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Steven E. Shreve
Stochastic Calculus for Finance I
Student’s Manual: Solutions to Selected
Exercises
December 14, 2004
Springer
Berlin Heidelberg NewYork
Hong Kong London
Milan Paris Tokyo
Contents
1 The Binomial No-Arbitrage Pricing Model . . . . . . . . . . . . . . . . 1
1.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Probability Theory on Coin Toss Space . . . . . . . . . . . . . . . . . . . . 7
2.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
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 sellsfor 1.20 at time zero. Consider an agent who begins with wealth X0 = 0and 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 witha cash position of −4∆0 − 1.20Γ0. If this is positive, it is invested in themoney market; if it is negative, it represents money borrowed from the moneymarket. At time one, the value of the agent’s portfolio of stock, option andmoney market is
X1 = ∆0S1 + Γ0(S1 − 5)+ − 5
4(4∆0 + 1.20Γ0) .
Assume that both H and T have positive probability of occurring. Show thatif there is a positive probability that X1 is positive, then there is a positiveprobability that X1 is negative. In other words, one cannot find an arbitragewhen 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, andutilizing the numbers given in Example 1.1.1, we can write:
X1(H) = 8∆0 + 3Γ0 −5
4(4∆0 + 1.20Γ0),
X1(T ) = 2∆0 + 0 · Γ0 −5
4(4∆0 + 1.20Γ0)
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
X1(H) = −X1(T ).
In other words, either X1(H) and X1(T ) are both zero, or they have oppositesigns. Taking into account that both p > 0 and q > 0, we conclude that ifthere is a positive probability that X1 is positive, then there is a positiveprobability that X1 is negative.
Exercise 1.6 (Hedging a long position - one period.). Consider a bankthat has a long position in the European call written on the stock price inFigure 1.1.2. The call expires at time one and has strike price K = 5. InSection 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 tossingturns out, the bank wants to have
5
4· 1.20 = 1.50
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 marketto accomplish this.
Solution. The trader should use the opposite of the replicating portfoliostrategy worked out in Example 1.1.1. In particular, she should short 1
2 shareof stock, which generates $2 income. She should invest this in the moneymarket. At time one, if the stock goes up in value, the bank has an optionworth $3, has $
(54 ·2)
= $2.50 in the money market, and must pay $4 to coverthe 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 bankhas 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 Example1.2.1, with S0 = 4, u = 2, d = 1
2 , and take the interest rate r = 14 , so that
p = q = 12 . For n = 0, 1, 2, 3, define Yn =
∑n
k=0 Sk to be the sum of thestock prices between times zero and n. Consider an Asian call option thatexpires at time three and has strike K = 4 (i.e., whose payoff at time three is(
14Y3 − 4
)+). This is like a European call, except the payoff of the option is
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,
v3(s, y) =(
14y − 4
)+.
(i) Develop an algorithm for computing vn recursively. In particular, write aformula for vn in terms of vn+1.
(ii) Apply the algorithm developed in (i) to compute v0(4, 4), the price of theAsian option at time zero.
1.7 Solutions to Selected Exercises 3
(iii) Provide a formula for δn(s, y), the number of shares of stock which shouldbe 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)-sttoss 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
1 + r[pvn+1(us, y + us) + qvn+1(ds, y + ds)],
δn(s, y) =vn+1(us, y + us) − vn+1(ds, y + ds)
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) =4
5
[1
2v3(32, 60) +
1
2v3(8, 36)
]= 6.40,
v2(4, 16) =4
5
[1
2v3(8, 24) +
1
2v3(2, 18)
]= 1,
v2(4, 10) =4
5
[1
2v3(8, 18) +
1
2v3(2, 12)
]= 0.20,
v2(1, 7) =4
5
[1
2v3(2, 9) +
1
2v3(.50, 7.50)
]= 0.
4 1 The Binomial No-Arbitrage Pricing Model
We use the algorithm again to compute the relevant values of v1:
v1(8, 12) = 45
[12v2(16, 28) + 1
2v2(4, 16)]
= 2.96,
v1(2, 6) = 45
[12v2(4, 10) + 1
2v2(1, 7)]
= 0.08.
Finally, we may now compute
v0(4, 4) =4
5
[1
2v1(8, 12) +
1
2v1(2, 6)
]= 1.216.
Exercise 1.9 (Stochastic volatility, random interest rate). Consider abinomial pricing model, but at each time n ≥ 1, the “up factor” un(ω1ω2 . . . ωn),the “down factor” dn(ω1ω2 . . . ωn), and the interest rate rn(ω1ω2 . . . ωn) areallowed to depend on n and on the first n coin tosses ω1ω2 . . . ωn. The initialup factor u0, the initial down factor d0, and the initial interest rate r0 are notrandom. More specifically, the stock price at time one is given by
S1(ω1) =
u0S0 if ω1 = H,d0S0 if ω1 = T,
and, for n ≥ 1, the stock price at time n + 1 is given by
Sn+1(ω1ω2 . . . ωnωn+1) =
un(ω1ω2 . . . ωn)Sn(ω1ω2 . . . ωn) if ωn+1 = H,dn(ω1ω2 . . . ωn)Sn(ω1ω2 . . . ωn) if ωn+1 = T.
One dollar invested in or borrowed from the money market at time zero growsto an investment or debt of 1 + r0 at time one, and, for n ≥ 1, one dollar in-vested in or borrowed from the money market at time n grows to an investmentor debt of 1 + rn(ω1ω2 . . . ωn) at time n + 1. We assume that for each n andfor all ω1ω2 . . . ωn, the no-arbitrage condition
0 < dn(ω1ω2 . . . ωn) < 1 + rn(ω1ω2 . . . ωn) < un(ω1ω2 . . . ωn)
holds. We also assume that 0 < d0 < 1 + r0 < u0.
(i) Let N be a positive integer. In the model just described, provide analgorithm for determining the price at time zero for a derivative securitythat at time N pays off a random amount VN depending on the result ofthe first N coin tosses.
(ii) Provide a formula for the number of shares of stock that should be held ateach time n (0 ≤ n ≤ N − 1) by a portfolio that replicates the derivativesecurity VN .
(iii) Suppose the initial stock price is S0 = 80, with each head the stock priceincreases by 10, and with each tail the stock price decreases by 10. Inother words, S1(H) = 90, S1(T ) = 70, S2(HH) = 100, etc. Assume theinterest 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 + r0 − d0
u0 − d0, q0 =
u0 − 1 − r0
u0 − d0,
and for each and and for all ω1ω2 . . . ωn,
pn(ω1ω2 . . . ωn) =1 + rn(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn)
un(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn),
qn(ω1ω2 . . . ωn) =un(ω1ω2 . . . ωn) − 1 − rn(ω1ω2 . . . ωn)
un(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn).
In place of (1.2.16), we define for n = N − 1, N − 2, . . . , 1,
Vn(ω1ω2 . . . ωn) =1
1 + r
[pn(ω1ω2 . . . ωn)Vn+1(ω1ω2 . . . ωnH)
+qn(ω1ω2 . . . ωn)Vn+1(ω1ω2 . . . ωnT )],
and for the the case n = 0 we adopt the definition
V0 =1
1 + r
[p0V1(H) + q0V1(T )
].
(ii) The number of shares of stock that should be held at time n is still givenby (1.2.17):
∆n(ω1 . . . ωn) =Vn+1(ω1 . . . ωnH) − Vn+1(ω1 . . . ωnT )
Sn+1(ω1 . . . ωnH) − Sn+1(ω1 . . . ωnT ).
The proof that this hedge works, i.e., that taking the position ∆n in thestock at time n and holding it until time n+1 results in a portfolio whosevalue at time n + 1 is Vn+1, is the same as the proof given for Theorem1.2.2.
(iii) If the stock price at a particular time n is x, then the stock price at thenext time is either x + 10 or x − 10. That means that the up factor isun = x+10
xand the down factor is dn = x−10
x. The corresponding risk-
neutral probabilities are
pn =1 − dn
un − dn
=1 − x−10
xx+10
x− x−10
x
=1
2,
qn =un − 1
un = dn
=x+10
x− 1
x+10x
− x−10x
=1
2.
Because these risk-neutral probabilities do not depend on the time n noron the coin tosses ω1 . . . ωn, we can easily compute the risk-neutral prob-
ability of an arbitrary sequence ω1ω2ω3ω4ω5 to be(
12
)5= 1
32 .
6 1 The Binomial No-Arbitrage Pricing Model
There are three ways for the call with strike 80 to expire in the moneyat time 5: either the five tosses result in five heads (S5 = 130), result infour heads and one tail (S5 = 110), or result in three heads and two tails(S5 = 90). The risk-neutral probability of five heads is 1
32 . If a tail occurs,it can occur on any toss, and so there are five sequences that have fourheads and one tail. Therefore, the risk-neutral probability of four headsand one tail is 5
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 risk-neutral probability of three heads and two tails is 10
32 . The time-zero priceof the call is
V0 =1
32· (130 − 80) +
5
32· (110 − 80) +
10
32· (90 − 80) = 9.375.
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 = 12 ,
q = 12 .
(ii) Compute ES1, ES2, and ES3. What is the average rate of growth of the
stock price under P?
(iii) Answer (i) and (ii) again under the actual probabilities p = 23 , q = 1
3 .
Solution.
(i) The distribution of S3 under the risk-neutral probabilities p and q is
32 8 2 .50p3 3p2q 3pq2 q3
With p = 12 , q = 1
2 , this becomes
32 8 2 .50.125 .375 .375 .125
(ii) By Theorem 2.4.4,
ES3
(1 + r)3= E
S2
(1 + r)2= E
S1
(1 + r)= ES0 = S0 = 4.
