swap option to receive fixed v hen the fixed rate in the so ap option is the loin ard swap rate. Problem 28.19. Suppose that zero rates are as in Problem 2‹S. 17. L'se DerivaGein to determine the value o/ an oplion to pay a (axed i-‹i/e oJ 6°X• and receive LIBOR an a five-year sn'ap starling in one 5'ear. Assutne that they princ ipal is SUB million, payments are excliangeJ semionnuall j, and the strap rate volatilit v is 21%‹. Wc choose the Caps and Swap Optitins workshcct of DerivaGem and choose Sw ap Option as the Underlying Type. We enter 100 as the Principal, 1 as the Start (Years), 6 as the End (Ycars), 6% as the Swap Rate, and Semiannual as the Settlement Frequency. We choose BlackEuropean as the pricing model, enter 21% as the Volatility and check the Pay fiixed button. We do not chcck the Imply Rreakcvcn Rate and imply Volatility boxes. The value of the swap option is 5.63. Problem 28.20. describe hour you would (a) cal‹:ulate cap flat volatiliti‹'s fi-om cay strol vol‹itilities and LI›) 221 0U2012 Pearson Educati‹3n

calculate cap spot volatilities from cap flat volatilities. (a) To calculate flat volatilities from spot volatilities we choose a strike rate and use the spot volatilities to calculate caplet prices. We then sum the caplet prices to obtain cap prices and imply flat volatilities trom Black’s model. The answer is slightly dependent on the strike price chosen. This procedure ignores any volatility smile in cap pricing.

(b)To calculate spot volatilities from flat volatilities the first step is usually to interpolate between the flat volatilities so that we have a flat volatility for each caplet payment date. We choose a strike price and use the flat volatilities to calculate cap prices. By subtracting successive cap prices we obtain caplet prices from which we can imply spot volatilities. The answer is slightly dependent on the strike price chosen. This procedure also ignores any volatility smile in caplet pricing.

222 '2012 Pearson Education

CHAPTER 29 Convexity, Timing, and Quanto Adjustments Problem 29.1. Explain how you would value a derivative that pays off 1 00/t in five years where R is the oneyear interest rate (annually compounJeJ) observed in]oiir years. What difference would it make if the payoff were in (a) 4 veor.s and th) 6 years'? The value of the derivative is 100s, ,P(0, 5) where P(0, i) is the value of a / -year zerocoupon bond today and fi

is the forward rate for the period between i, and /z , expressed

with annual compounding. If the payoff is made in four years the value is 100(fi, , c)P(0, 4) where ‹ is the convexity adjustment given by equation (29.2). The formula for the convexity adjustment is:

where o , is the volatility of the forw’ard rate between times /, and /, The expression l0u(fi, , c) is the expected payoff‘in a world that is forward risk neutral with respect to a zero-coupon bond maturing at time four years. If the payoff is wade in six 4p years, the value is from equation given6)by 100(fi,(29.4) , c)/’(0, exp —where is the correlation between the (4,5) and (4,6) forward rates. As an approximation we can assume that p — — 1 , o-, 5= o-4, and fi, , = R, , . Approximating the exponential function we then get the value of the derivative as 1oops,— c)P(ñ,6) . Problem 29.2. Explain whether any c‹›nvexity or timing adjustments are neces ary when

(a)

be wish to value o spt‘ead option that Paf's off every quarter the excess (if any) of the five-year swap rnte over the three-month LIBOR rate applied to a principal o/$l 00. The payoff occurs 90 days after lhe rates are observed. (b) We wish to value a derivative that pays off every quarter the three-month LIBOR rate mixv.y the three-month Treasury bill rate. The payoffoccurs 90 days after the rates are observed. (c) A convexity adjustment is neccssary for the swap rate (d) No convexity or timing adjustments are necessary. Problem 29.3. Suppo.se that in Example 2d.3 o] Section 2d.2 the payoff occurs after one year (i.e., when the interest rate is observed) rather than in 15 months. What difference does this make to the input.s to Black’s models? There are two differences. The discounting is done over a 1.0-year period instead of over a 1.25-year period. Also a convexity adjustment to the forward rate is necessary. From equation 223 â 2 0 l 2 Pearson Education

(29.2) the convexity adjustment is: 0.07’ x 0.2" 11.25 x

= 0.00001

I 0.25 x 0.07 or about halt a basis point. In the formula for thc caplet we set F, — — 0.07005 instead of 0.07 . This means that d, — — —0.5642 and d_, — — —0.7642 . With continuous compounding the l 5-month rate is 6.50o and the forward rate betw'een 12 and 15 months is It.94%›. The 12 month rate is therefore 6.39% The caplet price becomcs 0.25 x 10, 000e ' " " ” " [0.07005 ‘(—0.5642) — 0.08.\ (—0.7642)] = 5.29 or $5.29. Problem 29.4. The yield curves i.s flat at 1ñ% per annum i› ith annual coinJ›ounJing. Call ulale the value of an instrument i»liere, in five years ' time, ltte tit o-vear .s›i ap rate (»ith annual compounding) i.s receiveJ ‹and a fi.red rate out 0% is paiJ. Both are applied to a notional principal of S 100. A.ssume that the volatilit)' of the strap rate i5’29“o per annum. Explain why the value of the instrument is Jifferent fi-oni zero. The convexity adjustment discussed in Section 29.1 leads to the instrument being worth an amount slightly diltcrcnt from zero. Define G( ) as thc value as seen in five years of a twoyear bond vt ith a coupon of 1 O0o as a function of’its 1).1 ) ield. 1.1 ( "I + 1 + y (1 ' i) 0. I 2.2

It follows that U’(0.1) =— 1.735 5 and G“(0.1) = 4.6552 and the convexity adjustment that must be made for the tn o-year swap- rate is 4.6582

- 0.00268 i .7355 We can therefore value the insirumcnt on the assumption that the swap rate will be 10.268% in five years. The value tif the instrumcnt is 0.26_ .167 8 or $0.167. (L5 x 0.1’ x 0.2' x Sx

Problem 29.5. What difference doe.s it make in Problem 29.4 i[the .sn›aJ› rate is observeJ in fi ve years, hut the exe hang‹• of pa¡vnent.f takes place in (a) st.x vear.s, and (h) .seven years? Assume that the volatilities of all forward rate.f are 20’Xn. Assume also that flue fomvard s ’ap rate for the perioJ between years five and .yever has a correlation of 0.S i ’itll the %rward inlcre•st rate heRi ecu yeurs fi»e and six and a correlation o] !1. 95 worth the forward interest ralc betiv'ee•n year.s five anâ se yen, 224 2012 Pearson Education

In this case we have to make a timing adjustment as well as a convexity adjustment to the forward swap ratc. For (a) equation (29.4) ,hows that thc timing adjustment involves multiplying the swap rate by 0 ‘‘ 0 20 p0$20 x 0.l x 5 = 0.9856 exp so that it becomes 10.268 x 0.985ò = 10. 120 . The value of the instrument is 1.1‘ or $0.06b. For (b) equation (29.4) shows that the timing adjustment involves multiplying the swap rate by exp — 0.95 x 0.2 x 0.2 x 0. x 2 x = 0.9660 1+ 5 so that it becomes 10.268 0.966 = 9.919 . The0.1 value of thc instrument is now 0.081

1.1 = —0.042 ’

or —$0.042.

