Steven E. Shreve

Stochastic Calculus for Finance I
Student’s Manual: Solutions to Selected Exercises
December 14, 2004

Springer
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Preface

This document contains solutions to half the exercises appearing in Stochastic Calculus for Finance I: The Binomial Asset Pricing Model, Springer, 2003. Steven E. Shreve December 2004 Pittsburgh, Pennsylvania USA

Contents

1

The Binomial No-Arbitrage Pricing Model . . . . . . . . . . . . . . . . 1.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 7 7

2

Probability Theory on Coin Toss Space . . . . . . . . . . . . . . . . . . . . 2.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

State Prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4

American Derivative Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5

Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 5.8 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6

Interest-Rate-Dependent Assets . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 6.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1 The Binomial No-Arbitrage Pricing Model

1.7 Solutions to Selected Exercises
Exercise 1.2. Suppose in the situation of Example 1.1.1 that the option sells for 1.20 at time zero. Consider an agent who begins with wealth X0 = 0 and at time zero buys ∆0 shares of stock and Γ0 options. The numbers ∆0 and Γ0 can be either positive or negative or zero. This leaves the agent with a cash position of −4∆0 − 1.20Γ0 . If this is positive, it is invested in the money market; if it is negative, it represents money borrowed from the money market. At time one, the value of the agent’s portfolio of stock, option and money market is X1 = ∆0 S1 + Γ0 (S1 − 5)+ − 5 (4∆0 + 1.20Γ0 ) . 4

Assume that both H and T have positive probability of occurring. Show that if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. In other words, one cannot find an arbitrage when the time-zero price of the option is 1.20. Solution. Considering the cases of a head and of a tail on the first toss, and utilizing the numbers given in Example 1.1.1, we can write: 5 X1 (H) = 8∆0 + 3Γ0 − (4∆0 + 1.20Γ0 ), 4 5 X1 (T ) = 2∆0 + 0 · Γ0 − (4∆0 + 1.20Γ0 ) 4 Adding these, we get X1 (H) + X1 (T ) = 10∆0 + 3Γ0 − 10∆0 − 3Γ0 = 0, or, equivalently,

2

1 The Binomial No-Arbitrage Pricing Model

In other words, either X1 (H) and X1 (T ) are both zero, or they have opposite signs. Taking into account that both p > 0 and q > 0, we conclude that if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Exercise 1.6 (Hedging a long position - one period.). Consider a bank that has a long position in the European call written on the stock price in Figure 1.1.2. The call expires at time one and has strike price K = 5. In Section 1.1, we determined the time-zero price of this call to be V0 = 1.20. At time zero, the bank owns this option, which ties up capital V0 = 1.20. The bank wants to earn the interest rate 25% on this capital until time one, i.e., without investing any more money, and regardless of how the coin tossing turns out, the bank wants to have 5 · 1.20 = 1.50 4 at time one, after collecting the payoff from the option (if any) at time one. Specify how the bank’s trader should invest in the stock and money market to accomplish this. Solution. The trader should use the opposite of the replicating portfolio 1 strategy worked out in Example 1.1.1. In particular, she should short 2 share of stock, which generates $2 income. She should invest this in the money market. At time one, if the stock goes up in value, the bank has an option 5 worth $3, has $ 4 · 2 = $2.50 in the money market, and must pay $4 to cover the short position in the stock. This leaves the bank with $1.50, as desired. On the other hand, if the stock goes down in value, then at time one the bank has an option worth $0, still has $2.50 in the money market, and must pay $1 to cover the short position in stock. Again, the bank has $1.50, as desired. Exercise 1.8 (Asian option). Consider the three-period model of Example 1 1.2.1, with S0 = 4, u = 2, d = 1 , and take the interest rate r = 4 , so that 2 n 1 p = q = 2 . For n = 0, 1, 2, 3, define Yn = ˜ ˜ k=0 Sk to be the sum of the stock prices between times zero and n. Consider an Asian call option that expires at time three and has strike K = 4 (i.e., whose payoff at time three is + 1 ). This is like a European call, except the payoff of the option is 4 Y3 − 4 based on the average stock price rather than the final stock price. Let vn (s, y) denote the price of this option at time n if Sn = s and Yn = y. In particular, + 1 v3 (s, y) = 4 y − 4 . (i) Develop an algorithm for computing vn recursively. In particular, write a formula for vn in terms of vn+1 . (ii) Apply the algorithm developed in (i) to compute v0 (4, 4), the price of the Asian option at time zero.

X1 (H) = −X1 (T ).

1.7 Solutions to Selected Exercises

3

(iii) Provide a formula for δn (s, y), the number of shares of stock which should be held by the replicating portfolio at time n if Sn = s and Yn = y. Solution. (i), (iii) Assume that at time n, Sn = s and Yn = y. Then if the (n + 1)-st toss results in H, we have Sn+1 = us, Yn+1 = Yn + Sn+1 = y + us.

If the (n + 1)-st toss results in T , we have instead Sn+1 = ds, Yn+1 = Yn + Sn+1 = y + ds.

Therefore, formulas (1.2.16) and (1.2.17) take the form vn (s, y) = 1 [˜vn+1 (us, y + us) + q vn+1 (ds, y + ds)], p ˜ 1+r vn+1 (us, y + us) − vn+1 (ds, y + ds) δn (s, y) = . us − ds

(ii) We first list the relevant values of v3 , which are v3 (32, 60) = (60/4 − 4)+ = 11, v3 (8, 36) = (36/4 − 4)+ = 5, v3 (8, 24) = (24/4 − 4)+ = 2, v3 (8, 18) = (18/4 − 4)+ = 0.50, v3 (2, 18) = (18/4 − 4)+ = 0.50, v3 (2, 12) = (12/4 − 4)+ = 0, v3 (2, 9) = (9/4 − 4)+ = 0, v3 (.50, 7.50) = (7.50/4 − 4)+ = 0. We next use the algorithm from (i) to compute the relevant values of v2 : v2 (16, 28) = v2 (4, 16) = v2 (4, 10) = v2 (1, 7) = 4 1 1 v3 (32, 60) + v3 (8, 36) 5 2 2 4 1 1 v3 (8, 24) + v3 (2, 18) 5 2 2 1 4 1 v3 (8, 18) + v3 (2, 12) 5 2 2 = 6.40, = 1, = 0.20,

4 1 1 v3 (2, 9) + v3 (.50, 7.50) = 0. 5 2 2

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1 The Binomial No-Arbitrage Pricing Model

We use the algorithm again to compute the relevant values of v1 : v1 (8, 12) = v1 (2, 6) =
4 5 4 5 1 2 v2 (16, 28) 1 2 v2 (4, 10) 1 + 2 v2 (4, 16) = 2.96, 1 + 2 v2 (1, 7)

= 0.08.

Finally, we may now compute v0 (4, 4) = 1 4 1 v1 (8, 12) + v1 (2, 6) 5 2 2 = 1.216.

Exercise 1.9 (Stochastic volatility, random interest rate). Consider a binomial pricing model, but at each time n ≥ 1, the “up factor” un (ω1 ω2 . . . ωn ), the “down factor” dn (ω1 ω2 . . . ωn ), and the interest rate rn (ω1 ω2 . . . ωn ) are allowed to depend on n and on the first n coin tosses ω1 ω2 . . . ωn . The initial up factor u0 , the initial down factor d0 , and the initial interest rate r0 are not random. More specifically, the stock price at time one is given by S1 (ω1 ) = u0 S0 if ω1 = H, d0 S0 if ω1 = T,

and, for n ≥ 1, the stock price at time n + 1 is given by Sn+1 (ω1 ω2 . . . ωn ωn+1 ) = un (ω1 ω2 . . . ωn )Sn (ω1 ω2 . . . ωn ) if ωn+1 = H, dn (ω1 ω2 . . . ωn )Sn (ω1 ω2 . . . ωn ) if ωn+1 = T.

One dollar invested in or borrowed from the money market at time zero grows to an investment or debt of 1 + r0 at time one, and, for n ≥ 1, one dollar invested in or borrowed from the money market at time n grows to an investment or debt of 1 + rn (ω1 ω2 . . . ωn ) at time n + 1. We assume that for each n and for all ω1 ω2 . . . ωn , the no-arbitrage condition 0 < dn (ω1 ω2 . . . ωn ) < 1 + rn (ω1 ω2 . . . ωn ) < un (ω1 ω2 . . . ωn ) holds. We also assume that 0 < d0 < 1 + r0 < u0 . (i) Let N be a positive integer. In the model just described, provide an algorithm for determining the price at time zero for a derivative security that at time N pays off a random amount VN depending on the result of the first N coin tosses. (ii) Provide a formula for the number of shares of stock that should be held at each time n (0 ≤ n ≤ N − 1) by a portfolio that replicates the derivative security VN . (iii) Suppose the initial stock price is S0 = 80, with each head the stock price increases by 10, and with each tail the stock price decreases by 10. In other words, S1 (H) = 90, S1 (T ) = 70, S2 (HH) = 100, etc. Assume the interest rate is always zero. Consider a European call with strike price 80, expiring at time five. What is the price of this call at time zero?