Therefore,
8 2 Probability Theory on Coin Toss Space
ES1 = (1 + r)S0 = (1.25)(4) = 5,
ES2 = (1 + r)2S0 = (1.25)2(4) = 6.25,
ES3 = (1 + r)3S0 = (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 sameas the interest rate of the money market.
(iii) The distribution of S3 under the probabilities p and q is
32 8 2 .50p3 3p2q 3pq2 q3
With p = 23 , q = 1
3 , this becomes
32 8 2 .50.2963 .4444 .2222 .0371
To compute the average rate of growth, we reason as follows:
EnSn+1 = En
(Sn
Sn+1
Sn
)= SnEn
(Sn+1
Sn
)= (pu + qd)Sn.
In our case,
pu + qd =2
3· 2 +
1
3· 12 = 1.5.
In other words, the average rate of growth of the stock price under theactual probabilities is 50%. Finally, taking expectations, we have
ESn+1 = E(EnSn+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 submartin-gale. In other words, let M0,M1, . . . ,MN be a martingale and let ϕ be aconvex 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 martingaleproperty, we have
EnMn+1 = Mn,
2.9 Solutions to Selected Exercises 9
and henceϕ(EnMn+1) = ϕ(Mn).
On the other hand, by the conditional Jensen’s inequality, we have
Enϕ(Mn+1) ≥ ϕ(EnMn+1).
Combining these two, we get
Enϕ(Mn+1) ≥ ϕ(Mn),
and since n is arbitrary, this implies that the sequence of random variablesϕ(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. Definethe discrete-time stochastic integral (sometimes called a martingale transform)I0, I1, . . . , IN by setting I0 = 0 and
In =n−1∑
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 dependon 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 themartingale property.
Exercise 2.8. Consider an N -period binomial model.
(i) Let M0,M1, . . . ,MN and M ′0,M
′1, . . . ,M
′N be martingales under the risk-
neutral measure P. Show that if MN = M ′N (for every possible outcome
of the sequence of coin tosses), then, for each n between 0 and N , we haveMn = M ′
n (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 arandom variable that can depend on all N coin tosses. Define recursivelyVN−1, VN−2, . . . , V0 by the algorithm (1.2.16) of Chapter 1. Show that
V0,V1
1 + r, . . . ,
VN−1
(1 + r)N−1,
VN
(1 + r)N
is a martingale under P.
10 2 Probability Theory on Coin Toss Space
(iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define
V ′n = En
[VN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Show that
V ′0 ,
V ′1
1 + r, . . . ,
V ′N−1
(1 + r)N−1,
VN
(1 + r)N
is a martingale.
(iv) Conclude that Vn = V ′n 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 risk-neutral pricing formula (2.4.11) of Chapter 2).
Solution.
(i) We are given that Mn = M ′N . For n between 0 and N − 1, this equality
and the martingale property imply
Mn = En[MN ] = En[M ′N ] = Mn.
(ii) For n between 0 and N − 1, we compute the following conditional expec-tation:
En
[Vn+1
(1 + r)n+1
](ω1ω2 . . . ωn)
= pVn+1(ω1ω2 . . . ωnH)
(1 + r)n+1+ q
Vn+1(ω1ω2 . . . ωnT )
(1 + r)n+1
=Vn(ω1ω2 . . . ωn)
(1 + r)n,
where the second equality follows from (1.2.16). This is the martingalproperty for Vn
(1+r)n .
(iii) The martingale property forV ′
n
(1+r)n follows from the iterated conditioning
property (iii) of Theorem 2.3.2. According to this property, for n between0 and n − 1,
En
[V ′
n+1
(1 + r)n+1
]= En
[1
(1 + r)n+1En+1
[VN
(1 + r)N−(n+1)
]]
=1
(1 + r)nEn
[En+1
[VN
(1 + r)N−n
]]
=1
(1 + r)nEn
[VN
(1 + r)N−n
]
=V ′
n
(1 + r)n.
2.9 Solutions to Selected Exercises 11
(iv) Since the processes in (ii) and (iii) are martingales under the risk-neutralprobability measure and they agree at the final time N , they must agreeat all earlier times because of (i).
S0 = 4r0 = 1
4
S1(H) = 8
r1(H) = 1
4
S1(T ) = 2
r1(T ) = 1
2
S2(HH) = 12
S2(HT ) = 8
S2(TH) = 8
S2(TT ) = 2
Fig. 2.8.1. A stochastic volatility, random interest rate model.
Exercise 2.9 (Stochastic volatility, random interest rate). Considera two-period stochastic volatility, random interest rate model of the typedescribed in Exercise 1.9 of Chapter 1. The stock prices and interest rates areshown in Figure 2.8.1.
(i) Determine risk-neutral probabilities
P(HH), P(HT ), P(TH), P(TT ),
such that the time-zero value of an option that pays off V2 at time two isgiven by the risk-neutral pricing formula
V0 = E
[V2
(1 + r0)(1 + r1)
].
(ii) Let V2 = (S2 − 7)+. Compute V0, V1(H), and V1(T ).
(iii) Suppose an agent sells the option in (ii) for V0 at time zero. Compute theposition ∆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 valueof 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
12 2 Probability Theory on Coin Toss Space
of whether the second coin toss results in head or tail, the value of herportfolio at time two will be (S2 − 7)+?
Solution.
(i) For the first toss, the up factor is u0 = 2 and the down factor is d0 = 12 .
Therefore, the risk-neutral probability of a H on the first toss is
p0 =1 + r0 − d0
u0 − d0=
1 + 14 − 1
2
2 − 12
=1
2,
and the risk-neutral probability of T on the first toss is
q0 =u0 − 1 − r0
u0 − d0=
2 − 1 − 14
2 − 12
=1
2.
If the first toss results in H, then the up factor for the second toss is
u1(H) =S2(HH)
S1(H)=
12
8=
3
2,
and the down factor for the second toss is
d1(H) =S2(HT )
S1(H)=
8
8= 1.
It follows that the risk-neutral probability of getting a H on the secondtoss, given that the first toss is a H, is
p1(H) =1 + r1(H) − d1(H)
u1(H) − d1(H)=
1 + 14 − 1
32 − 1
=1
2,
and the risk-neutral probability of T on the second toss, given that thefirst toss is a H, is
q1(H) =u1(H) − 1 − r1(H)
u1(H) − d1(H)=
32 − 1 − 1
432 − 1
=1
2,
If the first toss results in T , then the up factor for the second toss is
u1(T ) =S2(TH)
S1(T )=
8
2= 4,
and the down factor for the second toss is
d1(H) =S2(TT )
S1(T )=
2
2= 1.
It follows that the risk-neutral probability of getting a H on the secondtoss, given that the first toss is a T , is
2.9 Solutions to Selected Exercises 13
p1(T ) =1 + r1(T ) − d1(T )
u1(T ) − d1(T )=
1 + 12 − 1
4 − 1=
1
6,
and the risk-neutral probability of T on the second toss, given that thefirst toss is a T , is
q1(T ) =u1(T ) − 1 − r1(T )
u1(T ) − d1(T )=
4 − 1 − 12
4 − 1=
5
6.
The risk-neutral probabilities are
P(HH) = p0p1(H) = 12 · 1
2 = 14 ,
P(HT ) = p0q1(H) = 12 · 1
2 = 14 ,
P(TH) = q0p1(T ) = 12 · 1
6 = 112 ,
P(TT ) = q0q1(T ) = 12 · 5
6 = 512 .
(ii) We compute
V1(H) =1
1 + r1(H)
[p1(H)V2(HH) + q1(H)V2(HT )
]
=4
5
[1
2· (12 − 7)+ +
1
2· (8 − 7)+
]
= 2.40,
V1(T ) =1
1 + r1(T )
[p1(T )V2(TH) + q1(T )V2(TT )
]
=2
3
[1
6· (8 − 7)+ +
5
6· (2 − 7)+
]
= 0.111111,
V0 =1
1 + r0[p0V1(H) + q0V1(T )]
=4
5
[1
2· 2.40 +
1
2· 0.1111
]
= 1.00444.
We can confirm this price by computing according to the risk-neutralpricing formula in part (i) of the exercise:
14 2 Probability Theory on Coin Toss Space
V0 = E
[V2
(1 + r0)(1 + r1)
]
=V2(HH)
(1 + r0)(1 + r1(H))· P(HH) +
V2(HT )
(1 + r0)(1 + r1(H))· P(HT )
+V2(TH)
(1 + r0)(1 + r1(T ))· P(TH) +
V2(TT )
(1 + r0)(1 + r1(T ))· P(TT )
=(12 − 7)+
(1 + 14 )(1 + 1
4 )· 1
4+
(8 − 7)+
(1 + 14 )(1 + 1
4 )· 1
4
+(8 − 7)+
(1 + 14 )(1 + 1
2 )· 1
12+
(2 − 7)+
(1 + 14 )(1 + 1
2 )· 5
12
= 0.80 + 0.16 + 0.04444 + 0
= 1.00444.
(iii) Formula (1.2.17) still applies and yields
∆0 =V1(H) − V1(T )
S1(H) − S1(T )=
2.40 − 0.111111
8 − 2= 0.381481.
(iv) Again we use formula (1.2.17), this time obtaining
∆1(H) =V2(HH) − V2(HT )
S2(HH) − S2(HT )=
(12 − 7)+ − (8 − 7)+
12 − 8= 1.