Problem 29.6. The yf-ice oJ et òonJ at time T , measured in terme of its yield, is G(»..) . Asenme geometric Brownian motion for lhe [oinvard bonJ vielJ, , in a ivorld that is forw’ard risl‹ neutral ii ith respect to a hond maturing at time T . Supyo.se that the growth i’atr o/ the jorwarJ bond field is r and its volatility a,. . (a) D.se Ito’s lemma to calciilate the proce.s.s for the]ooi ard bond Erie.e in terme of o:, p, . r , aHd 0 (1') .

(b) The forward hond price should follow a martinqale in the worlJ considered. k'se this fact to calctilate an expression for a. (c) Show lhat the expression jor a is, to a first approximation, consistent with equation (29.1). (d) The process for i is The forward bond price is C(j ) . From IU’s lemma, its process is d G( v) = [G'( )aJ +

G’(y),

›^] dt + G’( v)cr y d-

(b) Since thc expected growth rate of G(y ) is zero

(c) Assumint as an approximation that ,› always equals its initial vafue of v, , this shows that the growth rate of t is 225 C20 1 2 Pearson £iducation

1

’t›’‹

*2 ^'(y ) *’”

The variable ,. starts at v0and cnds as y . The convexity adjustment to y when we are calculating the expected value of yt in a world that is forward risk neutral with respect to a zero-coupon bond maturing at time T is approximately yzT times this or "(y› 2 G'(Jt)

This is consistent with equation f29.l). Problem 29.7. The variable S is an investment asset proviJing income at rate g ttlEastired in currency A. It follow.s fhe process d8 — — S ât + a S dz in the real world. defining new variables as necessary', give the proce.s.s followeJ by S , and the corresponding market price of risk, in (a) A woi-ld that is the traJitional risk-neutral n'orld [or curren‹’y A. (b) A world that i.y the traditional risk-neutral w'orld for currency B. (‹) A w'orld that i.s/ortvnrd risk meu/rs/ »ilh respect to a zero-coupon currency A bond (J) A +rorld that i.s forward risk neutral Frith respect to a zero coupon currency B bond maturing at time T . (c) In the traditional risk-neutral world the process followed by N is dS — - (r q)S dt + cr S d. wherer is the instantaneous risk-free rate. The market price or -risk is zero. (b) In the traditional risk-neutral world for currency B the process is dS — — (r — q y p , a a )S dt + a z S dz where is the exchange rate (units of A per unit of B), o-p is the volatility ot 9 and pq is the coefficient o1 correlation between 9 and S . The market pricc orñ -risk is (c) In a world that is foovard risk neutral with respect to a zero-coupon bond in currency A maturing at time T dS — — {r — q+ O a p)S dl mz S dz where29.8. cr is the bond price volatility. The market price of & -risk is cr, Problem (d) In option a worldprovides that is foe 'ard risk with respect bond currency A call a pavoff at neutral time T of max(S, — to x, 0)a zero-coupon ven, where S is the in dollar priceB maturing at time T 226 92012 Pearson Education ot F (units ofA per unit of B, and wherer is thc forward exchange rate, o-, is the volatility y„ is the correlation betweenr and .Y . Thc market price ofñ -risk is o-, y„ „ .

of gold at time T anJ r is the .strike price. Assuming that the storage co.st.s of gold are zero and defining other variables as necessaz, calculate the value o[the contract. Define P(t,I): Price in yen at time / of a bond paying 1 yen at time T E3.): Expectation in world that is forward risk neutral with respect to /'(/,T) N: Dollar forward price of gold for a contract maturing at time T to. Value ofr at time ai.: zero Forward rate (dollars per yen) G: Volatilityexchange of Volatility of G no: We assume that S, is lognormal, We can work in a world that is forward risk neutral with respect to Pvt,T) to get the value of the call as N(0, T)[E (5 )N(d, ) — N(d )] where d,

d The expected gold price in a world that is forward risk-neutral with respect to a zero-coupon dollar bond maturing at time T is F . It follows from equation (29.6) that Ez(Sz) - F (1 + pa z,mzT) Hence the option price. measured in yen, is PRO,T)[Fz(I + per, m T)N(d,—) KN(d,,)] where In[I,(1 + d,

Problem 29.9. Su¡›po.se that an index of Canadian s/ocky currently stands at 400. The Canadian dollar i.s currently worth 0.70 2’.S. Jollars. The risk-free interest rates in Canada and the U.S. are conston! at 6% an‹f 4°Xn, respectively. the d/ ripesc/ yieféf on the index is 3‘Xo. Hcfine Q as the number- oj CanaJian dollar.f per LI.S dollar anJ S as the value of the inde.x. The volatility of N i.$ 2K“Xo, the volatility of p is 6°Xu, and the correlation between S and Q i.s 0.4. U.se DerivaGert to determine the value of a two year Amei‘ican-style call option on the index if (a) It pays ojj in CunaJian dollars the amount by whi‹’h the index exceed,s 400. (b) It pays off in U.S. dollar.s the amount by’ which the index exceed.s 4ñ0.

227 92012 Pearson Education

(a) (b)

The value of the option can be calculated by setting S, — 400, K— —400, r = 0.06 ,

q = 0.03 , o = 0.2 , and T —- 2 . With 100 time steps the value (in Canadian dollars) is 52.92. The growth rate of‘ the index using the CDN numeraire is 0.06 — 0.03 or 3%. When we switch to the USD numeraire we increase thc growth rate of the index by 0.4 x 0.2s 0.06 or 0.45 % per year to 3.48%. The option can therefore be calculated using DerivaGein with S — — 400, K — — 400, i = 0.04 , q = 0.04 — 0.0345 = 0.0052 , o- = 0.2 , and T —- 2 . With 100 time steps DerivaGem gives the value as 57.51.

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CHAPTER 30 Interest Rate Derivatives: Models of the Short Rate Problem 30.1. Wluit is the Jifference het»een an equilibrium model and a no-arbitrage model? Equilibrium models usually start with assumptions about economic i-aiaables and derive the behavior of interest rates. Thc initial term structure is an output from the model. In a ioarbitrage model the initial term structure is an input. The bchavior of interest rates in a noarbitrage model is designed to be consistent with the initial terns structure.

In Vasicek’s model the standard dcs iation stays at 1‘%. tn the Rcnflleman and Bartter model the standard deviation is proportional to the level of the short rate. When the short rate incrcases from 4% to 8"% the standard deviation increases from 1% to 2%. In the Cox. Ingcrsoll, and Ross model the standard deviation of the short rate is proportional to the square root of the short rate. When the short rate increases from 4%â to 8% the standard deviation increases from 1 fio to 1.414% . Problem 30.3.

