hull teacher's manual

Preface

Notes for the Instructor Answers to Assignment Questions Course Organization Test Bank Questions Test Bank Answers Additional Questions Answers to Additional Questions Slides

Contents

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v

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115 120 142 144 149 151

Answers to Assignment Questions

Chapter 1: Introduction

1.28. The arbitrageur could borrow money to buy 100 ounces of gold today and short futures contracts on 100 ounces ounces of gold for delivery in one year. This means that gold is purchased for $500 per ounce and sold for $700 per ounce. The return (40% per annum) is far greater than the 10% cost of the borrowed funds. This is such a profitable opportunity that the arbitrageur should buy as many ounces of gold as possible and short futures contracts on the same number of ounces. Unfortunately arbitrage opportunities as profitable as this rarely arise in practice.

1.29. The investment in call options entails higher risks but can lead to higher returns. If the stock price stays at $94, an investor who buys call options loses $9400 whereas an investor who buys shares neither gains nor loses anything. If the stock price rises to $120, the investor who buys call options gains

2000 x (120 - 95) - 9400 = $40,600

An investor who buys shares gains

100 x (120 - 94) = $2,600

The strategies are equally profitable if the stock price rises to a level, 8, where

100 x (8 - 94) = 2000(8 - 95) - 9400

or 8= 100

The option strategy is therefore more profitable if the stock price rises above $100.

1.30. Suppose 8T is the price of oil at the bond's maturity. In addition to $1000 the standard oil bond pays:

ST < $25 $40> 8T > $25

8T > $40

o 170(8T - 25)

2,550

This is the payoff from 170 call options on oil with a strike price of 25 less the payoff from 170 call options on oil with a strike price of 40. The bond is therefore equivalent to a regular bond plus a long position in 170 call options on oil with a strike price of $25 plus a short position in 170 call options on oil with a strike price of $40. The investor has what is termed a bull spread on oil. This is discussed in Chapter 9.

1.31. The range forward contract is a portfolio consisting of a long position in a call option with a strike price of 1.4600 and a short position in a put option with a strike price

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of 1.4200. These options can be valued by DerivaGem. In the first worksheet, we set the exchange rate equal to 1.4454 (the average of the bid and offer quotes in Table 1.1 in the text), the volatility equal to 15%, the risk-free rate equal to 5.0%, the foreign risk-free rate equal to 6.4%, the time to exercise as 0.25 years. The value of a call option with a strike price of 1.4600 is shown by the software to be 0.03384. The value of a put option with a strike price of 1.4200 is shown by the software to be 0.03296. The value of the portfolio is therefore 0.03384 - 0.03296 = 0.00088. This is close to zero. We can make it exactly zero by increasing the strike price of the call option slightly. Trial and error shows that when the strike price of the call is increased to 1.4622, the value of the call is almost exactly the same as the value of the put.

1.32. Suppose that Fa is the one-year forward price of gold. If Fa is relatively high, the trader can borrow $250 at 6%, buy one ounce of gold and enter into a forward contract to sell gold in one year for Po. The profit made in one year is

Po - 250 x 1.06 = Fo - 265

If Po is relatively low, the trader can sell one ounce of gold for $249, invest the proceeds at 5.5%, and enter into a forward contract to buy the gold back for Po. The profit (relative to the position the trader would be in if the gold were held in the portfolio during the year) is

249 x 1.055 - Po = 262.695 - Po

This shows that there is no arbitrage opportunity if the forward price is between $262.695 and $265 per ounce.

1.33. ImportCo should buy three-month call options on £10 million with a strike price of 1.4600. ExportCo should buy three-month put options on £10 million with a strike price of 1.4200. In this case the foreign exchange rate is 1.4454 (the average of the bid and offer quotes in Table 1.1.), the (domestic) risk-free rate is 3%, the foreign risk-free rate is 4.4%, the volatility is 12%, and the time to exercise is 0.25. Using the Equity ..FXJndex.2utures worksheet in the DerivaGem Options Calculator select Currency as the underlying and Analytic European as the option type. The software shows that a call with a strike price of 1.46 is worth 0.0256 and a put with a strike of 1.42 is worth 0.0248. This means that the hedging would cost 0.0256 x 10,000,000 or $256,000 for ImportCo and $0.0248 x 10,000, 000 or $248,000 for ExportCo.

1.34. The trader has a long European call option with strike price K and a short European put option with strike price K. Suppose the price of the underlying asset at the maturity of the option is ST. If ST > K, the call option is exercised by the investor and the put option expires worthless. The payoff from the portfolio is ST - K. If ST < K, the call option expires worthless and the put option is exercised against the investor. The cost to the investor is K - ST. Alternatively we can say that the payoff to the investor is ST - K (a negative amount). In all cases, the payoff is ST - K, the same as the payoff from the forward contract. The trader's position is equivalent to a forward contract with delivery price K.

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Suppose that F is the forward price. If K = F, the forward contract that is created has zero value. Because the forward contract is equivalent to a long call and a short put, this shows that the price of a call equals the price of a put when the strike price is F.

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Answers to Assignment Questions Chapter 2: Mechanics of Futures Markets

2.26. There is a margin call if $1000 is lost on the contract. This will happen if the price of wheat futures rises by 20 cents from 250 cents to 270 cents per bushel. $1500 can be withdrawn if the futures price falls by 30 cents to 220 cents per bushel.

2.27. Keynes and Hicks argue that speculators on average make money from commodity futures trading and hedgers on average lose money from commodity futures trading. If speculators tend to have short positions in Sugar-World futures, the Keynes and Hicks argument implies that futures prices overstate expected future spot prices. For example, the expected future Sugar-World price in October 2002 is less than 7.73 cents per pound (which is the settlement price for the October 2002 contract). Table 2.2 shows Sugar-World futures prices decline fast. The Keynes and Hicks argument therefore implies a very fast decline for the expected price of Sugar-World over the 18 months following March 15, 2001 if speculators are short. To understand the meaning of the expected future price of a commodity, suppose that there are N different possible prices at a particular future time: PI, P2 , ••• , PN. Define qi as the (subjective) probability the price being Pi (with ql +q2+ ... +qN == 1). The expected future price is

i=1

Different people may have different expected future prices for the commodity. The expected future price in the market can be thought of as an average of the opinions of different market participants. Of course, in practice the actual price of the commodity at the future time may prove to be higher or lower than the expected price. Keynes and Hicks is not the last word on the relationship between futures prices and expected future spot prices. Chapter 3 discusses this issue from the perspective of the trade offs between risk and return.

2.28. The May 2001 settlement price for corn is 2101 cents per bushel. The May 2002 settlement price for corn is 254 cents per bushel. You could go long one May 2001 corn contract and short one May 2002 contract. In May 2001 you take delivery of the corn borrowing funds at 5% (or perhaps a little more than 5%). Assuming that the 20 cents for storage is paid up front, the borrowing requirement is 230.75 cents per bushel. At the end of a year 230.75 x 1.05 == 242.29 cents per bushel has to be repaid. The strategy therefore leads to a profit of 254 - 242.29 == 11.71 cents per bushel. Note that this profit is independent of the actual price of corn in May 2001 or May 2002. It will be slightly affected by the daily settlement procedures.

2.29. (a) For gold the standard deviation of daily changes is $2.77 per ounce or $277 per contract. For a 1% risk this means that the maintenance margin should be set at

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277 x v'2 x 2.33 = 912. For a 0.1% risk the maintenance margin should be set at 277 x v'2 x 3.09 = 1,210. For crude oil the standard deviation of daily changes is $0.31 per barrel or $310 per contract. For a 1% risk this means that the maintenance margin should be set at 310 x v'2 x 2.33 = 1,021. For a 0.1% risk the maintenance margin should be set at 310 x v'2 x 3.09 = 1,355. (b) The initial margin is set at 1,362 for crude oil. There are 151 margin calls and 7 times (out of 1201 days) where the investor is tempted to walk away. The initial margin is set at 1,217 for gold. There are 93 margin calls and 9 times (out of 823 days) when the investor is tempted to walk away. When the 0.1% risk level is used there are 3 times when the oil investor might walk away and 6 times when the gold investor might do so. These results suggest that extreme movements occur more often that the normal distribution would suggest. Here are some notes on how I handled the Excel calculations. Suppose that the initial margin is in cell Ql and the maintenance margin is in cell Q2. Suppose further that the change in the oil futures price is in column D of the spreadsheet and the margin balance is in column E. Consider cell E7. This is updated with an instruction of the form:

= IF(E6 < $Q$2, $Q$I, IF(E6 + D7 * 1000 > $Q$I, $Q$I, E6 + D7 * 1000»

Returning 1 in cell F7 if there has been a margin call and zero otherwise requires an instruction of the form:

= IF(E7 < $Q$2, 1,0)

Returning 1 in cell G7 if there has been an incentive to walk away and zero otherwise requires an instruction of the form:

= IF(E6 + D7 * 1000 < 0, 1,0)

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3.24.


Chapter 3: Determination of Forward and Futures Prices

(a) The present value, I, of the income from the security is given by:

1= e-O.08x2/12 + e-O.08x5/12 = 1.9540

From equation (3.6) the forward price, Po, is given by:

Po (50 - 1.9540)eO.08xO.5 = 50.01

or $50.01. The initial value of the forward contract is (by design) zero. The fact that the forward price is very close to the spot price should come as no surprise. When the compounding frequency is ignored the dividend yield on the stock equals the risk-free rate of interest.

(b) In three months: 1= e-O.08X2/12 = 0.9868

3.25 .

The delivery price, K, is 50.01. From equation (3.10) the value of the short forward contract, f, is given by

f = -(48 - 0.9868 - 50.01e-o.08X3/12) = 2.01

and the forward price is

(48 - 0.9868)eo.08X3/12 47.96

. (a) My explanation of this problem to students usually goes as follows. Suppose that the price of gold is $250 per ounce and the corporate client wants to borrow $250,000. The client has a choice between borrowing $250,000 in the usual way and borrowing 1,000 ounces of gold. If it borrows $250,000 in the usual way, an amount equal to 250,000 x 1.11 = $277,500 must be repaid. If it borrows 1,000 ounces of gold it must repay 1,020 ounces. In equation (3.16) r = 0.0925 and u = 0.005 so that the forward price is

250e(O.0925+0.005)Xl = 275.603

By buying 1,020 ounces of gold in the forward market the corporate client can ensure that the repayment of the gold loan costs

1,020 x 275.603 $281,115

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Clearly the cash loan is the better deal (281,115> 277,500). What is the correct rate of interest? Suppose that R is the rate of interest on the gold loan. The client must repay 1,000(1 + R) ounces of gold. When forward contracts are used the cost of this is

1,000(1 + R) x 275.603

This equals the $277,500 required on the cash loan when R = 0.688%. The rate of interest on the gold loan is too high by about 1.31%.

(b) In practice Central Banks do lend gold. the lease rate of 1.5% that is specified is typical. It redcues the forward price to

250e(O.0925+0.005-0.015)Xl = 271.500

The cost of buying 1,020 ounces in the forward market is now

1,020 x 271.500 = $276,930

The gold loan is now a better deal (276,930 < 277,500). The break-even rate of interest on the gold loan is 2.21%.

3.26. It is likely that the bank will price the product on assumption that the company chooses the delivery date least favorable to the bank. If the foreign interest rate is higher than the domestic interest rate then

1. The earliest delivery date will be assumed when the company has a long position. 2. The latest delivery date will be assumed when the company has a short position.

If the foreign interest rate is lower than the domestic interest rate then 1. The latest delivery date will be assumed when the company has a long position. 2. The earliest delivery date will be assumed when the company has a short position.

If the company chooses a delivery which, from a purely financial viewpoint, is suboptimal the bank makes a gain.

3.27. The daily marking to market of futures contracts ensures that the futures trader makes an immediate profit of $4,000. For the forward trader, the gain is not realized until the end of the forward contract (which is in three months). The gain made by the forward trader is therefore the present value of $4,000, rather than $4,000. Presumably the risk-free interest rate is about 10% per year or 2.5% per three months so that the impact of the discounting is to reduce the $4,000 gain to $3,900. The discounting effect can be seen in equation (3.8). This shows that when the forward price, Fo change by a certain amount, the value of a forward contract, f changes by the present value of that amount. The effect is symmetrical. If the forward and futures exchange rates had both gone down by 0.004, the futures trader would have taken an immediate loss of $4,000 while the forward trader would have taken a loss of only $3,900.

3.28. The value of the contract to the bank at time Tl is 81 - K l . The bank will choose K2 so that the new (rolled forward) contract has a value of 81 - Kl. This means that

81e-r/(T2-T l) - K2e-r(T2-Td = 81 - Kl

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where rand r I and the domestic and foreign risk-free rate observed at time Tl and applicable to the period between time Tl and T2. This means that

This equation shows that there are two components to K2. The first is the forward price at time T1 • The second is an adjustment to the forward price equal to the bank's gain on the first part of the contract compounded forward at the domestic risk-free rate.

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Chapter 4: Hedging Strategies Using Futures

4.22. Denote Xi and Yi by the i-th observation on the change in the futures price and the change in the spot price respectively.

An estimate of U F is

An estimate of Us is

An estimate of p is

LXi = 0.96 LYi = 1.30

LX~ = 2.4474 LY; = 2.3594

L XiYi = 2.352

)2.4474 _ 0.962 = 0 5116

9 10 x 9 .

2.3594 9

. 10 x 2.352 - 0.96 x 1.30 0 9 = .m

V(lD x 2.4474 - 0.962)(10 x 2.3594 - 1.302)

The minimum variance hedge ratio is

Us = 0 981 0.4933 = 0 946 P UF • x 0.5116 .

4.23. (a) The company should short

or 280 contracts.

(1.2 - 0.5) x 100,000,000 1000 x 250

(b) The company should take a long position in

or 120 contracts.

(1.5 - 1.2) x 100,000,000 1000 x 250

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4.24. To hedge the February 2003 purchase the company should take a long position in March 2003 contracts for the delivery of 800,000 pounds of copper. The total number of contracts required is 800,000/25,000 = 32. Similarly a long position in 32 September 2003 contracts is required to hedge the August 2003 purchase. For the February 2004 purchase the company could take a long position in 32 September 2003 contracts and roll them into March 2004 contracts during August 2003. (As an alternative, the company could hedge the February 2004 purchase by taking a long position in 32 March 2003 contracts and rolling them into March 2004 contracts.) For the August 2004 purchase the company could take a long position in 32 September 2003 and roll them into September 2004 contracts during August 2003. The strategy is therefore as follows

Oct. 2002: Enter into long position in 96 Sept. 2003 contracts Enter into a long position in 32 Mar. 2003 contracts

Feb. 2003: Close out 32 Mar. 2003 contracts Aug. 2003: Close out 96 Sept. 2003 contracts

Enter into long position in 32 Mar. 2004 contracts Enter into long position in 32 Sept. 2004 contracts

Feb. 2004: Close out 32 Mar. 2004 contracts Aug. 2004: Close out 32 Sept. 2004 contracts

With the market prices shown the company pays

69.00 + 0.8 x (72.30 - 69.10) = 71.56

for copper in February, 2003. It pays

65.00 + 0.8 x (72.80 - 64.80) 71.40

for copper in August 2003. As far as the February 2004 purchase "is concerned, it loses 72.80 - 64.80 = 8.00 on the September 2003 futures and gains 76.70 - 64.30 = 12.40 on the March 2004 futures. The net price paid is therfore

77.00 + 0.8 x 8.00 - 0.8 x 12.40 == 73.48

As far as the August 2004 purchase is concerned, it loses 72.80 - 64.80 = 8.00 on the September 2003 futures and gains 88.20 - 64.20 = 24.00 on the September 2004 futures. The net price paid is therfore

88.00 + 0.8 x 8.00 - 0.8 x 24.00 = 75.20

The hedging scheme succeeds in keeping the price paid in the range 71.40 to 75.20. In October 2002 the initial margin requirement on the 128 contracts is 128 x $2, 000 or $256,000. There is a margin call when the futures price drops by more than 2 cents. This happens to the March 2003 contract between October 2002 and February 2003, to the September 2003 contract between October 2002 and February 2003, and to the September 2003 contract between February 2003 and August 2003.

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4.25. (a) The theoretical futures price is

1250e(o.o6-o.03)XO.25 = 1259.41

(b) The number of contracts the fund manager should short is

50,000,000 0.87 x 1250 x 250 = 139.2

Rounding to the nearest whole number, 139 contracts should be shorted.

( c) The following table shows that the strategy has the effect of locking in a return of about $500,000. By hedging the portfolio manager earns 1% in two months. This is the risk-free interest rate. To illustrate the calculations in the table consider the first column. If the index in two months is 1,000, the futures price is 1000e(o.o6-o.o3)xo.oo8333 = 1002.50. The gain on the short futures position is therefore

(1259.41 - 1002.50) x 250 x 139 = $8,927,530

Ignoring the difference between continuous compounding and compounding every two months, the return on the index is 3/6=0.5% in the form of dividend and -250/1250 = -20% in the form of capital gains. The total return on the index is therefore -19.5%. The risk-free rate is 1% per two months. The return is therefore -20.5% in excess of the risk-free rate. From the capital asset pricing model we expect the return on the portfolio to be 0.87 x - 20.5% = -17.835% in excess of the risk-free rate. The portfolio return is therefore -16.835%. The loss on the portfolio is 0.16835 x 50,000,000 or $8,417,500. When this is combined with the gain on the futures the total gain is $510,030.

Index in Two months 1000 1100 1200 1300 1400

Futures Price ($) 1002.50 1102.75 1203.00 1303.25 1403.50 Gain on Futures ($) 8,927,530 5,443,830 1,960,130 -1,523,570 5,007,270 Index Return -19.5% -11.5% -3.5% 4.5% 12.5% Excess Ind. Return -20.5% -12.5% -4.5% 3.5% 11.5% Excess Port. Return -17.835% -10.875% -3.915% 3.045% 10.005% Port. Return -16.835% -9.875% -2.915% 4.045% 11.005% Port. Gain ($) -8,417,500 -4,937,500 -1,457,500 2,022,500 5,502,500 Total Gain ($) 510,030 5~,330 502,630 498,930 495,230

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Answers to Assignment Questions Chapter 5: Interest Rate Markets

5.33. The Eurodollar futures contract price of 89.5 means that the Eurodollar futures rate is 10.5% per annum with quarterly compounding and an actual/360 day count. This becomes 10.5 x 365/360 = 10.646% with an actual/actual day count. This is

41n(1 + 0.25 x 0.10646) = 0.1051

or 10.51% with continuous compounding. The forward rate given by the 91-day rate and the 182-day rate is 10.4% with continuous compounding. This suggests the following arbitrage opportunity:

1. Buy Eurodollar futures. 2. Borrow 182-day money. 3. Invest the borrowed money for 91 days.

5 .. 34. The U.S. Eurodollar futures contract maturing at time T enables an investor to lock in the forward rate for the period between T and T* where T* is three months later than T. If f is the forward rate, the U.S. dollar cash flows that can be locked in are

- Ae-r(T* -T) at time T +A at time T*

where A is the principal amount. To convert these to Canadian dollar cash flows, the Canadian company must enter into a short forward foreign exchange contract to sell Canadian dollars at time T and a long forward foreign exchange contract to buy Canadian dollars at time T*. Suppose F and F* are the forward exchange rates for contracts maturing at times T and T*. (These represent the number of Canadian dollars per U.S. dollar.) The Canadian dollars to be sold at time T are

Ae-r(T*-T) F

and the Canadian dollars to be purchased at time T* are

AF'"

The forward contracts convert the U.S. dollar cash flows to the following Canadian dollar cash flows:

_Ae-r(T*-T) F at time T + AF* at time T*

This is a Canadian dollar LIB OR futures contract where the principal amount is AF*.

5.35. (a) The duration of Portfolio A is

1 x 2000e-O.1x1 + 10 x 6000e-O.lxlO 01 = 5.95 2000e-O.1X1 + 6000e- . xlO

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Since this is also the duration of Portfolio B, the two portfolios do have the same duration.

(b) The value of Portfolio A is

2000e-0.1 + 6000e-0.1X10 = 4016.95

When yields increase by 10 basis points its value becomes

2000e -0.101 + 6000e -0.101 x 10 = 3993.18

The percentage decrease in value is

23.77 x 100 = 0 59~ 4016.95 . /0

The value of Portfolio B is

5000e-0.lX5.95 = 2757.81

When yields increase by 10 basis points its value becomes

5000e-0.l01x5.95 = 2741.45

The percentage decrease in value is

16.36 x 100 = 0 59~ 2757.81 . /0

The percentage changes in the values of the two portfolios for a 10 basis point increase in yields are therefore the same.

(c) When yields increase by 5% the value of Portfolio A becomes

2OO0e-0.15 + 6000e-0.15XlO = 3060.20

and the value of Portfolio B becomes

5000e-0.15x5.95 = 2048.15

The percentage reduction in the values of the two portfolios are: 956.75

Porftolio A: 4016.95 x 100 = 23.82

709.66 Porftolio B: 2757.81 x 100 = 25.73

Since the percentage decline in value of Portfolio A is less than that of Portfolio B, Portfolio A has a greater convexity (see Figure 5.3 in text).

5.36. (a) The zero rate for a maturity of six months, expressed with continuous compounding is 21n(1 + 2/98) = 4.0405%. The zero rate for a maturity of one year, expressed with continuous compounding is In(l + 5/95) = 5.1293. The 1.5-year rate is R where

3.1e-0.040405xo.5 + 3.1e-0.051293Xl + 103.1e-RX 1.5 = 101

The solution to this equation is R = 0.054429. The 2.O-year rate is R where 4e-0.040405xO.5 + 4e-0.051293Xl + 4e-0.054429X1.5 + 104e-Rx2 = 104

The solution to this equation is R = 0.058085. These results are shown in the table below

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Maturity Zero Forward Par Yield Par Yield (years) Rate (%) Rate (%) semi ann. (%) cont. compo

0.5 4.0405 4.0405 4.0816 4.0405 1.0 5.1293 6.2181 5.1813 5.1154 1.5 5.4429 6.0700 5.4986 5.4244 2.0 5.8085 6.9054 5.8620 5.7778

(b) The continuously compounded forward rates calculated using equation (5.1) are shown in the third column of the table

( c) The par yield, expressed with semiannual compounding, can be calculated from the formula in Section 5.3. It is shown in the fourth column of the table. In the fifth column of the table it is converted to continuous compounding

(d) The price of the bond is

3.5e-o.040405XO.5 + 3.5e-o.051293xl + 3.5e-o.054429X1.5 + 103.5e-o.o58o85x2 = 102.13

The yield on the bond, y satisfies

3.5e-1IXO.5 + 3.5e-1Ix1.0 + 3.5e-1Ix1.5 + 103.5e-yx2.o = 102.13

The solution to this equation is y = 0.057723. The bond yield is therefore 5.7723%.