Exercise 2.11 (Put–call parity). Consider a stock that pays no dividendin 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-neutralpricing formula (2.4.11):
Cn = En
[CN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Consider also a put with payoff PN = (K − SN )+ at time N , whose price atearlier times is
Pn = En
[PN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Finally, consider a forward contract to buy one share of stock at time N forK dollars. The price of this contract at time N is FN = SN −K, and its priceat earlier times is
Fn = En
[FN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
(Note that, unlike the call, the forward contract requires that the stock bepurchased 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 untilexpiration, explain why the payoff you receive is the same as the payoffof 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 forevery n.
(iii) Using the fact that the discounted stock price is a martingale under therisk-neutral measure, show that F0 = S0 − K
(1+r)N .
(iv) Suppose you begin at time zero with F0, buy one share of stock, borrowingmoney as necessary to do that, and make no further trades. Show thatat 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 forF0 at time zero, you can use this static replication to hedge your shortposition in the forward contract.)
(v) The forward price of the stock at time zero is defined to be that value ofK that causes the forward contract to have price zero at time zero. Theforward price in this model is (1 + r)NS0. Show that, at time zero, theprice of a call struck at the forward price is the same as the price of a putstruck at the forward price. This fact is called put–call parity.
(vi) If we choose K = (1 + r)NS0, we just saw in (v) that C0 = P0. Do wehave 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 ;
Case III: SN < K. Then CN = 0 and PN = K − SN = −FN .
In all three cases, we see that CN = FN + PN .
(ii)
Cn = En
[CN
(1 + r)N−n
]= En
[FN + PN
(1 + r)N−n
]
= En
[FN
(1 + r)N−n
]+ En
[PN
(1 + r)N−n
]= Fn + Pn.
(iii)
F0 = E
[FN
(1 + r)N
]= E
[SN − K
(1 + r)N
]
= E
[SN
(1 + r)N
]− E
[K
(1 + r)N
]= S0 −
K
(1 + r)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
(− K
(1 + r)N
)
= SN − K = FN .
(v) First of all, if K = (1 + r)NS0, 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. AnAsian option has a payoff based on the average stock price, i.e.,
VN = f
(1
N + 1
N∑
n=0
Sn
),
where the function f is determined by the contractual details of the option.
(i) Define Yn =∑n
k=0 Sk and use the Independence Lemma 2.5.3 to showthat 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 issome 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 computingvn(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
,
and whereas Sn and Yn depend only on the first n tosses, Sn+1
Sndepends
only on toss n + 1. According to the Independence Lemma 2.5.3, for anyfunction 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 · Sn+1
Sn
, Yn + Sn · Sn+1
Sn
)]
= hn(Sn, Yn),
where
hn(s, y) = Ehn+1
(s · Sn+1
Sn
, y + s · Sn+1
Sn
)
= phn+1(su, y + su) + qhn+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(
yN+1
). For n = N −1, . . . , 1, 0,
we have from the risk-neutral pricing formula (2.4.12) and (i) above that
Vn =1
1 + rEnVn+1 =
1
1 + rEnvn+1(Sn+1, Yn+1) = vn(Sn, Yn),
where
vn(s, y) =1
1 + r
[pvn+1(su, y + su) + qvn+1(sd, y + sd)
].
3
State Prices
3.7 Solutions to Selected Exercises
Exercise 3.1. Under the conditions of Theorem 3.1.1, show the followinganalogues of properties (i)–(iii) of that theorem:
(i′) P(
1Z
> 0)
= 1;
(ii′) E1Z
= 1;
(iii′) for any random variable Y ,
EY = E
[1
Z· Y]
.
In other words, 1Z
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
Z(ω)=
P(ω)
P(ω)
is defined and positive for every ω ∈ Ω.
(ii′) We compute
E1
Z=∑
ω∈Ω
1
Z(ω)P(ω) =
∑
ω∈Ω
P(ω)
P(ω)P(ω) =
∑
ω∈Ω
P(ω) = 1.
20 3 State Prices
(iii′) We compute
E
[1
Z· Y]
=∑
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 actualprobabilities p = 2
3 , q = 13 , define the estimates of S3 at various times by
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]:
M2(HH) = 23S3(HHH) + 1
2S3(HHT ) = 23 · 32 + 1
3 · 8 = 24,
M2(HT ) = 23S3(HTH) + 1
3S3(HTT ) = 23 · 8 + 1
3 · 2 = 6,
M2(TH) = 23S3(THH) + 1
3S3(THT ) = 23 · 8 + 1
3 · 2 = 6,
M2(TT ) = 23S3(THH) + 1
2S3(THT ) = 23 · 2 + 1
3 · 0.50 = 1.50.
We next compute M1 from the formula M1 = E1[S3]:
M1(H) =4
9S3(HHH) +
2
9S3(HHT ) +
2
9S3(HTH) +
1
9S3(HTT )
=4
9· 32 +
2
9· 8 +
2
9· 8 +
1
9· 2
= 18,
M1(T ) =4
9S3(THH) +
2
9S3(THT ) +
2
9S3(TTH) +
1
9S3(TTT )
=4
9· 8 +
2
9· 2 +
2
9· 2 +
1
9· 0.50
= 4.50.
Finally, we compute
M0 = E[S3]
=8
27S3(HHH) +
4
27S3(HHT ) +
4
27S3(HTH) +
4
27S3(THH)
+2
27S3(HTT ) +
2
27S3(THT ) +
2
27S3(TTH) +
1
27S3(TTT )
=8
27· 32 +
4
27· 8 +
4
27· 8 +
4
27· 8 +
2
27· 2 +
2
27· 2 +
2
27· 2 +
1
27· 0.50
= 13.50.
3.7 Solutions to Selected Exercises 21
ZZ
ZZ
ZZ
!!
!!
!!
aa
aa
aa
ZZ
!!
!!
!!
aa
aa
aa
M0 = 13.50
M1(T ) = 4.50
M1(H) = 18
M2(TT ) = 1.50
M2(HT ) = M2(TH) = 6
M2(HH) = 24
S3(HHH) = 32
S3(HHT ) = S3(HTH)
= S3(THH) = 8
S3(HTT ) = S3(THT )
= S3(TTH) = 2
S3(TTT ) = .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 thatM1 = E1[M2] and M0 = E0[M1] = E[M1], which we do below:
E1[M2](H) = 23M2(HH) + 1
3M2(HT ) = 23 · 24 + 1
3 · 6 = 18 = M1(H),
E1[M2](T ) = 12M2(TH) + 1
2M2(TT ) = 23 · 6 + 1
3 · 1.50 = 4.50 = M1(T ),
M0 = 23M1(H) + 1
3M1(T ) = 23 · 18 + 1
3 · 4.50 = 13.50 = M0.
Exercise 3.5 (Stochastic volatility, random interest rate). Considerthe model of Exercise 2.9 of Chapter 2. Assume that the actual probabilitymeasure is
P(HH) =4
9, P(HT ) =
2
9, P(TH) =
2
9, P(TT ) =
1
9.
The risk-neutral measure was computed in Exercise 2.9 of Chapter 2.
(i) Compute the Radon-Nikodym derivative Z(HH), Z(HT ), Z(TH) and
Z(TT ) of P with respect to P
(ii) The Radon-Nikodym derivative process Z0, Z1, Z2 satisfies Z2 = Z. Com-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 thismodel, which does not use the risk-neutral measure, is
22 3 State Prices
V1(H) =1 + r0
Z1(H)E1
[Z2
(1 + r0)(1 + r1)V2
](H)
=1
Z1(H)(1 + r1(H))E1[Z2V2](H),
V1(T ) =1 + r0
Z1(T )E1
[Z2
(1 + r0)(1 + r1)V2
](T )
=1
Z1(T )(1 + r1(T ))E1[Z2V2](T ),
V0 = E
[Z2
(1 + r0)(1 + r1)V2
].
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
4, P(HT ) =
1
4, P(TH) =
1
12, P(TT ) =
5
12.
Therefore, the Radon-Nikodym derivative is
Z(HH) =P(HH)
P(HH)=
1
4· 9
4=
9
16, Z(HT ) =
P(HT )
P(HT )=
1
4· 9
2=
9
8,
Z(TH) =P(TH)
P(TH)=
1
12· 9
2=
3
8, Z(TT ) =
P(TT )
P(TT )=
5
12· 9
1=
15
4,
(ii)
Z1(H) = E1[Z2](H)
= Z2(HH)Pω2 = H given that ω1 = H+Z2(HT )Pω2 = T given that ω1 = H
= Z2(HH)P(HH)
P(HH) + P(HT )+ Z2(HT )
P(HT )
P(HH) + P(HT )
=9
16·
49
49 + 2
9
+9
8·
29
49 + 2
9
=3
4,
3.7 Solutions to Selected Exercises 23
Z1(T ) = E1[Z2](T )
= Z2(TH)Pω2 = H given that ω1 = T+Z2(TT )Pω2 = T given that ω1 = T
= Z2(TH)P(TH)
P(TH) + P(TT )+ Z2(TT )
P(TT )
P(TH) + P(TT )
=3
8·
29
29 + 1
9
+15
4·
19
29 + 1
9
=3
2,
Z0 = E0[Z1]
= E[Z1]
= Z1(H)(P(HH) + P(HT )
)+ Z1(T )
(P(TH) + P(TT )
)
=3
4·(
4
9+
2
9
)+
3
2·(
2
9+
1
9
)
= 1.