11 the pricc of a traded security fol1o*'cd a mean-rei erting or path-depcndent process there z'ould be market inefficiency. Thc short-term interest rate is not thc price of a traded security. In other words we cannot trade something w hose price is all ays the short-term interest rate. There is therefore no market inct‘ficiency v hen the short-term iuterest rate follows a meat- reverting or path-dependent process. We can trade bonds and other instruments v hose prices do depend on the short rate. The prices of these instruments do not follow mean-reverting or pathdependent processes. Problem 30.4. Explain Ihe Jifjerence henveen a one-fac'tor and a twin-(actor inte•re•sl rate tnoJel. In a one-factor model there is one sourcc o1 uncertainty driving alI rates. This usually means that in any short period of time all rates move in the same direction Ibut not necessarily by the samc amount). In a two-factor model, there are two sources ot'uncertainty driving all rates. The first source of uncertainty usually gives rise to a roughly parallel shift in rates. The sccond gives rise to a twist whcrc long and short rates mov cs in opposite directions. Problem 30.5. Can the approach desci ibeJ in 3ection 30.4 jor decompos in,g an oytion on a coupon-hearing hond into a portfolio of option.s on zero-cotipon boiiñy lie u.red in conjun‹ tion ith a hi o229 ›2012 Pearson Educati on

fâ‹ tor model? Explain your ansv-er. No. The approach in Section 30.4 relies on the argument that, at any given time, all bond prices are moving in the same dircction. This is not true w hen there is more than one factor. Problem 30.6. Suppo.se that a H. I and h — — 0.1 in hoth the Vasicek and the Cox, Ingersoll, Ross model. In both inoJels, lhe• initial .short rate is 10’Zo and the initial .standard Jeviution of the short rate change in a short time At is 0 . 0 2 s . Compare the prices given bv the models for a zerocoipon bond that matures in year 10. In Vasicck’s model, n = 0.1, b — — 0.1 , and o- — 0.02 so that B(t,t ! 0) ' ( '' '°) = 6.32121 0 ' A(i, 1

2 . 1 10) = exp 6'3212 I — 10)(0.1 x 0— 0.0002) 0.01

0.0004 x 6.32121' 0.4

— — 0.71 557 Thc bond price is thereforc 0.715 b7e°’ “"'"'"' ' = 0.38046 In the Cox, lngersoll, and Ross model, o — 0. t , b — — v.I and cr = 0.02 /

= 0.0632. Also

y— Q = (y a)ie"“ — 1) + 2y = ' 0.92992

Define

B(l,t2 10)

2(e’'"

= 6.07650

= 0.1341 6 .4(t,t 10) =

2ye""”’ '"‘ °" — 0.69746

The bond price is thcrefore 0.69?46e ‘ """”’ ' = 0.37986 Problem 30.7. Suf›pose that a — — 0.1 , b — — 0.ñS, and o- — — ñ.015 in Vasicek’.i model with the initial value of the short rate being 5’X‹. Cal‹’ulate the price ofa one-year European call option on a zerocoipon bond with a principal oJ $100 lhat mature.f in three years w'hen the strike price is J87. Using the notation in the text, ,

3, T — — 1 , L = 100 , K — — b7 , and ^P 0.01 0.1 , P(0,3) = 0.55092 , and 6 — 1 .14277 so that equation From equation (30.6), P{H,1) = 0.94988 (30.20) gives the call price as call price is 1 00 x 0.55092 x .V(1.1427 7) — 57 x 0.94988 x A'(1.11688) = 2.59 230 ‹i’92012 Pearson Education

Problem 30.8. Repeat Problem 3ñ.7 valuing a European put option worth a s/rite oJ $87. What r.s the put—call parity relationship between the prices ofEuropean call and pul options? Show that the yt‹/ anâ call option prices satisfy put -c'all paritv in this case. As mentioned in the text, equation (30.20) for a call option is essentially the same as Black’s model. By analogy with Black’s formulas corresponding expression for a put option is x#(0,7)N(— h + u p) — fP(0, ) V(—/i) In this case the put price is 87 x 0.94988 x A(—1. 11688) — 100s 0.85092 x V(—1 . 14277) = 0.14 Since the underlying bond pays no coupon, putwall parity states that the put price plus the bond price should equal the call price plus the present value of the strikc price. The bond pricc is 85.09 and the present value of the strike price is 87 x 0.94985 = 52.G4 . Putwall parity is therefore satisfied: 82.64 + 2.59 = S5.09 + 0.14

Problem 30.9. Suppose that a — — 0.05, ñ = 0.0b, and a — — 0.015 in Vasicek’.s model with the initial short-term interest rate being 6“Xo. GHlCtllate the price o/a 2. 1-year European call option on a hond that will mature in three yc•ars. Suppose that the bond pays a c'oupon o[5% semiannually. The principal of the bond is 100 and the .strike prices of the option is 99. The .strike yrice is thr cash price (not the quoteJ price) that will he paid]or the bond. As explained in Section 30.4, the first stage is to calculate the value of r at time 2.1 years which is such that the value of the bond at that time is 99. Denoting this value of r by ’, we must solve 2.5 A(2.1, 2.5)e " 2 "2 '”’ + 102.5 N(2.1,3.0)e ” 2 ''’"'‘ 99 where the A and B functions are given by equations (30.7) and (30.8). The solution to this is r‘ = 0.066 . Since 2.5 A(2.1, 2.5)e°D'“ ’" 5 "’”’ = 2.43473 and . )0. = 96.56438 102.5 A(2. 1, 3.0)e 6(•. 3I,O "'

the call option on the coupon-bearing bond can be decomposed into a call option with a strike price of 2.43473 on a bond that pays off 2.5 at time 2.5 years and a call option with a stri ke price of 96.56438 on a bond that pays off 102.5 at time 3.0 ycars. Equation (30.20) shows that the value of the first option is 0.009085 and the valuc of the second option is 0.806143. The total value of the option is therefore 0.815238. Problem 30.10. Use the answer to Problem 30.9 anJ put—call parity arguments to calculate the price of a put option that has the same terms as the call option in Problem 30.9. Put-call parity shows that:

› -- c+ PV(K) —(B — 1) z here c is the call price, K is the strike price, is the present value or’ the coupons, and B, 231

is the bond price. In this case ‹ = 0.8152 , PV K) = 9ñ P(0. 2.1) = S7.1222 , B I— — 2.5 z P(0. 2.5) + 1.02.5 P(0, 3) = $7.4730 so that the put price is — 0.8152 + 8.7.1222 — 57.4730 = 0.4644 z Problem 30.11. In the Hull—White model, u — — 0.05 and a —- ñ.ñ I. C’alculate the price of a othe-yo•ar EtiroJlean c all of›tion on a zero-coupon hond that ii ill oia/tire in five years whcn the tenn .structure is flttl th I K‘fo, the principal o[the hond is 5100, and the .in-ike grit:e is $6b. —0.6065 . Also Using the notation in the text P[11,T) — e ' ' 0.9048 and P[0,3’) — e°’ ' —

—1 e°"’” ' = 0.0329 2x 0.08 and £ = —0.4 192 so that the call price is 100 x 0.6065.\'(h) — 68 x 0.9045.i (—/i rr„) = 0.439 Problem 30.12. Suppo.se that ‹i — — 0.05 nt d a ñ.ñ 1.5 in the Hull—White model ii ith the initial term strur.ture lieing jlat at 6% with serniunnucil comj›ounding. C"alculate the pi’ice oj a 2.1-year European call option on a bond tidal will mature in three br.ars. SupJ›o.se that the bond pay.s a coupon of J % per annum .seniiunnnollv. The J›rincif›al o} the bond i,s ION and the strike price of the oj tion is 99. The strike price i the ca.sl1 yric e (not //ie quoted price) that will be paid for the hond. The relevant parameters for the Hull—\Vhite nitidel are