5.37. (a) On the first day of the delivery month the bond has 15 years and 7 months to maturity. The value of the bond assuming it lasts 15.5 years and all rates are 6% per annum with semiannual compounding is

31 5 100 ~ 1.03i + 1.0331 = 140.00

The conversion factor is therefore 1.4000.

(b) On the first day of the delivery month the bond has 21 years and 4 months to maturity. The value of the bond assuming it lasts 21.25 years and all rates are 8% per annum with semiannual compounding is

1 [ 42 3.5 100 ] v'f.03 3.5 + ~ 1.03i + 1.0342 = 113.66

Subtracting the accrued interest of 1.75, this becomes 111.91. The conversion factor is therefore 1.1191.

(c) For the first bond, the quoted futures price times the conversion factor is

118.71825 x 1.4000 = 166.2056

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This is 2.7944 less than the quoted bond price. For the second bond, the quoted futures price times the conversion factor is

118.71825 x 1.1191 = 132.8576

This is 3.1424 less than the quoted bond price. The :first bond is therefore the cheapest to deliver.

(d) The price received for the bond is 166.2056 plus accrued interest. There are 176 days between January 1,2002 and June 25,2002. There are 181 days between January 1, 2002 and July 1, 2002. The accrued interest is therefore

176 5 x 181 = 4.8619

The cash price received for the bond is therefore 171.0675.

5.38. The number of short futures contracts required is

100, 000,000 x 4.0 = 364.3 122,000 x 9.0

Rounding to the nearest whole number 364 contracts should be shorted.

(a) This increases the number of contracts that should be shorted to

100,000,000 x 4.0 = 468.4 122,000 x 7.0

or 468 when we round to the nearest whole number.

(b) In this case the gain on the short futures position is likely to be less than the loss on the loss on the bond portfolio. This is because the gain on the short futures position depends on the size of the movement in long-term rates and the loss on the bond portfolio depends on the size of the movement in medium-term rates. Duration-based hedging assumes that the movements in the two rates are the' same.

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Answers to Assignment Questions Chapter 6: Swaps

6.18. The two-year swap rate implies that a two-year LIB OR bond with a coupon of 11 % sells for par. If R2 is the two-year zero rate

11e-O.10X I.O + 111e-R2x2.o = 100

so that R2 = 0.1046 The three-year swap rate implies that a three-year LIBOR bond with a coupon of 12% sells for par. If Ra is the three-year zero rate

12e-O.10X1.0 + 12e-O.l046x2.0 + 112e-R3xa.o = 100

so that Ra = 0.1146 The two and three year rates are therefore 10.46% and 11.46% with continuous compounding.

6.19. There is a 1% differential between the yield on sterling and doll~r 5-year bonds. The financial intermediary could use this differential when designing a swap. For example, it could (a) allow company X to borrow dollars at 1% per annum less than the rate offered on sterling funds, that is, at 11% per annum and (b) allow company Y to borrow sterling at 1% per annum more than the rate offered on dollar funds, that is, at 11!% per annum. However, as shown in Figure 6.1, the financial intermediary would not then earn a positive spread.

£: 12.0% £: n.5% £: 12.0% ... ...

Company ... Financia1 ... Company .. .... .. ~ X = Institution .. Y $: 10.5% ..

$: 11.0% $: 10.5%

Figure 6.1 First attempt at designing swap for Problem 6.19

To make 0.5% per annum, the financial intermediary could add 0.25% per annum, to the rates paid by each of X and Y. This means that X pays 11.25% per annum, for dollars and Y pays 11.75% per annum, for sterling and leads to the swap shown in Figure 6.2. The financial intermediary would be exposed to some foreign exchange risk in this swap. This could be hedged using forward contracts.

£: 12.0% £: 11.75% £: 12.0%

... Company ... Financia1 ... Company .. .... - ".

"" X .. Institution .. Y $: 10.5% .. .. $: 11.25% $: 10.5%

Figure 6.2 Final Swap for Problem 6.19

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6.20. The swap can be regarded as a long position in a floating-rate bond combined with a short position in a fixed-rate bond. The correct discount rate is 12% per annum with quarterly compounding or 11.82% per annum with continuous compounding. Immediately after the next payment the floating-rate bond will be worth $100 million. The next floating payment ($ million) is

0.118 x 100 x 0.25 = 2.95

The value of the floating-rate bond is therefore

102.95e-O.1l82X2/12 = 100.941

The value of the fixed-rate bond is

2.5e-O.1l82X2/12 + 2.5e-O.1l82X5/12 + 2.5e-O.1l82X8/12

+2.5e-O.1l82Xll/12 + 102.5e-O.1l82X14/12 = 98.678

The value of the swap is therefore

100.941 - 98.678 = $2.263 million

As an alternative approach we can value the swap as a series of forward rate agreements. The calculated value is

(2.95 - 2.5)e-O.1l82X2/12 + (3.0 _ 2.5)e-O.1182X5/12

+(3.0 - 2.5)eO.1l82X8/12 + (3.0 - 2.5)e-O.1l82Xll/12

+(3.0 - 2.5)e-O.1l82X14/12 = $2.263 million

which is in agreement with the answer obtained using the first approach.

6.21. The financial institution is long a USD bond and short an AUD bond. The value of the USD bond (in millions of dollars) is

0.48e-O.07Xl + 12.48e-o.o7x2 = 11.297

The value of the AUD bond (in millions of AUD) is

1.6e-O.09Xl + 21.6e-o.09x2 = 19.504

The value of the swap (in millions of dollars) is therefore

11.297 - 19.504 x 0.62 = -0.795

or -$795,000.

32

As an alternative we can value the swap as a series of forward foreign exchange contracts. The one-year forward exchange rate is 0.62e-0.02 = 0.6077. The two-year forward exchange rate is 0.62e-0.02x2 = 0.5957. The value of the swap in millions of dollars is therefore

(0.48 - 1.6 x 0.6077)e-0.07Xl + (12.48 - 21.6 x 0.5957)e-O.07x2 = -0.795

which is in agreement with the first calculation.

6.22. The spread between the interest rates offered to A and B is 0.4% (or 40 basis points) on sterling loans and 0.8% (or 80 basis points) on U.S. dollar loans. The total benefit to all parties from the swap is therefore

80 - 40 = 40 basis points

It is therefore possible to design a swap which will earn 10 basis points for the bank while making each of A and B 15 basis points better off than they would be by going directly to financial markets. One possible swap is shown in Figure 6.3. Company A borrows at an effective rate of 6.85% per annum in U.S. dollars. Company B borrows at an effective rate of 10.45% per annum in sterling. The bank earns a 10-basis-point spread. The way in which currency swaps such as this operate is as follows. Principal amounts in dollars and sterling that are roughly equivalent are chosen. These principal amounts flow in the opposite direction to the arrows at the time the swap is initiated. Interest payments then flow in the same direction as the arrows during the life of the swap and the principal amounts flow in the same direction as the arrows at the end of the life of the swap. Note that the bank is exposed to some exchange rate risk in the swap. It earns 65 basis points in U.S. dollars and pays 55 basis points in sterling. This exchange rate risk could be hedged using forward contracts.

£: 11.0% £: 10.45% £: 11.0% ~ .JI

Company - Financial Company ... .. .... ... A ... Institution ... B $:6.2% r r

$: 6.85% $:6.2%

Figure 6.3 One Possible Swap for Problem 6.22

33

Answers to Assignment Questions Chapter 7: Mechanics of Options Markets

1.15. (a) In this case the options are out of the money. No margin is required for the options. The shares can be purchased using a margin account with an initial outlay of only 50% of their cost: 0.5 x 500 x 28 = $7,000. Since $1,500 is received for the options, the investor's net initial outlay is $5,500. (b) In this case margin available for the purchase of the shares is reduced by the extent to which the options are in the money. The cost of the shares is $16,000. For the purposes of calculating the part of this that can be bought on margin, the share price is reduced to $30 so that funding of $7,500 is provided. In addition $1,500 can be funded from the sale of the options. The net cash investment is therefore 16,000 - 7,500 - 1,500 or $7,000.

1.16. Figure 7.1 shows the way in which the investor's profit varies with the stock price in the first case. For stock prices less than $30 there is a loss of $1,200. As the stock price increases from $30 to $50 the profit increases from -$1,200 to $800. Above $50 the profit is $800. Students may express surprise that a call which is $10 out of the money is less expensive than a put which is $10 out of the money. This could be because of dividends or the crashophobia phenomenon discussed in Chapter 15. Figure 7.2 shows the way in which the profit varies with stock price in the second case. In this case the profit pattern has a zigzag shape. The problem illustrates how many different patterns can be obtained by including calls, puts, and the underlying asset in a portfolio.

1.11. Executive stock option plans account for a high percentage of the total remuneration received by executives. When the market is rising fast (as it was for much of the 1990s) many corporate executives do very well out of their stock option plans - even when their company does worse than its competitors. Large institutional investors have argued that executive stock options should be structured so that the payoff depends how the company has performed relative to an appropriate industry index. In a regular executive stock option the strike price is the stock price at the time the option is issued. In the type of relative-performance stock option favored by institutional investors, the strike price at time t is SoIt/lo where So is the company's stock price at the time the option is issued, 10 is the value of an equity index for the industry in which the company operates at the time the option is issued, and It is the value of the index at time t. If the company's performance equals the performance of the industry, the options are always at-the-money. If the company outperforms the industry, the options become in the money. If the company underperforms the industry, the options become out of the money. Note that a relative performance stock option can provide a payoff when both the market and the company's stock price decline. Relative performance stock options clearly provide a better way of rewarding senior management for superior performance. Some companies have argued that, if they

34

2000

-=S1000 ... G..

0

·1000

·2000

.

. . ' . .

"

. -----.Loog Stock -LoogPut --Shot1Can -Tdal

Figure 7.1 Profit in first case considered Problem 7.16

2000 -01000 ... A.

80 ·1000 - . ---- Loog Stock

·2000 -Loog Put --Shot1Call -ldal

Figure 7.2 Profit for the second case considered Problem 7.16

introduce relative performance options when their competitors do not, they will lose some of their top management talent.

7.18. DerivaGem shows that the value of the option is 4.57. The option's intrinsic value is 32 - 30 = 2.00. The option's time value is therefore 4.57 - 2.00 = 2.57. A time value

35

of zero would indicate that it is optimal to exercise the option immediately. In this case the value of the option would equal its intrinsic value. When the stock price is 20, DerivaGem gives the value of the option as 12, which is its intrinsic value. When the stock price is 25, DerivaGem gives the value of the options as 7.54, indicating that the time value is still positive (= 0.54). Keeping the number of time steps equal to 50, trial and error indicates the time value disappears when the stock price is reduced to 21.6 or lower. (With 500 time steps this estimate of how low the stock price must become is reduced to 21.3.)

36


Chapter 8: Properties of Stock Options

8.21. If the call is worth $3, put-call parity shows that the put should be worth

3 + 20e-O.lxO.25 + e-O.lxO.08333 - 19 = 4.50

This is greater than $3. The put is therefore undervalued relative to the call. The correct arbitrage strategy is to buy the put, buy the stock, and short the ca.11.

8.22. Consider a portfolio that is long one option with strike price K 1 , long one option with strike price K3, and short two options with strike price K 2• The value of the portfolio can be worked out in four different situations

ST :5 K 1 : Portfolio Value = 0 Kl < ST :5 K 2 : Portfolio Value = ST - Kl K2 < ST < Ka: Portfolio Value = ST - Kl - 2(ST - K 2)

= K2 - Kl - (ST - K 2) > 0 ST > Ka: Portfolio Value = ST - Kl - 2(ST - K2) + ST - Ka

= K2 - Kl - (Ka - K2) =0

This shows that the value is always either zero or positive at the expiration of the option. In the absence of arbitrage possibilities it must be positive or zero today. This means that

or C2 :5 0.5(Cl + C3)

. Note that students often think they have proved this by writing down

2C2 < 2(So-K2e- rT)

Ca :5 So - Kae-rT

and subtracting the middle inequality from the sum of the other two. But they are deceiving themselves. Inequality relationships cannot be subtracted. For example, 9 > 8 and 5 > 2, but it is not true that 9 - 5 > 8 - 2

8.23. The corresponding result is P2 :5 0.5(P1 + Pa)

37

where PI, 112 and Pa are the prices of European put option with the same maturities and strike prices KI, K2 and Ka respectively. This can be proved from the result in Problem 8.22 using put-call parity. Alternatively we can consider a portfolio consisting of a long position in a put option with strike price KI, a long position in a put option with strike price Ka, and a short position in two put options with strike price K 2 •

The value of this portfolio in different situations is given as follows

ST < K 1 : Portfolio Value = KI - ST - 2(K2 - ST) + Ka - ST = Ka - K2 - (K2 - Kt) =0

KI < ST < K2: Portfolio Value = Ka - ST - 2(K2 - ST) = Ka - K2 - (K2 - ST) >0

K2 < ST < Ka: Portfolio Value = Ka - ST ST > Ka: Portfolio Value = 0

Since the portfolio value is always zero or positive at some future time the same must be true today. Hence

or

8.24. (a) Suppose V is the value of the company and D is the face value of the debt. The value of the manager's position in one year is

max(V - D, 0)

This is the payoff from a call option on V with strike price D. (b) The debt holders get

min(V, D)

= D - max(D - V, 0)

Since max(D - V, 0) is the payoff from a put option on V with strike price D, the debt holders have in effect made a risk-free loan (worth D at maturity with certainty) and written a put option on the value of the company with strike price D. The position of the debt holders in one year can also be characterized as

V - max(V - D, 0)

This is a long position in the assets of the company combined with a short position in a call option on the assets with a strike price of D. The equivalence of the two characterizations can be presented as an application of put-eall parity. (c) The manager can increase the value of his or her position by increasing the value of the call option in a). It follows that the manager should attempt to increase both V

38

and the volatility of V. To see why incre88ing the volatility of V is beneficial, imagine what happens when there are large changes in V. If V increases, the manager benefits to the full extent of the change. If V decreases, much of the downside is borne by the company's lenders.

8.25. DerivaGem shows that the price of the call option is 6.9686 and the price of the put option is 4.1244. In this case

c + D + Ke-rT = 6.9686 + 0.5e-O.06XO.5 + 40e-O.06xl = 45.1244

Also p+ S = 4.1244 + 41 = 45.1244

As the time to maturity becomes very large and there are no dividends, the price of the call option approaches the stock price of 41. (For example when T = 100 it is 40.94.) This is because the call option can be regarded as a position in the stock where the price does not have to be paid for a very long time. The present value of what has to be paid is close to zero. As the time to maturity becomes very large the price of the European put option becomes close to zero. (For example when T = 100 it is 0.04.) This is because the present value of what might be received from the put option becomes close to zero.

39

Answers to Assignment Questions Chapter 9: 'frading Strategies Involving Options

9.19. A butterfly spread is created by buying the $55 put, buying the $65 put and selling two of the $60 puts. This costs 3 + 8 - 2 x 5 = $1 initially. The following table shows the profit/loss from the strategy.

Stock Price Payoff Profit

ST ~ 65 0 -1 60 ~ ST < 65 65-ST 64-ST 55~ST<60 ST-55 ST-56 ST <55 0 -1

The butterfly spread leads to a loss when the final stock price is greater then $64 or less than $56.

9.20. There are two alternative profit patterns for part (a). These are shown in Figures 9.1 and 9.2. In Figure 9.1 the long maturity (high strike price) option is worth more than the short maturity (low strike price) option. In Figure 9.2 the reverse is true. There is no ambiguity about the profit pattern for part (b). This is shown in Figure 9.3.

Profit

Figure 9.1 Investor's Profit/Loss in Problem 9.20a when long maturity call is worth more than short maturity call

40

Profit

Figure 9.2 Investor's Profit/Loss in Problem 9.20a when short maturity call is worth more than long maturity call

Profit

Figure 9.3 Investor's Profit/Loss in Problem 9.20b

9.21. The variation of an investor's profit/loss with the terminal stock price for each of the four strategies is shown in Figure 9.4. In each case the dotted line shows the profits from the components of the investor's position and the solid line shows the total net profit.

41

9.22. (a) A call option with a strike price of 25 costs 7.90 and a call option with a strike price of 30 costs 4.18. The cost of the bull spread is therefore 7.90 - 4.18 = 3.72. The profits ignoring the impact of discounting are

Profit

Profit

(a)

(c)

Stock Price Range

ST :::; 25 25 < 8T < 30

8r > 30

Profit

Profit

(b)

(d)

Profit

-3.72 8T - 28.72

1.28

Figure 9.4 Investor's Profit/Loss in Problem 9.21

(b) A put option with a strike price of 25 costs 0.28 and a put option with a strike price of 30 costs 1.44. The cost of the bear spread is therefore 1.44 - 0.28 = 1.16. The profits ignoring the impact of discounting are

42

Stock Price Range

ST =5 25 25<ST<30

ST>30

Profit

+3.84 28.84 - ST

-1.16

(c) Call options with maturities of one year and strike prices of 25, 30, and 35 cost 8.92, 5.60, and 3.28, respectively. The cost of the butterfly spread is therefore 8.92 + 3.28 - 2 x 5.60 = 1.00. The profits ignoring the impact of discounting are

Stock Price Range

ST =5 25 25 < ST < 30 30~ST<35

ST >35

Profit

-1.00 ST - 26.00 34.00 - ST

-1.00

(d) Put options with maturities of one year and strike prices of 25, 30, and 35 cost 0.70,2.14,4.57, respectively. The cost of the butterfly spread is therefore 0.70+4.57-2 x 2.14 = 0.99. Allowing for rounding errors, this is the same as in (c). The profits are the same as in (c). (e) A call option with a strike price of 30 costs 4.18. A put option with a strike price of 30 costs 1.44. The cost of the straddle is therefore 4.18 + 1.44 = 5.62. The profits ignoring the impact of discounting are

Stock Price Range Profit

24.58 - ST ST - 35.62

(f) A six-month call option with a strike price of 35 costs 1.85. A six-month put option with a strike price of 25 costs 0.28. The cost of the strangle is therefore 1.85 + 0.28 = 2.13. The profits ignoring the impact of discounting are

Stock Price Range

ST < 25 25 < ST < 35 ST~ 35

43

Profit

22.87 - ST -2.13

ST - 37.13


Chapter 10: Introduction to Binomial Trees

10.14. At the end of six months the value of the option will be either $12 (if the stock price . is $60) or $0 (if the stock price is $42). Consider a. portfolio consisting of:

+~ : shares -1 : option

The value of the portfolio is either 42~ or 60~ - 12 in six months. If

42~ = 60~ -12

Le., ~ = 0.6667

the value of the portfolio is certain to be 28. For this value of ~ the portfolio is therefore riskless. The current value of the portfolio is:

0.6667 x 50 - !

where! is the value of the option. Since the portfolio must earn the risk~free ra.te of interest

(0.6667 x 50 - !)eO.12XO.5 = 28

i.e., ! = 6.96

The value of the option is therefore $6.96. This can also be calculated using risk~neutral valuation. Suppose that p is the pro~ ability of an upward stock price movement in a risk~neutral world. We must have

6Op+ 42(1- p) = 50 x eO.06

i.e., 18p = 11.09

or: p= 0.6161

The expected value of the option in a risk-neutral world is:

12 x 0.6161 + 0 x 0.3839 = 7.3932

This has a present value of 7.3932e-o.06 = 6.96

44

Hence the above answer is consistent with risk-neutral valuation.

10.15. (a) A tree describing the behavior of the stock price is shown in Figure 10.1. The risk-neutral probability of an up move, p, is given by

eO.25xO.12 - 0.90 p = 1.1 - 0.9 = 0.6523

Calculating the expected payoff and discounting, we obtain the value of the option as

[2.4 x 2 x 0.6523 x 0.3477 + 9.6 x 0.34772]e-O.12XO.5 = 2.118

The value of the European option is 2.118. This can also be calculated by working back through the tree as shown in Figure 10.1. The second number at each node is the value of the European option. (b) The value of the American option is shown as the third number at each node on the tree. It is 2.537. This is greater than the value of the European option because it is optimal to exercise early at node C.

48.400 0.000 0.000

39.600 2.400 2.400

32.400 9.600 9.600

Figure 10.1 'free to evaluate European and American put options in Problem 10.15. At each node, upper number is the stock price; next number is the European put price; final number is the American put price

10.16. 'frial and error shows that immediate early exercise is optimal when the strike price is above 43.2. This can be also shown to be true algebraically. Suppose the strike price increases by a relatively small amount e. This increases the value of being at node C by e and the value of being at node B by 0.3477e-o.03f = O.3374f. It therefore increases the value of being at node A by

(0.6523 X 0.3374f + 0.3477e)e-O.03 = 0.551f

45

For early exercise at node A we require 2.537 + 0.551€ < 2 + € or € > 1.196. This corresponds to the strike price being greater than 43.196.

10.17. If the option is a put its value is

[0.6523 x 0 + 0.3477 x 3]e-O.12XO.25 = 1.01227

In the real world the expected payoff from the put option is

0.7041 x 0 + 0.2959 x 3 = 0.8877

The expected return on the put is R where

0.8877 = 1.01227e-o.25R

This gives R = -0.525 or -52.5%. The call option has a highly positive beta and so earns a high positive expected return. The put option has a highly negative beta and so earns a high negative return.

10.18. This type of option is known as a power option. A tree describing the behavior of the stock price is shown in Figure 10.2. The risk-neutral probability of an up move, p, is given by

eO.05x2/12 - 0.9 p = 1.08 - 0.9 = 0.6020

Calculating the expected payoff and discounting, we obtain the value of the option 88

[0.7056 x 2 x 0.6020 x 0.3980 + 32.49 x 0.39802]e-O.05X4/12 = 5.394

The value of the European option is 5.394. This can also be calculated by working back through the tree as shown in Figure 10.2. The second number at each node is the value of the European option.

(b) Early exercise at node C would give 9.0 which is less than 13.2449. The option should therefore not be exercised early if it is American.

46

32.400 0.2785

F

34.922 0.000

29.160 0.7056

24.300 32.49

Figure 10.2 Tree to evaluate European power option in Problem 10.18. At each node, upper number is the stock price; next number is

the option price

10.19. (a) This problem is based on the material in Section 10.8. In this case 8t = 0.25 so that u = eO.30xv'O.25 = 1.1618, d = l/u = 0.8607, and

eO.04XO.25 - 0.8607 p = 1.1618 _ 0.8607 = 04959

(b) and (c) The value of the option using a two-step tree 88 given by DerivaGem is shown in Figure 10.3 to be 3.3739. To use DerivaGem choose the first worksheet, select Equity as the underlying type, and select Binomial European 88 the Option Type. After carrying out the calculations select Display Tree.