We may also check directly that EZ = 1, as follows:
EZ = Z(HH)P(HH) + Z(HT )P(HT ) + Z(TH)P(TH) + Z(TT )P(TT )
=9
16· 4
9+
9
8· 2
9+
3
8· 2
9+
15
4· 1
9
=1
4+
1
4+
1
12+
5
12= 1.
(iii) We recall that
V2(HH) = 5, V2(HT ) = 1, V2(TH) = 1, V2(TT ) = 0.
We computed in part (ii) that
Pω2 = H given that ω1 = H =49
49+ 2
9
=2
3,
Pω2 = T given that ω1 = H =29
49+ 2
9
=1
3,
Pω2 = H given that ω1 = T =29
29+ 1
9
=2
3,
Pω2 = T given that ω1 = T =19
29+ 1
9
=1
3.
Therefore,
24 3 State Prices
V1(H) =1
Z1(H)(1 + r1(H)
)E1[Z2V2](H)
=
(3
4· 5
4
)−1 [Z2(HH)V2(HH)Pω2 = H given that ω1 = H
+Z2(HT )V2(HT )Pω2 = T given that ω1 = H]
=16
15
[9
16· 5 · 2
3+
9
8· 1 · 1
3
]
= 2.40,
V1(T ) =1
Z1(T )(1 + r1(T )
)E1[Z2V2](T )
=
(3
2· 3
2
)−1 [Z2(TH)V2(TH)Pω2 = H given that ω1 = T
+Z2(TT )V2(TT )Pω2 = T given that ω1 = T]
=4
9
[3
8· 1 · 2
3+
15
4· 0 · 1
3
]
= 0.111111,
and
V0 = E
[Z2V2
(1 + r0)(1 + r1)
]
=Z2(HH)V2(HH)
(1 + r0)(1 + r1(H))P(HH) +
Z2(HT )V2(HT )
(1 + r0)(1 + r1(H))P(HT )
+Z2(TH)V2(TH)
(1 + r0)(1 + r1(T ))P(TH) +
Z2(TT )V2(TT )
(1 + r0)(1 + r1(T ))P(TT )
=
(5
4· 5
4
)−19
16· 5 · 4
9+
(5
4· 5
4
)−19
8· 1 · 2
9+
(5
4· 3
2
)−13
8· 1 · 2
9
+
(5
4· 3
2
)−115
4· 0 · 1
9
=16
25· 5
4+
16
25· 1
4+
8
15· 1
12= 1.00444.
Exercise 3.6. Consider Problem 3.3.1 in an N -period binomial model withthe utility function U(x) = ln x. Show that the optimal wealth process cor-responding to the optimal portfolio process is given by Xn = X0
ζn, n =
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
When U(x) = ln x, U ′(x) = 1x
and the inverse function of U ′ is I(y) = 1y.
Therefore,
XN =1
λζN
We must choose λ to satisfy (3.3.26), which in this case takes the form
X0 = E[ζNI(λζN )
]=
1
λ.
Substituting this into the previous equation, we obtain
XN =X0
ζN
.
Because Xn
(1+r)n is a martingale under the risk-neutral measure P, we have
Xn
(1 + r)n= En
[XN
(1 + r)N
]=
1
Zn
En
[ZNXN
(1 + r)N
]=
1
Zn
En [ζNXN ] =X0
Zn
.
Therefore,
Xn =(1 + r)nX0
Zn
=X0
ζn
.
Exercise 3.8. The Lagrange Multiplier Theorem used in the solution of Prob-lem 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 (p1ζ1, . . . , pmζm), is not the zero vector, thenthe 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 thatthere is an optimal solution; the solution to the Lagrange multiplier equa-tions may in fact minimize the expected utility. The solution could also beneither a maximizer nor a minimizer. Therefore, in this exercise, we outline adifferent 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 givenby (3.3.25) X∗
N rather than XN . In other words,
X∗N = I
(λ
(1 + r)NZ
), (3.6.1)
where λ is the solution of equation (3.3.26). This permits us to use the nota-tion XN for an arbitrary (not necessarily optimal) random variable satisfying(3.3.19). We must show that
EU(XN ) ≤ EU(X∗N ). (3.6.2)
26 3 State Prices
(i) Fix y > 0, and show that the function of x given by U(x)−yx is maximizedby 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
and replace the dummy variable y by the random variable λZ(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 functionof x, the difference U(x) − yx is a concave function of x. The derivativeof 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 atthe point where its derivative is zero. The inequality (3.6.2) is just thisstatement.
(ii) Making the suggested replacements, we obtain
U(XN ) − λZXN
(1 + r)N≤ U
(I
(λZ
(1 + r)N
))− λZ
(1 + r)NI
(λZ
(1 + r)N
).
Taking expectations under P and using the fact that Z is the Radon-Nikodym derivative of P with respect to P, we obtain
EU(XN ) − λEXN
(1 + r)N
≤ EU
(I
(λZ
(1 + r)N
))− λE
[Z
(1 + r)NI
(λZ
(1 + r)N
)].
From (3.3.19) and (3.3.26), we have
EXN
(1 + r)N= X0 = E
[Z
(1 + r)NI
(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 theinterest rate be r = 1
4 so the risk-neutral probabilities are p = q = 12 .
(i) Determine the price at time zero, denoted V P0 , 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 V C0 , 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 V S0 , of the American straddle
that expires at time three and has intrinsic value gS(s) = gP (s) + gC(s).
(iv) Explain why V S0 < V P
0 + V C0 .
Solution
(i) The payoff of the put at expiration time three is
V P3 (HHH) = (4 − 32)+ = 0,
V P3 (HHT ) = V P
3 (HTH) = V P3 (THH) = (4 − 8)+ = 0,
V P3 (HTT ) = V P
3 (THT ) = V P3 (TTH) = (4 − 2)+ = 2,
V P3 (TTT ) = (4 − 0.50)+ = 3.50.
Because 11+r
p = 11+r
q = 25 , the value of the put at time two is
28 4 American Derivative Securities
V P2 (HH) = max
(4 − S2(HH)
)+,2
5V P
3 (HHH) +2
5V P
3 (HHT )
= max
(4 − 16)+,
2
5· 0 +
2
5· 0
= max0, 0= 0,
V P2 (HT ) = max
(4 − S2(HT )
)+,2
5V P
3 (HTH) +2
5V P
3 (HTT )
= max
(4 − 4)+,
2
5· 0 +
2
5· 2
= max0, 0.80= 0.80,
V P2 (TH) = max
(4 − S2(TH)
)+,2
5V P
3 (THH) +2
5V P
3 (THT )
= max
(4 − 4)+,
2
5· 0 +
2
5· 2
= max0, 0.80= 0.80,
V P2 (TT ) = max
(4 − S2(TT )
)+,2
5V P
3 (TTH) +2
5V P
3 (TTT )
= max
(4 − 1)+,
2
5· 2 +
2
5· 3.50
= max3, 2.20= 3.
At time one the value of the put is
V P1 (H) = max
(4 − S1(H)
)+,2
5V P
2 (HH) +2
5V P
2 (HT )
= max
(4 − 8)+,
2
5· 0 +
2
5· 0.80
= max0, 0.32= 0.32,
V P1 (T ) = max
(4 − S1(T )
)+,2
5V P
2 (TH) +2
5V P
2 (TT )
= max
(4 − 2)+,
2
5· 0.80 +
2
5· 3
= max2, 1.52= 2.
The value of the put at time zero is
4.9 Solutions to Selected Exercises 29
V P0 = max
(4 − S0)
+,2
5V P
1 (H) +2
5V P
1 (T )
= max
(4 − 4)+,
2
5· 0.32 +
2
5· 2
= max0, 0.928= 0.928.
(ii) The payoff of the call at expiration time three is
V C3 (HHH) = (32 − 4)+ = 28,
V C3 (HHT ) = V C
3 (HTH) = V C3 (THH) = (8 − 4)+ = 4,
V C3 (HTT ) = V C
3 (THT ) = V C3 (TTH) = (2 − 4)+ = 0,
V C3 (TTT ) = (0.50 − 4)+ = 0.
Because 11+r
p = 11+r
q = 25 , the value of the call at time two is
V C2 (HH) = max
(S2(HH) − 4
)+,2
5V C
3 (HHH) +2
5V C
3 (HHT )
= max
(16 − 4)+,
2
5· 28 +
2
5· 4
= max12, 12.8= 12.8,
V C2 (HT ) = max
(S2(HT ) − 4
)+,2
5V C
3 (HTH) +2
5V C
3 (HTT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 0
= max0, 1.60= 1.60,
V C2 (TH) = max
(S2(TH) − 4
)+,2
5V C
3 (THH) +2
5V C
3 (THT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 0
= max0, 1.60= 1.60,
V C2 (TT ) = max
(S2(TT ) − 4
)+,2
5V C
3 (TTH) +2
5V C
3 (TTT )
= max
(1 − 4)+,
2
5· 0 +
2
5· 0
= max0, 0= 0.
At time one the value of the call is
30 4 American Derivative Securities
V C1 (H) = max
(S1(H) − 4
)+,2
5V C
2 (HH) +2
5V C
2 (HT )
= max
(8 − 4)+,
2
5· 12.8 +
2
5· 1.60
= max4, 5.76 = 5.76,
V C1 (T ) = max
(S1(T ) − 4
)+,2
5V C
2 (TH) +2
5V C
2 (TT )
= max
(2 − 4)+,
2
5· 1.60 +
2
5· 0
= max0, 0.64 = 0.64.