1.7 Solutions to Selected Exercises

5

Solution. (i) We adapt Theorem 1.2.2 to this case by defining p0 = ˜ 1 + r 0 − d0 , u0 − d 0 q0 = ˜ u0 − 1 − r 0 , u0 − d 0

and for each and and for all ω1 ω2 . . . ωn , pn (ω1 ω2 . . . ωn ) = ˜

1 + rn (ω1 ω2 . . . ωn ) − dn (ω1 ω2 . . . ωn ) , un (ω1 ω2 . . . ωn ) − dn (ω1 ω2 . . . ωn ) un (ω1 ω2 . . . ωn ) − 1 − rn (ω1 ω2 . . . ωn ) . qn (ω1 ω2 . . . ωn ) = ˜ un (ω1 ω2 . . . ωn ) − dn (ω1 ω2 . . . ωn ) 1 pn (ω1 ω2 . . . ωn )Vn+1 (ω1 ω2 . . . ωn H) ˜ 1+r +˜n (ω1 ω2 . . . ωn )Vn+1 (ω1 ω2 . . . ωn T ) , q

In place of (1.2.16), we define for n = N − 1, N − 2, . . . , 1, Vn (ω1 ω2 . . . ωn ) =

and for the the case n = 0 we adopt the definition V0 = 1 p0 V1 (H) + q0 V1 (T ) . ˜ ˜ 1+r

(ii) The number of shares of stock that should be held at time n is still given by (1.2.17): ∆n (ω1 . . . ωn ) = Vn+1 (ω1 . . . ωn H) − Vn+1 (ω1 . . . ωn T ) . Sn+1 (ω1 . . . ωn H) − Sn+1 (ω1 . . . ωn T )

The proof that this hedge works, i.e., that taking the position ∆n in the stock at time n and holding it until time n + 1 results in a portfolio whose value at time n + 1 is Vn+1 , is the same as the proof given for Theorem 1.2.2. (iii) If the stock price at a particular time n is x, then the stock price at the next time is either x + 10 or x − 10. That means that the up factor is un = x+10 and the down factor is dn = x−10 . The corresponding riskx x neutral probabilities are pn = ˜ qn = ˜ 1 − dn = un − d n un − 1 = un = d n x+10 x

1−

x−10 x − x−10 x

= =

1 , 2 1 . 2

x+10 x −1 x+10 x−10 x − x

Because these risk-neutral probabilities do not depend on the time n nor on the coin tosses ω1 . . . ωn , we can easily compute the risk-neutral prob1 1 5 ability of an arbitrary sequence ω1 ω2 ω3 ω4 ω5 to be 2 = 32 .

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1 The Binomial No-Arbitrage Pricing Model

There are three ways for the call with strike 80 to expire in the money at time 5: either the five tosses result in five heads (S5 = 130), result in four heads and one tail (S5 = 110), or result in three heads and two tails 1 (S5 = 90). The risk-neutral probability of five heads is 32 . If a tail occurs, it can occur on any toss, and so there are five sequences that have four heads and one tail. Therefore, the risk-neutral probability of four heads 5 and one tail is 32 . Finally, if there are two tails in a sequence of five tosses, there 10 ways to choose the two tosses that are tails. Therefore, the riskneutral probability of three heads and two tails is 10 . The time-zero price 32 of the call is V0 = 1 5 10 · (130 − 80) + · (110 − 80) + · (90 − 80) = 9.375. 32 32 32

2 Probability Theory on Coin Toss Space

2.9 Solutions to Selected Exercises
Exercise 2.2. Consider the stock price S3 in Figure 2.3.1. (i) What is the distribution of S3 under the risk-neutral probabilities p = 1 , ˜ 2 q = 2. ˜ 1 (ii) Compute ES1 , ES2 , and ES3 . What is the average rate of growth of the stock price under P?
2 (iii) Answer (i) and (ii) again under the actual probabilities p = 3 , q = 1 . 3

Solution. (i) The distribution of S3 under the risk-neutral probabilities p and q is ˜ ˜ 32 8 2 .50 p3 3˜2 q 3˜q 2 q 3 ˜ p ˜ p˜ ˜ With p = 2 , q = 2 , this becomes ˜ 1 ˜ 1 2 .50 32 8 .125 .375 .375 .125 (ii) By Theorem 2.4.4, E Therefore, S2 S1 S3 =E =E = ES0 = S0 = 4. 3 2 (1 + r) (1 + r) (1 + r)

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2 Probability Theory on Coin Toss Space

ES1 = (1 + r)S0 = (1.25)(4) = 5, ES2 = (1 + r)2 S0 = (1.25)2 (4) = 6.25, ES3 = (1 + r)3 S0 = (1.25)3 (4) = 7.8125. In particular, we see that ES3 = 1.25 · ES2 , ES2 = 1.25 · ES1 , ES1 = 1.25 · S0 . Thus, the average rate of growth of the stock price under P is the same as the interest rate of the money market. (iii) The distribution of S3 under the probabilities p and q is 32 8 2 .50 3 2 p 3p q 3pq 2 q 3
2 1 With p = 3 , q = 3 , this becomes

8 2 .50 32 .2963 .4444 .2222 .0371 To compute the average rate of growth, we reason as follows: En Sn+1 = En Sn In our case, Sn+1 Sn = S n En Sn+1 Sn = (pu + qd)Sn .

2 1 · 2 + · 12 = 1.5. 3 3 In other words, the average rate of growth of the stock price under the actual probabilities is 50%. Finally, taking expectations, we have pu + qd = ESn+1 = E En Sn+1 = 1.5 · ESn , so that ES1 = 1.5 · ES0 = 6,

ES2 = 1.5 · ES1 = 9, ES3 = 1.5 · ES2 = 13.50.

Exercise 2.3. Show that a convex function of a martingale is a submartingale. In other words, let M0 , M1 , . . . , MN be a martingale and let ϕ be a convex function. Show that ϕ(M0 ), ϕ(M1 ), . . . , ϕ(MN ) is a submartingale. Solution Let an arbitrary n with 0 ≤ n ≤ N − 1 be given. By the martingale property, we have En Mn+1 = Mn ,

2.9 Solutions to Selected Exercises

9

and hence ϕ(En Mn+1 ) = ϕ(Mn ). On the other hand, by the conditional Jensen’s inequality, we have En ϕ(Mn+1 ) ≥ ϕ(En Mn+1 ). Combining these two, we get En ϕ(Mn+1 ) ≥ ϕ(Mn ), and since n is arbitrary, this implies that the sequence of random variables ϕ(M0 ), ϕ(M1 ), . . . , ϕ(MN ) is a submartingale. Exercise 2.6 (Discrete-time stochastic integral). Suppose M0 , M1 , . . . , MN is a martingale, and let ∆0 , ∆1 , . . . , ∆N −1 be an adapted process. Define the discrete-time stochastic integral (sometimes called a martingale transform) I0 , I1 , . . . , IN by setting I0 = 0 and n−1 In = j=0 ∆j (Mj+1 − Mj ), n = 1, . . . , N.

Show that I0 , I1 , . . . , IN is a martingale. Solution. Because In+1 = In + ∆n (Mn+1 − Mn ) and In , ∆n and Mn depend on only the first n coin tosses, we may “take out what is known” to write En [In+1 ] = En In + ∆n (Mn+1 − Mn ) = In + ∆n En [Mn+1 ] − Mn . However, En [Mn+1 ] = Mn , and we conclude that En [In+1 ] = In , which is the martingale property. Exercise 2.8. Consider an N -period binomial model. (i) Let M0 , M1 , . . . , MN and M0 , M1 , . . . , MN be martingales under the riskneutral measure P. Show that if MN = MN (for every possible outcome of the sequence of coin tosses), then, for each n between 0 and N , we have Mn = Mn (for every possible outcome of the sequence of coin tosses). (ii) Let VN be the payoff at time N of some derivative security. This is a random variable that can depend on all N coin tosses. Define recursively VN −1 , VN −2 , . . . , V0 by the algorithm (1.2.16) of Chapter 1. Show that V0 , V1 VN −1 VN ,..., , 1+r (1 + r)N −1 (1 + r)N

is a martingale under P.

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2 Probability Theory on Coin Toss Space

(iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define Vn = En Show that V0 , is a martingale. (iv) Conclude that Vn = Vn for every n (i.e., the algorithm (1.2.16) of Theorem 1.2.2 of Chapter 1 gives the same derivative security prices as the riskneutral pricing formula (2.4.11) of Chapter 2). Solution. (i) We are given that Mn = MN . For n between 0 and N − 1, this equality and the martingale property imply Mn = En [MN ] = En [MN ] = Mn . (ii) For n between 0 and N − 1, we compute the following conditional expectation: En Vn+1 (ω1 ω2 . . . ωn ) (1 + r)n+1 Vn+1 (ω1 ω2 . . . ωn T ) Vn+1 (ω1 ω2 . . . ωn H) +q ˜ =p ˜ n+1 (1 + r) (1 + r)n+1 Vn (ω1 ω2 . . . ωn ) = , (1 + r)n VN , n = 0, 1, . . . , N − 1. (1 + r)N −n

VN −1 VN V1 ,..., , N −1 (1 + r)N 1+r (1 + r)

where the second equality follows from (1.2.16). This is the martingal Vn property for (1+r)n . n (iii) The martingale property for (1+r)n follows from the iterated conditioning property (iii) of Theorem 2.3.2. According to this property, for n between 0 and n − 1,

V

En

Vn+1 (1 + r)n+1

1 VN En+1 (1 + r)n+1 (1 + r)N −(n+1) 1 VN = En En+1 (1 + r)n (1 + r)N −n VN 1 En = (1 + r)n (1 + r)N −n Vn = . (1 + r)n = En

2.9 Solutions to Selected Exercises

11

(iv) Since the processes in (ii) and (iii) are martingales under the risk-neutral probability measure and they agree at the final time N , they must agree at all earlier times because of (i).

S2 (HH) = 12 S1 (H) = 8 1 r1 (H) = 4 S0 = 4 r0 = 1 4 S1 (T ) = 2 r1 (T ) = 1 2 S2 (T T ) = 2 S2 (HT ) = 8 S2 (T H) = 8

Fig. 2.8.1. A stochastic volatility, random interest rate model.