(d) With 5, 50, 100, and 500 time steps the value of the option is 3.9229, 3.7394, 3.7478, and 3.7545, respectively.

47

At each node: Upper value = Underlying Asset Price Lower value = Option Price

Values In red are a result olearly exercise.

Strike price = 40 Discount factor per step = 0.9900 Tim estep, dt = 0.2500 years, 91.25 days Growth factor per step, a = 1.0101 P robabil:ty of up move, p = 0.4959 Up step size, u = 1.1618 Down step size, d = 0.8607

Node Time: 0.0000 0.2500

'-5""3"'. 9""'9""'4~35~

0.5000

Figure 10.3 'free produced by DerivaGem to evaluate European option in Problem 10.19.

48


Chapter 11: Model of the Behavior of Stock Prices

11.12. With the notation in the text

In this case 8 = 50, Jl = 0.16, CT = 0.30 and ot = 1/365 = 0.00274. Hence

08 50 rv 4»(0.16 x 0.00274, 0.30 x v'0.00274)

= 4»(0.00044, 0.0157)

and 08 rv 4»(50 x 0.00044, 50 x 0.0157)

that is, 08 rv 4»(0.022, 0.785)

(a) The expected stock price at the end of the next day is therefore 50.022 (b) The standard deviation of the stock price at the end of the next day is 0.785 (c) 95% confidence limits for the stock price at the end of the next day are

50.022 - 1.96 x 0.785 and 50.022 + 1.96 x 0.785

i.e., 48.48 and 51.56

. Note that some students may consider one trading day rather than one calendar day. Then ot = 1/252 = 0.00397. The answer to (a) is then 50.032. The answer to (b) is 0.945. The answers to part (c) are 48.18 and 51.88.

11.13. (a) The probability distributions are:

4»(2.0 + 0.1, v'0.16) = 4»(2.1,0.4)

4»(2.0 + 0.6, v'0.16 x V6) = 4»(2.6, 0.98)

4»(2.0 + 1.2, v'0.16 x v'i2) = 4»(3.2, 1.39)

(b) The chance of a random sample from 4»(2.6,0.98) being negative is

N (-~~~) = N(-2.65)

49

where N(x) is the cumulative probability that a standardized normal variable [i.e., a variable with probability distribution ¢(O, 1)] is less than x. From normal distribution tables N( -2.65) = 0.0040. Hence the probability of a negative cash position at the end of six months is 0.40%. Similarly the probability of a negative cash position at the end of one year is

N ( - 1~:9) = N( -2.30) = 0.0107

or 1.07%. (c) In general the probability distribution of the cash position at the end of x months is

¢(2.0 + O.lx, 0.4y'x)

The probability of the cash position being negative is maximized when:

is minimized. Define

2.0+ O.lx 0.4y'x

= 2.0+0.1x =5 -, 025 ~ Y 0.4y'x x +. x

~! = -2.5x-! + 0.125x- i 3

= x-~ (-2.5 + 0.125x)

This is zero when x = 20 and it is easy to verify that lfly/dx2 > 0 for this value of x. It therefore gives a minimum value for y. Hence the probability of a negative cash position is greatest after 20 months.

11.14. The process followed by B, the bond price, is from Ito's lemma (see equation 11.2):

In this case

so that:

Hence

8B -0' Ot - , 8B 8x

1 B=

x

dB = -a Xo - x)- + -8 X - dt - -8xdz [ ( 1 1 2 22] 1 x2 2 x3 x2

[ 1 8

2] 8 = -a(xo-x)-+- dt--dz

x2 x x

50

The expected instantaneous rate at which capital gains are earned from the bond is therefore:

1 82

-a(xo - x)- + -X 2 X

The expected interest per unit time is 1. The total expected instantaneous return is therefore:

1 82

1 - a(xo - X) - + -X2 X

When expressed as a proportion of the bond price this is:

a 2 = X - - (Xo - X) + 8

X

11.15. (a) In this case 8y/8S = 2, 82y/8S2 = 0, and 8y/8t = 0 so that Ito's lemma gives

dy = 2/l's dt + 20B dz

or dy = p,ydt + aydz

(b) In this case 8y/8S = 28, 82y/8S2 = 2, and 8y/8t = 0 so that Ito's lemma gives

dy = (2p,S2 + a2 S2) dt + 2aS2 dz

or dy = (2p, + ( 2 )y dt + 2ay dz

(c) In this case 8y/8S = eS , 82y/8S2 = eS , and 8y/8t = 0 so that Ito's lemma gives

dy = (p,Ses + a2S 2eS /2) dt + uSes dz

or dy = [p,y In y + a 2y(ln y)2 /2] dt + ay In y dz

(d) In this case 8y/8S = _er(T-t) /82 = -y/S, 82y/8S2 = 2er(T-t) /S3 = 2y/S2, and ay/at = _rer(T-t) / S = -ry so that Ito's lemma gives

dy = (-ry - p,y + u2y) dt - uydz

or dy = -(r + p, - ( 2)ydt - uydz

11.16. The variable In ST is normally distributed with mean In So + (p, - u2/2)T and standard deviation uVT. In this case So = 50, p, = 0.12, T = 2, and u = 0.30 so that the mean

51

and standard deviation ofln ST are In 50+(0.12-0.32/2)2 = 4.062 and 0.3'\1'2 = 0.424, respectively. Also, In 80 = 4.382. The probability that ST > 80 is the same as the probability that In BT > 4.382. This is

1 - N (4.382 - 4.062) = 1 - N(O 754) 0.424 .

where N(x) is the probability that a normally distributed variable with mean zero and standard deviation 1 is less than x. From the tables at the back of the book N(0.754) = 0.775 so that the required probability is 0.225.

52


Chapter 12: The Black-Scholes Model

12.24. In this case So = 50, J.t = 0.18 and q = 0.30. The probability distribution of the stock price in two years, ST, is lognormal and is, from equation (12.3), given by:

InST t'.I 4>[ln50 + (0.18 - 0~9)2, 0.3V2J

i.e., In ST t'.I 4>(4.18, 0.42)

The mean stock price is from equation (12.4)

50e2XO.18 = 50eO.36 = 71.67

and the standard deviation is, from equation (12.5),

50e2XO.18v'eO.09X2 - 1 = 31.83

95% confidence intervals for In ST are

4.18 -1.96 x 0.42 and 4.18 + 1.96 x 0.42

i.e., 3.35 and 5.01

These correspond to 95% confidence limits for ST of

i.e., 28.52 and 150.44

12.25. The calculations are shown in the table below

LUi = 0.09471 L u~ = 0.01145

and an estimate of standard deviation of weekly returns is:

0.01145 _ 0.094712 = 0 02884 13 14 x 13 .

53

r

The volatility per annum is therefore 0.02884V52 = 0.2079 or 20.79%. The standard error of this estimate is

or 3.9% per annum.

0.2079 = 0.0393 v'2 x 14

Problem 12.25 Computation of Volatility

Week Closing Stock

Price ($) Price Relative

= Si/Si-l

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15

30.2 32.0 31.1 30.1 30.2 30.3 30.6 33.0 32.9 33.0 33.5 33.5 33.7 33.5 33.2

1.05960 0.97188 0.96785 1.00332 1.00331 1.00990 1.07843 0.99697 1.00304 1.01515 1.00000 1.00597 0.99407 0.99104

Daily Return Ui = In(Si/ Si-l)

0.05789 -0.02853 -0.03268

0.00332 0.00331 0.00985 0.07551

-0.00303 0.00303 0.01504 0.00000 0.00595

-0.00595 -0.00900

12.26. (a) The expected value of the security is E[(ST )2] From equations (12.4) and (12.5), at time t :

E(ST) = Sep,(T-t)

var(ST) = S2e2P,(T-t)[e0'2(T-t) - 1]

Since var(ST) = E[(ST )2] - [E(ST )]2, it follows that E[(ST )2] = var(ST) + [E(ST )]2 so that

E[(ST )2] = S2e2P,(T-t)[e0'2(T-t) - 1] + S2e2p,(T-t)

= ~e(2p,+0'2)(T-t)

In a risk-neutral world J.t = r so that

Using risk-neutral valuation, the value of the derivative security at time t is

= s2e(2r+0'2)(T-t)e-r (T-t)

= s2e(r+0'2)(T-t)

54

(b) If: 1 = s2e(r+u2 )(T-t)

81 = -s2(r + 0'2)e(r+u2)(T-t)

8t 81 = 2Se(r+u2 )(T-t) 8S

821 = 2e(r+u2 )(T-t)

8S2

The left-hand side of the Black-Scholes differential equation is:

-S2(r + 0'2)e(r+u2)(T-t) + 2rS2e<r+(2 )(T-t) + 0'2s2e(r+u2 )(T-t)

=rS2e(r+u2 )(T-t)

=rf

Hence the Black-Scholes equation is satisfied.

12.21. In this case So = 30, K = 29, r = 0.05,0' = 0.25 and T = 0.3333

d = In (30/29) + (0.05 + 0.252/2) x 0.3333 = 0.4225 1 0.25v'0.3333

d = In (30/29) + (0.05 - 0.252/2) x 0.3333 = 0.2782 2 0.25v'0.3333

N(0.4225) = 0.6637, N(0.2782) = 0.6096

N( -0.4225) = 0.3363, N(-0.2782) = 0.3904

(a) The European call price is

30 x 0.6637 - 2ge-0.05xO.3333 x 0.6096 = 2.52

or $2.52. (b) The American call price is the same as the European call price. It is $2.52. (c) The European put price is

2ge-0.05xO.3333 x 0.3904 - 30 x 0.3363 = 1.05

or $1.05. (d) Put-call parity states that:

p+ So = c+ Ke-rT

In this case c = 2.52, So = 30, K = 29, p 1.05 and e-rT = 0.9835 and it is easy to verify that the relationship is satisfied.

55

12.28. (a) The present value of the dividend must be subtracted from the stock price. This gives a new stock price of:

12.29.

30 - 0.5e-O.125xO.05 = 29.5031

and d = In (29.5031/29) + (0.05 + 0.252/2) x 0.3333 = 0.3068 1 0.25v'0.3333

d = In (29.5031/29) + (0.05 - 0.252/2) x 0.3333 = 0.1625 2 0.25v'0.3333

N(d1 ) = 0.6205; N(d2 ) = 0.5645

The price of the option is therefore

29.5031 x 0.6205 - 2ge-O.3333xO.05 x 0.5645 = 2.21

or $2.21. (b) Since

N(-d1) = 0.3795, N(-d2) = 0.4355

the value of the option when it is a European put is

2ge-O.3333XO.05 x 0.4355 - 29.5031 x 0.3795 = 1.22

or $1.22. (c) H tl denotes the time when the dividend is paid:

K[l - e-r(T-td ] = 29(1 - e-O.05XO.2083) = 0.3005

This is less than the dividend. Hence the option should be exercised immediately before the ex-dividend date for a sufficiently high value of the stock price.

We first value the option assuming that it is not exercised early, we set the time to maturity equal to 0.5. There is a dividend of 0.4 in 2 months and 5 months. Other parameters are So = 18, K = 20, r = 10%, q = 30%. DerivaGem gives the price as 0.7947. We next value the option assuming that it is exercised at the five-month point just before the final dividend. DerivaGem gives the price as 0.7668. The price given by Black's approximation is therefore 0.7947. DerivaGem also shows that the correct American option price calculated with 100 time steps is 0.8243. It is never optimal to exercise the option immediately before the first ex-dividend date when

where Dl is the size of the first dividend, and tl and t2 are the times of the first and second dividend respectively. Hence we must have:

that is, Dl < 0.494

56

It is never optimal to exercise the option immediately before the second ex-dividend date when:

where D2 is the size of the second dividend. Hence we must have:

that is, D2 ::;; 0.166

It follows that the dividend can be as high as 16.6 cents per share without the American option being worth more than the corresponding European option.

57


Chapter 13: Options on Stock Indices, Currencies, and FUtures

13.41. In this case the risk-neutral probability of an up move is

(1 - .875)/(1.125 - 0.875) = 0.5

The value of the option is

e -O.07xO.25 [0.5 x 3 + 0.5 x 01 = 1.474

13.42. In this case the risk-neutral probability of an up movement in one month is

e(r-q).o.t - d

u-d e(O.08-0.03) xO.25 - 0.9

1.1 _ 0.9 = 0.5629

The probability of a down movement is 0.4371. A tree of index movements together with the corresponding European and American option prices is shown in Figure 13.1. The European price is 11.88 while the American price is 13.56.

363.00 0.00 0.00

297.00 3.00 3.00

243.00 57.00 57.00

Figure 13.1 Tree to evaluate European and American put options in Problem 13.42. At each node, upper number is the stock price; next number is the European put price; final number is the American put price

58

13.43. In this case So = 0.75, K = 0.75, r = 0.07, rJ = 0.09, u = 0.04 and T = 0.75. The option can be valued using equation (13.9)

and

d1 = In (0.75/0.75) + (0.07 - 0.09 + 0.0016/2) x 0.75 = -0.4157

0.04JO.75

~ = d1 - 0.04JO.75 = -0.4503

N(dl) = 0.3388, N(~) = 0.3262

The value of the call, c, is given by

c = 0.75e-O.09xO.75 x 0.3388 - 0.75e-0.07xO.75 x 0.3262 = 0.0054

i.e., it is 0.54 cents. From put-call parity

so that c = 0.0054 + 0.75e-0.07xO.75 - 0.75e-o.09XO.75 = 0.0160

The option to buy 0.75 USD with 1.00 CDN is the same as the same as an option to sell one Canadian dollar for 0.75 USD. In other words it is -a put option on the Canadian dollar and its price is 0.0160 USD.

·13.44. In this case Fo = 525, K = 525, r = 0.06, T = 0.4167. We wish to find the value of u for which p = 20 where:

This must be done by trial and error. When u = 0.2, p = 26.36. When u = 0.15, p = 19.78. When u = 0.155, p = 20.44. When u = 0.152, p = 20.04. These calculations show that the implied volatility is approximately 15.2% per annum.

13.45. Suppose that K is the value of the fund at the beginning of the year and ST is the value of the fund at the end of the year. The salary of a fund manager is

a max(ST - K, 0)

where a is a constant. This shows that a fund manager has a call option on the value of the fund at the end of the year. All of the parameters determining the value of this call option are outside the control of the fund manager except the volatility of the fund. The fund manager has an incentive to make the fund as volatile as possible! This may not correspond with the desires of the investors. One way of making the fund highly volatile would

59

be by investing only in high-beta stocks. Another would be by using the whole fund to buy call options on a market index. It might be argued that a fund manager would not do this because of the risk which the manager faces. If the fund earns a negative return the manager's salary is zero. However, a fund manager could hedge the risk of a negative return by, on his or her own account, taking a short position in call options on a stock market index. The position could be chosen so that if the market goes up, the gain on salary more than offsets the losses on the call options. If the market goes down the fund manager ends up with the price received for the call options. It is easy to see that the strategy becomes more attractive as the riskiness of the fund's portfolio increases. To summarize, it is clear that the (superficially attractive) remuneration package is open to abuse and does not necessarily motivate the fund managers to act in the best interests of the fund's investors.

13.46. The prices of call options on May corn futures with strike prices of 190, 200, 210, 220, 230, and 240 are 201, 12, 5i, 2i, i, and !. The corresponding put prices are i, Ii, 4!, 11!, 20, and 29'2' The futures price for the May 2001 contract is, from Table 2.2, 210l There are 37 trading days to maturity~ Using DerivaGem with Fo = 210.75, r = 5%, T = 37/365, gives implied volatilities for calls with strike prices of 200, 210, 220, 230, and 240 as 18.78%, 18.26%, 20.97%, 22.81%, and 26.46%, respectively. The price of the call with a strike price of 190 indicates that it should be exercised immediately and an implied volatility cannot be calculated. The implied volatilities for puts with strike prices of 190, 200, 210, 220, 230, and 240 are 17.26%, 18.56%, 19.20%, 21.13%, 22.26%, and 24.17%, respectively. We do not expect put-call parity to hold exactly for American options and so there is no reason why the implied volatility of a call should be exactly the same as the implied volatility of a put. Nevertheless it is reassuring that they are close. There is a tendency for high strike price options to have a higher implied volatility. As explained in Chapter 15, this is an indication that the probability distribution for corn futures prices in the future has a fatter right tail and thinner left tail than the lognormal distribution.

13.47. Table 13.1 shows that the last trade in the June call option on DJX with a strike price of 100 was at a price of $5. The last trade in the June put on DJX with a strike price of 100 was at a price of $4.10. To calculate the implied volatility of the call we use the Equity..FXJndex..Futures_Options worksheet of DerivaGem, select the Underlying Type as Index and the Option Type as Analytic European. We then enter the Index Level as 100.31, the Risk-Free Rate as 4.5%, the Dividend Yield as 2%, and the Exercise Price as 100. Nothing need be entered for Volatility. The Price is entered as 5 in the second half of the Option Data table. There are 93 calendar days and 65 trading days to maturity. IT the Time to Exercise is measured in calendar days it is entered as 93/365=0.2548 years. If it measured in trading days, it is entered as 65/252=0.2579. The call and Implied Volatility buttons should be selected. Clicking on Calculate then causes the Implied Volatility to appear as the Volatility. When calendar days assumption is made the implied volatility is 22.58%. When the trading

60

days assumption is made it is 22.43%. For the put option we change the price to 4.10 and select Put instead of Call. When the calendar days assumption is made, the implied volatility of the put is 22.78%. When the trading days assumption is made, it is 22.67%. These results show that the implied volatility of the call is close to that of the put. As is explained formally in Chapter 15, put-call parity implies that the two implied volatilities should be equal.

61


Chapter 14: The Greek Letters

14.24. The price, delta, gamma, vega, theta, and rho of the option are 3.7008,0.6274,0.050, 0.1135, -0.00596, and 0.1512. When the stock price increases to 30.1, the option price increases to 3.7638. The change in the option price is 3.7638 - 3.7008 = 0.0630. Delta predicts a change in the option price of 0.6274 x 0.1 = 0.0627 which is very close. When the stock price increases to 30.1, delta increases to 0.6324. The size of the increase in delta is 0.6324 - 0.6274 = 0.005. Gamma predicts an increase of 0.050 x 0.1 = 0.005 which is the same. when the volatility increases from 25% to 26%, the option price increases by 0.1136 from 3.7008 to 3.8144. This is consistent with the vega value of 0.1135. When the time to maturity is changed from 1 to 1-1/365 the option price reduces by 0.006 from 3.7008 to 3.6948. This is consistent with a theta of -0.00596. Finally when the interest rate increases from 5% to 6% the value of the option increases by 0.1527 from 3.7008 to 3.8535. This is consistent with a rho of 0.1512.

14.25. The delta of the portfolio is

-1,000 x 0.50 - 500 x 0.80 - 2,000 x (-0.40) - 500 x 0.70 = -450

The gamma of the portfolio is

-1,000 x 2.2 - 500 x 0.6 - 2,000 x 1.3 - 500 x 1.8 = -6,000

The vega of the portfolio is

-1,000 x 1.8 - 500 x 0.2 - 2,000 x 0.7 - 500 x 1.4 = -4,000

(a) A long position in 4,000 traded options will give a gamma-neutral portfolio since the long position has a gamma of 4,000 x 1.5 = +6, 000. The delta of the whole portfolio (including traded options) is then:

4,000 x 0.6 - 450 = 1,950

Hence, in addition to the 4,000 traded options, a short position in £1,950 is necessary so that the portfolio is both gamma and delta neutral. (b) A long position in 5,000 traded options will give a vega-neutral portfolio since the long position has a vega of 5,000 x 0.8 = +4, 000. The delta of the whole portfolio (including traded options) is then

5,000 x 0.6 - 450 = 2,550

62

Hence, in addition to the 5,000 traded options, a short position in £2,550 is necessary so that the portfolio is both vega and delta neutral.

14.26. Let WI be the position in the first traded option and W2 be the position in the second traded option. We require:

6,000 = 1.5wI + 0.5W2

4,000 = 0.8WI + 0.6W2

The solution to these equations can easily be seen to be WI = 3, 200, W2 = 2, 400. The whole portfolio then has a delta. of

-450 + 3,200 x 0.6 + 2,400 x 0.1 = 1, 710

Therefore the portfolio can be made delta, gamma and vega neutral by taking a long position in 3,200 of the first traded option, a long position in 2,400 of the second traded option and a short position in £1,710.

14.27. The product provides a six-month return equal to

max (0, OAR)

where R is the return on the index. Suppose that So is the current value of the index and ST is the value in six months. When an amount A is invested, the return received at the end of six months is:

ST-SO A max (0, 004 So )

0.4A ( ) = aomax 0, ST - So

This is O.4A/So of at-the-money European call options on the index. With the usual notation, they have value:

O;OA [Soe-qT N(dd - Soe-rT N(d2 )]

= 0.4A[e-qT N(d1 ) - e-rT N(d2 )]

In this case r = 0.08, q = 0.25, T = 0.50 and q = 0.03

d (0.08 - 0.03 + 0.252/2) 0.50

1 = f'i'\"iFn = 0.2298 0.25yO.50

d2 = d1 - 0.25v'0.50 = 0.0530

N(d1 ) = 0.5909; N(d2 ) = 0.5212

63

The value of the European call options being offered is

0.4A(e-O.03XO.5 x 0.5909 - e-O.08xO.5 x 0.5212)

= 0.0325A

This is the present value of the payoff from the product. If an investor buys the product he or she avoids having to pay 0.0325A at time zero for the underlying option. The cash flows to the investor are therefore Time 0: -A + 0.0325A = 0.9675A After six months: + A The return with continuous compounding is 2In(1/0.9675) = 0.066 or 6.6% per annum. The product is therefore slightly less attractive than a risk-free investment.

14.28. (a)

KN'(d2) = KN'(dl - uVT) = ~e-(dV2)+dll1"';T-a2T/2 v'27r

Because dl uy'T = In(F / K) + u2T /2 the second equation reduces to

KN'(d2) = ~e-(dV2)+ln(F/K) = ~e-dV2 v'27r v'21r

The result follows. (b)

Because

8e = e-rT N(d ) + e-rT F N'(d ) 8dl _ e-rT N'(d ) 8d2 8F 1 1 8F 2 8F

it follows from the result in (a) that

(c)

8e = e-rTN(dd 8F

8e =e-rTFN'(dl)8dl _e-rTKN'(d2)8d2 8u 8u 8u

Because dl = d2 + uy'T

From the result in (a) it follows that

8e = e-rTFN'(dl)VT 8u

64

(d) Rho is given by Be T Br = -Te-r [FN(d1 ) - KN(d2 )]]

or -cT. Because q = r in the case of a futures option there are two components to rho. One arises from differentiation with respect to r, the other from differentiation with respect to q.