= 0.05 and o- = 0.015 . Sctting

A = 0.4

(2.1, 3) =

B(2.1, 2.5) Also from equation (30.26), A(2. 1. s) = 0.99925 The first stage is to calculate the value of r at time 2.1 years which is such that the › alue of the bond at that timc is 99. Denoting this value of R by R’, +'c must solve 2.5e ’!”’ + 102.5 (2.1, 3)e°" 2 ' ”’ = 99 The solution to this for fi’ turns out to be 6.626%. The option on the coupon bond is decomposed into an opti‹in with a strike price of 9ti.565 on a zero-coupon bond w ith a principal of 102.5 and an option with a strike price of 2.435 on a zero-coupon bond with a principal of 2.5. The first option is worth 0.0103 and the second option is worth 0.9343. The total value of thc option is therefore 0.9446. (Note that the initial short rate with continuous compounding is 5.91 fio.) Problem 30.13. L".se a ch‹ingc• of numeraire orgii/nen/ to show that the relationship hetw een the futures rate and fomvard rate for the Ho—Lee moJel is u.s .shown in Section 6.3. k'se rite relationship to v'erifl’ the exprrs 'ion for 8(t) given for Ihr he—Lee model in equation (:10. 11). (Hint. The ]iitiire.s J›rice is u martingale ii’lien the n7cfi-Tel price ofri.sk /.y zero. Thr]orw'ard price is a martii›gale ›i hen lli‹• market price of risk is a Piero-roupon honJ matw-ing at the .same time as the jorwai d coil tracl.) Wc will consider instantaneous forward and futures rates. (A more general result involving 232 .^.2012 Pearson Education

the forward and futures rate applying to a period of time between T and F is proved in Technical Note I on the author’s site.) F(t,T) -

— ST

Qn P(t, T)

Because P(t,T) — — A(t,T)e "" " F(t,T) —

Qn A(t,T) — r[T — t) Qn A(t,T) — r(T c°F(t,T) or ›T — t) ör [ In A(t,T) — r(T- t)“] — —1 From Itö’s lemma

dF(0, T) — — ... a dThe instantaneous foiv'ard rate with maturity / has a drift of zero in a world that is forw'ard risk neutral with respect to P(t,T) . This is a world where the market price of risk is —a[T — /) . When we move to a world where the market price of risk is zero the drift of the forward rate increases to o’(T —/) . lntegrating this between = 0 and / = T we see that the forward rate grows by a total of a-"T" 1 2 between time 0 and time T in a world where the market price of risk is zero. The futures rate has zero growth rate in this world. At time T the forward rate equals the futures rate. It fol lows that the futures rate must exceed the forward rate by o-2Fz / 2 at time zero. This is consistent with the formula in Section 6.4. Define G(0, i) as the instantaneous futures rate for maturity so that G(0, i) — F{0, i) = o't' / 2 and G, (0, /) — N, (0, /) = ° 2 r

In the traditional risk-neutral world the expected value of r at time i is the futures rate, G(0, i) . This means that the expected growth in r at time i must be 6,(0, i) so that 8[t) — — G,(0, f) . It follows that This is equation (30.11). Problem 30.14. Use a .similar approach to that in Proble m 50.13 to derive the relationship between the fixtures rate anJ the forward rife [or the Hull-White model. H.se the relationship to verifv the expression]or B(i) given for the Uu//—life model in equation (30.14)

Because P t. T) = A(l,T)e° B'' "’

F(t,T) —-

ST

In P(t,T)

233 N20l2 Pearson Education

[ln A(t,F—) rB{t,T) [ln Art,T) — rB{t,T) O

mln Art,T—)

From Itö’s lemma

rB[t,T) — — e°""°" d (0, T) — — , T,.+ ene°""°" dz• This has drift of zero in a world that is forward risk ncutral with respect to Pat,T) . This is a world where the market price of risk is —oB(i, T) . When wc move to a world where the market price of risk is zero the drift of F(0,T) increases to o-ze "' ' B(t,T) . Integrating this bemeen / = 0 and i — T we see that the forward rate grows by a total of

between time 0 and time T in a world where the market price of risk is zero. The matures price has zero growth rate in this world. At time T the forward price equals the futures price. lt follows that the futures price must exceed the forward price by

at time zero.' Define G(0,/) as the instantaneous futures rate for maturity i so that 2

G(0, i) — F(0,

und

2o’

' To produce a result relating the futures rate for the period between timcs rI and T,qto the forward rate for this period we can proceed as in Technical Note 1 on the author’s web site. The drift of the forward rate is m B(t,T,, )—2 2( B(t, Ti ) z

=

2a (T — T,)

[e“ (—2e°"" + 2e "' ) + e’ [e ’"

rr' )’ 2

e "‘" ]

lntegrating between time 0 and time T we get e 2a’(T — T,› [' ’

1

2e

( ”’ )— e

2

‘"') / a * (e2 "‘ 1)(e '°"° — e “”' ) / (2s)]

[4(1 — e " ) —(1 — e ' ” )(1 e " )] B(TE,T, )

2 T„ —[B(T T ,T› )(—1 e°”'"' ) + 2aB(0, I )I ] 4a This is the amount by w hich the futures rate exceeds the forward rate at time zero.

234 2012 Pearson Education

In the traditional risk-neutral world the expected value of r at time i is the futures rate, G(0, i) . This means that the expected growth in r at time i must be Ci, (0, i) — a{i — G(0, t) so that J(/) — nr — G,(0, — / ) u[r — U(0./)] . lt folloivs that

This proves equation (30.14). Problem 30.15. Suppose a — — 0.ñ S a— — 0.015 , and the term striictut-e is fiot ml 10°Z‹ . Construct o trinomial tree for the Hull—White model wtiere there are Evo time s/eyJi, each one year in length. The time step, fil , is 1 so that Ar — 0.01 5 0.0259b . Also y„„, = 4 showing that the branching method should chans* ftiur steps from thc center of the tree. With only three steps we nev cr reach the point u here the branching changes. The trcc is shown in Figurc 530.1.

Nodc r p, /›„

A ÏÙ.00% 0.1667 0.6666

B l 2.61* 0.1429 0.6542

C 10.01% 0.1667 0.6666

p

0.1667

0.1929

0.IG67 0.1429

D 7.41 0.I929 0.6642

E 15.J4 o 0.1217 0.6567

F 12.64% 11.1429 0.6642

G 10.04% 0.lô 67

H 7.44‘?c ().1929 0.6642

1 4.84% 0.2217 0.6567

0.2217

0.1929

ti.16f*7

0.1429

0.1217

Figure 530.1 Tree for Problcm 30 15 Problem 30.16. Calciilnte the price of a ivan-i car z•ero-coupon hoiid jrom the tree in Figure 38.6. A two-year zcro-coup‹›n bond pays off $100 at the ends of the final branches. At node B it is 235 201 2 Pearson éducation