Exercise 2.9 (Stochastic volatility, random interest rate). Consider a two-period stochastic volatility, random interest rate model of the type described in Exercise 1.9 of Chapter 1. The stock prices and interest rates are shown in Figure 2.8.1. (i) Determine risk-neutral probabilities P(HH), P(HT ), P(T H), P(T T ), such that the time-zero value of an option that pays off V2 at time two is given by the risk-neutral pricing formula V0 = E V2 . (1 + r0 )(1 + r1 )

(iii) Suppose an agent sells the option in (ii) for V0 at time zero. Compute the position ∆0 she should take in the stock at time zero so that at time one, regardless of whether the first coin toss results in head or tail, the value of her portfolio is V1 . (iv) Suppose in (iii) that the first coin toss results in head. What position ∆1 (H) should the agent now take in the stock to be sure that, regardless

(ii) Let V2 = (S2 − 7)+ . Compute V0 , V1 (H), and V1 (T ).

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2 Probability Theory on Coin Toss Space

of whether the second coin toss results in head or tail, the value of her portfolio at time two will be (S2 − 7)+ ? Solution.
1 (i) For the first toss, the up factor is u0 = 2 and the down factor is d0 = 2 . Therefore, the risk-neutral probability of a H on the first toss is

p0 = ˜

1+ 1 − 1 + r 0 − d0 4 = 1 u0 − d 0 2− 2 2−1− u0 − 1 − r 0 = 1 u0 − d 0 2− 2

1 2

=

1 , 2

and the risk-neutral probability of T on the first toss is q0 = ˜
1 4

=

1 . 2

If the first toss results in H, then the up factor for the second toss is u1 (H) = 12 3 S2 (HH) = = , S1 (H) 8 2

and the down factor for the second toss is d1 (H) = S2 (HT ) 8 = = 1. S1 (H) 8

It follows that the risk-neutral probability of getting a H on the second toss, given that the first toss is a H, is p1 (H) = ˜ 1+ 1 −1 1 1 + r1 (H) − d1 (H) = 3 4 = , u1 (H) − d1 (H) 2 2 −1

and the risk-neutral probability of T on the second toss, given that the first toss is a H, is q1 (H) = ˜ u1 (H) − 1 − r1 (H) = u1 (H) − d1 (H)
3 2

−1− 3 2 −1

1 4

=

1 , 2

If the first toss results in T , then the up factor for the second toss is u1 (T ) = S2 (T H) 8 = = 4, S1 (T ) 2

and the down factor for the second toss is d1 (H) = 2 S2 (T T ) = = 1. S1 (T ) 2

It follows that the risk-neutral probability of getting a H on the second toss, given that the first toss is a T , is

2.9 Solutions to Selected Exercises

13

p1 (T ) = ˜

1+ 1 −1 1 1 + r1 (T ) − d1 (T ) 2 = = , u1 (T ) − d1 (T ) 4−1 6

and the risk-neutral probability of T on the second toss, given that the first toss is a T , is q1 (T ) = ˜ 4−1− u1 (T ) − 1 − r1 (T ) = u1 (T ) − d1 (T ) 4−1
1 2

=

5 . 6

The risk-neutral probabilities are P(HH) = p0 p1 (H) = ˜ ˜ P(HT ) = p0 q1 (H) = ˜ ˜ P(T H) = q0 p1 (T ) = ˜ ˜ P(T T ) = q0 q1 (T ) = ˜˜ (ii) We compute V1 (H) = 1 p1 (H)V2 (HH) + q1 (H)V2 (HT ) ˜ ˜ 1 + r1 (H) 1 4 1 · (12 − 7)+ + · (8 − 7)+ = 5 2 2 = 2.40, 1 V1 (T ) = p1 (T )V2 (T H) + q1 (T )V2 (T T ) ˜ ˜ 1 + r1 (T ) 5 2 1 · (8 − 7)+ + · (2 − 7)+ = 3 6 6 = 0.111111, 1 V0 = [˜0 V1 (H) + q0 V1 (T )] p ˜ 1 + r0 4 1 1 = · 2.40 + · 0.1111 5 2 2 = 1.00444.
1 2 1 2 1 2 1 2

· · · ·

1 2 1 2 1 6 5 6

1 = 4, 1 = 4,

= =

1 12 , 5 12 .

We can confirm this price by computing according to the risk-neutral pricing formula in part (i) of the exercise:

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2 Probability Theory on Coin Toss Space

V0 = E

V2 (1 + r0 )(1 + r1 ) V2 (HH) V2 (HT ) = · P(HH) + · P(HT ) (1 + r0 )(1 + r1 (H)) (1 + r0 )(1 + r1 (H)) V2 (T H) V2 (T T ) + · P(T H) + · P(T T ) (1 + r0 )(1 + r1 (T )) (1 + r0 )(1 + r1 (T )) (8 − 7)+ 1 1 (12 − 7)+ · + = 1 1 1 1 · 4 (1 + 4 )(1 + 4 ) 4 (1 + 4 )(1 + 4 ) (2 − 7)+ (8 − 7)+ 1 5 · 1 1 1 · 12 + (1 + 4 )(1 + 2 ) (1 + 4 )(1 + 1 ) 12 2 = 0.80 + 0.16 + 0.04444 + 0 = 1.00444. +

(iii) Formula (1.2.17) still applies and yields ∆0 = V1 (H) − V1 (T ) 2.40 − 0.111111 = = 0.381481. S1 (H) − S1 (T ) 8−2

(iv) Again we use formula (1.2.17), this time obtaining ∆1 (H) = (12 − 7)+ − (8 − 7)+ V2 (HH) − V2 (HT ) = = 1. S2 (HH) − S2 (HT ) 12 − 8

Exercise 2.11 (Put–call parity). Consider a stock that pays no dividend in an N -period binomial model. A European call has payoff CN = (SN − K)+ at time N . The price Cn of this call at earlier times is given by the risk-neutral pricing formula (2.4.11): Cn = En CN , n = 0, 1, . . . , N − 1. (1 + r)N −n

Consider also a put with payoff PN = (K − SN )+ at time N , whose price at earlier times is Pn = E n PN , n = 0, 1, . . . , N − 1. (1 + r)N −n

Finally, consider a forward contract to buy one share of stock at time N for K dollars. The price of this contract at time N is FN = SN − K, and its price at earlier times is Fn = E n FN , n = 0, 1, . . . , N − 1. (1 + r)N −n

(Note that, unlike the call, the forward contract requires that the stock be purchased at time N for K dollars and has a negative payoff if SN < K.)

2.9 Solutions to Selected Exercises

15

(i) If at time zero you buy a forward contract and a put, and hold them until expiration, explain why the payoff you receive is the same as the payoff of a call; i.e., explain why CN = FN + PN . (ii) Using the risk-neutral pricing formulas given above for Cn , Pn , and Fn and the linearity of conditional expectations, show that Cn = Fn + Pn for every n. (iii) Using the fact that the discounted stock price is a martingale under the K risk-neutral measure, show that F0 = S0 − (1+r)N . (iv) Suppose you begin at time zero with F0 , buy one share of stock, borrowing money as necessary to do that, and make no further trades. Show that at time N you have a portfolio valued at FN . (This is called a static replication of the forward contract. If you sell the forward contract for F0 at time zero, you can use this static replication to hedge your short position in the forward contract.) (v) The forward price of the stock at time zero is defined to be that value of K that causes the forward contract to have price zero at time zero. The forward price in this model is (1 + r)N S0 . Show that, at time zero, the price of a call struck at the forward price is the same as the price of a put struck at the forward price. This fact is called put–call parity. (vi) If we choose K = (1 + r)N S0 , we just saw in (v) that C0 = P0 . Do we have Cn = Pn for every n? Solution (i) Consider three cases: Case I: SN = K. Then CN = PN = FN = 0; Case II: SN > K. Then PN = 0 and CN = SN − K = FN ; In all three cases, we see that CN = FN + PN . Case III: SN < K. Then CN = 0 and PN = K − SN = −FN .

(ii) Cn = En = En (iii) F0 = E SN − K FN =E N (1 + r) (1 + r)N SN K K . =E −E = S0 − N N (1 + r) (1 + r) (1 + r)N FN + P N CN = En N −n (1 + r) (1 + r)N −n FN PN + En = Fn + Pn . N −n (1 + r) (1 + r)N −n

16

2 Probability Theory on Coin Toss Space

(iv) At time zero, your portfolio value is F0 = S0 + (F0 − S0 ). At time N , the value of the portfolio is SN + (1 + r)N (F0 − S0 ) = SN + (1 + r)N = SN − K = F N . (v) First of all, if K = (1 + r)N S0 , then, by (iii), F0 = 0. Further, if F0 = 0, then, by (ii), C0 = F0 + P0 = P0 . (vi) No. This would mean, in particular, that CN = PN , and hence (SN − K)+ = (K − SN )+ , which in turn would imply that SN (ω) = K for all ω, which is not the case for most values of ω. Exercise 2.13 (Asian option). Consider an N -period binomial model. An Asian option has a payoff based on the average stock price, i.e., VN = f 1 Sn N + 1 n=0
N

−

K (1 + r)N

,

where the function f is determined by the contractual details of the option. (i) Define Yn = k=0 Sk and use the Independence Lemma 2.5.3 to show that the two-dimensional process (Sn , Yn ), n = 0, 1, . . . , N is Markov. (ii) According to Theorem 2.5.8, the price Vn of the Asian option at time n is some function vn of Sn and Yn ; i.e., Vn = vn (Sn , Yn ), n = 0, 1, . . . , N. Give a formula for vN (s, y), and provide an algorithm for computing vn (s, y) in terms of vn+1 . Solution (i) Note first that Sn+1 = Sn · Sn+1 , Sn Yn+1 = Yn + Sn · Sn+1 , Sn n n+1 and whereas Sn and Yn depend only on the first n tosses, SSn depends only on toss n + 1. According to the Independence Lemma 2.5.3, for any function hn+1 (s, y) of dummy variables s and y, we have

2.9 Solutions to Selected Exercises

17

En [hn+1 (Sn+1 , Yn+1 )] = En hn+1 Sn · = hn (Sn , Yn ), where hn (s, y) = Ehn+1 s ·