14.29. For the option considered in Section 14.1, So = 49, K = 50, r = 0.05, 0' = 0.20, and T = 20/52. DerivaGem shows that e = -0.011795 x 365 = -4.305, a = 0.5216, r = 0.065544, n = 2.4005. The left hand side of equation (14.7)

1 -4.305 + 0.05 x 49 x 0.5216 + 2' x 0.22 X 492 x 0.065~44 = 0.120

The right hand side is 0.05 x 2.4005 = 0.120

This shows that the result in equation (14.7) is satisfied.

14.30. Consider the first week. The portfolio consists of a short position in 100,000 options and a long position in 52,200 shares. The value of the option changes from $240,053 at the beginning of the week to $188,760 at the end of the week for a gain of $51,293. The value of the shares change from 52,200 x 49 = $2,557,800 to 52,200 x 48.12 = $2,511,864 for a loss of $45,936. The net gain is 51,293 - 45,936 = $5,357. The gamma and theta (per year) of the portfolio are -6554.4 and 430,533 so that equation (14.6) predicts the gain as

1 1 430533 x 52 - 2' x 6554.4 x (48.12 - 49)2 = 5742

The results for all 20 weeks are shown in the following table. Week Actual Gain Predicted Gain

1 5,357 5,742 2 5,689 6,093 3 -19,742 -21,084 4 1,941 1,572 5 3,706 3,652 6 9,320 9,191 7 6,249 5,936 8 9,491 9,259 9 961 870

10 -23,380 -18,992 11 1,643 2,497 12 2,645 1,356 13 11,365 10,923 14 -2,876 -3,342 15 12,936 12,302 16 7,566 8,815 17 -3,880 -2,763 18 6,764 6,899 19 4,295 5,205 20 4,804 4,805

65


Chapter 15: Volatility Smiles

15.16. In liquidation the company's stock price must be at least 300,000/100,000 = $3. The company's stock price should therefore always be at least $3. This means that the stock price distribution that has a less heavy left tail and heavier right tail than the lognormal distribution. An upward sloping volatility smile can be expected.

15.17. (a) If p is the risk-neutral probability of a positive outcome (stock price rises to $24), we must have

24p + 18(1 - p) = 20eO.08xO.0833

so that p = 0.356 (b) The price of a call option with strike price K is (24 - K)pe-O.08XO.08333 when K < 24. Call options with strike prices of 19, 20, 21, 22, and 23 therefore have prices 1. 766, 1.413, 1.060, 0.707, and 0.353, respectively. (c) From Dei-ivaGem the implied volatilities of the options with strike prices of 19, 20, 21, 22, and 23 are 49.8%, 58.7%, 61.7%, 60.2%, and 53.4%, respectively. The volatility smile is therefore a "frown" with the volatilities for deep-out-of-the-money and deep-in-the-money options being lower than those for close-to-the-money options. (d) The price of a put option with strike price K is (K _18)(1_p)e-O.08XO.08333. Put options with strike prices of 19, 20, 21, 22, and 23 therefore have prices of 0.640, 1.280, 1.920, 2.560, and 3.200. DerivaGem gives the implied volatilities as 49.81%, 58.68%, 61.69%, 60.21%, and 53.38%. Allowing for rounding errors these are the same as the implied volatilities for put options.

15.18. The calculations are shown in the following table. For example, when the strike price is 34, the price of a call option with a volatility of 10% is 5.926, and the price of a call option when the volatility is 30% is 6.312. When there is a 60% chance of the first volatility and 40% of the second, the price is 0.6 x 5.926 + 0.4 x 6.312 = 6.080. The implied volatility given by this price is 23.21. The table shows that the uncertainty about volatility leads to a classic volatility smile similar to that in Figure 15.1 of the text. In general when volatility is stochastic with the stock price and volatility uncorrelated we get a pattern of implied volatilities similar to that observed for currency options.

66

Call Option Price Call Option Price Implied Strike Price 10% Volatility 30% Volatility Weighted Price Volatility (%)

34 5.926 6.312 6.080 23.21 36 3.962 4.749 4.277 21.03 38 2.128 3.423 2.646 18.88 40 0.788 2.362 1.418 18.00 42 0.177 1.560 0.730 18.80 44 0.023 0.988 0.409 20.61 46 0.002 0.601 0.242 22.43

15.19. The following table shows the percentage of daily returns greater than 1, 2, 3, 4, 5, and 6 standard deviations for each currency. The pattern is similar to that in Table 15.1.

>lsd > 2sd > 3sd > 4sd > 5sd > 6sd

AUD 24.8 5.3 1.3 0.2 0.1 0.0 BEF 24.3 5.7 1.3 0.6 0.2 0.0 CHF 26.1 4.2 1.3 0.6 0.1 0.0 DEM 23.9 5.0 1.4 0.6 0.1 0.0 DKK 26.7 5.8 1.3 0.3 0.0 0.0 ESP 28.2 5.1 0.9 0.3 0.1 0.0 FRF 26.0 5.4 1.4 0.2 0.0 0.0 GBP 23.9 6.4 1.1 0.4 0.1 0.0 ITL 25.4 6.6 1.1 0.2 0.0 0;0 NLG 25.6 5.7 1.7 0.2 0.0 0.0 SEK 28.2 5.2 1.0 0.0 0.0 0.0

Normal 31.7 4.6 0.3 0.0 0.0 0.0

67

15.20. The results are shown in the table below.

> 3sd down > 3sd up

TSE 0.88 0.22 S&P 0.55 0.44

FTSE 0.55 0.66 CAC 0.33 0.33

Nikkei 0.55 0.66

Total 0.57 0.46

15.21 Define Cl and PI as the values of the call and the put when the volatility is 0'1. Define c:z and P2 as the values of the call and the put when the volatility is 0'2. From put-call parity

If follows that

PI + Soe-qT = Cl + Ke-rT

P2 + Soe-qT = c:z + Ke-rT

PI-P2 =CI- C2

68


Chapter 16: Value at Risk

16.16. The change in the value of the portfolio for a small change 6y in the yield is approximately - D B6y where D is the duration and B is the value of the portfolio. It follows that the standard deviation of the daily change in the value of the bond portfolio equals DBuy where uy is the standard deviation of the daily change in the yield. In this case D = 5.2, B = 6,000,000, and u y = 0.0009 so that the standard deviation of the daily change in the value of the bond portfolio is

5.2 x 6,000,000 x 0.0009 = 28,080

The 20-day 90% VaH for the portfolio is 1.282 x 28,080 x v'2O = 160,990 or $160,990. This approach assumes that only parallel shifts in the term structure can take place. Equivalently it assumes that all rates are perfectly correlated or that only one factor drives term structure movements. Alternative more accurate approaches described in the chapter are (a) cash flow mapping and (b) a principal components analysis.

16.17. The variance of the portfolio (in thousands of dollars) is

0.0182 x 3002 + 0.012 X 5002 + 2 x 300 x 500 x 0.6 x 0.018 x 0.012 = 104.04

The standard deviation is )104.04 = 10.2. Since N( -1.96) = 0.025, the I-day 97.5% VaH is 10.2 x 1.96 = 19.99 and the 100day 97.5% VaH is JW x 19.99 = 63.22. The lO-day 97.5% VaH is therefore $63,220. The 100day 97.5% value at risk for the gold investment is 5,400 x v'ill x 1.96 = 33,470. The 10-day 97.5% value at risk for the silver investment is 6,000 x v'ill x 1.96 = 37,188. The diversification benefit is

33,470 + 37,188 - 63,220 = $7,438

16.18. An approximate relationship between the daily change in the value of the portfolio, 6P and the proportional daily change in the value of the asset 6x is

6P = 10 x 126x = 1206x

The standard deviation of 6x is 0.02. It follows that the standard deviation of 6P is 2.4. The I-day 95% VaH is 2.4 x 1.65 = $3.96.

16.19. Using the same notation as in Problem 16.18, the quadratic relationship is

6P = 10 x 126x + 0.5 x 102 x (-2.6)6x2

or 6P = 1206x - 1306x2

69

(a) From Appendix 16B, the first three moments of 6P are -1300"2, 12020"2 + 3 X

13020"4, and -9 x 1202 x 130u4 - 15 X 13030"6 where 0" is the standard deviation of ox. Substituting 0" = 0.02, the first three moments are -0.052, 5.768, and -2.698. (b) The first two moments imply that the mean and standard deviation of 6P is -0.052 and 2.402. The 5% fractile of the distribution is -0.052 - 2.402 x 1.65 = -4.02. The I-day 95% VaR is therefore $4.02. (c) The skewness of the distribution, f,p, is

1 2.4023 [-2.698 + 3 x 5.768 x 0.052 - 2 x 0.0522] = -0.13

We can use the Cornish-Fisher expansion in Appendix 16B setting q = 0.05 to obtain

1 Wq = -1.65 - 6(1.652

- 1) x 0.13 = -1.687

so that the 5% fractile of the distribution is

-0.052 - 2.402 x 1.687 = 4.10

This is the I-day 95% VaH..

16.20. The cash flows are as follows

Year 1

2-yr Bond 5 3-yr Bond 5 5-yr Bond -5

Total 5 Present Value 4.756 Impact of 1 bp change -0.0005

2 3

105 0 5 105

-5 -5

105 100 95.008 86.071

-0.0190 -0.0258

4 5

0 0 0 0

-5 -105

-5 -105 -4.094 -81.774 0.0016 0.0409

The duration relationship is used to calculate the last row of the table. When the one-year rate increases by one basis point, the value of the cash flow in year 1 decreases by 1 x 0.0001 x 4.756 = 0.0005; when the two year rate increases by one basis point, the value of the cash flow in year 2 decreases by 2 x 0.0001 x 95.008 = 0.0190; and so on.

The sensitivity to the first factor is

-0.0005 x 0.32 - 0.0190 x 0.35 - 0.0258 x 0.36 + 0.0016 x 0.36 + 0.0409 x 0.36

or -0.00081. The sensitivity to the second factor is

-0.0005 x (-0.32) - 0.0190 x (-0.10) - 0.0258 x 0.02 + 0.0016 x 0.14 + 0.0409 x 0.17

or 0.0087 The sensitivity to the third factor is

-0.0005 x (-0.37) - 0.0190 x (-0.38) - 0.0258 x (-0.30) + 0.0016 x (-0.12)

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+0.0409 x (-0.04)

or 0.0133. Assuming one factor, the standard deviation of the one-day change in the portfolio value is 0.00081 x 17.49 = 0.0142. The 20-day 95% VaR is therefore 0.0142 x 1.645v'20 = 0.104 Assuming two factors, the standard deviation of the one-day change in the portfolio value is

.)0.000812 X 17.492 + 0.00872 X 6.052 = 0.0545

The 20-day 95% VaR is therefore 0.0545 x 1.645v'20 = 0.401 Assuming three factors, the standard deviation of the one-day change in the portfolio value is

.)0.000812 X 17.492 + 0.00872 X 6.052 + 0.01332 X 3.102 = 0.0683

The 20-day 95% VaR is therefore 0.0683 x 1.645v'20 = 0.502. In this case the second and third factor are important in calculating VaR.

16.21. This assignment is useful for consolidating students' understanding of alternative approaches to calculating VaR, but it is calculation intensive. Realistically students need some programming skills to make the assignment feasible. My answer follows the usual practice of assuming that the 1O-day 99% value at risk is vro times the 1-day 99% value at risk. Some students may try to calculate a 10-day VaR directly, which is fine. (a) From DerivaGem, the values of the two option positions are -5.413 and -1.014. The deltas are -0.589 and 0.284, respectively. An approximate linear model relating the change in the portfolio value to proportional change, OXl, in the first stock price and the proportional change, OX2, in the second stock price is

of = -0.589 x 500Xl + 0.284 X 200X2

or of = -29.450xl + 5.68oX2

The daily volatility of the two stocks are 0.28/ V252 = 0.0176 and 0.25/ V252 = 0.0157, respectively. The one-day variance of of is

29.452 X 0.01762 + 5.682 X 0.01572

- 2 x 29.45 x 0.0176 x 5.68 x 0.0157 x 004 = 0.2396

The one day standard deviation is, therefore, 0.4895 and the 10-day 99% VaR is 2.33 x ViO x 004895 = 3.61. (b) The gammas of the two option positions are -0.0321 and -0.0678. The quadratic approximation is

of = -0.589 x 500Xl + 0.284 X 200X2 - 0.5 x 0.0321 x 502ox~ - 0.5 x 0.0678 x 2Q2ox~

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or 8P = -29.4548xl + 5.6868x2 - 40.1018x~ - 13.5648x~

In the notation of Appendix 16B, n = 2, 0:1 = -29.454, 0:2 = 5.686, Pu = -40.101, 1322 = -13.564, P12 = 0, 0'11 = 0.01762 = 0.000311, 0"22 = 0.01572 = 0.000248, and 0"12 = 0.4 x 0.0176 x 0.0157 = 0.000111. It follows that

E(8P) = -40.101 x 0.000311 + 13.564 x 0.000248 = -0.0158

A1so

and E(8p3 ) = -2.927

The mean, standard deviation, and skewness can be calculated as -0.01584, 0.4910, and -0.1505, respectively. The estimate of the one percentile point of the probability distribution when only the first two moments are used is

-0.01584 - 2.33 x 0.4910 = -1.1598

The 10-day 99% value at risk when the first two moments are used is, therefore, v'iO x 1.1598 = 3.67. (c) Using the notation of Appendix 16B

1 2 Wq = -2.33 - 6(2.33 - 1) x 0.1505 = -2.441

The 1 percentile point of the probability distribution when the first three moments are used is

-0.01584 - 2.441 x 0.4910 = 1.2144

The 10-day 99% value at risk when the first three moments are used is, therefore, v'iO x 1.2144 = 3.84. (d) In the partial simulation approach, we simulate changes in the stock prices over a one-day period (building in the correlation) and then use the quadratic approximation to calculate the change in the portfolio value on each simulation trial. The one percentile point of the probability distribution of portfolio value changes turns out to be 1.22. The 10-day 99% value at risk is, therefore, 1.22v'iO or about 3.86. (e) In the full simulation approach, we simulate changes in the stock price over oneday (building in the correlation) and revalue the portfolio on each simulation trial. The results are very similar to (d) and the estimate of the 10-day 99% value at risk is about 3.86.

16.22. We can create a portfolio with zero delta in Sample Application E by changing the position in the stock from 1,000 to 513.58. (This reduces delta by 1,000- 513.58 = 486.42.) In this case the true VaH. is 48.86; the VaH. given by the linear model is 0.00; and the VaH. given by the quadratic model is -35.71.

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Other zero-delta examples can be created by changing the option portfolio and then zeroing out delta by adjusting the position in the underlying asset. The results are similar. The software shows that neither the linear model nor the quadratic model gives good answers when delta is zero. The linear model always gives a VaR of zero because the model assumes that the portfolio has no risk. (For example, in the case of one underlying asset 6P = 6,68.) When there are no cross gammas the quadratic model assumes that 6P is always positive. (For example, in the case of one underlying asset 6P = 0.51'(68)2.) This gives a negative VaR. In practice many portfolios do have deltas close to zero because of the hedging activities described in Chapter 14. This has led many financial institutions to prefer historical simulation to the model building approach.

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Chapter 17: Estimating Volatilities and Correlations

17.15. The proportional change in the price of gold is -2/300 = -0.00667. Using the EWMA model the variance is updated to

0.94 X 0.0132 + 0.06 X 0.006672 = 0.00016153

so that the new daily volatility is v'0.00016153 = 0.01271 or 1.271% per day. Using GARCH (1,1) the variance is updated to

0.000002 + 0.94 x 0.0132 + 0.04 X 0.006672 = 0.00016264

so that the new daily volatility is v'0.00016264 = 0.1275 or 1.275% per day.

17.16. The proportional change in the price of silver is zero. Using the EWMA model the variance is updated to

0.94 X 0.0152 + 0.06 x 0 = 0.0002115

so that the new daily volatility is v'0.0002115 = 0.01454 or 1.454% per day. Using GARCH (1,1) the variance is updated to

0.000002 + 0.94 x 0.0152 + 0.04 x 0 = 0.0002135

so that the new daily volatility is v'0.0002135 = 0.01461 or 1.461% per day. The initial covariance is 0.8 x 0.013 x 0.015 = 0.000156 Using EWMA the covariance is updated to

0.94 x 0.000156 + 0.06 x 0 = 0.00014664

so that the new correlation is 0.00014664/(0.01454 x 0.01271) = 0.7935 Using GARCH (1,1) the covariance is updated to

0.000002 + 0.94 x 0.000156 + 0.04 x 0 = 0.00014864

so that the new correlation is 0.00014864/(0.01461 x 0.01275) = 0.7979. The w pa..rameter defines the long run average value of a variance or a covariance. There is no reason why we should expect the long run average daily variance for gold and silver should be the same. There is also no reason why we should expect the long run average covariance between gold and silver to be the same as the long run average variance of gold or the long run average variance of silver. In practice, therefore, we are likely to want to allow w in a GARCH(I,I) model to depend to vary from market variable to market variable. (Some instructors may want to use this problem as a lead in to multivariate GARCH models.

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17.17. My results give "best" values for ,\ higher than the 0.94 used by lliskMetrics. For AUD, BEF, CHF, DEM, DKK, ESP, FRF, GBP, ITL, NLG, and SEK they are 0.983, 0.967, 0.968, 0.960, 0.971, 0.983, 0.965, 0.977, 0.939, 0.962, and 0.989, respectively. For TSE, S&P, FTSE, CAC, and Nikkei, they are 0.991,0.989,0.958,0.974, and 0.961, respectively.

17.18. (a) The long·run average variance, VL, is

w = 0.000002 = 0.0001 1- a - /3 0.02

The long run average volatility is v'0.0001 = 0.01 or 1% per day (b) From equation (17.13) the expected variance in 20 days is

0.0001 + 0.9820(0.0152 - 0.0001) = 0.000183

The expected volatility is therefore v'0.000183 = 0.0135 or 1.35%. Similarly the expected volatilities in 40 and 60 days are 1.25% and 1.17%, respectively. (c) The average of the expected variances at the ends of days 0, 1, 2, ... ,19 is 0.000204. The volatility that should be used for a 20.day option is therefore v'0.000204 = 0.01428 or 1.428% per day. This is 1.4280252) = 22.67% per year. Similarly, the volatilities that should be used for 40. and 60-day options are 21.69% and 20.89%, respectively. (d) When today's volatility increases to 2% the expected volatilities in 20, 40, and 60 days are 1.73%, 1.53% and 1.38% per day, respectively. (e) When today's volatility increases to 2% the volatilities used to price 20, 40, and 60 day options increase to 29.67%, 27.85%, and 26.35%, respectively.

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Answers to Assignment Questions Chapter 18: Numerical Procedures

18.25. The tree is shown in Figure 18.1. The value of the option is estimated as 0.0207. and its delta is estimated as

0.006221 - 0.041153 = -0404 0.858142 - 0.764559 .

At each node: Upper value = Underlying Asset Price Lower value = Option Price

Shaded values are a result of early exercise.

Strike price", 0.8 Discount factor per step", 0.9802

TIme step. dt", 0.3333 years, 121.67 days Growth factor per step, a = 1.0101

Probability of up move, p '" 0.5726 Up step size, u '" 1.0594 Down step size, d '" 0.9439

'-:0:-::. 9~09~1~45:-f

Node Tlme: 0.0000 0,3333

o

0,6667 1,0000

Figure 18.1 Thee for Problem 18.25

18.26. In this case Fo = 8.5, K = 9, r = 0.12, T = 1, a = 0.25, and 6t parameters for the tree are

u = ecrVlf; = eO.25VO.25 = 1.1331

d = .!. = 0.8825 u

a=l

a - d 1 - 0.8825 p = u - d = 1.1331- 0.8825 = 0.469

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0.25. The

The tree output by DerivaGem for the American option is shown in Figure 18.2. The estimated value of the option is $0.596. The tree produced by DerivaGem for the European version of the option is shown in Figure 18.3. The estimated value of the option is $0.586. The Black-Scholes price of the option is $0.570. The control variate estimate of the price of the option is therefore

0.596 + 0.570 - 0.586 = 0.580

At each node: Upper value = Underlying Asset Price lower value = Option Price

Shaded values are a result ot early exercise.

Strike price = 9 Discount factor per step = 0.9704 Time step. dt = 0.2600 years. 91.26 days Growth factor per step. a = 1.0000 Probability of up move. p .. 0.4688 Up step size, u" 1.1331

Node Time: 0.0000 0.2500 0.5000 0.7500 1.0000

Figure 18.2 Tree for American option in Problem 18.26

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At each node: Upper value. Underlying Anet Price Lower value. Opllon Price

Shaded values are a rnult of early exercise.

S trike price .. 9 Discount factor per step = 0.9704 Time slep, dt = 0.2500 years. 91.25 days Growth factor per slep, a .. 1.0000 Probability of up mow, p .. 0.4688 Up step size. u .. 1.1331 Down step size, d ,. 0.8825

Node Time: 0.0000 0.2500

f":':':'~"::'l:'f

0.5000 0.7500 1.0000

Figure 18.3 Tree for European option in Problem 18.26

18.27. DerivaGem gives the value of the option as 0.989. Black's approximation sets the price of the American call option equal to the maximum of two European options. The first lasts the full six months. The second expires just before the final ex-dividend date. In this case the software shows that the first European option is worth 0.957 and the second is worth 0.997. Black's model therefore estimates the value of the American option as 0.997. This is close to the tree value of 0.989.

18.28. In this case equation (18.32) becomes

where

aj = 1 +1rot [-4(r - rf )jot + 40'2j20t]

b~ = 1 (1 _ 0'2j20t) 3 1 + rot

cj = 1 +lrot [4(r - rf)jot + 40'2iot]

The parameters are r = 0.06, r f = 0.09, 0' = 0.15, 8 = 1.60, K = 1.62, T = 1, ot = 0.25, 08 = 0.2 and we obtain the table shown below. The option price is $0.062.