The value of the call at time zero is
V C0 = max
(S0 − 4)+,
2
5V C
1 (H) +2
5V C
1 (T )
= max
(4 − 4)+,
2
5· 5.76 +
2
5· 0.64
= max0, 2.56= 2.56.
(iii) Note that gS(s) = |s − 4|. The payoff of the straddle at expiration timethree is
V S3 (HHH) = |32 − 4| = 28,
V S3 (HHT ) = V S
3 (HTH) = V S3 (THH) = |8 − 4| = 4,
V S3 (HTT ) = V S
3 (THT ) = V S3 (TTH) = |2 − 4| = 2,
V S3 (TTT ) = |0.50 − 4| = 3.50.
We see that the payoff of the straddle is the payoff of the put given inthe solution to (i) plus the payoff of the call given in the solution to (ii).Because 1
1+rp = 1
1+rq = 2
5 , the value of the straddle at time two is
V S2 (HH) = max
∣∣S2(HH) − 4∣∣, 2
5V S
3 (HHH) +2
5V S
3 (HHT )
= max
|16 − 4|, 2
5· 28 +
2
5· 4
= max12, 12.8= 12.8,
V S2 (HT ) = max
∣∣S2(HT ) − 4∣∣+,
2
5V S
3 (HTH) +2
5V S
3 (HTT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 2
= max0, 2.40= 2.40,
4.9 Solutions to Selected Exercises 31
V S2 (TH) = max
(S2(TH) − 4
)+,2
5V S
3 (THH) +2
5V S
3 (THT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 2
= max0, 2.40= 2.40,
V S2 (TT ) = max
∣∣S2(TT ) − 4∣∣, 2
5V S
3 (TTH) +2
5V S
3 (TTT )
= max
|1 − 4|, 2
5· 2 +
2
5· 3.50
= max3, 2.20= 3.
One can verify in every case that V S2 = V P
2 + V C2 . At time one the value
of the straddle is
V S1 (H) = max
∣∣S1(H) − 4∣∣, 2
5V S
2 (HH) +2
5V S
2 (HT )
= max
|8 − 4|, 2
5· 12.8 +
2
5· 2.40
= max4, 6.08= 6.08,
V S1 (T ) = max
∣∣S1(T ) − 4∣∣, 2
5V S
2 (TH) +2
5V S
2 (TT )
= max
|2 − 4|, 2
5· 2.40 +
2
5· 3
= max2, 2.16= 2.16.
We have V S1 (H) = 6.08 = 0.32 + 5.76 = V P
1 (H) + V C1 (H), but V S
1 (T ) =2.16 < 2 + 0.64 = V P
1 (T ) + V C1 (T ).
The value of the straddle at time zero is
V S0 = max
|S0 − 4|, 2
5V S
1 (H) +2
5V S
1 (T )
= max
|4 − 4|, 2
5· 6.08 +
2
5· 2.16
= max0, 3.296= 3.296.
We have V S0 = 3.296 < 0.928 + 2.56 = V P
0 + V C0 .
32 4 American Derivative Securities
(iv) For the put, if there is a tail on the first toss, it is optimal to exercise attime one. This can be seen from the equation
V P1 (T ) = max
(4 − S1(T )
)+,2
5V P
2 (TH) +2
5V P
2 (TT )
= max
(4 − 2)+,
2
5· 0.80 +
2
5· 3
= max2, 1.52= 2,
which shows that the intrinsic value at time one if the first toss resultsin T is greater than the value of continuing. On the other hand, for thecall 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 onthe first toss.
The straddle has the intrinsic value of a put plus a call. When it is ex-ercised, both parts of the payoff are received. In other words, it is notan American put plus an American call, because these can be exercisedat different times whereas the exercise of a straddle requires both theput payoff and the call payoff to be received. In the computation of thestraddle price
V S1 (T ) = max
∣∣S1(T ) − 4∣∣, 2
5V S
2 (TH) +2
5V S
2 (TT )
= max
|2 − 4|, 2
5· 2.40 +
2
5· 3
= max2, 2.16= 2.16,
we see that it is not optimal to exercise the straddle at time one if the firsttoss results in T . It would be optimal to exercise the put part, but not thecall part, and the straddle cannot exercise one part without exercising theother. Greater value is achieved by not exercising both parts than wouldbe achieved by exercising both. However, this value is less than would beachieved if one could exercise the put part and let the call part continue,and thus V S
1 (T ) < V P1 (T ) + V C
1 (T ). This loss of value at time one resultsin a similar loss of value at the earlier time zero: V S
0 < V P0 + V C
0 .
Exercise 4.3. In the three-period model of Figure 1.2.2 of Chapter 1, let theinterest rate be r = 1
4 so the risk-neutral probabilities are p = q = 12 . Find
the time-zero price and optimal exercise policy (optimal stopping time) forthe path-dependent American derivative security whose intrinsic value at each
time n, n = 0, 1, 2, 3, is(4 − 1
n+1
∑n
j=0 Sj
)+
. This intrinsic value is a put on
the average stock price between time zero and time n.
4.9 Solutions to Selected Exercises 33
Solution The intrinsic value process for this option is
G0 = (4 − S0)+
= (4 − 4)+ = 0,
G1(H) =(4 − S0+S1(H)
2
)+
=(4 − 4+8
2
)+= 0,
G1(T ) =(4 − S0+S1(T )
2
)+
=(4 − 4+2
2
)+= 1,
G2(HH) =(4 − S0+S1(H)+S2(HH)
3
)+
=(4 − 4+8+16
3
)+= 0,
G2(HT ) =(4 − S0+S1(H)+S2(HT )
3
)+
=(4 − 4+8+4
3
)= 0,
G2(TH) =(4 − S0+S1(T )+S2(TH)
3
)+
=(4 − 4+2+4
3
)+= 0.6667,
G2(TT ) =(4 − S0+S1(T )+S2(TT )
3
)+
=(r − 4+2+1
3
)= 1.6667.
At time three, the intrinsic value G3 agrees with the option value V3. In otherwords,
V3(HHH) = G3(HHH)
=
(4 − S0 + S1(H) + S2(HH) + S3(HHH)
4
)+
=
(4 − 4 + 8 + 16 + 32
4
)+
= 0,
V3(HHT ) = G3(HHT )
=
(4 − S0 + S1(H) + S2(HH) + S3(HHT )
4
)+
=
(4 − 4 + 8 + 16 + 8
4
)+
= 0,
V3(HTH) = G3(HTH)
=
(4 − S0 + S1(H) + S2(HT ) + S3(HTH)
4
)+
=
(4 − 4 + 8 + 4 + 8
4
)+
= 0,
V3(HTT ) = G3(HTT )
=
(4 − S0 + S1(H) + S2(HT ) + S3(HTT )
4
)+
=
(4 − 4 + 8 + 4 + 2
4
)+
= 0,
34 4 American Derivative Securities
V3(THH) = G3(THH)
=
(4 − S0 + S1(T ) + S2(TH) + S3(THH)
4
)+
=
(4 − 4 + 2 + 4 + 8
4
)+
= 0,
V3(THT ) = G3(THT )
=
(4 − S0 + S1(T ) + S2(TH) + S3(THT )
4
)+
=
(4 − 4 + 2 + 4 + 2
4
)+
= 1,
V3(TTH) = G3(TTH)
=
(4 − S0 + S1(T ) + S2(TT ) + S3(TTH)
4
)+
=
(4 − 4 + 2 + 1 + 2
4
)+
= 1.75,
V3(TTT ) = G3(TTT )
=
(4 − S0 + S1(T ) + S2(TT ) + S3(TTT )
4
)+
=
(4 − 4 + 2 + 1 + 0.50
4
)+
= 2.125.
We use the algorithm of Theorem 4.4.3, noting that p1+r
= q1+r
= 25 , to obtain
V2(HH) = max
G2(HH),
2
5V3(HHH) +
2
5V3(HHT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
V2(HT ) = max
G2(HT ),
2
5V3(HTH) +
2
5V3(HTT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
4.9 Solutions to Selected Exercises 35
V2(TH) = max
G2(TH),
2
5V3(THH) +
2
5V3(THT )
= max
0.6667,
2
5· 0 +
2
5· 1
= max0.6667, 0.40= 0.6667,
V2(TT ) = max
G2(TT ),
2
5V3(TTH) +
2
5V3(TTT )
= max
1.6667,
2
5· 1.75 +
2
5· 2.125
= max1.6667, 1.55= 1.6667.
Continuing, we have
V1(H) = max
G1(H),
2
5V2(HH) +
2
5V2(HT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
V1(T ) = max
G1(T ),
2
5V2(TH) +
2
5V2(TT )
= max
1,
2
5· 0.6667 +
2
5· 1.6667
= max1, 0.9334= 1,
V0 = max
G0,
2
5V1(H) +
2
5V1(T )
= max
0,
2
5· 0 +
2
5· 1
= max0, 0.40= 0.40.
To find the optimal exercise time, we work forward. Since V0 > G0, oneshould not exercise at time zero. However, V1(T ) = G1(T ), so it is optimal toexercise at time one if there is a T on the first toss. If the first toss results inH, 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
exercise rule we choose in this case. If the payoff were(4 − 1
n+1
∑n
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 stop-ping times in S0. List all the stopping times in S0 (there are 26), and fromamong those, list the stopping times that never exercise when the option isout of the money (there are 11). For each stopping time τ in the latter set,compute E
[Iτ≤2
(45
)τGτ
]. Verify that the largest value for this quantity is
given by the stopping time of (4.4.6), the one which makes this quantity equalto the 1.36 computed in (4.4.7).