Sn+1 Sn+1 , Yn + Sn · Sn Sn

Sn+1 Sn+1 ,y + s · Sn Sn = phn+1 (su, y + su) + q hn+1 (sd, y + sd). ˜ ˜

Because En [hn+1 (Sn+1 , Yn+1 )] can be written as a function of (Sn , Yn ), the two-dimensional process (Sn , Yn ), n = 0, 1, . . . , N , is a Markov process. (ii) We have the final condition VN (s, y) = f y N +1

we have from the risk-neutral pricing formula (2.4.12) and (i) above that Vn = where vn (s, y) = 1 pvn+1 (su, y + su) + q vn+1 (sd, y + sd) . ˜ ˜ 1+r 1 1 En Vn+1 = En vn+1 (Sn+1 , Yn+1 ) = vn (Sn , Yn ), 1+r 1+r

. For n = N − 1, . . . , 1, 0,

3 State Prices

3.7 Solutions to Selected Exercises
Exercise 3.1. Under the conditions of Theorem 3.1.1, show the following analogues of properties (i)–(iii) of that theorem: (i ) P
1 Z

> 0 = 1;

1 (ii ) E Z = 1;

(iii ) for any random variable Y , EY = E 1 ·Y . Z

1 In other words, Z facilitates the switch from E to E in the same way Z facilitates the switch from E to E.

Solution (i ) Because P(ω) > 0 and P(ω) > 0 for every ω ∈ Ω, the ratio 1 P(ω) = Z(ω) P(ω) is defined and positive for every ω ∈ Ω. 1 = Z 1 P(ω) = Z(ω) P(ω) ω∈Ω (ii ) We compute E

ω∈Ω

P(ω)

P(ω) = ω∈Ω P(ω) = 1.

20

3 State Prices

(iii ) We compute E 1 ·Y Z = ω∈Ω P(ω) P(ω)

Y (ω)P(ω) = ω∈Ω Y (ω)P(ω) = EY.

Exercise 3.3. Using the stock price model of Figure 3.1.1 and the actual 2 probabilities p = 3 , q = 1 , define the estimates of S3 at various times by 3 Mn = En [S3 ], n = 0, 1, 2, 3. Fill in the values of Mn in a tree like that of Figure 3.1.1. Verify that Mn , n = 0, 1, 2, 3, is a martingale. Solution We note that M3 = S3 . We compute M2 from the formula M2 = E2 [S3 ]:
2 M2 (HH) = 3 S3 (HHH) + 1 S3 (HHT ) = 2 2 3 S3 (HT H) 2 3 S3 (T HH) 2 3 S3 (T HH) 1 3 S3 (HT T ) 1 3 S3 (T HT ) 1 2 S3 (T HT ) 2 1 3 · 32 + 3 · 8 2 1 3 ·8+ 3 ·2 2 1 3 ·8+ 3 ·2 2 1 3 · 2 + 3 · 0.50

= 24, = 6, = 6, = 1.50.

M2 (HT ) = M2 (T H) = M2 (T T ) =

+ + +

= = =

We next compute M1 from the formula M1 = E1 [S3 ]: M1 (H) = 4 2 2 1 S3 (HHH) + S3 (HHT ) + S3 (HT H) + S3 (HT T ) 9 9 9 9 4 2 2 1 = · 32 + · 8 + · 8 + · 2 9 9 9 9 = 18, 2 2 1 4 M1 (T ) = S3 (T HH) + S3 (T HT ) + S3 (T T H) + S3 (T T T ) 9 9 9 9 2 2 1 4 = · 8 + · 2 + · 2 + · 0.50 9 9 9 9 = 4.50.

Finally, we compute M0 = E[S3 ] 8 4 4 4 = S3 (HHH) + S3 (HHT ) + S3 (HT H) + S3 (T HH) 27 27 27 27 2 2 1 2 + S3 (HT T ) + S3 (T HT ) + S3 (T T H) + S3 (T T T ) 27 27 27 27 8 4 4 4 2 2 2 1 = · 32 + ·8+ ·8+ ·8+ ·2+ ·2+ ·2+ · 0.50 27 27 27 27 27 27 27 27 = 13.50.

3.7 Solutions to Selected Exercises

21

!!    
M0 = 13.50 M1 (H) = 18 M2 (HH) = 24

! !! aa

!

S3 (HHH) = 32

aa

Z Z  
M2 (HT ) = M2 (T H) = 6

aa S3 (HHT ) = S3 (HT H)  = S3 (T HH) = 8  Z Z S (HT T ) = S3 (T HT ) !! 3
= S3 (T T H) = 2

Z Z

M1 (T ) = 4.50

Z Z

!!
M2 (T T ) = 1.50

!! aa

aa

aa S3 (T T T ) = .50

Fig. 3.7.1. An estimation martingale.

We verify the martingale property. We have M2 = E2 [M3 ] because M3 = S3 and we used the formula M2 = E2 [S3 ] to compute M2 . We must check that M1 = E1 [M2 ] and M0 = E0 [M1 ] = E[M1 ], which we do below:
1 2 E1 [M2 ](H) = 3 M2 (HH) + 3 M2 (HT ) = 2 1 3 · 24 + 3 · 6 2 1 3 · 6 + 3 · 1.50 2 1 3 · 18 + 3 · 4.50

= 18 = M1 (H), = 4.50 = M1 (T ), = 13.50 = M0 .

E1 [M2 ](T ) = M0 =

1 2 M2 (T H) 2 3 M1 (H)

+ +

1 2 M2 (T T ) 1 3 M1 (T )

= =

Exercise 3.5 (Stochastic volatility, random interest rate). Consider the model of Exercise 2.9 of Chapter 2. Assume that the actual probability measure is P(HH) = 2 2 1 4 , P(HT ) = , P(T H) = , P(T T ) = . 9 9 9 9

The risk-neutral measure was computed in Exercise 2.9 of Chapter 2. (i) Compute the Radon-Nikod´m derivative Z(HH), Z(HT ), Z(T H) and y Z(T T ) of P with respect to P (ii) The Radon-Nikod´m derivative process Z0 , Z1 , Z2 satisfies Z2 = Z. Comy pute Z1 (H), Z1 (T ) and Z0 . Note that Z0 = EZ = 1. (iii) The version of the risk-neutral pricing formula (3.2.6) appropriate for this model, which does not use the risk-neutral measure, is

22

3 State Prices

V1 (H) = = V1 (T ) = = V0 =

Z2 1 + r0 E1 V2 (H) Z1 (H) (1 + r0 )(1 + r1 ) 1 E1 [Z2 V2 ](H), Z1 (H)(1 + r1 (H)) Z2 1 + r0 E1 V2 (T ) Z1 (T ) (1 + r0 )(1 + r1 ) 1 E1 [Z2 V2 ](T ), Z1 (T )(1 + r1 (T )) Z2 V2 . E (1 + r0 )(1 + r1 )

Use this formula to compute V1 (H), V1 (T ) and V0 when V2 = (S2 − 7)+ . Compare to your answers in Exercise 2.6(ii) of Chapter 2. Solution (i) In Exercise 2.9 of Chapter 2, the risk-neutral probabilities are P(HH) = 1 1 1 5 , P(HT ) = , P(T H) = , P(T T ) = . 4 4 12 12

Therefore, the Radon-Nikod´m derivative is y Z(HH) = 1 9 9 P(HH) = · = , P(HH) 4 4 16 P(T H) 1 9 3 = · = , P(T H) 12 2 8 Z(HT ) = P(HT ) 1 9 9 = · = , P(HT ) 4 2 8 P(T T ) 5 9 15 = · = , P(T T ) 12 1 4

Z(T H) = (ii)

Z(T T ) =

Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P{ω2 = H given that ω1 = H} +Z2 (HT )P{ω2 = T given that ω1 = H} P(HT ) P(HH) + Z2 (HT ) = Z2 (HH) P(HH) + P(HT ) P(HH) + P(HT ) = 9 · 16 3 = , 4
4 9 4 9

+

2 9

+

9 · 8

2 9 4 9

+

2 9

3.7 Solutions to Selected Exercises

23

Z1 (T ) = E1 [Z2 ](T ) = Z2 (T H)P{ω2 = H given that ω1 = T }

+Z2 (T T )P{ω2 = T given that ω1 = T } P(T T ) P(T H) + Z2 (T T ) = Z2 (T H) P(T H) + P(T T ) P(T H) + P(T T )
1 9

2 3 · 2 9 8 9+ 3 = , 2 Z0 = E0 [Z1 ]

=

+

15 · 4

1 9 2 9

+

1 9

= E[Z1 ] = Z1 (H) P(HH) + P(HT ) + Z1 (T ) P(T H) + P(T T ) 3 · 4 = 1. = 4 2 + 9 9 + 3 · 2 2 1 + 9 9

We may also check directly that EZ = 1, as follows: EZ = Z(HH)P(HH) + Z(HT )P(HT ) + Z(T H)P(T H) + Z(T T )P(T T ) 9 4 9 2 3 2 15 1 · + · + · + · = 16 9 8 9 8 9 4 9 1 5 1 1 + = 1. = + + 4 4 12 12 (iii) We recall that V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1, V2 (T T ) = 0. We computed in part (ii) that P{ω2 = H given that ω1 = H} = P{ω2 = T given that ω1 = H} = P{ω2 = H given that ω1 = T } = P{ω2 = T given that ω1 = T } = Therefore,
4 9 4 +2 9 9 2 9 4 +2 9 9 2 9 2 +1 9 9 1 9 2 1 9+9

2 , 3 1 = , 3 2 = , 3 1 = . 3 =

24

3 State Prices

V1 (H) = =

1 E1 [Z2 V2 ](H) Z1 (H) 1 + r1 (H) 3 5 · 4 4
−1

Z2 (HH)V2 (HH)P{ω2 = H given that ω1 = H} +Z2 (HT )V2 (HT )P{ω2 = T given that ω1 = H}