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Stock Price Time To Maturity (Months) 12 9 6 3 0

2.40 0.780 0.780 0.780 0.780 0.780 2.20 0.580 0.580 0.580 0.580 0.580 2.00 0.380 0.380 0.380 0.380 0.380 1.80 0.180 0.180 0.180 0.180 0.180 1.60 0.062 0.054 0.043 0.027 0.000 1.40 0.011 0.007 0.003 0.000 0.000 1.20 0.001 0.000 0.000 0.000 0.000 1.00 0.000 0.000 0.000 0.000 0.000 0.80 0.000 0.000 0.000 0.000 0.000

18.29. (a) For the binomial model in Section 18.5 there are two equally likely changes in the logarithm of the stock price in a time step of length bt. These are (r - a2/2)bt + aV'li and (r - a2/2)bt - aV'li. The expected change in the logarithm of the stock price is

0.5[(r - a2/2)bt + o-vTt] + 0.5[(r - 0-2/2)bt - o-vTt] = (r - 0-2/2)bt

This is correct. [See equation (12.3).] The variance of the change in the logarithm of the stock price is

0.50-2bt + 0.5u2bt = u2bt

This is correct. [See equation (12.3).] (b) For the trinomial tree model in Section 18.5, the change in the logarithm of the stock price in a time step of length bt is +av'3bt, 0, and -o-V3bt with probabilities

2

3' _ /it (r _ 0-

2) + .!.

V~ 2 6

The expected change is

It variance is 0-2bt plus a term of order (bt)2. These are correct. (c) To get the expected change in the logarithm of the stock price in time bt correct we require

1 2 1 (a 2) -(Inu) + -(Inm) + -(lnd) = r - - bt

6 3 6 2

The relationship m2 = ud implies In m = 0.5(In u + In d) so that the requirement becomes

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or

The expected change in In S is In m. To get the variance of the change in the logarithm of the stock price in time at correct we require

1 1 -(Inu -In m)2 + -(1nd -In m)2 = u20t 6 6

Because In u - In m = - (In d - In m) it follows that

Inu = lnm+ uV30t

Ind = Inm - uV30t

These results imply that

18.30. See chart

80.00

75.00

70.00

65.00

80.00

55.00

50.00 2

Problem 18.30.

6 m M a ~ ~ ~ M ~ ~ ~ 50

nSlap.

80

90.00

88.00

88.00

84.00

82.00

90.00

78.00

78.00

74.00

72.00

70.00 2 8 10

90.00

88.00

88.00

84.00

82.00

80.00

78.00

78.00

74.00

72.00

70.00 2 8 10

14 18

14

Problem 18.30b

18

81

22 28 nSlap.

30

Problem 18.300

22 28 nSlap.

30

38 42 48 50

38 48 50

23.0%

22.0%

21.0%

20.0%

2 10 14 18

82

Problem 18.3Od

22 26

nStepe 30 34 38 42


Chapter 19: Exotic Options

19.23. It is instructive to consider two different ways of valuing this instrument. From the perspective of a sterling investor it is a cash or nothing put. The variables are So = 1/1.48 = 0.6757, K = 1/1.50 = 0.6667, r = 0.08, q = 0.04, (}' = 0.12, and T = 1. The derivative pays off if the exchange rate is less than 0.6667. The value of the derivative is 10, OOON( _d2)e-O.08xl where

d = In(0.6757/0.6667) + (0.08 - 0.04 - 0.122/2) = 0 3852 2 0.12 .

Since N( -d2 ) = 0.3501, the value of the derivative is 10,000 x 0.3501 x e-0.08 = 3,231 or 3,231. In dollars this is 3,231 x 1.48 = $4782 FrOJ;Il the perspective of a dollar investor the derivative is an asset or nothing call. The variables are So = 1.48, K = 1.50, r = 0.04, q = 0.08, (}' = 0.12 and T = 1. The value is 10,000N(dl)e-0.08xl where

d = In(1.48/1.50) + (0.04 - 0.08 + 0.122 /2) = -03852 1 Q12 .

N(d1 ) = 0.3500 and the value of the derivative is as before 10,000 x 1.48 x 0.3500 x e-O•08 = 4782 or $4,782.

19,24. The price of the option is 3.528. (a) The option price is a humped function of the stock price with the maximum option price occuring for a stock price of about $57. If you could choose the stock price there would be a trade off. High stock prices give a high probability that the option will be knocked out. Low stock prices give a low potential payoff. For a stock price less than $57 delta is positive (as it is for a regular call option); for a stock price greater that $57, delta is negative. (b) Delta increases up to a stock price of about 45 and then decreases. This shows that gamma can be positive or negative. (c) The option price is a humped function of the time to maturity with the maximum option price occurring for a time to maturity of 0.5 years. This is because too long a time to maturity means that the option has a high probability of being knocked out; too short a time to maturity means that the option has a low potential payoff. For a time to maturity less than 0.5 years theta is negative (as it is for a regular call option); for a time to maturity greater than 0.5 years the theta of the option is positive. (d) The option price is also a humped function of volatility with the maximum option price being obtained for a volatility of about 20%. This is because too high a volatility means that the option has a high probability of being knocked out; too Iowa volatility

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means that the option has a low potential payoff. For volatilities less than 20% vega is positive (as it is for a regular option); for volatilities above 20% vega is negative.

19.25. (a) Both approaches use a one call option with a strike price of 50 and a maturity of 0.75. In the first approach the other 15 call options have strike prices of 60 and equally spaced times to maturity. In the second approach the other 15 call options have strike prices of 60, but the spacing between the times to maturity decreases as the maturity of the barrier option is approached. The second approach to setting times to maturity produces a better hedge. This is because the chance of the barrier being hit at time t is an increasing function of t. AB t increases it therefore becomes more important to replicate the barrier at time t. (b) By using either trial and error or the Solver tool we see that we come closest to matching the price of the barrier option when the maturities of the third and fourth options are changed from 0.25 and 0.5 to 0.39 and 0.65. (c) To calculate delta for the two 16-option hedge strategies it is necessary to change the last argument of EPortfolio from 0 to 1 in cells L42 and X42. To calculate delta for the barrier option it is necessary to change the last argument of BarrierOption in cell F12 from 0 to 1. To calculate gamma and vega the arguments must be changed to 2 and 3, respectively. The delta, gamma, and vega of the barrier option are -0.0221, -0.0035, and -0.0254. The delta, gamma, and vega of the :first 16-option portfolio are -0.0262, -0.0045, and -0.1470. The delta, gamma, and vega of the second 16-option portfolio are -0.0199, -0.0037, and -0.1449. The second of the two 16-option portfolios provides Greek letters that are closest to the Greek letters of the barrier option. Interestingly neither of the two portfolios does particularly well on vega.

19.26 A natural approach is to attempt to replicate the option with positions in: (a) A European call option with strike price 1.00 maturing in two years (b) A European put option with strike price 0.80 maturing in two years (c) A European put option with strike price 0.80 maturing in 1.5 years (d) A European put option with strike price 0.80 maturing in 1.0 years (e) A European put option with strike price 0.80 maturing in 0.5 years The first option can be used to match the value of the down-and-out-call for t = 2 and S > 1.00. the others can be used to match it at the following (t, S) points: (1.5, 0.80) (1.0, 0.80), (0.5, 0.80), (0.0, 0.80). Following the procedure in the text, we find that the required positions in the options are as shown in the following table. The value of the portfolio initially is 0.482. This is only a little less than the value of the down-and-out-option which is 0.488. This example is different from the example in the text in a number of ways. Put options and call options are used in the replicating portfolio. The value of the replicating portfolio converges to the value of the option from below rather than from above. Also, even with relatively few options, the value of the replicating portfolio is close to the value of the down-and-out option.

19.27. In this case Ml = 917.07 and M2 = 904028.7 so that the option can be valued as an option on futures where the futures price is 917.07 and volatility is y'ln(904028. 7/917.072

or 26.88%. The value of the option is 100.74. DerivaGem gives the price as 86.77 (set

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Option Strike Maturity Type Price (years) Position

Call 1.00 2.00 +1.0000 Put 0.80 2.00 -0.1255 Put 0.80 1.50 -0.1758 Put 0.80 1.00 -0.0956 Put 0.80 0.50 -0.0547

option type =Asian). The higher price for the first option arises because the average is calculated from prices at times 0.25, 0.50, 0.75, and 1.00. The mean of these times is 0.625. By contrast the corresponding mean when the price is observed continuously is 0.50. The later a price is observed the more uncertain it is and the more it contributes to the value of the option.

19.28. For the regular option the theoretical price is 239,599. For the average price option the theoretical price is 115,259. My 20 simulation runs ( 40 outcomes because of the antithetic calculations) gave results as shown in the following table.

Ave. Hedging Cost S.D. Hedging Cost Ave Trading Vol (20 wks) Ave Trading Vol (last 10 wks)

Regular Call

247,628 17,833

412,440 187,074

Ave Price Call

114,837 12,123

291,237 51,658

These results show that the standard deviation of using delta hedging for an average price option is lower than that for a regular option. However, using the criterion in Chapter 14 (standard deviation divided by value of option) hedge performance is better for the regular option. Hedging the average price option requires less trading, particularly in the last 10 weeks. This is because we become progressively more certain about the average price as the maturity of the option is approached.

19.29 The value of the option is 1093. It is necessary to change cells F20 and F46 to 0.67. Cells G20 to G39 and G46 to G65 must be changed to calculate delta of the compound option. Cells H20 to H39 and H46 to H65 must be changed to calculate gamma of the compound option. Cells 120 to 140 and 146 to 166 must be changed to calculate the Black-Scholes price of the call option expiring in 40 weeks. Similarly cells J20 to J40 and J46 to J66 must be changed to calculate the delta of this option; cells K20 to K40 and K46 to K66 must be changed to calculate the gamma of the option.

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The payoffs in cells N9 and NIO must be calculated as MAX(I40-0.015,0)*lOOOOO and MAX(I66-0.015,O) * 100000. Delta plus gamma hedging works relatively poorly for the compound option. On 20 simulation runs the cost of writing and hedging the option ranged from 200 to 2500.

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Chapter 20: More on Models and Numerical Procedures

20.11. To use the approach in Section 20.6 we construct a tree for Y(t) = G(t)/S(t) where G(t) is the minimum value of the index to date and S(t) is the value of the index at time t. The tree parameters are a = 1.0050, p = 0.5000, 'U = 1.1052, and d = 0.9048. We use the tree in Figure 20.1 to value the option in units of the stock index. This means that we value an instrument that pays off 1-Y(t). The tree shows that the value of the option is 0.1019 units of the stock index or 400 x 0.1019 = 40.47 dollars. DerivaGem shows that the value given by the analytic formula is 53.38. This is higher than the value given by the tree because the tree assumes that the stock price is observed only three times when the minimum is calculated.

Figure 20.1 Tree for Problem 20.17.

20.18. (a) The six-month call option with a strike price of 1.05 should be valued with a volatility of 13.4% is worth 0.01829. The call option with a strike price of 1.10 should be valued with a volatility of 14.3% and is worth 0.00959. The bull spread is therefore worth 0.01829 - 0.00959 = 0.00870. (b)We now ask what volatility if used to value both options gives this price. Using the DerivaGem Application Builder in conjunction with Goal Seek we find that the answer is 11.42%. (c) Yes, this does support the contention at the beginning of the chapter that the correct volatility for valuing exotic options can be counterintuitive. We might reasonably expect the volatility to be between 13.4% (the volatility used to value the first option) and 14.3% (the volatility used to value the second option). 11.42% is well

87

outside this range. The reason why the volatility is relatively low is as follows. The option provides the same payoff as a regular option with a 1.05 strike price when the asset price is between 1.05 and 1.10 and a lower payoff when the asset price is over 1.10. The implied probability distribution of the asset price (see Figure 15.2) is less heavy ~han the lognormal distribution in the 1.05 to 1.10 range and more heavy than the lognormal distribution in the> 1.10 range. This means that using a volatility of 13.4% (which is the implied volatility of a regular option with a strike price of 1.05) will give a price than is too high. (d) The bull spread provides a payoff at only one time. It is therefore correctly valued by the IVF model.

20.19. Consider first the least squares approach. At the two-year point the option is in the money for paths 1, 3, 4, 6, and 7. The five observations on 8 are 1.08, 1.07, 0.97, 0.77, and 0.84. The five continuation values are 0, 0.10e-0 .06 , 0.21e-0 .06 , 0.23e-0.06,

0.12e-0.06 • The best fit continuation value is

-1.394 + 3.7958 - 2.27682

The best fit continuation values for the five paths are 0.0495, 0.0605, 0.1454, 0.1785, and 0.1876. These show that the option should be exercised at the two-year point for all five paths. There are six paths at the one-year point for which the option is in the money. These are paths 1,4, 5, 6, 7, and 8. The six observations on 8 are 1.09, 0.93, 1.11, 0.76, 0.92, and 0.88. The six continuation values are 0.05e-0 .06 , 0.16e-0 .06 , 0, 0.36e-0 .06 , 0.2ge-0 .06 , and O. The best fit continuation value is

2.055 - 3.3178 + 1.34182

The best fit continuation values for the six paths are 0.0327, 0.1301, 0.0253, 0.3088, 0.1385, and 0.1746. These show that the option should be exercised at the one-year point for paths 1, 4, 6, 7, and 8 The value of the option if not exercised at time zero is therefore

!(0.04e-0.06 + 0 + 0.06e-0 .012 + 0.20e-0.06 + 0 + 0.37e-0 .06 + 0.21e-0 .06 + 0.25e-0 .06) 8

or 0.133. Exercising at time zero would yield 0.13. The option should therefore not be exercised at time zero and its value is 0.133. Consider next the exercise boundary parametrization approach. At time two years it is optimal to exercise when the stock price is 0.84 or below. At time one year it is optimal to exercise whenever the option is in the money. The value of the option assuming no early exercise at time zero is therefore

! (0.04e -0.06 +0+0.10e -0.018 +0.20e -0.06 +0.02e -0.06 +0.37 e -0.06 +0.21e -0.06 +0.25e -0.06)i 8

or 0.139. Exercising at time zero would yield 0.13. The option should therefore not be exercised at time zero. the value at time zero is 0.139. However, this may be biased

88

upward. As explained in the text, once the early exercise boundary has been obtained we carry out a new Monte carlo simulation using the early exercise boundary to value the option.

20.20. The price of the option using Merton's model can be calculated using the first 20 terms in the formula in section 20.2. For strike prices of 80, 90, 100, 110, and 120, the option prices are 22.64, 14.17, 7.67, 3.86, and 2.04 respectively. The implied volatilities are 26.00%, 23.64%, 22.85%, 23.65%, and 25.42%, respectively. The smile is similar to that for foreign currencies in Chapter 15. the probability distribution of the asset price price in six months has heavier tails than the lognormal distribution.

89


Chapter 21: Martingales and Measures

21.16 (a) The value of the option can be calculated by setting So = 400, K = 400, r = 0.06, q = 0.03, U = 0.2, and T = 2. With 100 time steps the value (in Canadian dollars) is 52.92. (b) 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 the growth rate of the index by 0.4 x 0.2 x 0.06 or 0.48% per year to 3.48%. The option can therefore be calculated using DerivaGem with So = 400, K = 400, r = 0.04, q = 0.04 - 0.0348 = 0.0052, U = 0.2, and T = 2. With 100 time steps DerivaGem gives the value as 57.51.

21.18. (a) The no-arbitrage arguments in Chapter 3 show that

J(t) F(t) = P(t, T)

(b) From Ito's lemma:

Therefore

dIn P = (p,p - u~/2) dt + Up dz

dInJ = (p,J - u'/2)dt + uJ dz

dIn ~ = d(lnJ -lnP) = (1'1 - u'/2 - p,p + u~/2)dt + (uJ - up) dz

so that

or dF = (p,J - p,p + u~ - ulup)Fdt + (uJ - up)Fdz

In a world that is forward risk neutral with respect to pet, T), F has zero drift. The process for F is

dF = (ul - up)Fdz

(c) In the traditional risk-neutral world, 1'1 = p,p = r where r is the short-term risk-free rate and

dF = (u~ - uJup)Fdt + (ul - up)Fdz

Note that the answers to parts (b) and (c) are consistent with the market price of risk being zero in (c) and Up in (b). When the market price of risk is Up, 1'1 = r + uJup and p,p = r+u~.

90

(d) In a world that is forward risk~neutral with respect to a bond maturing at time T*, p,p = r + o-pO'p and f.tl = r + O'pO'I so that

or dF = (0'1 - O'P)(O'p - O'p)Fdt + (0'1 - O'p)Fdz

21.19. (a) We require the expected value of the Nikkei index in a dollar risk-neutral world. In a yen risk-neutral world the expected value of the index is 20,000e(O.02-0.01)X2 20,404.03. In a dollar risk-neutral world the analysis in Section 21.8 shows that this becomes

20,404.03eo.3XO.20XO.12x2 = 20,699.97

The value of the instrument is therefore

20,699.97e-o.04X2 = 19,108.48

(b) An amount 8Q yen is invested in the Nikkei. Its value in yen changes to

In dollars this is worth 8Q1 +68/8

Q+6Q

where 6Q is the increase in Q. When terms of order two and higher are ignored, the dollar value becomes

8(1 + 68/8 - 6Q/Q)

The gain on the Nikkei position is therefore 68 - 86Q/Q When 8Q yen are shorted the gain in dollars is

This equals 86Q/Q when terms of order two and higher are ignored. The gain on the whole position is therefore 68 as required. (c) In this case the investor invests $20,000 in the Nikkei. The investor converts the funds to yen and buys 100 times the index. The index rises to 20,050 so that the investment becomes worth 2,005,000 yen or

2,005, 000 = 20 110 33 99.7 ,.

dollars. The investor therefore gains $110.33. The investor also shorts 2,000,000 yen. The value of the yen changes from $0.0100 to $0.01003. The investor therefore loses

91

0.00003 x 2,000,000 = 60 dollars on the short position. The net gain is 50.33 dollars. This is close to the required gain of $50. (d) Suppose that the value of the instrument is V. When the index changes by 68 yen the value of the instrument changes by

8V~8 88

dollars. We can calculate 8V /88. Part (b) of this question shows how to manufacture an instrument that changes by 68 dollars. This enables us to delta-hedge our exposure to the index.

92


Chapter 22: Interest Rate Derivatives: The Standard Market Models

22.27. The present value of the coupon payment is

35e-o.08xO.25 = 34.31

Equation (22.2) can therefore be used with Fo = (910 - 34.31)eo.08X8/12 = 923.66, r = 0.08, U = 0.10 and T = 0.6667. When the strike price is a cash price, K = 900 and

d = In(923.66/900) + 0.005 x 0.6667 = 0.3587 1 0.1,",,0.6667

d2 = d1 - 0.1 ,",,0.6667 = 0.2770

The option price is therefore

900e-O.08xO.6667 N( -0.2770) - 875.69N( -0.3587) = 18.34

or $18.34. When the strike price is a quoted price 5 months of accrued interest must be added to 900 to get the cash strike price. The cash strike price is 900 + 35 x 0.8333 = 929.17. In this case

d = In(923.66/929.17) + 0.005 x 0.6667 = -0.0319 1 0.1 ,",,0.6667

d2 = d1 - 0.1 ,",,0.6667 = -0.1136

and the option price is

929. 17e-o.o8xo.6667 N(0.1136) - 875.69N(0.0319) = 31.22

or $31.22.

22.28. The quoted futures price corresponds to a forward rate is 8% per annum with quarterly compounding and actual/360. The parameters for Black's model are therefore: Fk = 0.08, K = 0.08, R = 0.075, Uk = 0.15, tk = 0.75, and P(O, tk+l) = e-O.075xl = 0.9277

d1

= 0.5 X 0.152

x 0.75 = 0.0650 0.15,",,0.75

d2 = 0.5 X 0.152

x 0.75 = -0.0650 0.15,",,0.75

and the call price, c, is given by

c = 0.25 x 1,000 x 0.9277 (0.08N(0.0650) - 0.08N( -0.0.650)} = 0.96

93

22.29. The payoff from the swaption is a series of five cash flows equal to ma.x[0.076 - BT, O} in million of dollars where BT is the five-year swap rate in four years. The value of an annuity that provides $1 per year at the end of years 5, 6, 7,8, and 9 is

9 1 ~ 1.08i = 2.9348

The value of the swaption in millions of dollars is therefore

2.9348[0.076N( -d2 ) - 0.08N( -ddJ

where d = In(0.08/0.076) + 0.252

X 4/2 = 0.3526 1 0.25V4

d = In(0.08/0.076) - 0.252

X 4/2 = -0.1474 2 0.25v4

and

The value of the swaption is

2.9348[0.076N(0.1474) - 0.08N( -0.3526)} = 0.039554

or $39,554. This is the same answer as that given by DerivaGem. Note that for the purposes of using DerivaGem the zero rate is 7.696% continuously compounded for all maturities.

22.30. We use the Caps and Swap Options worksheet of DerivaGem. To set the zero curve as flat at 6% with continuous compounding, we need only enter 6% for one maturity. To value the cap we select Cap/Floor as the Underlying Type, enter Quarterly for the Settlement Frequency, 100 for the Principal, 0 for the Start (Years), 5 for the End (Years), 7% for the Cap/Floor Rate, and 20% for the Volatility. We select BlackEuropean as the Pricing Model and choose the Cap button. We do not check the Imply Breakeven Rate and Imply Volatility boxes. The value of the cap is 1.565. To value the floor we change the Cap/Floor Rate to 5% and select the Floor button rather than the Cap button. The value is 1.072. The collar is a long position in the cap and a short position in the floor. The value of the collar is therefore

1.565 - 1.072 = 0.493

22.31. We choose the third worksheet of DerivaGem and choose Swap Option as the Underlying Type. We enter 100 as the Principal, 2 as the Start (Years), 7 as the End (Years), 6% as the Swap Rate, and Semiannual as the Settlement Frequency. We also enter the zero curve information. We choose Black-European as the pricing model, enter 15% as the Volatility and check the Pay Fixed button. We do not check the

94

Imply Breakeven Rate and Imply Volatility boxes. The value of the swap option is 4.606. For a company that expects to borrow at LIBOR plus 50 basis points in two years and then enter into a swap to convert to five-year fixed-rate borrowings, the swap guarantees that its effective fixed rate will not be more than 6.5%. The swap option is the same as an option to sell a five-year 6% coupon bond for par in two years.