worth 100e°' ” i = 85.69 . At node C it is z orth 100s "" = 90.45 . At node D it is w orth 100c ' "" = 92.31 . It follows that at node A the bond is worth (88.69 x 0.25 -F 90.48 x 0.5 92.31 x 0.25)e "‘ = 81.88 or $81.88 Problem 30.17. Calculate the price o} u nio-year- zero-coupon bond h-cm the tree in Figure 30.9 and very that ii agree.s v ith the initial term .structui e. A two-year zero-coupon bond pays toff S100 at time Evo years. At node B it is worth 1 00e“ "’“' = 93.30 . At node C it is worth 1 00e " ' “ ' = 94.93 . At node D it is worth 100e “ ”" = 96.59. It follows that at node A the bond is worth (93.30 x 0.167 * 94.93 x 0.666 + 96.59 x 0. J67)e """ = 91.37 or $9.1 .37. Because 91.37 = 100c "" ” , the price of the t o-ycar bond agrees with thc initial tenn structure. Problem 30.18. Calculate the price o] an 18-month zero-coupon hond from the /ree lit Figure 30.10 and verf that it agrees ›i'ith the initial term structure. An 18-mtinth zero-coupon b‹ind pays of1 $100 at the final nodes of the tree. At node E it is w orth 100c "‘" = 95.70 . At node F it is worth 100e """ " = 96.81 . At node Ct it is worth 100c " "”" = 97.64. At node H it is orth 100s°’ "" = 98.2G . At node I it is worth 100c’"”" ° = 98.71 . At node B it is worth (0.118 x 95.70 + 0.654 x 9f›.81 + 0.228 x 97.64)c ’“‘ ' = 94.17 Similarly at nodes C and D it is w orth 95.60 and 96.6b. The value at node A is therefore (0.167 x 94.17 0.66é' x 95.60 0.167 x 9 €.€ Y)e ”’”‘ ‘ = 93.92 The IS-month zero rate is tL05 — 0.05e“‘ '“ ’ = 0.0418 . This gives the price of the 18-month zero-coupon bond as I 00e*'“"“ ° = 93.92 shtiwing that the tree agrees u’it1i the initial term structure. Problem 30.19. What does th‹• calibration oJ a one-factor t€Tltl E frffc/@e itlOdel involve? The calibration of a one-factor interest rate model involves determining its volatility parameters so that it matches the market prims of actively traded interest rate options as closely as possible. Problem 30.20. Usr. the DerivaGem softw'arc to value i 4 , 2 x 3, 3 x 2, nnd 4 • i European sump oj›tions to receive fixed and yay floating. Assume that the one, ni o, thrre, four, and five year interest rate.s are 6%, 5.5“ o, d‘o, 6.ñ%‹›, anJ 7’Xo, I‘esJ›ecti+ el y. The puyment frequency on the s»up is semiannual and the fixed rate i.s 6’X‹ per annum with .seminnni.io/ c ompounJitlg. Use the Hull— White moJel Frith a — — 3°Xv and a — — 1 % . Calculate the volatility implied h y Black’s model]or each option. The tiption prices are 0.1302, 0.0814, 0.0580, and 0.0274. The implied Black volatilities are 14.28%, 13.64%, 13.24%, and 12.81 o 236 'â“20 12 Pearson Education

Problem 30.21. Prove emotion.s (3ñ.25), (3ñ.26), and (3ñ. 7). From equation (30.15) P{t,t + L I) = A(1, f -I At )e‘""’ D ’ ”’ AI SO

so that c *“”' = Aft,t + it)e ’' '"'” “"

Hence equation (30.25) is true with B(t,T) =

B(t, F)A/ B(t,t + it

and A(t,T) — A{t,t + i t— ) " ' " ' B'' "' Aft,T) B(t,T)

In (i, F) = In Aft.T) —B(t, I + it)

Problem 30.22 (a) What is the second partial derivative o[P(t,T) with respec! to r in the Va.sicek and CIR moJels? In Section 30.2, kis presented as an alternative to the standard duration measure D. What is a similar alternative to the convexi5' measure in Section 4.9? (c) Mat is (or P(t,T).’! Hoax would you calculate jor a coupon-hearing bond“.! (d) Give a Taylor Serie.f expansion]or IP(I,T) in terms of Mr anJ fAr)’ for Va.sicek and CIR. (a)

= B(t,T)' A(t,T)e ^''’"" — — B(r.T) 2 P{i, Z)

(b) A corresponding definition for is

(c) When Q—P(t, I), ?’— — B(i,I)' . For a coupon-bearing bond C is a wcighted average of

the ’s for the constituent zero-coupon bonds 1'here w eights are prtiportional to bond prices.

237 fi2012 Pearson Education

(d) " 'T 6P(t. T) — —“ âr

Mr — —B{t,T) P(t,T)kr +

{i,T)'’ P(i,T) Mr‘" + . ..

Problem 30.23 Suppose that the short rate r is 4“Xo and i/.s real-world proces.s i.s cm = 0.1 [0.05— r} dt * 0.0 Jc

n'hile the risk-neutral proce.s.s i.s dr — — 0.1 [0.1 I — r] eg + 0.01 dz (a) What is the market price of interest rate ri.sk?

tb) What re the expected return and volatility for a 5-year zero-couyon hond in the riskneutral world? (c) What is the expected return and volatili5 [or a 5-year zero-coupon hond in the real world? (a) The risk neutral process Yot r has a drift rate w hich is 0.006/r higher than the real world process. The volatility is 0.01/r. This means that the market price of interest rate risk is —0.006/0.01 or —0.6. (b)The expected return on the bond in the risk-neutral world is the risk free rate of 4%. The volatility is 0.01 •B(0,5) where B('1 5)'

i — ¿ —0.

5

o. i

' ’ 935

(c) In thethe realvolatility world the i.e., is volatility 3.935%. of the bond is also 3.93500. The expected return is

higher by 0.6•0.03935 or 2.361%. It is 6.361%.

238 2012 Pearson Education

CHAPTER 31 Interest Rate Derivatives: HJM and LMM Problem 31.1. Explain the difference beuveen a Markov anJ a non-Markov model o/ the short rate. In a Markov model the expected change and volatility of the short rate at time / depend only on the value of the short ratc at time . In a non-Markov model they depend on the history of the short rate prior to time t . Problem 31.2. Prove the relationship between the driJt unJ volatility of the forward rate for the multifactor ver.sion ofHJM in equation (31.6). Equation (31.1) becomes dP(t,T) — r(t)P{t, T) dt Z

vy (I,T, A1, )P(/, r) dc,(r)

so that d ln[P[t, T, )] =

and

—

" "’

From equation (31.2) # "*

d f!(',T ,T 2)

Putting T — — T and T, — — T + it and taking limits as di tends to zero this becomes dF(t,T)' Using v (/, i, D,) = 0

›, (/, T,M

â T

dz (t)

The result in equation (31.6) follows by substituting . (I,T, D, ) = Problem 31.3. “When the forwarJ rate volatilit)’ s(t.T) in HJw is ‹ onstant, the Ho—Lee model results. ” Verify that flits is true by .showing that HJ.w gives a process for bonJ prices that i.s con.sistent Xi ffh the He Lee model in Chapte•r 30. Using the notation in Section 31.1, when .s is constant, v, (t,T) — .s v„ (t,T) — —0 Integrating v (/,T) 239 rT20 12 Pearson

'(I,T) — — s a(t) for some function o . Using the fact thatTv[T,T) 0 , e must have v(l,T) = s(T t) Using the notation from Chapter 30, in Ho—Lee P(i, T) - A{t.T)e "'"” ' . The standard deviation o1 the short rate is constant. lt foi lows from Itô’s lcmma that the standard deviation of the bond price is a constant times the bond price times T — t . The volatility of tlic bond price is thercfore constant times — / . This shows that Hz-Lee is consistent with a constant Problem 31.4. “When the forward rate volatili h, . (i,T) in H i l l i.y oe “'" " he Hull—White moJel results.” Ve r y that thi.f is true by short ing that HIV gi yes u proces.s for bonJ prim es that i.s consistent with the hull—White model in L."luipter 30. Integrating vz(t,T) v(t,T)