16 9 2 9 1 ·5· + ·1· 15 16 3 8 3 = 2.40, 1 V1 (T ) = E1 [Z2 V2 ](T ) Z1 (T ) 1 + r1 (T ) = = 3 3 · 2 2
−1

Z2 (T H)V2 (T H)P{ω2 = H given that ω1 = T } +Z2 (T T )V2 (T T )P{ω2 = T given that ω1 = T }

4 3 2 15 1 = ·1· + ·0· 9 8 3 4 3 = 0.111111, and V0 = E

Z 2 V2 (1 + r0 )(1 + r1 ) Z2 (HT )V2 (HT ) Z2 (HH)V2 (HH) P(HH) + P(HT ) = (1 + r0 )(1 + r1 (H)) (1 + r0 )(1 + r1 (H)) Z2 (T T )V2 (T T ) Z2 (T H)V2 (T H) P(T H) + P(T T ) + (1 + r0 )(1 + r1 (T )) (1 + r0 )(1 + r1 (T )) 5 5 · 4 4 +
−1

=

9 4 ·5· + 16 9
−1

5 5 · 4 4

−1

9 2 ·1· + 8 9

5 3 · 4 2

−1

3 2 ·1· 8 9

5 3 15 1 · ·0· 4 2 4 9 8 1 16 5 16 1 · + · + · = 25 4 25 4 15 12 = 1.00444. Exercise 3.6. Consider Problem 3.3.1 in an N -period binomial model with the utility function U (x) = ln x. Show that the optimal wealth process cor0 responding to the optimal portfolio process is given by Xn = Xn , n = ζ 0, 1, . . . , N , where ζn is the state price density process defined in (3.2.7). Solution From (3.3.25) we have XN = I λZ (1 + r)N = I(λζN ).

3.7 Solutions to Selected Exercises

25
1 y.

When U (x) = ln x, U (x) = Therefore,

1 x

and the inverse function of U is I(y) = XN = 1 λζN

We must choose λ to satisfy (3.3.26), which in this case takes the form X0 = E ζN I(λζN ) = 1 . λ

Substituting this into the previous equation, we obtain XN = Because
Xn (1+r)n

X0 . ζN

is a martingale under the risk-neutral measure P, we have = 1 Z N XN En Zn (1 + r)N = 1 X0 En [ζN XN ] = . Zn Zn

Xn XN = En (1 + r)n (1 + r)N Therefore,

Xn =

(1 + r)n X0 X0 = . Zn ζn

Exercise 3.8. The Lagrange Multiplier Theorem used in the solution of Problem 3.3.5 has hypotheses that we did not verify in the solution of that problem. In particular, the theorem states that if the gradient of the constraint function, which in this case is the vector (p1 ζ1 , . . . , pm ζm ), is not the zero vector, then the optimal solution must satisfy the Lagrange multiplier equations (3.3.22). This gradient is not the zero vector, so this hypothesis is satisfied. However, even when this hypothesis is satisfied, the theorem does not guarantee that there is an optimal solution; the solution to the Lagrange multiplier equations may in fact minimize the expected utility. The solution could also be neither a maximizer nor a minimizer. Therefore, in this exercise, we outline a different method for verifying that the random variable XN given by (3.3.25) maximizes the expected utility. We begin by changing the notation, calling the random variable given ∗ by (3.3.25) XN rather than XN . In other words,
∗ XN = I

λ Z , (1 + r)N

(3.6.1)

where λ is the solution of equation (3.3.26). This permits us to use the notation XN for an arbitrary (not necessarily optimal) random variable satisfying (3.3.19). We must show that
∗ EU (XN ) ≤ EU (XN ).

(3.6.2)

26

3 State Prices

(i) Fix y > 0, and show that the function of x given by U (x)−yx is maximized by y = I(x). Conclude that U (x) − yx ≤ U (I(y)) − yI(y) for every x. (3.6.3)

(ii) In (3.6.3), replace the dummy variable x by the random variable XN λZ and replace the dummy variable y by the random variable (1+r)N . Take expectations of both sides and use (3.3.19) and (3.3.26) to conclude that (3.6.2) holds. Solution (i) Because U (x) is concave, and for each fixed y > 0, yx is a linear function of x, the difference U (x) − yx is a concave function of x. The derivative of this function is U (x) − y, and this is zero if and only if U (x) = y, which is equivalent to x = I(y). A concave function has its maximum at the point where its derivative is zero. The inequality (3.6.2) is just this statement. (ii) Making the suggested replacements, we obtain U (XN ) − λZXN ≤U (1 + r)N I λZ (1 + r)N − λZ I (1 + r)N λZ (1 + r)N .

Taking expectations under P and using the fact that Z is the RadonNikod´m derivative of P with respect to P, we obtain y EU (XN ) − λE ≤ EU XN (1 + r)N I λZ (1 + r)N − λE Z I (1 + r)N λZ (1 + r)N .

From (3.3.19) and (3.3.26), we have E XN Z = X0 = E I (1 + r)N (1 + r)N λZ (1 + r)N .

Cancelling these terms on the left- and right-hand sides of the above equation, we obtain (3.6.2).

4 American Derivative Securities

4.9 Solutions to Selected Exercises
Exercise 4.1. In the three-period model of Figure 1.2.2 of Chapter 1, let the 1 interest rate be r = 4 so the risk-neutral probabilities are p = q = 1 . ˜ ˜ 2 (i) Determine the price at time zero, denoted V0P , of the American put that expires at time three and has intrinsic value gP (s) = (4 − s)+ . (ii) Determine the price at time zero, denoted V0C , of the American call that expires at time three and has intrinsic value gC (s) = (s − 4)+ .

(iii) Determine the price at time zero, denoted V0S , of the American straddle that expires at time three and has intrinsic value gS (s) = gP (s) + gC (s). (iv) Explain why V0S < V0P + V0C . Solution (i) The payoff of the put at expiration time three is V3P (HHH) = (4 − 32)+ V3P (HHT ) = V3P (HT H) V3P (HT T ) = V3P (T HT ) V3P (T T T ) = (4 − 0.50)+ Because
1 1 ˜ ˜ 2 1+r p = 1+r q = 5 ,

= 0, = V3P (T HH) = (4 − 8)+ = 0, = V3P (T T H) = (4 − 2)+ = 2, = 3.50.

the value of the put at time two is

28

4 American Derivative Securities

V2P (HH) = max

2 2 , V3P (HHH) + V3P (HHT ) 5 5 2 2 = max (4 − 16)+ , · 0 + · 0 5 5 = max{0, 0} = 0, 2 + 2 V2P (HT ) = max 4 − S2 (HT ) , V3P (HT H) + V3P (HT T ) 5 5 2 2 = max 4 − 4)+ , · 0 + · 2 5 5 = max{0, 0.80} = 0.80, 2 + 2 V2P (T H) = max 4 − S2 (T H) , V3P (T HH) + V3P (T HT ) 5 5 2 2 = max 4 − 4)+ , · 0 + · 2 5 5 = max{0, 0.80} 4 − S2 (HH)
+

= 0.80, V2P (T T ) = max 2 2 , V3P (T T H) + V3P (T T T ) 5 5 2 2 = max (4 − 1)+ , · 2 + · 3.50 5 5 = max{3, 2.20} 4 − S2 (T T )
+

= 3. At time one the value of the put is V1P (H) = max 2 2 , V2P (HH) + V2P (HT ) 5 5 2 2 = max (4 − 8)+ , · 0 + · 0.80 5 5 = max{0, 0.32} = 0.32, 2 + 2 V1P (T ) = max 4 − S1 (T ) , V2P (T H) + V2P (T T ) 5 5 2 2 = max (4 − 2)+ , · 0.80 + · 3 5 5 = max{2, 1.52} 4 − S1 (H)
+

= 2. The value of the put at time zero is

4.9 Solutions to Selected Exercises

29

2 2 V0P = max (4 − S0 )+ , V1P (H) + V1P (T ) 5 5 2 2 = max (4 − 4)+ , · 0.32 + · 2 5 5 = max{0, 0.928} = 0.928. (ii) The payoff of the call at expiration time three is V3C (HHH) = (32 − 4)+ V3C (HHT ) = V3C (HT H) V3C (HT T ) = V3C (T HT ) V3C (T T T ) = (0.50 − 4)+ Because
1 1 ˜ ˜ 2 1+r p = 1+r q = 5 ,

= 28, = V3C (T HH) = (8 − 4)+ = 4, = V3C (T T H) = (2 − 4)+ = 0, = 0.

the value of the call at time two is
+

V2C (HH) = max

2 2 , V3C (HHH) + V3C (HHT ) 5 5 2 2 = max (16 − 4)+ , · 28 + · 4 5 5 = max{12, 12.8} S2 (HH) − 4 = 12.8,
+

V2C (HT ) = max

2 2 , V3C (HT H) + V3C (HT T ) 5 5 2 2 = max 4 − 4)+ , · 4 + · 0 5 5 = max{0, 1.60} S2 (HT ) − 4 = 1.60,
+

V2C (T H) = max

2 2 , V3C (T HH) + V3C (T HT ) 5 5 2 2 = max 4 − 4)+ , · 4 + · 0 5 5 = max{0, 1.60} = 1.60, 2 + 2 V2C (T T ) = max S2 (T T ) − 4 , V3C (T T H) + V3C (T T T ) 5 5 2 2 = max (1 − 4)+ , · 0 + · 0 5 5 = max{0, 0} S2 (T H) − 4 = 0.