22.32. To calculate the convexity adjustment for the five-year rate define the price of a five year bond, as a function of its yield as

The convexity adjustment is

G(y) = e-5y

G' (y) = -5e-5y

Gil (V) = 25e -5y

0.5 X 0.082 x 0.252 X 4 x 5 = 0.004

Similarly for the two year rate the convexity adjustment is

0.5 X 0.082 x 0.252 X 4 x 2 = 0.0016

We can therefore value the derivative by assuming that the five year rate is 8.4% and the two-year rate is 8.16%. The value of the derivative is

0.24e-o.08X4 = 0.174

If the payoff occurs in five years rather than four years it is necessary to make a timing adjustment. From equation (22.18) this involves multiplying the forward rate by

exp [_ 1 x 0.25 x O'~~O; 0.08 x 4 x 1] = 0.98165

The value of the derivative is

0.24 x 0.98165e-o.o8x5 = 0.158

22.33. (a) In this case we must make a convexity adjustment to the forward swap rate. Define

6 4 100 G(y) = ~ (1 + V/2)i + (1 + V/2)6

95

so that , 6 2i 300

G (y) = - ~ (1 + y/2)i+l + (1 + y/2)1

" ~ i(i + 1) 1050 G (y) = £;t (1 + y/2)i+2 + (1 + y/2)8

G'(0.08) = -262.11 and G"(O.08) = 853.29 so that the convexity adjustment is

1 2 2 853.29 2 x 0.08 x 0.18 x 10 x 262.11 = 0.00338

The adjusted forward swap rate is 0.08 + 0.00338 = 0.08338 and the value of the derivative in millions of dollars is

0.08338 x 100 = 3 805 1.0420 •

(b) When the swap rate is applied to a yen principal we must make a quanto adjustment in addition to the convexity adjustment. From equation (21.36) this involves multiplying the forward swap rate by e-0.25XO.12xO.18xIO = 0.9474. (Note that the correlation is the correlation between the dollar per yen exchange rate and the swap rate. It is therefore -0.25 rather than +0.25.) The value of the derivative in millions of yen is

0.08338 x 0.9474 x 100 = 5865 1.01520 •

22.34. No adjustment is necessary for the forward rate applying to the period between years seven and eight. Using this, we can deduce from equation (22.18) that the forward rate applying to the period between years five and six must be multiplied by

exp [_ 1 x 0.16 x O·~~O; 0.06 x 5 x 2] = 0.9856

Similarly the forward rate applying to the period between year six and year seven must be multiplied by

exp [ _ 1 x 0.16 x O·~~O~ 0.06 x 6 x 1] = 0.9913

Similarly the forward rate applying to the period between year eight and nine must be multiplied by

exp [1 x 0.16 x O·~~O~ 0.06 x 8 x 1] = 1.0117

The adjusted forward average interest rate is therefore

0.25 x (0.08 x 0.9856 + 0.08 x 0.9913 + 0.08 + 0.08 x 1.0117) = 0.07977

The value of the derivative is

0.7977 x 1000 x 1.06-8 = 50.05

96


Chapter 23: Interest Rate Derivatives: Models of the Short Rate

23.24. The tree is shown in Figure 23.1. The probability on each upper branch is 1/6; the probability on each middle branch is 2/3; the probability on each lower branch is 1/6. The six month bond prices nodes E, F, G, H, I are 100e-0.1442XO.5, 100e-0.1197XO.5, 100e-0.0952XO.5, 100e-0.0707xo.5, and 100e-0.o462 x 0.5 ,respectively. These are 93.04, 94.19, 95.35, 96.53, and 97.71. The payoffs from the option at nodes E, F, G, H, and I are therefore 1.96,0.81, 0, 0, and O. The value at node B is (0.1667 x 1.96 + 0.6667 x 0.81)e-O.l095XO.5 = 0.8380. The value at node C is 0.1667 x 0.81 x e-O.0851XO.5 = 0.1294. The value at node D is zero. The value at node A is

(0.1667 x 0.8380 + 0.6667 x 0.1294)e-o.o750XO.5 = 0.217

The answer given by DerivaGem using the analytic approach is 0.213.

23.25. In the convexity adjustment formula, equation (23.32), h = 10, t2 = 10.25 so that

B(O, tl) = .!.(1 - e-IOa) = 6.3212 a

The convexity adjustment is therefore

or 81 basis points. The futures rate is 8% with quarterly compounding or 7.92% with continuous compounding. The forward rate is, therefore, 7.92-0.81=7.11% per annum with continuous compounding.

97

Figure 23.1 'free for Problem 23.24

23.26. (a) The implied value of u is 1.12%. (b) The value of the American option is 0.593 (c) The implied value of u is 18.49% and the value of the American option is 0.593. The two models give the same answer providing they are both calibrated to the same European price. (d) We get a negative interest rate if there are 10 down moves. The probability of this is 0.16667 x 0.16418 x 0.16172 x 0.15928 x 0.15687 x 0.15448 x 0.15218 x 0.14978 x 0.14747 x 0.14518 = 8.3 x 10-9

( e) The calculation is

0.164179 x 1.705358 x e-O.052571xO.l = 0.278516

23.27. The values of the four European swap options are 1.72, 1.73, 1.30, and 0.65, respectively. The implied Black volatilities are 13.37%, 13.34%, 13.43%, and 13.42%, respectively.

23.28. With 100 time steps the lognormal model gives prices of 5.569, 2.433, and 0.699 for strike prices of 95, 100, and 105. With 100 time steps the normal model gives prices of 5.493,2.511, and 0.890 for the three strike prices respectively. The normal model gives a heavier left tail and a less heavy right tail than the lognormal model for interest

98

rates. This translates into a less heavy left tail and a heavier right tail for bond prices. The arguments in Chapter 15 show that we expect the normal model to give higher option prices for high strike prices and lower option prices for low strike. This is indeed what we find.

23.29. The results are shown in Figure 23.2.

0.4850 0.4800 0.4750 G.4700 0.4650 0.4800 0.4550 0.4500 0.4450 0.4400

Trinomial ConYIKgIIIICII :-Tree I

I:-:-~

0.4350 .J.---..----..------,,------,

0.'700 0.11850

0.9800

o 20

(a)

40 ....... eo

Trtnomhll Convergence I--Treel

80

0"

750

1

;::~--~--~--~-~ o 40

."po

(c)

eo eo

0.8000 0.5800 0.5800 0.5700 0.5800 0.5800 0.5400 0.5300 0.~.J.--___,----~--~--~

o 20 40 . ..,. (b)

80

TI1nomIIII ConVII"". 1--Treel

1.

03S0

1 lE .:~ 1.0050 .J.-----..---.---..,.-----,--.

o 40 . ..,. (d)

80 80


99


Chapter 24: Interest Rate Derivatives: More Advanced Models

24.16. (a) The cumulative variances for one, two, three, and five years are 0.182 x 1 = 0.0324, 0.22 x 2 = 0.08, 0.222 x 3 == 0.1452, and 0.22 x 5 = 0.2, respectively. If the required forward rate volatilities are AI, A2, Aa, and A4 , we must have

A~ = 0.0324

A~ x 1 = 0.08 - 0.0324

A~ x 1 = 0.1452 - 0.08

A~ x 2 = 0.2 - 0.1452

It follows that Al = 0.18, A2 = 0.218, Aa = 0.255, and ~ = 0.166 (b) First we interpolate to obtain the spot volatility for the four-year caplet as 21%. The yield curve is flat at 4.879% with continuous compounding. We use DerivaGem to calculate the prices of caplets with a strike price of 5% where the underlying option matures in one, two, three, four, and five years. The results are 0.3252, 0.4857, 0.6216, 0.6516, and 0.6602, respectively. This means that the prices two-, three-, four-, five-, and six-year caps are 0.3252, 0.8109, 1.4325, 2.0841, and 2.7443. We use DerivaGem again to imply flat volatilities from these prices. The flat volatilities for two-, three-, four-, five-, and six-year caps are 18%, 19.14%, 20.28%, 20.49%, and 20.37%, respectively

24.17. The two types of flexi caps mentioned are more difficult to value than the fiexi cap considered in Section 24.3. There are two reasons for this. (i) They are American-style. (The holder gets to choose whether a caplet is exercised.) This makes the use of Monte Carlo simulation difficult. (ii) They are path dependent. In (a) the decision on whether to exercise a caplet is liable to depend on the number of caplets exercised so far. In (b) the exercise of a caplet is liable to depend on a decision taken some time earlier. In practice, fiexi caps are sometimes valued using a one-factor model of the short rate in conjunction with the techniques described in Section 20.5 for handling pathdependent derivatives. The flexi cap in (b) is worth more than the fiexi cap considered in Section 24.3. This is because the holder of the flexi cap in (b) has all the options of the holder of the flexi cap in the text and more. Similarly the flexi cap in (a) is worth more than the fiexi cap in (b). This is because the holder of the fiexi cap in (a) has all the options of the holder of the fiexi cap in (b) and more. We therefore expect the fiexi cap in (a) to be the most expensive and the fiexi cap considered in section 24.3 to be the least expensive.

100


Chapter 25: Swaps Revisited

25.9. The fixed side consists of four payments of usn 0.9 million. The present value in millions of dollars is

~9 ~9 ~9 ~9 1.05 + 1.052 + 1.053 + 1.054 = 2.85

The forward Australian LIB OR rate is 10% with annual compounding. In equation (25.4) Fi = 0.10, O'G,i = 0.15, O'F,i = 0.25, and Pi = 0.7 so that the quanto adjustment to the forward rate should be

0.1 x 0.7 x 0.15 x 0.25ti = 0.002625ti

The value of the floating payments is therefore

_1_ 1.02625 1.0525 1.07875 _ 3 28 1.05 + 1.052 + 1.053 + 1.054 - .

The value of the swap is 3.28 - 2.85 = 0.43.

25.10. When the CP rate is 6.5% and Treasury rates are 6% with semiannual compounding, the CMT% is 6% and an Excel spreadsheet can be used to show that the price of a 30-year bond with a 6.25% coupon is about 103.46. The spread is zero and the rate paid by P&G is 5.75%. When the CP rate is 7.5% and Treasury rates are 7% with semiannual compounding, the CMT% is 7% and the price of a 30-year bond with a 6.25% coupon is about 90.65. The spread is therefore

max[O, (98.5 x 7/5.78 - 90.65)/100]

or 28.64%. The rate paid by P&G is 33.89%.

25.11. You should be paying fixed and receiving floating. The counterparty will value the floating payments less than you because it does not make a convexity adjustemnt increasing forward rates. The size of the convexity adjustment for a forward rate increases with the forward rate, the forward rate volatility, the time between resets, and the time until the forward rate is observed. We therefore maximize the impact of the convexity adjustment by choosing long swaps involving high interest rate currencies, where the interest rate volatility is high and there is a long time between resets. The convexity adjustment for the payment at time ti is

0.12 X 0.182 X 1 X ti

1.1

101

25.12.

This is 0.000295ti' For a five year LIBOR-in-arrears swap the value of the convexity adjustment is

5 1 0000 000 """ 0.000295 , , L.J IIi

i=l .

or $3137.7. Similarly the value of the convexity adjustments for 10 and 20 year swaps is $8,552.4 and $18,827.5.

(a) Because the LIBOR zero curve is flat at 5% with annual compounding, the five-year swap rate for an annual-pay swap is also 5%. (As explained in Chapter 6 swap rates are par yields.) A swap where 5% is paid and LIBOR is receive would therefore be worth zero. A swap where 6% is paid and LIBOR is received has the same value as an instrument that pays 1% per year. Its value in millions of dollars is therefore

1 1 1 1 1 - 1.05 - 1.052 - 1.053 - 1.054 - 1.055 = -4.33

(b) In this case company X has, in addition to the swap in (a), a European swap option to enter into a two-year swap in three years. The swap gives company X the right to receive 6% and pay LIB OR. We value this in DerivaGem by using the Caps and Swap Options worksheet. We choose Swap Option as the Underling Type, set the Principal to 100, the Settlement Frequency to Annual, the Swap Rate to 6%, and the Volatility to 20%. The Start (Years) is 3 and the End (Years) is 5. The Pricing Model is Black-European. We choose Rec Fixed and do not check the Imply Volatility or Imply Breakeven Rate boxes. All zero rates are 4.879% with continuous compounding. We therefore need only enter 4.879% for one maturity. The value of the swap option is given as 2.18. The value of the swap with the cancelation option is therefore

-4.33 + 2.18 = -2.15

(c) In this case company X has, in addition to the swap in (a), granted an option to the counterparty. The option gives the counterparty the right to pay 6% and receive LIBOR on a two-year swap in three years. We can value this in DerivaGem using the same inputs as in (b) but with the Pay Fixed instead of the Rec Fixed being chosen. The value of the swap option is 0.57. The value of the swap to company X is

-4.33 - 0.57 = -4.90

(d) In this case company X is long the Rec Fixed option and short the Pay Fixed option. The value of the swap is therefore

-4.33 + 2.18 - 0.57 = -2.72

It is certain that one of the two sides will exercise its option to cancel in three years. The swap is therefore to all intents and purposes a three-year swap with no embedded options. Its value can also be calculated as

1 1 1 -1.05 - 1.052 - 1.053 = -2.72

102

The value of the fixed side of the swap in millions of dollars is

4 4 4 4 4 + IT + 1.12 + 1.13 + 1.14 = 16.68

Each forward rate is 10%. The convexity adjustment is from equation (25.2)

0.102

X 0.182

X 1 X ti = 0.OOO2945t. 1 + 0.10 xl'

The value of the floating side of the swap is in millions of dollars

5 5.01472 5.02945 5.05890 - 20 96 + 1.1 + 1.12 + 5.044173 + 1.14 - .

The value of the swap is therefore 16.68 - 20.96 = -4.28

103

Answers to Assignment Questions Chapter 26: Credit Risk

26.14. In this case Eo = 4, O'E = 0.60, D = 15, r = 0.06. Setting up the data in Excel, we can solve equations (23.4) and (23.5) by using Solver to search for the values of Va and O'y that minimize the square of the difference between the left hand side and right hand side in equation (23.4) plus the square of the difference between the left hand side and the right hand side in equation (23.5). The solution to the equations proves to be Va = 17.084 and O'y = 0.1576. The probability of default is N( -d2) or 15.61 %. The market value of the debt is 17.084 - 4 = 13.084. The present value of the promised payment on the debt is 15e-O•06x2 = 13.304. The expected loss on the debt is, therefore, (13.304 - 13.084)/13.304 or 1.65% of its no-default value. The expected recovery rate in the event of default is therefore (15.61 - 1.65)/15.61 or about 89%. The reason the recovery rate is so high is as follows. There is a default if the value of the assets moves from 17.08 to below 15. A value for the assets significantly below 15 is unlikely. Conditional on a default, the expected value of the assets is, therefore, not a huge amount below 15.

26.15. Equation (26.3) gives j-l

G j - B j = LPillij + Pjlljj

i=1

We assume that lljj is always positive (i.e. a bond holder never makes a gain from a terminal default). Because Pj > 0

so that

Also because r'

so that

j-l

G j - B j ~ LPillij

i=1

j-l

B j ~ G j - LPiG:ij

i=1

i=1

j-l

Pj ~ 1- LPi i=1

104

or

This proves the required result. When a 20-year bond is added in Table 26.4, there are 7 bonds. Equation (26.5) can be used to determine a17 = 92.13, a27 = 86.98, a37 = 82.07, a47 = 77.40 aS7 = 72.95, a67 = 53.73, and a77 = 26.98. Using the results in the final column of Table 26.5 it follows that

j-l

LPiaij = 19.85 i=l

Also E{:i Pi = 0.300 so that

j-l

a77(l - LPi) = 18.89 i=l

The price of a risk-free 20 year bond, G7 , is 125.10. It follows that the price of the 20-year corporate bond, B7, must lie between 125.10 - 19.85 = 105.25 and 125.10-19.85 - 18.89 = 86.36. From this it can be calculated that the yield on the 20 year bond must lie between 8.42% and 6.53%.

26.16. In this case 1

au = 1.06 (106 - 0.4 x 106) = 60

1 al2 = 1.06 (106 - 0.4 x 106) = 60

1 al3 = 1.06 (106 - 0.4 x 106) = 60

1 a22 = 1.062 (106 - 0.4 x 106) = 56.60

1 a23 = 1.062 (106 - 0.4 x 106) = 56.60

1 a33 = 1.063 (106 - 0.4 x 106) = 53.40

Also Gl = G2 = G3 = 100 while Bl = 98.8806, B2 = 97.4827, B3 = 95.8467 so that

PI = 100 - 98.8806 = 0.0187 60

= 100 - 97.4827 - 0.0187 x 60 = 00247 P2 56.60 .

= 100 - 95.84 - 0.0187 x 60 - 0.0247 x 56.60 = 00306 P3 53.60 .

105


Chapter 27: Credit Derivatives

27.20. In this case A(ti) = 0.04 and e(~) = 0 for all i. Also, V(tl) = 0.9418, V(t2) = 0.8869, veta) = 0.8353, and V(t4) = 0.7866 while U(tl) = 0.9561, U(t2) = 1.8565, u(ta) = 2.7045, and U(t4) = 3.5031. Also PI = 0.01, 1'2 = 0.015, Pa = 0.02, P4 = 0.025, 1r = 0.93, and 1- R - A(ti)R = 0.792

0.792(0.01 x 0.9418 + 0.015 x 0.8869 + 0.02 x 0.8353 + 0.025 x 0.7866) s=------~----------~--------~~~---~--~~--------~--0.01 x 0.9561 + 0.015 x 1.8565 + 0.02 x 2.7045 + 0.025 x 3.5031 + 0.93 x 3.5031

or 0.0136. The credit default swap spread is 136 basis points. This means that 0.5 x 1.36 = 0.68% of the notional principal would be paid every six months. If the swap were a binary swap we would not multiply by 0.792 in the numerator and the swap spread would be 172 basis points.

27.21. (a) Define Ui as the expected percentage of the no-default value lost from defaults at time i years and Vi as the exposure at time i years. The expected loss from defaults is

In this case UI = 1 - e-O.0020XI = 0.0020

U2 = e-O.0020XI - e-O.0040X2 = 0.0060

Ua = e-O.OO40x2 - e-O.Oo55xa = 0.0084

U4 = e-O.Oo55xa - e-O.0065x4 = 0.0093

U5 = e-O.0065X4 - e-O.0075XS = 0.0111

The parameter Vi is the value of an i year European option to enter into a swap with life 5 - i when fixed is received and floating is paid. From DerivaGem (in millions of dollars), VI = 0.9114, V2 = 0.9852, Va = 0.9060, V4 = 0.5455, and Vs = O. The expected loss from defaults is 0.020. (b) In this case VI = 2.47, V2 = 2.50, Va = 1.74, V4 = 0.91, and Vs = O. The Ui'S are the same. The expected loss from defaults is 0.0424. This is greater than the expected loss from defaults when fixed is received because the term structure is upward sloping. Floating payments are expected to increase. A swap where fixed is received is likely to move out of the money (lowering the credit risk) and a swap where floating is received is likely to move in the money (increasing the credit risk).

106

(c) The value of an annuity paying $1 per year for five years is

e-O.04xl + e-O.048x2 + e-O.053x3 + e-O.055X4 + e-O.056x5 = 4.28

The spread required to compensate for credit risk (in million of dollars) is therefore (0.020 + 0.0424)/4.28 = 0.0149 per year. This is about 1.5 basis points. (d) The payment made on the day of the default depends on the one-year LIBOR rate one year earlier. A swap option that includes this payment is therefore path dependent. It cannot be evaluated in a straightforward way. Monte Carlo simulation must be used.

27.22. Real world default probabilities are the true probabilities of defaults. They can be estimated from historical data. Risk-neutral default probabilities are the probabilities of defaults in a world where all market participants are risk neutral. They can be estimated from bond prices. Risk-neutral default probabilities are higher. This means that returns in the risk-neutral world are lower. From Table 26.8 the probability of a company moving from A to BBB or lower in one year is 5.92%. An estimate of the value of the derivative is therefore 0.0592 x 100 x e-O.05Xl = 5.63. The approximation in this is that we are using the real-world probability of a downgrade. To value the derivative correctly we should use the risk-neutral probability of a downgrade. Since the risk-neutral probability of a default is higher than the real-world probability, it seems likely that the same is true of a downgrade. This means that 5.63 is likely to be too low as an estimate of the value of the derivative.

27.23. In this case 0' = 0.25, bt = 1, r = 0.05 and the tree parameters are u = 1.2840, d = 0.7788, a = 1.0513, p = 0.5394, and 1 - p = 0.4606. The tree is shown in Figure 27.1. The upper number at each node is the stock price, the second number is the equity component of the value, the third number is the debt component of the value, and the fourth number is the total value. The convertible is valued at each node immediately after the coupon payment (a) The instrument cannot be converted at the final nodes. The value at each of these nodes is therefore 100. At the two-year point the instrument is worth (100+5)e-o.o7 = 97.90 if it is not converted. The instrument should be converted at node D since conversion will improve the value to 41.22 x 3.5 = 144.27. At node C the value of the instrument, if not converted, is (97.90 + 5)e-O.7 = 95.94. Because 19.47x3.6 is less than this, the instrument should not be converted at node C. AtnodeB, assuming no conversion, the equity value is 144.27xO.5394xe-o.o5 = 74.02. The debt value is

(97.90 x 0.4606 + 5)e-O•07 = 46.71

The total value is 120.73. If converted the instrument is worth 32.10 x 3.6 = ll5.56. The instrument should not therefore be converted. At node A, the equity component of the value is 74.02 x 0.5394 x e-O•05 = 37.98. The debt component is

(46.71 x 0.5394 + 95.94 x 0.4606 + 5)e-O.07 = 69.36

107

The value of the instrument is therefore 107.34. (b) Without the conversion option the instrument is worth (95.94 + 5)e-O•07 = 94.12. The value of the conversion option is, therefore, 107.34 - 94.12 = 13.22. (C) The bond would be called at node A. It would then be converted and the value at node A would be 115.56. The value of the equity components at node A would then be 115.56 x 0.5394 x e-O•05 = 59.29. The value of the debt component would be (95.94 x 0.4606 + 5)e-O•07 = 45.86. The total value of the instrument would be reduced to 105.15. The value of the conversion option (net of the call option) would be reduced to 105.15 - 94.12 = 11.03. (d) The tree would be constructed to reflect the dividends. This would involve subtracting the present value of the dividends from the initial stock price, constructing the tree in the usual way, and then adding in the present value of future dividends at each node. When calculating a value-if-converted number at a node, the present value of the dividend six months later would be taken into account.