Solution A stopping time is a random variable, and we can specify a stoppingtime by listing its values τ(HH), τ(HT ), τ(TH), and τ(TT ). The stoppingtime property requires that τ(HH) = 0 if and only if τ(HT ) = τ(TH) =τ(TT ) = 0. Similarly, τ(HH) = 1 if and only if τ(HT ) = 1 and τ(TH) = 1if and only if τ(TT ) = 1. The 26 stopping times in the two-period binomialmodel are tabulated below.
StoppingTime HH HT TH TT
τ1 0 0 0 0τ2 1 1 1 1τ3 1 1 2 2τ4 1 1 2 ∞τ5 1 1 ∞ 2τ6 1 1 ∞ ∞τ7 2 2 1 1τ8 2 2 2 2τ9 2 2 2 ∞τ10 2 2 ∞ 2τ11 2 2 ∞ ∞τ12 2 ∞ 1 1τ13 2 ∞ 2 2τ14 2 ∞ 2 ∞τ15 2 ∞ ∞ 2τ16 2 ∞ ∞ ∞τ17 ∞ 2 1 1τ18 ∞ 2 2 2τ19 ∞ 2 2 ∞τ20 ∞ 2 ∞ 2τ21 ∞ 2 ∞ ∞τ22 ∞ ∞ 1 1τ23 ∞ ∞ 2 2τ24 ∞ ∞ 2 ∞τ25 ∞ ∞ ∞ 2τ26 ∞ ∞ ∞ ∞
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(TH) = 1, G2(TT ) = 4.
The stopping times that take the value 1 when there is an H on the first tossare 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 thefirst 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 theoption is in the money. This leaves us with τ1 and the ten stopping timesτ17–τ26. We evaluate the risk-neutral expected payoff of these eleven stoppingtimes.
E
[Iτ1≤2
(4
5
)τ
Gτ1
]= G0 = 1,
E
[Iτ17≤2
(4
5
)τ
Gτ17
]=
1
4· 16
26G2(HT ) +
1
2· 4
5G1(T )
=4
25· 1 +
2
5· 3 = 1.36,
E
[Iτ18≤2
(4
5
)τ
Gτ18
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TH) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 1 +
4
25· 4 = 0.96,
E
[Iτ19≤2
(4
5
)τ
Gτ19
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TH)
=4
25· 1 +
4
25· 1 = 0.32,
E
[Iτ20≤2
(4
5
)τ
Gτ20
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 4 = 0.80,
E
[Iτ21≤2
(4
5
)τ
Gτ21
]=
1
4· 16
26G2(HT )
=4
25· 1 = 0.16,
E
[Iτ22≤2
(4
5
)τ
Gτ22
]=
1
2· 4
5G1(T )
=2
5· 3 = 1.20,
E
[Iτ23≤2
(4
5
)τ
Gτ23
]= +
1
4· 16
25G2(TH) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 4 = 0.80,
38 4 American Derivative Securities
E
[Iτ24≤2
(4
5
)τ
Gτ24
]= +
1
4· 16
25G2(TH)
=4
25· 1 = 0.16,
E
[Iτ25≤2
(4
5
)τ
Gτ25
]= +
1
4· 16
25G2(TT )
= +4
25· 4 = 0.64,
E
[Iτ26≤2
(4
5
)τ
Gτ26
]= 0.
The largest value, 1.36, is obtained by the stopping time τ17.
Exercise 4.7. For the class of derivative securities described in Exercise 4.6whose time-zero price is given by (4.8.3), let Gn = Sn − K. This derivativesecurity permits its owner to buy one share of stock in exchange for a paymentof K at any time up to the expiration time N . If the purchase has not beenmade at time N , it must be made then. Determine the time-zero value andoptimal exercise policy for this derivative security. (Assume r ≥ 0.)
Solution Set Yn = 1(1+r)n (Sn − K), n = 0, 1, . . . , N . We assume r ≥ 0.
Because the discounted stock price is a martingale under the risk-neutralmeasure and K
(1+r)n+1 is not random, we have
En[Yn+1] = En
[Sn+1
(1 + r)n+1
]− En
[K
(1 + r)n+1
]
=Sn
(1 + r)n− K
(1 + r)n+1
≥ Sn
(1 + r)n− K
(1 + r)n.
This shows that Yn, n = 0, 1, . . . , N , is a submartingale. According to Theorem4.3.3 (Optional Sampling—Part II), EYN∧τ ≤ EYN whenever τ is a stoppingtime. If τ is a stopping time satisfying τ(ω) ≤ N for every sequence of coin
tosses ω, this becomes EYτ ≤ EYN , or equivalently,
E
[1
(1 + r)τGτ
]≤ E
[1
(1 + r)NGN
].
Therefore,
V0 = maxτ∈S0,τ≤N
E
[1
(1 + r)τGτ
]≤ E
[1
(1 + r)NGN
].
On the other hand, because the stopping time that is equal to N regardlessof the outcome of the coin tossing is in the set of stopping times over whichthe above maximum is taken, we must in fact have equality:
4.9 Solutions to Selected Exercises 39
V0 = maxτ∈S0,τ≤N
E
[1
(1 + r)τGτ
]= E
[1
(1 + r)NGN
].
Hence, it is optimal to exercise at the final time N regardless of the outcomeof the coin tossing.
5
Random Walk
5.8 Solutions to Selected Exercises
Exercise 5.2. (First passage time for random walk with upwarddrift) Consider the asymmetric random walk with probability p for an upstep and probability q = 1 − p for a down step, where 1
2 < p < 1 so that0 < q < 1
2 . In the notation of (5.2.1), let τ1 be the first time the random walkstarting from level 0 reaches the level 1. If the random walk never reaches thislevel, 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
(1
f(σ)
)n
is a martingale.
(iii) Show that for σ > 0,
e−σ = E
[Iτ1<∞
(1
f(σ)
)τ1]
.
Conclude that Pτ1 < ∞ = 1.
(iv) Compute Eατ1 for α ∈ (0, 1).
(v) Compute Eτ1.
Solution.
(i) The function f(σ) satisfies f(0) = 1 and f ′(σ) = peσ−qe−σ, f ′′(σ) = f(σ).Since f is always positive, f ′′ is always positive and f is convex. Underthe assumption p > 1/2 > q, we have f ′(0) = p−q > 0, and the convexityof f implies that f(σ) > 1 for all σ > 0.
42 5 Random Walk
(ii) We compute
En[Sn+1] =
(1
f(σ)
)n+1
En
[eσ(Mn+Xn+1)
]
=
(1
f(σ)
)n+1
eσMnE[eσXn+1
]
=
(1
f(σ)
)n+1
eσMn(peσ + qe−σ
)
= eσMn
(1
f(σ)
)n
= Sn.
(iii) Because Sn∧τ1is 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,
limn→∞
eσMn∧τ1
(1
f(σ)
)n∧τ1
= Iτ1<∞eσ
(1
f(σ)
)τ
.
Letting n → ∞ in (5.8.1), we obtain
1 = E
[Iτ1<∞e
σ
(1
f(σ)
)τ1]
,
or equivalently,
e−σ = E
[Iτ1<∞
(1
f(σ)
)τ1]
. (5.8.2)
This equation holds for all positive σ. We let σ ↓ 0 to obtain the formula
1 = EIτ1<∞ = Pτ1 < ∞.
(iv) We now introduce α ∈ (0, 1) and solve the equation α = 1f(σ) for e−σ.
This equation can be written as
αpeσ + αqe−σ = 1,
which may be rewritten as
αq(e−σ
)2 − e−σ + αp = 0,
and the quadratic formula gives
5.8 Solutions to Selected Exercises 43
e−σ =1 ±
√1 − 4α2pq
2αq. (5.8.3)
We want σ to be positive, so we need e−σ to be less than 1. Hence wetake the negative sign, obtaining
e−σ =1 −
√1 − 4α2pq
2αq.
Substituting this into (5.8.2), we obtain the formula
Eατ1 =1 −
√1 − 4α2pq
2αq. (5.8.4)
(v) Differentiating (5.8.4) with respect to α leads to
Eτατ−1 =1 −
√1 − 4α2pq
2α2q√
1 − 4α2pq,
and letting α ↑ 1, we obtain
Eτ1 =1 −√
1 − 4pq
2q√
1 − 4pq=
1 −√
(1 − 2q)2
2q√
(1 − 2q)2=
1 − (1 − 2q)
2q(1 − 2q)=
1
1 − 2q=
1
p − q.
Exercise 5.4 (Distribution of τ2). Consider the symmetric random walk,and let τ2 be the first time the random walk, starting from level 0, reachesthe level 2. According to Theorem 5.2.3,
Eατ2 =
(1 −
√1 − α2
α
)2
for all α ∈ (0, 1).
Using the power series (5.2.21), we may write the right-hand side as
(1 −
√1 − α2
α
)2
=2
α· 1 −
√1 − α2
α− 1
= −1 +
∞∑
j=1
(α
2
)2j−2 (2j − 2)!
j!(j − 1)!
=
∞∑
j=2
(α
2
)2j−2 (2j − 2)!
j!(j − 1)!
=
∞∑
k=1
(α
2
)2k (2k)!
(k + 1)!k!.
(i) Use the above power series to determine Pτ2 = 2k, k = 1, 2, . . . .
44 5 Random Walk
(ii) Use the reflection principle to determine Pτ2 = 2k, k = 1, 2, . . . .