At time one the value of the call is

30

4 American Derivative Securities

V1C (H) = max

2 2 , V2C (HH) + V2C (HT ) 5 5 2 2 = max (8 − 4)+ , · 12.8 + · 1.60 5 5 = max{4, 5.76} = 5.76, 2 + 2 V1C (T ) = max S1 (T ) − 4 , V2C (T H) + V2C (T T ) 5 5 2 2 = max (2 − 4)+ , · 1.60 + · 0 5 5 = max{0, 0.64} = 0.64. S1 (H) − 4
+

The value of the call at time zero is 2 2 V0C = max (S0 − 4)+ , V1C (H) + V1C (T ) 5 5 2 2 = max (4 − 4)+ , · 5.76 + · 0.64 5 5 = max{0, 2.56} = 2.56. (iii) Note that gS (s) = |s − 4|. The payoff of the straddle at expiration time three is V3S (HHH) = |32 − 4| = 28, V3S (HHT ) = V3S (HT H) = V3S (T HH) = |8 − 4| = 4, V3S (HT T ) = V3S (T HT ) = V3S (T T H) = |2 − 4| = 2, V3S (T T T ) = |0.50 − 4| = 3.50. We see that the payoff of the straddle is the payoff of the put given in the solution to (i) plus the payoff of the call given in the solution to (ii). 1 1 Because 1+r p = 1+r q = 2 , the value of the straddle at time two is ˜ ˜ 5 V2S (HH) = max 2 2 S2 (HH) − 4 , V3S (HHH) + V3S (HHT ) 5 5 2 2 = max |16 − 4|, · 28 + · 4 5 5 = max{12, 12.8} = 12.8, V2S (HT ) = max 2 2 , V3S (HT H) + V3S (HT T ) 5 5 2 2 = max 4 − 4)+ , · 4 + · 2 5 5 = max{0, 2.40} S2 (HT ) − 4
+

= 2.40,

4.9 Solutions to Selected Exercises

31

V2S (T H) = max

2 2 , V3S (T HH) + V3S (T HT ) 5 5 2 2 = max 4 − 4)+ , · 4 + · 2 5 5 = max{0, 2.40} = 2.40, 2 2 V2S (T T ) = max S2 (T T ) − 4 , V3S (T T H) + V3S (T T T ) 5 5 2 2 = max |1 − 4|, · 2 + · 3.50 5 5 = max{3, 2.20} = 3. S2 (T H) − 4
+

One can verify in every case that V2S = V2P + V2C . At time one the value of the straddle is V1S (H) = max 2 2 S1 (H) − 4 , V2S (HH) + V2S (HT ) 5 5 2 2 = max |8 − 4|, · 12.8 + · 2.40 5 5 = max{4, 6.08} = 6.08, 2 2 V1S (T ) = max S1 (T ) − 4 , V2S (T H) + V2S (T T ) 5 5 2 2 = max |2 − 4|, · 2.40 + · 3 5 5 = max{2, 2.16} = 2.16. We have V1S (H) = 6.08 = 0.32 + 5.76 = V1P (H) + V1C (H), but V1S (T ) = 2.16 < 2 + 0.64 = V1P (T ) + V1C (T ). The value of the straddle at time zero is 2 2 V0S = max |S0 − 4|, V1S (H) + V1S (T ) 5 5 2 2 = max |4 − 4|, · 6.08 + · 2.16 5 5 = max{0, 3.296} = 3.296. We have V0S = 3.296 < 0.928 + 2.56 = V0P + V0C .

32

4 American Derivative Securities

(iv) For the put, if there is a tail on the first toss, it is optimal to exercise at time one. This can be seen from the equation V1P (T ) = max 2 2 , V2P (T H) + V2P (T T ) 5 5 2 2 = max (4 − 2)+ , · 0.80 + · 3 5 5 = max{2, 1.52} = 2, 4 − S1 (T )
+

which shows that the intrinsic value at time one if the first toss results in T is greater than the value of continuing. On the other hand, for the call the intrinsic value at time one if there is a tail on the first toss is (S1 (T ) − 4)+ = (2 − 4)+ = 0, whereas the value of continuing is 0.64. Therefore, the call should not be exercised at time one if there is a tail on the first toss. The straddle has the intrinsic value of a put plus a call. When it is exercised, both parts of the payoff are received. In other words, it is not an American put plus an American call, because these can be exercised at different times whereas the exercise of a straddle requires both the put payoff and the call payoff to be received. In the computation of the straddle price V1S (T ) = max 2 2 S1 (T ) − 4 , V2S (T H) + V2S (T T ) 5 5 2 2 = max |2 − 4|, · 2.40 + · 3 5 5 = max{2, 2.16} = 2.16,

we see that it is not optimal to exercise the straddle at time one if the first toss results in T . It would be optimal to exercise the put part, but not the call part, and the straddle cannot exercise one part without exercising the other. Greater value is achieved by not exercising both parts than would be achieved by exercising both. However, this value is less than would be achieved if one could exercise the put part and let the call part continue, and thus V1S (T ) < V1P (T ) + V1C (T ). This loss of value at time one results in a similar loss of value at the earlier time zero: V0S < V0P + V0C . Exercise 4.3. In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be r = 1 so the risk-neutral probabilities are p = q = 1 . Find ˜ ˜ 4 2 the time-zero price and optimal exercise policy (optimal stopping time) for the path-dependent American derivative security whose intrinsic value at each
1 time n, n = 0, 1, 2, 3, is 4 − n+1 j=0 Sj . This intrinsic value is a put on the average stock price between time zero and time n. n +

4.9 Solutions to Selected Exercises

33

Solution The intrinsic value process for this option is G0 = G1 (H) = G1 (T ) = G2 (HH) = 4 − G2 (HT ) = 4 − G2 (T H) = G2 (T T ) = 4− 4− (4 − S0 )
+ + + + + + +

= = =

4−

S0 +S1 (H) 2 S0 +S1 (T ) 2

4− 4− 4− r−

(4 − 4)+

= 0, = 0, = 1, = 0, = 0, = 0.6667, = 1.6667.

4−

4+8 + 2 4+2 + 2

S0 +S1 (H)+S2 (HH) 3 S0 +S1 (H)+S2 (HT ) 3 S0 +S1 (T )+S2 (T H) 3 S0 +S1 (T )+S2 (T T ) 3

= 4− = = 4− =

4+8+16 + 3 4+8+4 3 4+2+4 + 3 4+2+1 3

At time three, the intrinsic value G3 agrees with the option value V3 . In other words, V3 (HHH) = G3 (HHH) = = 4− 4− S0 + S1 (H) + S2 (HH) + S3 (HHH) 4 4 + 8 + 16 + 32 4
+ +

= 0, V3 (HHT ) = G3 (HHT ) = = 4− 4− S0 + S1 (H) + S2 (HH) + S3 (HHT ) 4 4 + 8 + 16 + 8 4
+ +

= 0, V3 (HT H) = G3 (HT H) = = 4− 4− S0 + S1 (H) + S2 (HT ) + S3 (HT H) 4 4+8+4+8 4
+ +

= 0, V3 (HT T ) = G3 (HT T ) = = 4− 4− S0 + S1 (H) + S2 (HT ) + S3 (HT T ) 4 4+8+4+2 4
+ +

= 0,

34

4 American Derivative Securities

V3 (T HH) = G3 (T HH) = = 4− 4− S0 + S1 (T ) + S2 (T H) + S3 (T HH) 4 4+2+4+8 4
+ +

= 0, V3 (T HT ) = G3 (T HT ) = = 4− 4− S0 + S1 (T ) + S2 (T H) + S3 (T HT ) 4 4+2+4+2 4
+ +

= 1, V3 (T T H) = G3 (T T H) = = 4− 4− S0 + S1 (T ) + S2 (T T ) + S3 (T T H) 4 4+2+1+2 4
+ +

= 1.75, V3 (T T T ) = G3 (T T T ) = = 4− 4− S0 + S1 (T ) + S2 (T T ) + S3 (T T T ) 4 4 + 2 + 1 + 0.50 4
+ +

= 2.125. We use the algorithm of Theorem 4.4.3, noting that p ˜ 1+r

=

q ˜ 1+r

2 = 5 , to obtain

2 2 V2 (HH) = max G2 (HH), V3 (HHH) + V3 (HHT ) 5 5 2 2 = max 0, · 0 + · 0 5 5 = 0, 2 2 V2 (HT ) = max G2 (HT ), V3 (HT H) + V3 (HT T ) 5 5 2 2 = max 0, · 0 + · 0 5 5 = 0,

4.9 Solutions to Selected Exercises

35

2 2 V2 (T H) = max G2 (T H), V3 (T HH) + V3 (T HT ) 5 5 2 2 = max 0.6667, · 0 + · 1 5 5 = max{0.6667, 0.40} = 0.6667, 2 2 V2 (T T ) = max G2 (T T ), V3 (T T H) + V3 (T T T ) 5 5 2 2 = max 1.6667, · 1.75 + · 2.125 5 5 = max{1.6667, 1.55} = 1.6667. Continuing, we have 2 2 V1 (H) = max G1 (H), V2 (HH) + V2 (HT ) 5 5 2 2 = max 0, · 0 + · 0 5 5 = 0, 2 2 V1 (T ) = max G1 (T ), V2 (T H) + V2 (T T ) 5 5 2 2 = max 1, · 0.6667 + · 1.6667 5 5 = max{1, 0.9334} = 1, 2 2 V0 = max G0 , V1 (H) + V1 (T ) 5 5 2 2 = max 0, · 0 + · 1 5 5 = max{0, 0.40} = 0.40. To find the optimal exercise time, we work forward. Since V0 > G0 , one should not exercise at time zero. However, V1 (T ) = G1 (T ), so it is optimal to exercise at time one if there is a T on the first toss. If the first toss results in H, the option is destined always be out of the money. With the intrinsic value Gn = 4 −
1 n+1 n j=1

Sj

+

defined in the exercise, it does not matter what n 1 exercise rule we choose in this case. If the payoff were 4 − n+1 j=1 Sj , so that exercising out of the money is costly (as one would expect in practice), then one should allow the option to expire unexercised.