25.00 37.98 69.36

107.34


52.92 o

100.00 100.00

32.10 o

100.00 100.00

19.47 o

100.00 100.00

11.81 o

100.00 100.00

27.24. We can value this using Black's model. The payoff for each basis point by which the swap spread exceeds 100 basis points is the value of an annuity where payments of 0.5 x 0.0001 x 100,000,000 = $5000 are made every six months for a five-year period with the first payment at time 1.5 years. The present value of this annuity is

5000(e-O.05X1.5 + e-O.05x2 + ... + e-O.o5xe = 41,558

108

The value of the instrument is therefore

41, 588[FoN(d1 ) - KN(d:a)J

where

d In(Fo/ K) + (12/2

1 = (1vT

d In(Fo/ K) - (12/2

2 = (1vT

and Fo = 90, K = 100, (1 = 0.15, and T = 1. This is

41,588 x 2.0217 = 84,078

The instrument ceases to exist is there is a default in the first year. We therefore multiply this valuation by the probability of no default in the first year to get a final valuation of 0.985 x 84, 078 = $82,817

109


Chapter 28: Real Options

28.9. In this case a = 0.05 and u = 0.15. We first define a variable X that follows the process

dX = -adt + udz

A tree for X constructed in the way described in Chapter 23 is shown in Figure 28.1. We now displace nodes so that the tree models In S in a risk-neutral world where S is the price of wheat. The displacements are chosen so that the initial price of wheat is 250 cents and the expected prices at the ends of the first and second time steps are 260 and 270 cents, respectively. Suppose that the displacement to give In S at the end of the first time step is Ql. Then

0.1667eQl +0.1837 + 0.6666eQl + 0.1667eQI-0.1837 = 260

so that Ql = 5.5551 The probabilities of nodes E, F, G, H, and I being reached are 0.0257, 0.2221, 0.5043, 0.2221, and 0.0257, respectively. Suppose that the displacement to give In S at the end of the second step is Q2. Then

so that Q2 = 5.5874. This leads to the tree for the price of wheat shown in Figure 28.2. Using risk-neutral valuation the value of the project (in thousands of dollars) is

-10 - 90e-O.05xO.5 + 2.70 x 40e-O.05xl = 4.94

This shows that the project is worth undertaking. Figure 28.3 shows the value of the project on a tree. The project should be abandoned at node D for a saving of 2.41. Figure 28.4 shows that the abandonment option is worth 0.39.

28.10. Figure 28.5 shows what Figure 28.4 in the text becomes if the abandonment options costs $3 million. The value of the abandonment option reduces from 1.94 to 1.21. Similarly Figure 28.6 below shows what Figure 28.5 in the text becomes if the expansion option costs $5 million. The value of the expansion option reduces from 1.06 to 0.40.

110

A 0.1867 0.6666 0.1867

B 0.1545 0.6660 0.1795

c 0.1867 0.6666 0.1867

F 0.1837

o 0.1795 0.6660 0.1545

Figure 28.1 Tree for X in Problem 28.9

A 0.1867 0.6666 0.1867

B 0.1545 0.6660 0.1795

c 0.1867 0.6666 0.1667

o 0.1795 0.6660 0.1545

Figure 28.2 Thee for price of wheat in Problem 28.9

111

Node A 0.1667 0.6666 0.1667

B 0.1545 0.6660 0.1795

c 0.1667 0.6666 0.1667

F 128.36

o 0.1795 0.6660 0.1545

Figure 28.3 Tree for value of project in Problem 28.9

Node A 0.1667 0.6666 0.1667

B 0.1545 0.6660 0.1795

c 0.1667 0.6666 0.1667

o 0.1795 0.6660 0.1545

Figure 28.4 Tree for abandonment option in Problem 28.10

112

Figure 28.5 Tree for expansion option in Problem 28.10

Figure 28.6 Tree for abandonment option in Problem 28.9

113


Chapter 29: Insurance, Weather, and Energy Derivatives

29.11 (a) The losses in mi1lions of dollars are approximately

<1>(150,50)

The reinsurance contract would payout 60% of the losses. The payout from the reinsurance contract is therefore

<1>(90,30)

The cost of the reinsurance is the expected layout in a risk-neutral world, discounted at the risk-free rate. In this case, the expected layout is the same in a risk-neutral world as it is in the real world. The value of the reinsurance contract in millions of dollars is therefore

90e-O.05Xl = 85.61

(b) The probability that losses will be greater than $200 million is the probability that a normally distributed variable is greater than one standard deviation above the mean. This is 0.1587. The expected payoff in millions of dollars is therefore 0.1587 x 100 = 15.87 and the value of the contract is

15.87e-o.05Xl = 15.10

114

Test Bank Questions

Pages 121 to 141 contain test bank questions for Chapters 1 to 18. Some questions are multiple choice; others have numerical answers that can easily be computed using a calcula.tor. In each of the seventeen tests, there are a total of ten answers that can be quickly graded as correct or incorrect.

The questions can be used at the beginning of a class to provide a quick test of whether students understand the material covered in the previous class. Alternatively, they can be used to provide instructors with ideas for midterm or final exam questions. The answers are on pages 142 to 143.

120

CHAPTER 1 Test Bank

1. List three types of traders in futures, forward, and options markets (i)

(ii) (iii)

2. Which of the following is not true (circle one) (a) When a OBOE option on IBM is exercised, IBM issues more stock (b) An American option can be exercised at any time during its life (c) An call option will always be exercised at maturity if the underlying asset price

is greater than the strike price (d) A put option will always be exercised at maturity if the strike price is greater

than the underlying asset price.

3. A trader enters into a one-year short forward contract to sell an asset for $60 when the spot price is $58. The spot price in one year proves to be $63. What is the traders

. I? goon or oss. . ..

4. A trader buys 100 European call options with a strike price of $20 and a time to maturity of one year. The cost of each option is $2. The price of the underlying asset proves to be $25 in one year. What is the trader's gain or loss? ...

5. A trader sells 100 European put options with a strike price of $50 and a time to maturity of six months. The cost of each option is $4. The price of the underlying asset is $41 in one year. What is the trader's gain or loss? ...

6. The price of a stock is $36 and the price of a three-month call option on the stock with a strike price of $36 is $3.60. Suppose a trader has $3,600 to invest and is trying to choose between buying 1,000 options and 100 stock. How high does the stock price have to rise for an investment in options to be as profitable as an investment in the stock? ...

7. A one-year call option on a stock with a strike price of $30 costs $3; a one-year put option on the stock with a strike price of $30 costs $4. Suppose that a trader buys two call options and one put option. (i) What is the breakeven stock price, above which the trader makes a profit?

(ii) What is the breakeven stock price below which the trader makes a profit?

121

CHAPTER 2 Test Bank

1. Which of the following is true (circle one) (a) Both forward and futures contracts are traded on exchanges. (b) Forward contracts are traded on exchanges, but futures contracts are not. (c) Futures contracts are traded on exchanges, but forward contracts are not. (d) Neither futures contracts nor forward contracts are traded on exchanges.

2. Which of the following is not true (circle one) (a) Futures contracts nearly always last longer than forward (b) Futures contracts are standardized; forward contracts are not. (c) Delivery or final cash settlement usually takes place with forward contracts; the

same is not true of futures contracts. (d) Forward contract usually have one specified delivery date; futures contract often

have a range of delivery dates.

3. In the corn futures contract a number of different types of corn can be delivered (with price adjustments specified by the exchange) and there are a number of different delivery locations. Which of the following is true (circle one) (a) This flexibility tends increase the futures price. (b) This flexibility tends decrease the futures price. (c) This flexibility may increase and may decrease the futures price. (d) This has no effect on the futures price

4. A company enters into a short futures contract to sell 50,000 pounds of cotton for 70 cents per pound. The initial margin is $4,000 and the maintenance margin is $3,000. What price of cotton futures will trigger a margin call? ...

5. A company enters into a long futures contract to buy 1,000 barrels of oil for $20 per barrel. The initial margin is $6,000 and the maintenance margin is $4,000. What price of oil futures will allow $2,000 to be withdrawn from the margin account? ...

6. On the floor of a futures exchange one futures contract is traded where both the long and short parties are closing out existing positions. What is the resultant change in the open interest? ...

122

7. Who determines when delivery will take place in a futures contract (circle one) (a) The party with the long position (b) The party with the short position (c) Either party can specify a delivery date (d) The exchange specifies the exact delivery date.

8. Keynes and Hicks argue ( circle one) (a) Speculators on average make profits from futures (b) Hedgers on average make profits from futures (c) Speculators on average make losses from futures (d) Neither hedgers nor speculators make profits from futures on average.

9. A hedger takes a long position in an oil futures contract on November 1, 1999 to hedge an exposure on March 1, 2000. The initial futures price is $20. On December

. 31, 1999 the futures price is $21. On March 1, 2000 it is $24. The contract is closed out on March 1, 2000. What gain is recognized in the accounting year January 1 to December 31, 2oo0? Each contract is on 1000 barrels of oil. . ..

10. What is your answer to question 9 if the trader is a speculator rather than a hedger?

123

CHAPTER 3

Test Bank

1. An interest rate is 15% per annum when expressed with annual compounding. What is the equivalent rate with continuous compounding? ...

2. An interest rate is 8% per annum when expressed with continuous compounding. What is the equivalent rate with semiannual compounding? ...

3. An interest rate is 12% when expressed with quarterly compounding. What is the equivalent rate with semiannual compounding? ...

4. An investor shorts 100 shares when the share price is $50 and closes out the position six months later when the share price is $43. The shares pay a dividend of $3 per share during the six months. What is the investor's profit or loss? ...

5. The spot price of an investment asset that provides no income is $30 and the risk-free rate for all maturities (with continuous compounding) is 10%. What is the three-year forward price? ...

6. Repeat question 5 on the assumption that the asset provides an income of $2 at the end of the first year and at the end of the second year. . ..

7. An exchange rate is 0.7000 and the six-month domestic and foreign risk-free interest rates are 5% and 7% (both expressed with continuous compounding). What is the six-month forward rate? ...

8. A short forward contract that was negotiated some time ago will expire in three months and has a delivery price of $40. The current forward price for three-month forward contract is $42. The three month risk-free interest rate (with continuous compounding) is 8%. What is the value of the short forward contract? ...

9. The spot price of an asset is positively correlated with the market. Which of the following would you expect to be true (circle one) (a) The forward price equals the expected future spot price (b) The forward price is greater than the expected future spot price. (c) The forward price is less than the expected future spot price. (d) The forward price is sometimes greater and sometimes less than the expected

future spot price.

10. The one-year Canadian dollar forward exchange rate is quoted as 1.4000. What is the corresponding futures quote? ...

124

CHAPTER 4 Test Bank

1. The basis strengthens unexpectedly. Which of the following is true (circle one) (a) a short hedger's position improves. (b) a short hedger's position worsens. (c) a short hedger's position sometimes worsens and sometimes improves. (d) a short hedger's position stays the same.

2. On March 1 the price of oil is $20 and the July futures price is $19. On June 1 the price of oil is $24 and the July futures price is $23.50. A company entered into a futures contracts on March 1 to hedge the purchase of oil on June 1. It closed out its position on June 1. What is the effective price paid by the company for the oil?

3. On March 1 the price of gold is $300 and the December futures price is $315. On November 1 the price of gold is $280 and the December futures price is $281. A gold producer entered into a December futures contracts on March 1 to hedge the sale of gold on November 1. It closed out its position on November 1. What is the effective price received by the producer for the gold? ...

4. Suppose that the standard deviation of monthly changes in the price of commodity A is $2. The standard deviation of monthly changes in a futures price for a contract on commodity B (which is similar to commodity A) is $3. The correlation between the futures price and the commodity price is 0.9. What hedge ratio should be used when hedging a one month exposure to the price of commodity A? ...

5. A company has a $36 million portfolio with a beta of 1.2. The S&P index is currently standing at 900. Futures contracts on $250 times the index can be traded. What trade is necessary to achieve the following. (Indicate the number of contracts that should be traded and whether the position is long or short.) (i) Eliminate all systematic risk in the portfolio

(ii) Reduce the beta to 0.9 ... (iii) Increase beta to 1.8 ...

6. A futures contracts trade with all delivery months. A compnay is hedging the purchase of the underlying asset on June 15. Which futures contract should it use (circle one) (a) The June contract (b) The July contract ( c) The May contract (d) The August contract

125

7. Which of the following is true (circle one) (a) The optimal hedge ratio is the slope of the best fit line when the spot price (on

the y-axis) is regressed against the futures price (on the x-axis). (b) The optimal hedge ratio is the slope of the best fit line when the futures price

(on the y-axis) is regressed against the spot price (on the x-axis). (c) The optimal hedge ratio is the slope of the best fit line when the change in the

spot price (on the y-axis) is regressed against the change in the futures price (on the x-axis).

(d) The optimal hedge ratio is the slope of the best fit line when the change in the futures price (on the y-axis) is regressed against the change in the spot price (on the x-axis).

8. Metallgesellschaft ran into problems because (circle one) (a) The price of oil rose (b) Margin calls created short term cash flow problems (c) Rolling hedges forward is a very dangerous strategy (d) All of the above

126

CHAPTER 5 Test Bank

1. The three-year zero rate is 7% and the four-year zero rate is 7.5% (both continuously compounded. What is the forward rate for the fourth year? ...

2. The six-month zero rate is 8% with semiannual compounding. The price of a oneyear bond that provides a coupon of 6% per annum semiannually is 97. What is the one-year continuously compounded zero rate? ...

3. The yield curve is flat at 6% per annum with semiannual compounding. What is the value of an FRA where the holder receives interest at the rate of 8% per annum for a six-month period on a principal of $1,000 starting in two years? ...

4. Under liquidity preference theory, which of the following is always true (circle one) (a) The forward rate for a certain maturity is higher than the spot rate with that

maturity. (b) Forward rates are unbiased predictors of expected future spot rates. (c) The spot rate for a certain maturity is higher than the par yield for that maturity. (d) Forward rates are higher than expected future spot rates.

5. It is May 1. The quoted price of a 12% Treasury bond with a face value of 100 that pays coupons on April 1 and October 1 is 105. What is the cash price? ...

6. What difference does it make to your answer to question 5 if the bond is a corporate bond? ...

7. The quoted futures price is 103.5. Which of the following four bonds is cheapest to deliver (circle one) (a) Quoted price = 110; conversion factor = 1.0400. (b) Quoted price = 160; conversion factor = 1.5200. (c) Quoted price =131; conversion factor = 1.2500. (d) Quoted price = 143; conversion factor = 1.3500.

8. A trader enters into a long position in one Eurodollar futures contract. How much does the trader gain or lose when the futures price quote increases by 6 basis points.

9. The six-month zero rate with continuous compounding is 5%. The Eurodollar futures quote for a contract maturing in six months is 94. Estimate the nine month zero rate with continuous compounding. (Ignore convexity adjustments and day count issues.)

127

10. A company invests $1,000 in a five-year zero-coupon bond and $4,000 in a ten-year zero-coupon bond. What is the duration of the portfolio? ...

128

CHAPTER 6 Test Bank

1. Suppose that the yield curve is flat at 5% per annum with continuous compounding. A swap with a notional principal of $100 million in which 6% is received and sixmonth LIBOR is paid will last another 15 months. Payments are exchanged every six months. The six-month LIBOR rate at the last reset date (three months ago) was 7%.

(i) What is the value of the fixed-rate bond underlying the swap? ...

(ii) What is the value of the floating-rate bond underlying the swap? ...

(iii) What is the value of the payment that will be exchanged in 3 months?

(iv) What is the value of the payment that will be exchanged in 9 months?

(v) What is the value of the payment that will be exchanged in 15 months?

(vi) What is the value of the swap? ...

2. A company can invest funds for five years at 6% per annum with semiannual compounding. The five-year swap rate is 6.3%. What floating rate of interest can the company earn? ...

3. Which of the following is true (circle one) ( a) Principals are not usually exchanged in a currency swap (b) The principal amounts usually flow in the opposite direction to interest payments

at the beginning of a currency swap and in the same direction as interest payments at the end of the swap.

(c) The principal amounts usually flow in the same direction as interest payments at the beginning of a currency swap and in the opposite direction to interest payments at the end of the swap.

(d) Principals are not usually specified in a currency swap

4. Suppose you enter into an interest rate swap where you are receiving floating and paying fixed. Which of the following is true. ( circle two) (a) Your credit risk is greater when the term structure is upward sloping than when

it is downward sloping. (b) Your credit risk is greater when the term structure is downward sloping than

when it is upward sloping. ( c) Your credit risk exposure increases when interest rates decline unexpectedly. (d) Your credit risk exposure increases when interest rates increase unexpectedly.

129

CHAPTER 7 Test Bank

1. Consider an exchange traded put option to sell 100 shares for $20. Give (a) the strike price and (b) the number of shares that can be sold after (i) A 5 for 1 stock split (a)... (b) .. .

(ii) A 25% stock dividend (a)... (b) .. .

(iii) A $5 cash dividend (a)... (b) ...

2. A trader writes two naked put option contracts. The option price is $3, the strike price is $ 40 and the stock price is $42. What is the initial margin? ...

3. Which of the following lead to mM issuing more shares (circle three) ( a) Some executive stock options are exercised (b) Some exchange-traded put options are exercised (c) Some exchange-traded call options are exercised (d) Some warrants on IBM are exercised (e) Some of ffiM's convertible debt is converted to equity.

130

CHAPTER 8 Test Bank

1. Which of the following are always positively related to the price of a European call option on a stock ( circle three) (a) The stock price (b) The strike price (c) The time to expiration (d) The volatility (e) The risk-free rate (f) The magnitude of dividends anticipated during the life of the option

2. What is the lower bound for the price of a two-year European call option on a stock when the stock price is $20, the strike price is $15, and the risk-free interest rate is 5% and there are no dividends? ...

3. How does your answer to question 2 change if the option is American?

4. What is the lower bound for the price of a six-month European put option on a stock when the stock price is $40, the strike price is $46 and the risk-free interest rate is 6%? ...

5. How does your answer to question 4 change if the option is American?

6. The price of a European call option on a non-dividend-paying stock with a strike price of $50 is $6. The stock price is $51, the risk-free rate (all maturities) is 6% and the time to maturity is one year. What is the price of a one-year European put option on the stock with a strike price of $50? ...

7. What difference does it make to your answer to question 6 if a dividend of $1 is expected in six months? ...

8. A call and a put on a stock have the same strike price and time to maturity. At 1O:ooam on a certain day, the price of the call is $3 and the price of the put is $4. At 10:01am news reaches the market that has no affect on the stock price, but increases its volatility. As a result the price of the call changes to $4.50. What would you expect the price of the put to change to? ...

131

CHAPTER 9 Test Bank

1. Six-month call options with strike prices of $35 and $40 cost $6 and $4, respectively. (i) What is the maximum gain when a bull spread is created from the calls?

(ii) What is the maximum loss when a bull spread is created from the calls? ...

(iii) What is the maximum gain when a bear spread is created from the calls?

(iv) What is the maximum loss when a bear spread is created from the calls?

2. Three-month European put options with strike prices of $50, $55, and $60 cost $2, $4, and $7, respectively. (i) What is the maximum gain when a butterfly spread is created from the put

options?

(ii) What is the maximum loss when a butterfly spread is created from the put options? ...

(iii) For what two values of ST does the holder of the butterfly spread breakeven with a profit of zero, where ST is the stock price in three months? . .. and ...

3. A three-month call with a strike price of $25 costs $2. A three-month put with a strike price of $20 and costs $3. A trader uses the options to create a strangle. For what two values of ST does the trader breakeven with a profit of zero, where ST is the stock price in three months? ... and ...

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CHAPTER 10 Test Bank

1. Consider a six month put option on a stock with a strike price of $32. The current stock price is $30 and over the next six months it is expected to rise to $36 or fall to $27. The risk-free interest rate is 6%. (i) What is the risk-neutral probability of the stock rising to $36? ...

(ii) What position in the stock is necessary to hedge a long position in 1 put option?

(iii) What is the value of the put option? ...

2. A power option pays off max[(8T - X), 0]2 at time T where 8T is the stock price at time T and X is the strike price. Consider the situation where X = 26 and T is one year. The stock price is currently $24 and at the end of one year it will be either $30 or $18. The risk-free interest rate is 5% (i) What is the risk-neutral probability of the stock rising to $30? ...

(ii) What position in the stock is necessary to hedge a short position in one power t · ? op Ion. . ..

(iii) What is the value of the power option? ...

3. A stock price is currently $100. Over each of the next two three-month periods it is expected to increase by 10% or fall by 10%. Consider a six-month European put option with a strike price of $95. The risk-free interest rate is 8% per annum (i) What is the risk-neutral probability of a 10% rise in each quarter?

(ii) What is the value of the option? ...

(iii) What is the value of the option if it is American?

(iv) What is the value of the option if it is a call rather than a put?

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1. A variable, x, starts at 10 and follows a generalized Wiener process

dS = adt+ bdz

where a = 2, b = 3, and dz is a Wiener process. (i) What is the mean value of the variable after three years?

(ii) What is the standard deviation of the value of the variable after three years?

(iii) What is the mean value of the variable after six months? ...

(iv) What is the standard deviation of the value of the variable after six months?

2. A variable, x, starts at 10 and follows a generalized Wiener process

dS=adt+bdz

During the first two years a = 4 and b = 3. During the following three years a = 6 and b = 4. (i) What is the mean value of the variable at the end of the five years? ...

(ii) What is the standard deviation of the variable at the end of the five years?

3. A stock price has a expected return of 12% per annum and a volatility of 25% per annum. Currently the stock price is $40. Assume 252 days per day. (i) What is the standard deviation of the stock price at the end of one day? ...

(ii) What is the width of the 95% confidence interval for the stock price at the end of one day? ...

4. The process followed by a stock price is

dS = J,tSdt + q Sdz

(i) What is the coefficient of dt in the process followed by Q = S3? Express your answer in terms of Q ...

(ii) What is the coefficient of dz in the process followed by Q = S3? Express your answer in terms of Q ...

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CHAPTER 12 . Test Bank

1. A stock price is $30, the expected return is 18% per annum and the volatility is 20% per annum. (i) What is the mean of the logarithm of the stock price in two years? ...

(ii) What is the standard deviation of the logarithm of the stock price in two years?

(iii) What are the lower and upper 95% confidence limits for the logarithm of stock . . t ? d PrIce In wo years. . .. an ...

(iv) What are the lower and upper 95% confidence limits for the stock price in two years? . .. and ...

2. For a call option on a non-dividend-paying stock, the stock price is $30, the strike price is $29, the risk-free interest rate is 6% per annum, the volatility is 20% per annum and the time to maturity is three months. Expressed in terms of the cumulative normal function, N(x), (i) What is the price of the option? ...

(Ii) What is the price of the option if it is a put?

(iii) What is the price of the option if a dividend of $2 is expected in two months?

3. In a six month American call option a dividend is expected at the end of five months. The strike price is $30 and the risk-free interest rate is 10%. How high does the dividend have to be for there to be some chance of early exercise? ...

4. A stock price is observed weekly with Sj being the ith observation. Define Ui = In(Si/Si-l)' Suppose that there are 40 observations on the u, and EUi = 0.18 while E u? = 0.06. Estimate the stock price volatility per annum. . ..

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1. Consider a European put option on a index. The index level is 1,000, the strike price is 1050, the time to maturity is six months, the risk-free rate is 4% per annum, and the dividend yield on the index is 2% per annum. What is a lower bound to the option

. ? prIce. . ..