1

for some iünction a . Using the fact that v(T,T) — — û , we must have r'(/, T) — —[I — e "’ ” ]= m0(t,T) Using thc notation from Chapter 30, in Hull—White I(i,T) — — .4(t,T)e° '" ' . Thc standard deviation ot the short rate is constant, o. It follows from Itñ’s lemma that the standard deviation of the bond pricc is mf(i.T)B(i,T) . The volatility of the bond pricc is therefore oB[I,T) . This shows that Hull-White is consistent 1'ith ,(i.T) — — cre°"'" “ . Problem 31.5. What is the ad antage o]!LMM over- H.IM? LMM is a similar model to HJM. It has the add antage over I JJ M that it ink olvcs forward rates that are readily observablc. HJM involves instantancous forw ard rates. Problem 31.6. Provide an intuitive exJ›lanation oj n’hy r/ ratchet cap increos eS i0 value a.s the number o] factor.s in‹’rease. A ratchet cap tends to provide relatively low payoffs if a high (low) interest rate at one rcsct date is ftillowed by a high (low ) interest rate at the next reset date. High payoffs occur when a low interest rate is follow ed by a high intercst rate. As the number o1 factors increase, the correlation between successive fonv-ard rates dcclines and there is a greater chance that a low interest rate will be followed by a high interest rate.

Equation (31.10 ) can bc written 24() !201 2 Pearson Education

As #, tends to zero, ,(/)J(/) becomes the standard dcviation of the instantaneous maturity forward rate at time / . Using the notation of Section 31.1 this is .i(/, i„ C,) . As #, tends to zero

tends to . s( /, r, A1 ) d Equation (31.10) therefore bectimes JF (/) — s(/. /, . D, ) ’ s(t, r,G, ) J z dt * s (/, /, , k1,) dv This is the HIM result. Problem 31.8. Explain w'hv a sticky' cap is more exyen.five than a .similar rate heI ‹map. In a ratchet cap, the cap ratc equals the previous reset rate, R , plus a spread. In the notation of the text it is + s . In a sticky cap the cap rate equal the previous capped rate plus a spread. In the notation ot the text it is min(fi , V ) + . . The cap rate in a ratchet cap is always at least a great as that in a sticky cap. Since the value of a cap is a decreasing function of the cap rate, it follows that a sticky cap is mtire expensive. Problem 31.9. Explain hy IOs and POs have o¡›posite sensitivities fo the rate of J›repayment.s When piepaymcnts increase, the principal is received sooner. This increases the value of a PO. When prcpayments increase, less interest is received. This decreases the value of an 10. Problem 31.10. “An oj›tion adfu.sted spread i ' analogous to the241 ield on a hond.” Explain Ihis sh/emeof. '8.'201 2 Pearson Education

A bond yield is the discount rate that causes the bond’s price to equal the markct pricc. The same discount rate is used for all maturities. An OAS is the parallel shift to thc Trcasury zero curve that causes the price ol an instrument such as a mortgage-backed security to equal its

Equation coefficients of d , in In P(i, i, ) — In P(t,t„, ) — ln[1 ñ, F,(/)] Equation (31.9) therefore becomes

Equation (31.15) follows. Problem 31.12. Prove the formula for the variance, K(P) , of the swap rate in equation (31.17).

From the equations in the text

T,+, ) P{i,P(l,T„ ,c ,— P(t,Tz) )

and

P(t,T,) P(/, F )

1 J =0

so that

(We employ the convention that empty sums equal zero and empty products equal one.) -quivalently

or

so that

where

From Ito’s lemm a the q IL COmp

The variance rate of .i(/) is therefore one nt

.

of the volatil ity of .i(/) is

.\ 0:

I + r, G (/)

242

C20l 2 Pearson Education

Problem 31.13. Prove equation (31.19).

so that

. — If we assume that G t) = G, to) for the purposes of calculating the swap volatility we see z from equation (31.17) that the volatility becomes Equating coefficients of di

(') $o

This is equation (31.19).

243 ñ*2012 Pearson Education

dt

CHAPTER 32

Swaps Revisited Problem 32.1. Calculate all the jixeJ cash flow.s and their exact timing for the sii’ap in Businc.s.s Snap.shot 32.1. Assume /ñof //ye dav count conventions are applieJ using target payment date.s rather than actual pa»ment Jate•s. Results are as follows Target payment date

Day ofii’eek

.4ctual par ment date

Days from pres ions to current turget pmt Statcs

Jul 1 1, 2010

Sunday

Jul 12, 2010

lS1

2,975,342

Jan 11, 2011 .Iul 11, 2011 Jan 11, 2012 Jul 11, 201 2 Jan 11, 2013 Jul 11, 2013 Jan 13, 2014 Jul 11, 2014 .lan 12, 2015

184 181 1 b4 182 184 lS1 1 54 181 184

3,024,65 S 2,975,342 3,024,658 2,991,751 3,024,fi5S 2,975,342 3,024,658 2,975,342 3,024 fi5S

Jan 11, 2011 Jul II, 2011 Jan 11. 2012 Jul 11, 2012 Jan 11, 2013 Jul 11, 2013 Jan 11, 2014 Jul 11, 2tl14 Jan 11, 2015

Tucsday Monday Wednesday Wednesday Friday Thursday Saturday Friday Sunday

Fixed Payment û$)

The fixed rate day count convention is Actua1/365. There are lb1 days between January 11, 2010 and July 11, 2010. This means that the fixed payments on July 11, 2010 is ' x 0.06 x 100 000 000 — .52 975 342 365 Other fixed payments arc calculated similarly. Problem 32.2. Suy tOS € that a .swaf9 SF**ities that u jixeJ rale is e•xchanged for tw ice the LIBOR rate, Can the sii’ap be• valued u.sing the “assume forward rute•s are realized” rule? Ycs. The swap is the same as one on twicc the principal where half the fixed rate is exchanged for the LIBOR ratc. Problem 32.3. What i.s the vulue oj a ti+'o-year fixed-for-floating coUipGunJin] 3wa} where the firinci¡›al i.s S I OH inillion and puvmcms ure made .seniiaonunf/y? Fixed interest is received und floating i,s paid. The fixed rate is 8‘Xo ctnd ii is compounded at R.3% tboth semiannilall compoundeJ). The fioating rate is LIBOR plus IN basis points and it i.s compounded at LIBOR plus 20 ba›is points. The LIBOR zero cwve is flail O/ 8“No with .semiannual compounding. The final fixed payment is in millions o1 dollars: [(4 x 1.0415 + 4) x 1.0415 + 4] x 1 .041 5 -F 4 = 17.0238 The final floating payment assuming forward rates are realized is