36

4 American Derivative Securities

Exercise 4.5. In equation (4.4.5), the maximum is computed over all stopping times in S0 . List all the stopping times in S0 (there are 26), and from among those, list the stopping times that never exercise when the option is out of the money (there are 11). For each stopping time τ in the latter set, 4 τ compute E I{τ ≤2} 5 Gτ . Verify that the largest value for this quantity is given by the stopping time of (4.4.6), the one which makes this quantity equal to the 1.36 computed in (4.4.7). Solution A stopping time is a random variable, and we can specify a stopping time by listing its values τ (HH), τ (HT ), τ (T H), and τ (T T ). The stopping time property requires that τ (HH) = 0 if and only if τ (HT ) = τ (T H) = τ (T T ) = 0. Similarly, τ (HH) = 1 if and only if τ (HT ) = 1 and τ (T H) = 1 if and only if τ (T T ) = 1. The 26 stopping times in the two-period binomial model are tabulated below. Stopping Time HH HT TH TT τ1 τ2 τ3 τ4 τ5 τ6 τ7 τ8 τ9 τ10 τ11 τ12 τ13 τ14 τ15 τ16 τ17 τ18 τ19 τ20 τ21 τ22 τ23 τ24 τ25 τ26 0 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ 0 1 1 1 1 1 2 2 2 2 2 ∞ ∞ ∞ ∞ ∞ 2 2 2 2 2 ∞ ∞ ∞ ∞ ∞ 0 1 2 2 ∞ ∞ 1 2 2 ∞ ∞ 1 2 2 ∞ ∞ 1 2 2 ∞ ∞ 1 2 2 ∞ ∞ 0 1 2 ∞ 2 ∞ 1 2 ∞ 2 ∞ 1 2 ∞ 2 ∞ 1 2 ∞ 2 ∞ 1 2 ∞ 2 ∞

The intrinsic value process for this option is given by

4.9 Solutions to Selected Exercises

37

G0 = 1, G1 (H) = −3, G1 (T ) = 3, G2 (HH) = −11, G2 (HT ) = G2 (T H) = 1, G2 (T T ) = 4. The stopping times that take the value 1 when there is an H on the first toss are mandating an exercise out of the money (G1 (H) = −3). This rules out τ2 – τ6 . Also, the stopping times that take the value 2 when there is an HH on the first two tosses are mandating an exercise out of the money (G2 (HH) = −11). This rules out τ7 –τ16 . For all other exercise situations, G is positive, so the option is in the money. This leaves us with τ1 and the ten stopping times τ17 –τ26 . We evaluate the risk-neutral expected payoff of these eleven stopping times. E I{τ1 ≤2} E I{τ17 ≤2} 4 5 4 5 τ Gτ1 = G0 = 1, τ Gτ17 =

E I{τ18 ≤2}

4 5

τ

Gτ18

E I{τ19 ≤2}

4 5

τ

Gτ19

E I{τ20 ≤2}

4 5

τ

Gτ20

E I{τ21 ≤2}

4 5

τ

Gτ21

E I{τ22 ≤2}

4 5

τ

Gτ22

E I{τ23 ≤2}

4 5

τ

Gτ23

1 16 1 4 · G2 (HT ) + · G1 (T ) 4 26 2 5 4 2 = · 1 + · 3 = 1.36, 25 5 1 16 1 16 1 16 = · G2 (HT ) + · G2 (T H) + · G2 (T T ) 4 26 4 25 4 25 4 4 4 ·1+ ·1+ · 4 = 0.96, = 25 25 25 1 16 1 16 = · G2 (HT ) + · G2 (T H) 4 26 4 25 4 4 = ·1+ · 1 = 0.32, 25 25 1 16 1 16 = · G2 (HT ) + · G2 (T T ) 4 26 4 25 4 4 = ·1+ · 4 = 0.80, 25 25 1 16 = · G2 (HT ) 4 26 4 = · 1 = 0.16, 25 1 4 = · G1 (T ) 2 5 2 = · 3 = 1.20, 5 1 16 1 16 = + · G2 (T H) + · G2 (T T ) 4 25 4 25 4 4 = ·1+ · 4 = 0.80, 25 25

38

4 American Derivative Securities

E I{τ24 ≤2}

E I{τ25 ≤2}

E I{τ26 ≤2}

1 16 Gτ24 = + · G2 (T H) 4 25 4 · 1 = 0.16, = 25 τ 4 1 16 Gτ25 = + · G2 (T T ) 5 4 25 4 = + · 4 = 0.64, 25 τ 4 Gτ26 = 0. 5 4 5

τ

The largest value, 1.36, is obtained by the stopping time τ17 . Exercise 4.7. For the class of derivative securities described in Exercise 4.6 whose time-zero price is given by (4.8.3), let Gn = Sn − K. This derivative security permits its owner to buy one share of stock in exchange for a payment of K at any time up to the expiration time N . If the purchase has not been made at time N , it must be made then. Determine the time-zero value and optimal exercise policy for this derivative security. (Assume r ≥ 0.)
1 Solution Set Yn = (1+r)n (Sn − K), n = 0, 1, . . . , N . We assume r ≥ 0. Because the discounted stock price is a martingale under the risk-neutral K measure and (1+r)n+1 is not random, we have

En [Yn+1 ] = En

Sn+1 K − En n+1 (1 + r) (1 + r)n+1 Sn K = − (1 + r)n (1 + r)n+1 K Sn − . ≥ (1 + r)n (1 + r)n

This shows that Yn , n = 0, 1, . . . , N , is a submartingale. According to Theorem 4.3.3 (Optional Sampling—Part II), EYN ∧τ ≤ EYN whenever τ is a stopping time. If τ is a stopping time satisfying τ (ω) ≤ N for every sequence of coin tosses ω, this becomes EYτ ≤ EYN , or equivalently, E Therefore, V0 = τ ∈S0 ,τ ≤N

1 1 G ≤E GN . τ τ (1 + r) (1 + r)N 1 1 G ≤E GN . τ τ (1 + r) (1 + r)N

max

E

On the other hand, because the stopping time that is equal to N regardless of the outcome of the coin tossing is in the set of stopping times over which the above maximum is taken, we must in fact have equality:

4.9 Solutions to Selected Exercises

39

V0 =

τ ∈S0 ,τ ≤N

max

E

1 1 G =E GN . τ τ (1 + r) (1 + r)N

Hence, it is optimal to exercise at the final time N regardless of the outcome of the coin tossing.

5 Random Walk

5.8 Solutions to Selected Exercises
Exercise 5.2. (First passage time for random walk with upward drift) Consider the asymmetric random walk with probability p for an up step and probability q = 1 − p for a down step, where 1 < p < 1 so that 2 1 0 < q < 2 . In the notation of (5.2.1), let τ1 be the first time the random walk starting from level 0 reaches the level 1. If the random walk never reaches this level, then τ1 = ∞. (i) Define f (σ) = peσ + qe−σ . Show that f (σ) > 1 for all σ > 0. (ii) Show that when σ > 0, the process Sn = eσMn is a martingale. (iii) Show that for σ > 0, e−σ = E I{τ1 1/2 > q, we have f (0) = p − q > 0, and the convexity of f implies that f (σ) > 1 for all σ > 0.

42

5 Random Walk

(ii) We compute En [Sn+1 ] = = = 1 f (σ) 1 f (σ) 1 f (σ) n+1 En eσ(Mn +Xn+1 ) n+1 eσMn E eσXn+1 n+1 eσMn peσ + qe−σ 1 f (σ) n = eσMn

= Sn .

(iii) Because Sn∧τ1 is a martingale starting at 1, we have 1 = ESn∧τ1 = EeσMn∧τ1 1 f (σ) n∧τ1 .

(5.8.1)

For σ > 0, the positive random variable eσMn∧τ1 is bounded above by eσ , and 0 < 1/f (σ) < 1. Therefore, lim eσMn∧τ1 1 f (σ) n∧τ1 n→∞

= I{τ1 sB , this B derivative is vB (s) = − s2 . Show that the best value of B for the option owner makes the derivative of vB (s) continuous at s = sB (i.e., the two formulas for vB (s) give the same answer at s = sB ).

50

5 Random Walk

Solution (i) With v(s) = sp , (5.7.5) becomes sp = Mulitplication by
2p sp

2 p p 2 1 p ·2 s + · ps , 5 5 2

leads to 2p = 2 p 2 2 (2 ) + , 5 5

and we may rewrite this as (2p ) −
2

5 p · 2 + 1 = 0, 2

a quadratic equation in 2p . The solution to this equation is 2p = 1 2 5 ± 2 25 −4 4 = 1 2 5 3 ± 2 2 .

(ii) With v(s) = As + B , we have lims→∞ v(s) = ±∞ unless A = 0. The s boundary condition lims→∞ v(s) = 0 implies therefore that A is zero. (iii) Since v(s) is positive, we must have B > 0. We note that lims↓0 fB (s) = lims→∞ fB (s) = ∞. Therefore, fB (s) takes the value zero for some s ∈ (0, ∞) if and only if its minimium over (0, ∞) is less than or equal to zero. To find the minimizing value of s, we set the derivative of fB (s) equal to zero: B − 2 + 1 = 0. s √ This results in the critical point sc = B. We note that the second derivative, 2B , is positive on (0, ∞), so the function is convex, and hence fB s3 attains a minimum at sc . The minimal value of fB on (0, ∞) is √ fB (sc ) = 2 B − 4. This is positive if B > 4, in which case fB (s) = 0 has no solution in (0, ∞). If B = 4, then fB (sc ) = 0 and sc is the only solution to the equation fB (s) = 0 in (0, ∞). If 0 < B < 4, then fB (sc ) < 0 and the equation fB (s) = 0 has two solutions in (0, ∞).