2. Consider a European call option on a currency. The exchange rate is 1.0000, the strike price is 0.9100, the time to maturity is one year, the domestic risk-free rate is 5% per annum, and the foreign risk-free rate is 3% per annum. What is a lower bound to the

t ' .? op Ion prIce. . ..

3. Suppose that a futures price is $20, the strike price of a nine-month European calIon the futures is $19, the risk-free rate is 8% per annum, and the value of the option is $3. What is the value of a put option with the same strike price and time to maturity?

4. An exchange rate is currently 0.8. It is expected to move up to 0.84 or down to 0.76 in the next three months. The risk-free interest rates in the domestic and foreign currency are 4% and 6%. (i) What is the probability of an up movement in a risk-neutral world? ...

(ii) What is the value of a three month call option with a strike price of 0.821

5. A futures price is currently 40 cents. It is expected to move up to 44 cents or down to 34 cents in the next six months. The risk-free interest rates is 6%. (i) What is the probability of an up movement in a risk-neutral world? ...

(ii) What is the value of a six-month put option with a strike price of 37 cents?

6. An index is currently 1,000, its volatility is 20% per annum, the risk-free rate is 4% per annum, and the dividend yield on the index is 2% per annum. Expressing your answers in terms of the cumulative normal distribution, N(x), (i) What is the value of a one-year call option with a strike price of 950?

(ii) What is the value of a one-year put option with a strike price of 950?

7. What is the value of a European call option on a futures contract when the futures price is $20, the strike price is $20, the risk-free interest rate is 10% per annum, the volatility is 40% per annum, and the time to maturity is one year. Express your answer in terms of a single cumulative normal distribution function, N(x). . ..

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1. A trader uses a stop-loss strategy to hedge a short position in a three-month call option with a strike price of 0.7000 on an exchange rate. The trader covers the option when the exchange rate is 0.7005 and assumes a naked position when the exchange rate is 0.6995. The value of the option is 0.1. Estimate the expected number of times the trader covers the position. . ..

2. A trader in the United States has a portfolio of derivatives on the Australian dollar with a delta of 456. The USD and AUD risk free interest rates are 5% and 8%. (i) What position in the Australian dollar creates a delta-neutral position? ...

(ii) What position in a one-year futures contract on the Australian dollar creates a delta-neutral position? ...

(iii) What position in a one-year forward contract on the Australian dollar creates a delta-neutral position? ...

3. A portfolio of derivatives on a stock has a delta of 2400 and a gamma of -100. (i) What position in the stock would create a delta-neutral portfolio? ...

(ii) An option on the stock with a delta of 0.6 and a gamma of 0.04 can be traded. What position in the option and the stock creates a portfolio that is both gamma and delta neutral? ...

4. A portfolio of derivatives on an asset is worth $10,000 and the risk-free interest rate is 5%. The delta and gamma of the portfolio is zero. What is the theta? ...

5. The delta of a European call option on a non-dividend-paying stock is 0.6, its gamma is 0.04 and its vega is 0.1 (i) What is the delta of a European put option with the same strike price and time

to maturity as the call option? ...

(ii) What is the gamma of a European put option with the same strike price and time to maturity as the call option? ...

(iii) What is the vega of a European put option with the same strike price and time to maturity as the call option? ...

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1. In a volatility smile diagram (i) What is plotted on the horizontal axis?

(ii) What is plotted on the vertical axis? ...

2. Indicate whether the tails mentioned below are heavier or less heavy than the tails for a lognormal distribution based on the at-the-money volatility. (i) Left tail of foreign currency implied distribution ...

(ii) Right tail of foreign currency implied distribution (iii) Left tail of stock index implied distribution ... (iv) Right tail of stock index implied distribution

3. Which of the following is true (circle one) (a) Volatility smile for European puts is the same as that for European calls (b) Volatility smile for European puts is the same as that for American puts (c) Volatility smile for European calls is the same as that for American calls (d) Volatility smile for American puts is the same as that for American calls

4. A volatility smile such as that seen for foreign currency options can be caused by ( circle two) (a) The fact that currencies are traded in different countries at different times of the

day (b) The fact that volatility is not constant (c) The fact that the activities of central banks causes occasional jumps in the ex

change rate (d) The fact that interest rates may be different in the two countries

5. Which of the following is true (circle one) (a) The volatility skew for equities is much more pronounced now than it was in 1985. (b) The volatility skew for equities was much more pronounced in 1985 than it is

now. (c) The volatility skew for equities is consistent with the Black-Scholes model. (d) The volatility skew for equities is similar to that for foreign currencies.

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1. The volatility of an asset is 25% per year and there are 252 days per year. (i) What is the volatility per day? ...

(ii) What is the volatility per 10 days? ...

2. Stock A has a daily volatility of 1.2% and stock B has a daily volatility of 1.8%. The correlation between the two stock price returns is 0.2. (i) What is the 97.5% 5-day VaR for a 1 million dollar investment in stock A? .. .

(ii) What is the 97.5% 5-day VaR for a 1 million dollar investment in stock B? .. .

(iii) What is the 97.5% 5-day VaR for a 1 million dollar investment in stock A and a 1 million dollar investment in stock B? ...

(iv) What is the benefit of diversification to the 97.5% 5-day VaR?

3. Consider a position in a single option on a stock. The position has a delta and gamma of 12 and 4, respectively. The stock price is 10. What is the model relating the change in the portfolio value, oP, to the return on the stock in one day, OX, when (i) Only delta is used. . ..

(ii) Delta and gamma are used. . ..

4. A bond portfolio worth 5 million has a modified duration of 6 years and the yield curve is flat at 7%. Assume only parallel shifts in rates can happen. (i) What.is the standard deviation of the change in the value of the bond portfolio

in one day if the standard deviation of the change in interest rates in one day is 0.1%? ...

(ii) What is the standard of the change in the value of the bond portfolio in one day if the volatility of interest rates is 1 % per day? ...

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CHAPTER 17

Test Bank

1. The estimate of the volatility of an asset made at the end of Tuesday is 1% per day. On Wednesday the return realized on the asset is -1.5%. Update the volatility estimate (i) Use the EWMA model with A = 0.9 ...

(ii) Use GARCH (1,1) with w = 0.000003, a = 0.05 and f3 = 0.94 ...

2. At the end of Thursday, the volatility of asset A is 2% per day and the volatility of asset B is 1% per day. Also the covariance between the assets is 0.0001. During Friday asset A produces a return of 3% and asset B produces a return of zero. An EWMA model with A = 0.9 is used. (i) What is the estimate of the correlation between the asset returns at the end of

Thursday? ...

(ii) What is the estimate of the volatility of asset A at the end of Friday?

(iii) What is the estimate of the volatility of asset B at the end of Friday?

(iv) What is an estimate of the correlation between asset A and asset B at the end of Friday? ...

3. The parameters in a GARCH (1,1) model are w = 0.000002, a = 0.04, and f3 = 0.95. The current volatility level is 1 % per day. (i) What is the long run average variance rate? ...

(ii) What is the expected variance in 20 days?

(iii) What is the expected variance in 50 days?

(iv) What is the expected variance in 20 days if the EWMA model is used with A = 0.95? ...

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1. An exchange rate has a volatility of 12%. The domestic and foreign risk-free interest rates are 6% and 4%, respectively. The time step on a binomial tree is three months. (i) What are the p, u and d parameters for a Cox, Ross, and Rubinstein tree?

p= ... u= ... d= ...

(ii) What are the u and d parameters for the alternative tree where probabilities are always 0.5?

u= ... d= ...

2. A stock price is initially $30. A dividend of $2 is expected in 1.5 months. The volatility of the stock price is 30% and the risk-free rate is 8%. Suppose the procedure described in the text for constructing a two-step binomial tree, with each step being one month, is used. What are the stock prices at the three final nodes?

... and and ...

3. Which of the following cannot be calculated directly from a binomial tree (Circle two) (a) vega (b) delta (c) rho (d) gamma (e) theta

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Test Bank Answers

Chapter 1 1. hedgers, speculators, arbitrageurs (c); 2. (a); 3. $3 loss; 4. $300 gain; 5. $500 loss; 6. $40; 7. (i) $35; (ii) $20.

Chapter 2 1. (c); 2. (a); 3. (b); 4. 72 cents; 5. $22; 6. down 1; 7. (b); 8. (a); 9. $4,000; 10. $3,000.

Chapter 3 1. 13.98%; 2. 8.16%; 3. 12.18%; 4. Gain = $400; 5. $40.50; 6. $35.84; 7. 0.6930; 8. -$1.96; 9. (c); 10. 0.7143.

Chapter 4 1. (a); 2. $19.50; 3. $314; 4. 0.6; 5. (i) 192 short; 48 short; 96 long; 6. (b); 7. (c); 8. (b).

Chapter 5 1. 9%; 2. 9.02%; 3. $8.61; 4. (d); 5. 105.98; 6. 106.00; 7. (c); 8. Gain = $150; 9. 5.32%; 10. 9 years.

Chapter 6 1. (i) 102.61; (ii) 102.21; (iii) -0.49; (iv) 0.45; (v) 0.44; (vi) 0.40; 2. LIBOR-0.3%; 3. (b); 4. (a) and (d).

Chapter 7 1. (i) $4 and 500; (ii) $16 and 125; (iii) $20 and 100; 2. $1,880; 3. (a), (d) and (e).

Chapter 8 1. (a), (d), and (e)i 2. $6.43; 3. $6.43 (no change); 4. $4.64; 5. $6; 6. $2.09; 7. $3.06; 8. $5.50.

Chapter 9 1. (i) $3; (ii) $2; (iii) $2; (iv) $3; 2. (i) $4; (ii) $1; (iii) $51 and $59; 3. $15 and $30.

Chapter 10 1. (i) 0.435; (ii) Long position in 0.555 shares; (iii) 2.74; 2. (i) 0.603; (ii) Long position in 1.333 shares; (iii) 9.17; 3. (i) 0.601; (ii) 2.14; iii) 2.14; (iv) 10.87.

Chapter 11 1. (i) 12; (ii) 5.20; (iii) 11; (iv) 2.12; 2. (i) 36; (ii) 8.12; 3. (i) 0.63; (ii) 2.47; 4. (i) (31' + ( 2 )Q; (ii) 3uQ.

Chapter 12 1. (i) 3.721; (ii) 0.283; (iii) 3.166 and 4.276 (iv) 23.71 and 71.95; 2. (i) 30N(O.5390) -28.57N(0.4390); (ii) 28.57N( -0.4390)-30N( -0.5390); (iii) 28.02N( -0. 1438)-28.57N( -0.2438);: 3. $0.25; 4. 28.09%.

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Chapter 13 1. 39.16; 2. 0.1048; 3. 2.06; 4. (i) 0.4501; (ii) 0.0089; 5. (i) 0.6; (ii) 1.16 cents 6. 980.20N(0.4565) - 912.75N(0.2565; (ii) 912.75N( -0.2565) - 980. 20N ( -0.4565); 7. 18.10[2N(0.2) - I}.

Chapter 14 1. 100; 2. (i) Short 456; (ii) Short 442.5; (iii) Short 494.0; 3. (i) Short 2,400; (ii) Long 2,500 options; Short 3,900 shares; 4. 500; 5. (i) -0.4; (ii) 0.04; (iii) 0.1.

Chapter 15 1. (i) Strike price; (ii) Implied volatility; 2. (i) heavier; (li) heavier; (iii) heavier; (iv) less heavy; 3. (a); 4. (b) and (c); 5. (a).

Chapter 16 1. (i) 1.57%; (ii) 4.98%; 2. (i) $52,592; (li) $78,888; (iii) $103,194; (iv) $28,286; 3. (i) 6P = 1206xj (ii) 6P = 1206x + 2oo6x2; 4. (i) $30,000; (ii) $21,000.

Chapter 17 1. (i) 1.06; (ii) 1.04; 2. (i) 0.5; (ii) 2.12%; (iii) 0.95%; (iv) 0.45; 3. (i) 0.0002; (ii) 0.000118; (iii) 0.000139; (iv) 0.0001.

Chapter 18 1. (i) 1.£ = 1.0618; d = 0.9418; p = 0.5267; (ii) 1.£ = 1.0652; d = 1.0584; 2. 23.56; 28.02; and 33.32; 3. (a) and (c).

143

Additional Questions

Pages 145 to 148 contain some additional questions that instructors might find useful. Summary answers are on pages 149 to 150.

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1. A box spread is a combination of a bull spread composed of two call options with strike prices Xl and X l and a bear spread composed of two put options with the same two strike

prices. a) Describe the payoff from a box spread on the expiration date of the options. b) What would be a fair price for the box spread today? Defme variables as necessary. c) Under what circumstances might an investor choose to construct a box spread? d) What sort of investor do you think is most likely to invest in such an option

combination, i.e. a hedger, speculator or arbitrageur? Explain your answer.

2. Form a long butterfly spread using the three call options in the table below.

Cl C2 C3 X=$90 X=$l00 X=$110

T= 180 days T= 180 days T=180days Price 16.3300 10.3000 6.0600 DELTA 0.7860 0.6151 0.4365 GAMMA 0.0138 0.0181 0.0187 THETA -11.2054 -12.2607 -11.4208 VEGA 20.4619 26.8416 27.6602 RHO 30.7085 25.2515 18.5394

a) What does it cost to establish the butterfly spread? b) Calculate each of the Greek measures for this butterfly spread position and explain how

each can be interpreted. c) How would you make this option portfolio delta neutral? What would be achieved by

doing so? d) Suppose that tomorrow the price of C1 falls to $12.18 while the prices of C2 and C3

remain the same. Does this create an arbitrage opportunity? Explain.

3. Consider a six month American put option on index futures where the current futures price is 450, the exercise price is 450, the risk-free rate of interest is 7 percent per annum, the continuous dividend yield of the index is 3 percent, and the volatility of the index is 30 percent per annum. The futures contract underlying the option matures in seven months. Using a three-step binomial tree, calculate a) the price of the American put option now, b) the delta of the option with respect to the futures price, c) the delta of the option with respect to the index level, and d) the price of the corresponding European put option on index futures. e) Apply the control variate technique to improve your estimate of the American option

price and of the delta of the option with respect to the futures price. Note that the Black-Scholes price of the European put option is $36.704 and the delta with respect to the futures price given by Black-Scholes is -0.442.

4. A financial institution trades swaps where 12 month LIBOR is exchanged for a fixed rate of interest. Payments are made once a year. The one-year swap rate (i.e., the rate that would be exchanged for 12 month LIB OR in a new one-year swap) is 6 percent. Similarly the two-year swap rate is 6.5 percent.

a) Use this swap data to calculate the one and two year LmOR zero rates, expressing the rates with continuous compounding.

145

b) What is the value of an existing swap with a notional principal of $10 million that has two years to go and is such that financial institution pays 7 percent and receives 12 month LlBOR? Payments are made once a year.

c) What is the value of a forward rate agreement where a rate of 8 percent will be received on a principaJ of $1 million for the period between one year and two years?

Note: All rates given in this question are expressed with annual compounding.

5. The term structure is flat at 5% per annum with continuous compounding. Some time ago a financial institution entered into a 5-year swap with a principal of $100 million in which every year it pays 12-month LIBOR and receives 6%. The swap now has two years eight months to run. Four months ago 12-month LIBOR was 4% (with annual compounding). What is the value of the swap today? What is the financial institution's credit exposure on the swap?

6. An American put option to sell a Swiss franc for USD has a strike price of 0.80 and a time to maturity of 1 year. The volatility of the Swiss franc is 10%. the USD interest rate is 6%, and the Swiss franc interest rate is 3% (both interest rates continuously compounded). The current exchange rate is 0.81. Use a three time step tree to value the option.

7. A European call option on a certain stock has a strike price of $30, a time to maturity of one year and an implied volatility of 30%. A put option on the same stock has a strike price of $30. a time to maturity of one year and an implied volatility of 33%. What is the arbitrage opportunity open to a trader. Does the opportunity work only when the lognormal assumption underlying Black-Scholes holds. Explain the reasons for your answer carefully.

8. A put option on the S&P 500 has an exercise price of 500 and a time to maturity of one year. The risk free rate is 7% and the dividend yield on the index is 3%. The volatility of the index is 20% per annum and the current level of the index is 500. A financial institution has a short position in the option.

a) Calculate the delta, gamma, and vega of the position. Explain how they can be interpreted.

b) How can the position be made delta neutral?

c) Suppose that one week later the index has increased to 515. How can delta neutrality be preserved?

9. A certain stock is selling for $4. Analysts consider the break up value of the company to be such that the stock price cannot fall below $3 per share. What are the likely differences between the option prices produced by Black-Scholes and

146

option prices in the market? Consider out-of-the-money and in-the-money calls and puts.

10. An interest rate swap with a principal of $100 million involves the exchange of 5% per annum (semiannually compounded) for 6-month LIBOR. The remaining life is 14 months. Interest is exchanged every six months. The 2 month, 8 month and 14 month rates are 4.5%.5%, and 5.4% with continuous compounding. Sixmonth LIBOR was 5.5% four months ago. 'What is the value of the swap?

11. The Deutschemark-Canadian dollar exchange rate is currently 1.0000. At the end of 6 months it will be either 1.1000 or 0.9000. What is the value of a 6 month option to sell one million Canadian dollars for 1.05 million deutschemarks. Verify that the answer given by risk neutral valuation is the same as that given by noarbitrage arguments. Is the option the same as one to buy 1.05 million deutschemarks for 1 million Canadian dollars? Assume that risk-free interest rates in Canada and Gennany are 8% and 6% per annum respectively.

12. An American put futures option has a strike price of 0.55 and a time to maturity of 1 year. The current futures price is 0.60. The volatility of the futures price is 25% and the interest rate(continuously compounded) is 6% per annum. Use a four time step tree to value the option.

13. Is it ever optimal to exercise early an American call option on a) the spot price of gold, b) the spot price of copper, c) the futures price of gold, and d) the average price of gold measured between time zero and the current time. Explain your answers.

14. The future probability distribution of a stock price has a fatter right tail and thinner left tail than the 10gnonnal distribution. Describe the effect of this on the prices of in-the-money and out-of-the-money calls and puts. What is the volatility smile that would be observed?

15. A bank has just sold a call option on 500,000 shares of a stock. The strike price is 40; the stock price is 40; the risk-free rate is 5%; the volatility is 30%; and the time to maturity is 3 months.

a) What position should the company take in the stock for delta neutrality?

b) Suppose that the bank does set up a delta neutral position as soon as the option has been sold and the stock price jumps to 42 within the first hour of trading. What trade is necessary to maintain delta neutrality? Explain whether the bank has gained or lost money in this situation. (You do not need to calculate the exact amount gained or lost.)

c) Repeat part b) on the assumption that the stock jumps to 38 instead of 42

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16. A bank has sold a product that offers investors the total return (excluding dividends) on the Toronto 300 index over a one year period. The return is capped at 20%. If the index goes down the original investment of the investor is returned.

a) What option position is equivalent to the product

b) Write down the fonnulas you would use to value the product and explain in detail how you would decide whether it is a good deal to the investor

17. Use a three step tree to value a three month American put option on wheat futures. The current futures price is is 380 cents, the strike price is 370 cents, the risk-free rate is 5% per annum, and the volatility is 25% per annum. Explain carefully what happens if the investor exercises the option after two months. Suppose that the futures price at the time of exercise is 362 and the most recent settlement price is 360.

18. a) A bank's assets and liabilities both have a duration of 5 years. Is the bank hedged against interest rate movements? Explain carefully any limitations of the hedging scheme it has chosen.

b) Explain what is meant by basis risk in the situation where a company knows it will be purchasing a certain asset in two months and uses a three-month futures contract to hedge its risk.

19. a) Give an example of how a swap might be used by a portfolio manager.

b) Explain the nature of the credit risks to a financial institution in a swap agreement

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ANSWERS

1. (a) The box spread pays off X2-XI in all circumstances (b) It should be worth the present value of X2-XI today, c) and d) An arbitrageur might invest in a box spread if it is mispriced in the market today.

2. (a) 1.79 b) Greek letters are -0.0077, -0.0037, etc c) For delta neutrality we buy 0.0077 of the underlying asset. Small changes in the price of the underlying asset then have very little effect on the value of the whole portfolio d) Yes. We have a positive cash flow when we set up the butterfly spread today and a zero or positive cash in 180 days

3. a) 40.13, b) -0.449, c) Fo=Soeo.04*71l2 or So=O.9769Fo so that delta with respect to

the index level is =0.439, d) 39.81, e) American option price becomes 40.13+36.704-39.81=37.02. Delta becomes -0.449 -0.444+0.442=-0.451

4. a) One year rate is 5.827%, two-year rate is 6.313%, b) -$91,000, c) $8,500 5. 3.50. This is also the credit exposure. 6. 0.021 7. Put is priced too high relative to call. Sell put and buy call. This works regardless

of whether the assumptions underlying Black-Scholes hold 8. a) 0.371, -0.0038, -1.85, b) Sell 0.371 of index for each option sold, c) Delta

changes to 0.317 so 0.054 of index must be bought back 9. This is a thin left taillfat right tail situation (opposite to what is usually

encountered for equity). Black-Scholes is likely to overprice out of the money puts and in the money calls. Similarly it is likely to underprice out of the money calls and in the money puts.

to. $841,000 assuming floating is received. 11. Assume that the exchange rate is OM per $. p is then 0.450 and the value of the

option is about 80,000 OM. Yes the two options are the same. 12.0.036 13. a) no b)yes c)yes d)yes 14. This will lead to a smile where volatility increases with strike price. this is the

opposite of what is usuaUy observed. 15. a) Delta of long position in one option is 0.563. Bank should buy 281,500 shares

b) Delta changes to 0.686. Bank should buy a further 61,500 shares. The bank has a negative gamma and so is likely to have lost money from the big move, c) Delta changes to 0.427. The bank should sell 68,000 shares. It will have lost money in this situation as well

16. Suppose that So is invested in the product where So is the index level today. The value of the investment in one year is So plus the payoff from a bull spread. The bull spread is created from a long call option with strike price So and a short call option with strike price 1.2So. the interest earned can be calculated by valuing the options. This can be compared with other market opportunities.

17. Value is 15.14 cents. Total gain from exercise after 2 months is 8 cents. there would be a 10 cent cash payoff and a short futures position worth -2 cents.

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18. a) limitations relate to possibility of non-parallel shifts in the term structure and the possibility of large movements b) baSis risk arises from the difference between spot and futures price in 2 months

19. a) A swap could be used to change an asset earning a fixed rate of interest to one earning a floating rate. b) credit risks arise from the possibility of a default when the swap has a positive value and the counterparty defaults.

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