Solution
(i) Since the random walk can reach the level 2 only on even-numbered steps,Pτ2 = 2k − 1 = 0 for k = 1, 2, . . . . Therefore,
Eατ2 =∞∑
k=1
α2kPτ2 = 2k =
∞∑
k=1
(α
2
)2k (2k)!
(k + 1)!k!.
Equating coefficients, we conclude that
Pτ = 2k =(2k)!
(k + 1)!k!
(1
2
)2k
, k = 1, 2, . . . .
(ii) For k = 1, 2, . . . , the number of paths that reach or exceed level 2 by time2k is equal to the number of paths that are at level 2 at time 2k plus thenumber of paths that exceed level 2 at time 2k plus the number of pathsthat reach the level 2 before time 2k but are below level 2 at time 2k. Apath is of the last type if and only if its reflected path exceeds level 2 attime 2k. Thus, the number of paths that reach or exeed level 2 by time 2kis equal to the number of paths that are at level 2 at time 2k plus twicethe number of paths that exeed level 2 at time 2k. For the symmetricrandom walk, every path has the same probability, and hence
Pτ2 ≤ 2k = PM2k = 2 + 2PM2k ≥ 4= PM2k = 2 + PM2k ≥ 4 + PM2k ≤ −4= 1 − PM2k = 0 − PM2k = −2
= 1 − (2k)!
k!k!
(1
2
)2k
− (2k)!
(k + 1)!(k − 1)!
(1
2
)2k
.
Consequently,
Pτ2 = 2k= Pτ2 ≤ 2k − Pτ2 ≤ 2k − 2
=
(1
2
)2k−2 [(2k − 2)!
(k − 1)!(k − 1)!+
(2k − 2)!
k!(k − 2)!
]
−(
1
2
)2k [(2k)!
k!k!+
(2k)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [4(2k − 2)!
(k − 1)!(k − 1)!− (2k)!
k!k!
]
+
(1
2
)2k [4(2k − 2)!
k!(k − 2)!− (2k)!
(k + 1)!(k − 1)!
]
5.8 Solutions to Selected Exercises 45
=
(1
2
)2k [2k · 2k(2k − 2)! − 2k(2k − 1)(2k − 2)!
k!k!
]
+
(1
2
)2k [(2k + 2)(2k − 2)(2k − 2)! − 2k(2k − 1)(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [4k2(2k − 2)! − (4k2 − 2k)(2k − 2)!
k!k!
]
+
(1
2
)2k [(4k2 − 4)(2k − 2)! − (4k2 − 2k))(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [2k(2k − 2)!
k!k!+
2(k − 2)(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k(2k − 2)!
(k + 1)!k!
[(2k(k + 1) + 2(k − 2)k
]
=
(1
2
)2k(2k − 2)!
(k + 1)!k!2k(2k − 1)
=
(1
2
)2k(2k)!
(k + 1)!k!.
Exercise 5.7. (Hedging a short position in the perpetual Americanput). Suppose you have sold the perpetual American put of Section 5.4 andare hedging the short position in this put. Suppose at the current time thestock price is s and the value of your hedging portfolio is v(s). Your hedge isto first consume the amount
c(s) = v(s) − 4
5
[1
2v(2s) +
1
2v(s
2
)](5.7.3)
and then take a position
δ(s) =v(2s) − v
(s2
)
2s − s2
(5.7.4)
in the stock. (See Theorem 4.2.2 of Chapter 4. The processes Cn and ∆n inthat theorem are obtained by replacing the dummy variable s by the stockprice Sn in (5.7.3) and (5.7.4); i.e., Cn = c(Sn) and ∆n = δ(Sn).) If you hedgethis way, then regardless of whether the stock goes up or down on the nextstep, the value of your hedging portfolio should agree with the value of theperpetual American put.
(i) Compute c(s) when s = 2j for the three cases j ≤ 0, j = 1 and j ≥ 2.
(ii) Compute δ(s) when s = 2j for the three cases j ≤ 0, j = 1 and j ≥ 2.
46 5 Random Walk
(iii) Verify in each of the three cases s = 2j for j ≤ 0, j = 1 and j ≥ 2 thatthe hedge works (i.e., regardless of whether the stock goes up or down,the value of your hedging portfolio at the next time is equal to the valueof the perpetual American put at that time).
Solution
(i) Throughout the following computations, we use (5.4.6):
v(2j) =
4 − 2j , if j ≤ 1,4
2j, if j ≥ 1.
For j ≤ 0, we have
c(2j) = v(2j) − 4
5
[1
2v(2j+1) +
1
2v(2j−1)
]
= 4 − 2j − 2
5
[4 − 2j+1 + 4 − 2j−1
]
= 4 − 2j − 2
5
[8 − 5 · 2j−1
]
=4
5.
For j = 1, we have
c(2) = v(2) − 4
5
[1
2v(4) +
1
2v(1)
]
= 2 − 2
5[1 + 3]
=2
5.
Finally, for j ≥ 2,
c(2j) = v(2j) − 4
5
[1
2v(2j+1) +
1
2v(2j−1)
]
=4
2j− 2
5
[4
2j+1+
4
2j−1
]
=4
2j− 2
5
[4
2j+1+
16
2j+1
]
= 0.
(ii) For j ≤ 0, we have
δ(2j) =v(2j+1) − v(2j−1)
2j+1 − 2j−1=
(4 − 2j+1
)−(4 − 2j−1
)
2j+1 − 2j−1= −1.
5.8 Solutions to Selected Exercises 47
For j = 1, we have
δ(2) =v(4) − v(1)
4 − 1=
1 − 3
4 − 1= −2
3.
Finally, for j ≥ 2,
δ(2j) =v(2j+1) − v(2j−1)
2j+1 − 2j−1
=
4
2j+1− 4
2j−1
2j+1 − 2j−1
=4 − 16
2j+1· 1
(4 − 1)2j−1
= − 4
22j.
(iii) If the stock price is 2j , where j ≤ 0, then we have a portfolio valuedat v(2j) = 4 − 2j . We consume c(2j) = 4
5 and short one share of stock(δ(2j) = −1). This creates a cash position of
4 − 2j − 4
5+ 2j =
16
5,
which is invested in the money market. When we go to the next period,the money market investment grows to 5
4 · 165 = 4. If the stock goes up to
2j+1, our short stock position has value −2j+1, so our portfolio is valuedat 4− 2j+1, which is the option value in this case. If the stock goes downto 2j−1, our portfolio is valued at 4− 2j−1, which is the option value alsoin this case.
When the stock price is 2j with j ≤ 0, the owner of the option shouldexercise, and we are prepared for that by being short one share of stock.When she exercises, she will deliver the share of stock, which will coverour short position. But when she exercises, we must pay her 4, so we aremaintaining a cash position of 4. At the beginning of any period in whichshe fails to exercise, we get to consume the present value of the interestthat will accrue to this cash position over the next period.
If the stock price is 2, then we have a portfolio valued at v(2) = 2. Weconsume c(2) = 2
5 and short 23 of a share of stock (δ(2) = − 2
3 ). Thiscreates a cash position of
2 − 2
5+
2
3· 2 =
44
15,
which is invested in the money market. When we go to the next period,the money market investment grows to 5
4 · 4415 = 11
3 . If the stock goes up
48 5 Random Walk
to 4, the short stock position has value − 23 · 4 = − 8
3 , so the portfolio isvalued at 11
3 − 83 = 1, which is the option value in this case. If the stock
goes down to 1, the short stock position has value − 23 · 1 = − 2
3 , so theportfolio is valued at 11
3 − 23 = 3, which is the option value also in this
case.
Finally, if the stock price is 2j , where j ≥ 2, then we have a portfolio valuedat 4
2j . We consume c(2j) = 0 and short 422j shares of stock (δ(2j) = − 4
22j ).This creates a cash position of
4
2j+
4
22j· 2j =
8
2j,
which is invested in the money market. When we go to the next period,the money market investment grows to
5
4· 8
2j=
10
2j.
If the stock goes up to 2j+1, the short stock position has value
− 4
22j· 2j+1 = − 8
2j,
so the portfolio is valued at
10
2j− 8
2j=
2
2j=
4
2j+1,
which is the option value in this case. If the stock goes down to 2j−1, theshort stock position has value
− 4
22j· 2j−1 = − 2
2j,
so the portfolio is valued at
10
2j− 2
2j=
8
2j=
4
2j−1,
which is the option value also in this case.
Exercise 5.9. (Provided by Irene Villegas.) Here is a method for solvingequation (5.4.13) for the value of the perpetual American put in Section 5.4.
(i) We first determine v(s) for large values of s. When s is large, it is notoptimal to exercise the put, so the maximum in (5.4.13) will be given bythe second term,
4
5
[1
2v(2s) +
1
2v(s
2
)]=
2
5v(2s) +
2
5v(s
2
).
5.8 Solutions to Selected Exercises 49
We thus seek solutions to the equation
v(s) =2
5v(2s) +
2
5v(s
2
). (5.7.5)
All such solutions are of the form sp for some constant p, or linear com-binations of functions of this form. Substitute sp into (5.7.5), obtain aquadratic equation for 2p, and solve to obtain 2p = 2 or 2p = 1
2 . Thisleads to the values p = 1 and p = −1, i.e., v1(s) = s and v2(s) = 1
sare
solutions to (5.7.5).
(ii) The general solution to (5.7.5) is a linear combination of v1(s) and v2(s),i.e.,
v(s) = As +B
s. (5.7.6)
For large values of s, the value of the perpetual American put must begiven by (5.7.6). It remains to evaluate A and B. Using the second bound-ary condition in (5.4.15), show that A must be zero.