We thus have either 2p = 2 or 2p = 1 , and hence either p = 1 or p = −1. 2

(iv) Since vB (s) = B for all large values of s, we maximize this by choosing s B as large as possible, i.e., B = 4. For values of B < 4, the curve B lies s below the curve 4 (see Figure 5.8.1), and values of B > 4 are not possible s because of part (iii).

5.8 Solutions to Selected Exercises

51

(iv) We see from the tangency of the curve y = 4 with the intrinsic value s y = 4 − s at the point (2,2) in Figure 5.8.1 that y = 4 and y = 4 − s have s the same derivative at s = 2. Indeed, d 4 ds s =− 4 s2 = −1,

s=2

s=2

and as noted in the statement of the exercise, d (4 − s) = −1. ds y 4

(2, 2)

y=

4 s

y =4−s 1 2 3
B s

y=

3 s

4 for B = 3 and B = 4.

s

Fig. 5.8.1. The curve y =

6 Interest-Rate-Dependent Assets

6.9 Solutions to Selected Exercises
Exercise 6.1. Prove parts (i), (ii), (iii) and (v) of Theorem 2.3.2 when conditional expectation is defined by Definition 6.2.2. (Part (iv) is not true in the form stated in Theorem 2.3.2 when the coin tosses are not independent.) Solution We take each of parts (i), (ii), (iii) and (v) of Theorem 2.3.2 in turn. (i) Linearity of conditional expectations. According to Definition 6.2.2, En [c1 X + c2 Y ](ω 1 . . . ω n ) = ω n+1 ,...,ω N

c1 X(ω 1 . . . ω n ω n+1 . . . ω N ) + c2 Y (ω 1 . . . ω n ω n+1 . . . ω N ) ×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } X(ω 1 . . . ω n ω n+1 . . . ω N ) ω n+1 ,...,ω N

= c1

+c2

×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } Y (ω 1 . . . ω n ω n+1 . . . ω N ) ω n+1 ,...,ω N

= c1 En [X](ω 1 . . . ω n ) + c2 En [Y ](ω 1 . . . ω n ).

×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }

(ii) Taking out what is known. If X depends only on the first n coin tosses, then

54

6 Interest-Rate-Dependent Assets

En [XY ](ω 1 . . . ω n ) = ω n+1 ,...,ω N

X(ω 1 . . . ω n )Y (ω 1 . . . ω n ω n+1 . . . ω N ) ×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } Y (ω 1 . . . ω n ω n+1 . . . ω N ) ω n+1 ,...,ω N

= X(ω 1 . . . ω n )

= X(ω 1 . . . ω n )En [Y ](ω 1 . . . ω n )

×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }

(iii) Iterated conditioning. If 0 ≤ n ≤ m ≤ N , then because Em [X] depends only on the first m coin tosses and P{ωn+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N ω m+1 ,...,ω N

|ω1 = ω 1 , . . . , ωn = ω n } we have En Em [X] (ω 1 . . . ω n ) = ω n+1 ,...,ω N

= P{ωn+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n },

Em [X](ω 1 . . . ω m ) ×P{ωn+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } Em [X](ω 1 . . . ω m ) ω n+1 ,...,ω m ω m+1 ,...,ω N

=

×P{ωn+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N = ω n+1 ,...ω m

|ω1 = ω 1 , . . . , ωn = ω n } Em [X](ω 1 . . . ω m )

=

×P{ωn+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n } X(ω 1 . . . ω m ω m+1 . . . ω N ) ω n+1 ,...,ω m ω m+1 ,...,ω N

×P{ωn+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n }, where we have used the definition

×P{ωm+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m }

6.9 Solutions to Selected Exercises

55

Em [X](ω 1 . . . ω m ) = ω m+1 ,...,ω N

X(ω 1 . . . ω m ω m+1 . . . ω N ) ×P{ωm+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m }

in the last step. Using the fact that P{ωm+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m } = P{ωn+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } = P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n },

×P{ωn+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n }

we may write the last term in the above formula for En Em [X] (ω 1 . . . ω n ) as En Em [X] (ω 1 . . . ω n ) = ω n+1 ,...,ω m ω m+1 ,...,ω N

X(ω 1 . . . ω m ω m+1 . . . ω N ) ×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n } X(ω 1 . . . ω N ) ω n+1 ,...,ω N

=

= En [X](ω 1 . . . ω n ).

×P{ωn+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }

(iv) Conditional Jensen’s inequality. This follows from part (i) just like the proof of part (v) in Appendix A. Exercise 6.2. Verify that the discounted value of the static hedging portfolio constructed in the proof of Theorem 6.3.2 is a martingale under P. Solution The static hedging portfolio in Theorem 6.3.2 is, at time n, to short Sn Bn,m zero coupon bonds maturing at time m and to hold one share of the asset with price Sn . The value of this portfolio at time k, where n ≤ k ≤ m, is Sn Xk = S k − Bk,m , k = n, n + 1, . . . , m. Bn,m For n ≤ k ≤ m − 1, we have Ek [Dk+1 Xk+1 ] = Ek [Dk+1 Sk+1 ] − Sn Ek [Dk+1 Bk+1,m ] . Bn,m

56

6 Interest-Rate-Dependent Assets

Using the fact that the discounted asset price is a martingale under the risk-neutral measure and also using (6.2.5) first in the form Dk+1 Bk+1,m = Ek+1 [Dm ] and then in the form Dk Bk,m = Ek [Dm ], we may rewrite this as Ek [Dk+1 Xk+1 ] = Dk Sk − Sn Ek Ek+1 [Dm ] Bn,m Sn = D k Sk − Ek [Dm ] Bn,m Sn = D k Sk − Dk Bk,m Bn,m = D k Xk .

This is the martingale property. Exercise 6.4. Using the data in Example 6.3.9, this exercise constructs a 1 hedge for a short position in the caplet paying (R2 − 3 )+ at time three. We observe from the second table in Example 6.3.9 that the payoff at time three of this caplet is 2 , V3 (HT ) = V3 (T H) = V3 (T T ) = 0. 3 Since this payoff depends on only the first two coin tosses, the price of the caplet at time two can be determined by discounting: V3 (HH) = V2 (HH) = 1 1 V3 (HH) = , 1 + R2 (HH) 3 V2 (HT ) = V2 (T H) = V2 (T T ) = 0.

Indeed, if one is hedging a short position in the caplet and has a portfolio valued at 1 at time two in the event ω1 = H, ω2 = H, then one can simply 3 1 invest this 3 in the money market in order to have the 2 required to pay off 3 the caplet at time three.
2 21

In Example 6.3.9, the time-zero price of the caplet is determined to be (see (6.3.10)).

(i) Determine V1 (H) and V1 (T ), the price at time one of the caplet in the events ω1 = H and ω1 = T , respectively.
2 (ii) Show how to begin with 21 at time zero and invest in the money market and the maturity two bond in order to have a portfolio value X1 at time one that agrees with V1 , regardless of the outcome of the first coin toss. Why do we invest in the maturity two bond rather than the maturity three bond to do this?

(iii) Show how to take the portfolio value X1 at time one and invest in the money market and the maturity three bond in order to have a portfolio value X2 at time two that agrees with V2 , regardless of the outcome of the first two coin tosses. Why do we invest in the maturity three bond rather than the maturity two bond to do this?

6.9 Solutions to Selected Exercises

57

Solution. (i) We determine V1 by the risk-neutral pricing formula. In particular, V1 (H) = 1 E1 [D2 V2 ](H) D1 (H)

= P{ω2 = H|ω1 = H}D2 (HH)V2 (HH) +P{ω2 = T |ω1 = H}D2 (HT )V2 (HT ) 2 6 1 1 6 4 = · · + · ·0 = , 3 7 3 3 7 21 1 E1 [D2 V2 ](T ) = 0. V1 (T ) = D1 (T ) (ii) We compute the number of shares of the time-two maturity bond by the usual formula: ∆0 = V1 (H) − V1 (T ) = B1,2 (H) − B1,2 (T )
4 21 − 0 6 5 7 − 7

=

4 . 3
2 21

It is straight-forward to verify that this works. Set X0 = V0 =

and compute

X1 (H) = ∆0 B1,2 (H) + (1 + R0 )(X0 − ∆0 B0,2 ) 2 4 11 4 4 6 − · = = V1 (H), = · + 3 7 21 3 14 21 X1 (T ) = ∆0 B1,2 (T ) + (1 + R0 )(X0 − ∆0 B0,2 ) 4 5 2 4 11 = · + − · = 0 = V1 (T ). 3 7 21 3 14 We do not use the maturity three bond because B1,3 (H) = B1,3 (T ), and this bond therefore provides no hedge against the first coin toss. (iii) In the event of a T on the first coin toss, the caplet price is zero, the hedging portfolio has zero value, and no further hedging is required. In the event of a H on the first toss, we hedge by taking in the maturity three bond the position ∆1 (H) = V2 (HH) − V2 (HT ) = B2,3 (HH) − B2,3 (HT )
1 3 1 2

−0 2 =− . 3 −1

It is straight-forward to verify that this works. We compute X2 (HH) = ∆1 (H)B2,3 (HH) + (1 + R1 (H))(X1 (H) − ∆1 (H)B1,3 (H)) 2 4 2 1 7 4 1 + · = V2 (HH), =− · + = 3 2 6 21 3 7 3 X2 (HT ) = ∆1 (H)B2,3 (HT ) + (1 + R1 (H))(X1 (H) − ∆1 (H)B1,3 (H)) 2 7 4 2 4 = − ·1+ + · = 0 = V2 (HT ). 3 6 21 3 7

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6 Interest-Rate-Dependent Assets

We do not use the maturity two bond because B2,2 (HH) = B2,2 (HT ), and this bond therefore provides no hedge against the second coin toss.