[(4.05 x 1.04 I + 4.05) x l .041 * 4.05] 1 .041* 4.05 = 1.7.223 S The › alue of the so ap is therefore —0.2000 / (1.04‘) = —0.17 1 0 or —$171,000. Problem 32.4. What is the value o[a five- vear s w'op ›rhere LIBOR is paid in the u.dual ivav and in return LIBOR compounded at LIBOR is receiveJ on the other side? The j›rincipal on hoth sides is $ 100 million. Pavme•nt dales on the pay siJe and comf›ounding dates on the receive 'ide are everv six months and the ielJ ctrl e is ilat at 5% 'ith .sew iannval compounding. The value is zero. The reccive side is the same as the pay side with the cash flows compounded forward at LIBOR. Compounding cash flows forw'ard at LIBOR does not change their value. Problem 32.5. E.xplain carefully 'hy a hatjk might choose to Jisc'ount ‹cash flows on a cm-i-ency .en eye at a rate slightlv different from LIBELS. In theory, a new floating-for-floating swap should involve exchanging LIBOR in one currency for LIBOR in another currcncy (with no spreads added). In practice, macr‹ieconomic effects give rise to spreads. Financial institutions often adjust the discount rates they use to allow for this. Suppose that USD LIBOR is always exchanged Sz iss franc LIBOR plus 15 basis points. Financial institutions u tiuld discount US G cash flows at USD LI BOR and Swiss franc cash flow s at LIBOR plus 15 basis points. This would ensure that the floating-for-tloating swap is valued consistently with the market. Problem 32.6. Call ulate the total convexity/timing adjustment in Example 32.3 of Section 32.4 if all cap volatilities are 1R% in.stead o[2O“>o cf/7cf volGlililies for- all options on five-year swum.y are 13%› in.stead of 15 o L. What .slioillJ thr Jivr ear s›rup rate in three yearns ’ time he a.s.hunted for In this case, = 0.05, , . = 0.13 , T, — ñ. j , I — — 0.05, o, — 0.1s . and p. = 0.7 for all i . It is still true that U ( y ) = —437.603 and G ( i ) = 2261.23 . Equation (32.2) gives the total convexity/timing adjustment as 0.0000b92/ or 0.892 basis points per year until the swap rate is observed. The swap rate in three years shuuld be assumed to be 5.026b'/ . The value of the Problem 32.7. Expl‹iin wh a plain vanilla intere.st rates .swap anJ //ie conlpoiiiiJing s ap in See'tion 32.2 ‹:an h‹• i’alueJ using the “assume forw’cii-d rate.s area realized” rule, buI u LIBOR-in-arrears swap In a plain vanilla swap we can enter into a series of FRAs to exchange the floating cash flows for their values if the “assume forward rates arc rcalizc‹t rule” is used. In the case of a compounding svmp Section 32.2 shows that we are able to enter into a series of FRAs that exchange the final floating rate cash flow for its value w hen the “assume forward rates are realized rule” is used. There is no way ot entering into FRAS so that the floating-rate cash flow s in a LI hOIt-in-arrears sv ap aic exchanged for their values when the “assume forward rates are realized rule” is used. 245 ›2fJ 4.2 Pgarson Education

Problem 32.8. In the accrual swap discussed in the text, the hxeJ siJe accrues only ritten the floating reference rate lies helow a certain level. Discuss how the analysis can be extended to copc with a situation where thc fixed.side accrue.f only when the floating i eference rnte is ago ve one level and below another. Suppose that the fixed rate accrues only whcn the floating reference rate is below fi,. and above P,. where fi,. < R, . In this case the swap is a regular swap plus two serics of binary options, one for each day of the lifc ot the swap. Using the notation in the text, the riskneutral probability that LIBOR will be above fi,. on day i is N(d_,) where d, —

In(I. I R ) — cr‘/,““ 1 2

The probability that it will be below fi, where ñ, < R, is v (—d’,) where from the viewpoint of the party paying fixed, the swap is a regular swap plus binary options. The binary options corresponding to day i have a total value of

(This ignores the small timing adjustment mentioned in Section 32.6.)

246 2012 Pearson Education

CHAPTER 33 Energy and Commodity Derivatives Problem 33.1. What i.s meant hy HDD and CDD'? A day’s HDD is max(0, 65 — A) and a day’s CDD is max(0, N — 65) where A is the average of the highest and lowest temperaturc during the day at a specified weather station, measured in degrees Fahrenheit. Problem 33.2. How is a typical natural gn.y [orward contract structured'? It is an agreement by one side to deliver a specified amount of gas at a roughly uniform rate during a month to a particular hub for a specified price. Problem 33.3. Distinguish between the historical data anJ the risk-neutral approach to valuing a derivative. Under what circumstances do they give the same answer? The historical data approach to valuing an option involves calculating the expected payoff using historical data and discounting the payoff at the risk-free rate. The risk-neutral approach involves calculating the expected payoff in a risk-neutral world and discounting at the risk-free rate. The two approaches give the same answer when percentage changes in the underlying market variables have zero correlation with stock market returns. (In these circumstances all risks can be diversified away.) Problem 33.4. Suppose that each day during July the minimum temperature is 68 Fahrenheit and the maximum temperature is 52’ Fahrenheit. What i.y the payoff from a call option on the cumulative CDD during July u ith a strike oJ 250 anJ a payment rate o[85,000 per degree day." The average temperature cach day is 75 . The CDD each day is therefore 10 and the cumulative CDD for the month is 10 x 31 = 310 . The payoff from the call option is therefore (310 — 250) x 5, 000 = $300, 000 . Problem 33.5. Why is the price o[electricity more volatile• than that of other energy sources? Unlike most commodities electricity cannot be stored easily. If the demand for electricity exceeds the supply, as it sometimes does during the air conditioning season, the price of electricity in a deregulated environment will skyrocket. When supply and demand become matched again the price will return to former levels. Problem 33.6. Why is he historical data approach apJ›ropriate for pricing a w’eather derivatives contract 247 2012 Pearson Education

and a CAT bond."! There is no systematic risk (i.e., risk that is priced by the market) in weather derivatives and CAT bonds. Problem 33.7. “HDD and CDD can he regarJeJ as payo/fs from options on temperature. ” Expl‹iin this statement. HDD is max(6—5 A,0) where .‹ is the average or the maximum and minimum temperature during the day. This is the payoff from a put option on A with a strike price of 65. CDD is max{A — 65, 0) . This is the payoff from call option on A with a strike price of 65. Problem 33.8. Suppose that you have 50 years o] temperature data at your disposal. Explain carefully the analyses you would carry out to vo/tie a Jorw’ard contract on the cumulative CDD%r a particular month. It would be useful to calculate the cumulative CDD for July of each year of the last 50 years. A linear regression relationship C DD — — a + 6/ + e

could then be estimated where n and é are constants, i is the time in years measured from the start of the 50 years, and e is the error. This relationship allows for linear trends in temperature through time. The expected CDD for next year (year 51) is then a 5 lb . This could be used as an estimatc of the forward CDD. Problem 33.9. Wauld vow expect the volatility of the one-year forward prfce o/ozf to be greater than or less than the volatility o/ the spot price? Explain vonr answer. The volatility of the one-year forward price will be less than the volatility of the spot price. This is because, when the spot price changes by a certain amount, mean reversion will cause the forward price will change by a lcsser amount. Problem 33.10. What are the chara‹ teristics of an energy how ce »here the prit’e has a very high volatility and a very' high rate of mean reyers ion.! Give an e.xample o/ such an energv source. The price of the energy sourcc will show big changes, but will be pulled back to its long-run average level fast. Electricity is an example of an energy source with these characteristics. Problem 33.11. /-Amy can an ene•rgy producer u.se derivative markets to hedge risks'! The energy producer faces quantity risks and price risks. lt can use weather derivatives to hedge the quantity risks and eneryy derry atives to hcdgc against the price risks. Problem 33.12. Explain how a 3 z Yoption conlt-act]or ,day 2009 on electricity with Jailv exercise w'orks. Explain how a 5 x Z opiion contract for .day 2ñ09 on clef tric'i5' with monthly exerci.se worlds. 248 c20l2 Pearson Education

Which is w'orth more.'" A 5 8 contract for May, 2009 is a contract to provide electricity for tive days per week during the off-peak period (11PM to 7AM). When daily exercise is specified, thc holder of the option is able to choose each weekday iv hether he or she will buy electricity at thc strike price at the agreed rate. When there is monthly exercise, he or she chooses once at the beginning of the month whether electricity is to be bought at the strike price at the agreed rate for the whole month. The option with daily exercise is worth more. Problem 33.13. Explain how C.4 T bonds n.'orI:. CAT bonds (catastrophc bonds) arc an altcrnativc to reinsurance for an insurance company that has taken on a certain catastrophic risk (e.g., the risk o1‘ a hurricane or an earthquake) and wants to get rid of it. CAT bonds are issued by the insurance company. Thcy provide a higher rate of interest than government bonds. However, the bondholders agree to forego interest, and possibly principal, to meet any claims against the insurance company that are within a prespecified range. Problem 13.14.

ConsiJer two bonds that have the same coupon, time to maturity and price. One i.s a B-rateJ corporato• bond. The othc•r i s a CAT bond. An analy.st.s Isa.sed on hi.storical data .show.s that the rxperted losses on the two honds in each year o/ their li[e is the same. Which Idend woiilJ you aJvise a portfolio manager in buy and z'hv.’• Thc CAT bond has ›'ery little systematic risk. Whether a particular type of catastrophe occurs is indcpcndcnt of the return on the market. The risks in the CAT bond are likely to be largely “diversified away” by the other in› estments in the portfolio. A B-rated bond does have systematic risk that cannot be diversified away. It is likcly therefore that the CAT bond is a better addition to the portfolio.

249 2012 Pearson Education

CHAPTER 34 Real Options Problem 34.1. Explain the difference betw'een the net pre.sent valtie aj›proach and the ri,sk-neutral valuation approach %r valuing a new capital inve.student of›portunitv. What are the acl»antage.s of the ri.sk-neutral valuation aJ pu-oach for valuing real option.s? In the net present value approach, ciish flow s are estimated in the real u orld and discounted at a risk-adjusted discount rate. In the risk-neutral valuation approach, cash flows are estimated in the risk-neutral world and discounted at the risk-free interest ratc. The risk- neutral valuation approach is arguably more appropriate for valuing real options because it is very difficult to determine the appropriate risk-adjusted discount ratc when options arc valued. Problem 34.2. The market pri‹’e of risk for copper is 0.5, the volatilii) of copper price.s i.s 2 H‘Xo yer annum, the spot price is 80 cents per pounJ, and the six-month future.s J›rice i.s 75 cent.s per j›ound. What is the e.xpe‹ teJ percentage growth rate in copyer price.s over the next .sis month.s? In a risk-neutral world the expected price of clipper in six months is 75 cents. This corresponds to an expected growth rate ot 21n(75 /' 80) — = 12.9*^ per annum. The decrcasc in the growth rate when we move from the real z'orld to the risk-neutral orld is the volatility of copper times its market price of risk. This is 0.2 x 0.5 = 0.1 or 10oZ» per amiuin. It follow s that the expected growth ratc of the price of copper in the real world is —2.9* . Problem 34.3. Consider a commoJily v ith ‹ onstant volatili / r m anJ an expected growth rate that i.s a function .solely of time. Short’ that, in the traditional risk-ne•ulral »orld,

where Sz is the value of the cominoJi3› at time r , I (t) i.s the [utui-e.s pt-ice at time I) fot- a conii act maturing at time I , and p(m, v) is u normal Jisli-ibution with mean m anJ variance In this casc or d ln S [y(/) — cr' ,' 2] J/ + cr Jc so that ln S is normal with mean

and standard deviation oT

In 3 —o'T / 2 . Section 33.4 show s that 250 L^20l2 Pearson Education

Jt

//(/) =

[In N(f)]

so that Since

(0) = I, the result follows.

Problem 34.4. Dierive o relation.ship l›ehi een the convenience yield of a cmmmcdi 5' and its market price of + isk. We explained the ctincept tif a convenience yield for a commodity in Chapter S It is a measure of the benefits realized from ownershi of the physical commodity that are not real zed by the holders of a iiiturcs contract. It r is the con›'cnicnce yicld and « is the storage cost, equation (5.17) show's that thc commodity bchaves likc an invcstmcnt asset that provides a return cqual to r „ . In a r1k-neutra)l wo rld s growth is, therefore,

The convenience yield of a commodity can be related to its market price o1 risk. From Section 34.2, the expcctcd growth of the commodity pricc in a risk-ncutral world is m —ds , where m is its expected growth in the real world, › its volatility, and is its market price of risk. It follows that

Problem 34.5. The correlation hen›’een a company’’s gross re»enue and the market index i.s 0.2. The excess return of the market over the risk-free rate is 6"Xu and the volatility of the market index is I b“Xo. What is the market price ofri,sk for the company’.s t-evenue'? In equation (34.2) of risk lambda is

— 0.2 , /i —r — — 0.06, and cr = 0.18. It follo s that the market price 0.2 x 0.Of› = 0.067 0.18

Problem 34.6. A company can but an option for the Jelivei of one million units of a commoJity in three years at $25 per unit. The three year jiitures price is $24. The risk-free interest rate is 5“Xoper annum with continuous compounding ailJ the vc*/u/i/ify of the futures yrice is 20’Xo per annum. How inns h is the option worth? The option can be i alued using Black’s model. In this case — 24, A = 2fi , r = 0.05 , = 0.2 . and T — — 3. The value of an option to purchase one unit at $25 is c°" [I} \' (r/, ) — K.h"[ 1 )]

whcre

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This is 2.459. The ›'a1ue of the option to purchase one million units is therefore $2,489,000. Problem 34.7. A Jri ver entering into a car lease agreement can obtain the right to huy the car in four year.s for $10,000. The curryI'tt VffJHC o] fl1€ Cir is $30,OOH. The value of the c‹ir, S , is expected to Jñ/low' the ¡Proce.s.s JS — — is dt mS dwhere — — —0.25 . = 0.15 , unJ ‹lz is a Wiener-proc ess. Thr market price oj risk for the car pt ice i.s estimateJ to be —0.1 . What is the value of the option? Assuse that fate risk-free rate for all maturities is 6"Xo. The expcctcd growth rate of the car price in a risk-neutral »'or1d is —0.25 — (—0. 1 0.1 5) = —0. 235 The cxpcctcd value of the car in a risk-neuual world in four years, (S ) , is therefore 30, 000e Chapter 14 the value of the option is

’ ' = â11, 719 . Using thc rcsult in thc appendix to

e '[E(s.› ( ,) — •' )] »'here d, — —

ln( . (S, ) / K)+ cr'T / 2 cr T

i = II.06 , rr = 0.15 , T — — 4 , anâ K — — 10, 000 . It is $1,532.

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