(iii) We have thus established that for large values of s, v(s) = Bs
for someconstant B still to be determined. For small values of s, the value of theput is its intrinsic value 4 − s. We must choose B so these two functionscoincide at some point, i.e., we must find a value for B so that, for somes > 0,
fB(s) =B
s− (4 − s)
equals zero. Show that, when B > 4, this function does not take the value0 for any s > 0, but, when B ≤ 4, the equation fB(s) = 0 has a solution.
(iv) Let B be less than or equal to 4 and let sB be a solution of the equationfB(s) = 0. Suppose sB is a stock price which can be attained in the model(i.e., sB = 2j for some integer j). Suppose further that the owner of theperpetual American put exercises the first time the stock price is sB orsmaller. Then the discounted risk-neutral expected payoff of the put isvB(S0), where vB(s) is given by the formula
vB(s) =
4 − s, if s ≤ sB ,Bs, if s ≥ sB .
(5.7.7)
Which values of B and sB give the owner the largest option value?
(v) For s < sB , the derivative of vB(s) is v′B(s) = −1. For s > sB , this
derivative is v′B(s) = − B
s2 . Show that the best value of B for the optionowner makes the derivative of vB(s) continuous at s = sB (i.e., the twoformulas for v′
B(s) give the same answer at s = sB).
50 5 Random Walk
Solution
(i) With v(s) = sp, (5.7.5) becomes
sp =2
5· 2psp +
2
5· 1
2psp,
Mulitplication by 2p
sp leads to
2p =2
5(2p)
2+
2
5,
and we may rewrite this as
(2p)2 − 5
2· 2p + 1 = 0,
a quadratic equation in 2p. The solution to this equation is
2p =1
2
(5
2±√
25
4− 4
)=
1
2
(5
2± 3
2
).
We thus have either 2p = 2 or 2p = 12 , and hence either p = 1 or p = −1.
(ii) With v(s) = As + Bs, we have lims→∞ v(s) = ±∞ unless A = 0. The
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 tozero:
−B
s2+ 1 = 0.
This results in the critical point sc =√
B. We note that the second deriva-tive, 2B
s3 , is positive on (0,∞), so the function is convex, and hence fB
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 theequation fB(s) = 0 in (0,∞). If 0 < B < 4, then fB(sc) < 0 and theequation fB(s) = 0 has two solutions in (0,∞).
(iv) Since vB(s) = Bs
for all large values of s, we maximize this by choosing
B as large as possible, i.e., B = 4. For values of B < 4, the curve Bs
liesbelow the curve 4
s(see Figure 5.8.1), and values of B > 4 are not possible
because of part (iii).
5.8 Solutions to Selected Exercises 51
(iv) We see from the tangency of the curve y = 4s
with the intrinsic valuey = 4− s at the point (2,2) in Figure 5.8.1 that y = 4
sand y = 4− s have
the same derivative at s = 2. Indeed,
d
ds
4
s
∣∣∣∣s=2
= − 4
s2
∣∣∣∣s=2
= −1,
and as noted in the statement of the exercise,
d
ds(4 − s) = −1.
1 2 3
y
s
4
4
y = 4
s
y = 3
sy = 4− s
(2, 2)
Fig. 5.8.1. The curve y = B
sfor B = 3 and B = 4.
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 con-ditional expectation is defined by Definition 6.2.2. (Part (iv) is not true in theform 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[c1X + c2Y ](ω1 . . . ωn)
=∑
ωn+1,...,ωN
(c1X(ω1 . . . ωnωn+1 . . . ωN ) + c2Y (ω1 . . . ωnωn+1 . . . ωN )
)
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= c1
∑
ωn+1,...,ωN
X(ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn+c2
∑
ωn+1,...,ωN
Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= c1En[X](ω1 . . . ωn) + c2En[Y ](ω1 . . . ωn).
(ii) Taking out what is known. If X depends only on the first n cointosses, then
54 6 Interest-Rate-Dependent Assets
En[XY ](ω1 . . . ωn)
=∑
ωn+1,...,ωN
X(ω1 . . . ωn)Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= X(ω1 . . . ωn)
∑
ωn+1,...,ωN
Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= X(ω1 . . . ωn)En[Y ](ω1 . . . ωn)
(iii) Iterated conditioning. If 0 ≤ n ≤ m ≤ N , then because Em[X] de-pends only on the first m coin tosses and
∑
ωm+1,...,ωN
Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn= Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn,
we have
En
[Em[X]
](ω1 . . . ωn)
=∑
ωn+1,...,ωN
Em[X](ω1 . . . ωm)
×Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn=
∑
ωn+1,...,ωm
Em[X](ω1 . . . ωm)∑
ωm+1,...,ωN
×Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn=
∑
ωn+1,...ωm
Em[X](ω1 . . . ωm)
×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn=
∑
ωn+1,...,ωm
∑
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn,
where we have used the definition
6.9 Solutions to Selected Exercises 55
Em[X](ω1 . . . ωm)
=∑
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm
in the last step. Using the fact that
Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn
= Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn= Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn,
we may write the last term in the above formula for En
[Em[X]
](ω1 . . . ωn)
as
En
[Em[X]
](ω1 . . . ωn)
=∑
ωn+1,...,ωm
∑
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn=
∑
ωn+1,...,ωN
X(ω1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= En[X](ω1 . . . ωn).
(iv) Conditional Jensen’s inequality. This follows from part (i) just likethe proof of part (v) in Appendix A.
Exercise 6.2. Verify that the discounted value of the static hedging portfolioconstructed 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 shortSn
Bn,mzero 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
Xk = Sk − Sn
Bn,m
Bk,m, k = n, n + 1, . . . ,m.
For n ≤ k ≤ m − 1, we have
Ek [Dk+1Xk+1] = Ek[Dk+1Sk+1] −Sn
Bn,m
Ek [Dk+1Bk+1,m] .
56 6 Interest-Rate-Dependent Assets
Using the fact that the discounted asset price is a martingale under therisk-neutral measure and also using (6.2.5) first in the form Dk+1Bk+1,m =
Ek+1[Dm] and then in the form DkBk,m = Ek[Dm], we may rewrite this as
Ek[Dk+1Xk+1] = DkSk − Sn
Bn,m
Ek
[Ek+1[Dm]
]
= DkSk − Sn
Bn,m
Ek[Dm]
= DkSk − Sn
Bn,m
DkBk,m
= DkXk.
This is the martingale property.
Exercise 6.4. Using the data in Example 6.3.9, this exercise constructs ahedge for a short position in the caplet paying (R2 − 1
3 )+ at time three. Weobserve from the second table in Example 6.3.9 that the payoff at time threeof this caplet is
V3(HH) =2
3, V3(HT ) = V3(TH) = V3(TT ) = 0.
Since this payoff depends on only the first two coin tosses, the price of thecaplet at time two can be determined by discounting:
V2(HH) =1
1 + R2(HH)V3(HH) =
1
3, V2(HT ) = V2(TH) = V2(TT ) = 0.
Indeed, if one is hedging a short position in the caplet and has a portfoliovalued at 1
3 at time two in the event ω1 = H,ω2 = H, then one can simplyinvest this 1
3 in the money market in order to have the 23 required to pay off
the caplet at time three.
In Example 6.3.9, the time-zero price of the caplet is determined to be221 (see (6.3.10)).
(i) Determine V1(H) and V1(T ), the price at time one of the caplet in theevents ω1 = H and ω1 = T , respectively.
(ii) Show how to begin with 221 at time zero and invest in the money market
and the maturity two bond in order to have a portfolio value X1 at timeone 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 maturitythree bond to do this?
(iii) Show how to take the portfolio value X1 at time one and invest in themoney market and the maturity three bond in order to have a portfoliovalue X2 at time two that agrees with V2, regardless of the outcome of thefirst two coin tosses. Why do we invest in the maturity three bond ratherthan 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
D1(H)E1[D2V2](H)
= Pω2 = H|ω1 = HD2(HH)V2(HH)
+Pω2 = T |ω1 = HD2(HT )V2(HT )
=2
3· 6
7· 1
3+
1
3· 6
7· 0 =
4
21,
V1(T ) =1
D1(T )E1[D2V2](T ) = 0.
(ii) We compute the number of shares of the time-two maturity bond by theusual formula:
∆0 =V1(H) − V1(T )
B1,2(H) − B1,2(T )=
421 − 067 − 5
7
=4
3.
It is straight-forward to verify that this works. Set X0 = V0 = 221 and compute
X1(H) = ∆0B1,2(H) + (1 + R0)(X0 − ∆0B0,2)
=4
3· 6
7+
2
21− 4
3· 11
14=
4
21= V1(H),
X1(T ) = ∆0B1,2(T ) + (1 + R0)(X0 − ∆0B0,2)
=4
3· 5
7+
2
21− 4
3· 11
14= 0 = V1(T ).
We do not use the maturity three bond because B1,3(H) = B1,3(T ), and thisbond 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, thehedging portfolio has zero value, and no further hedging is required. In theevent of a H on the first toss, we hedge by taking in the maturity three bondthe position
∆1(H) =V2(HH) − V2(HT )
B2,3(HH) − B2,3(HT )=
13 − 012 − 1
= −2
3.
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
3· 1
2+
7
6
(4
21+
2
3· 4
7
)=
1
3= V2(HH),
X2(HT ) = ∆1(H)B2,3(HT ) + (1 + R1(H))(X1(H) − ∆1(H)B1,3(H))
= −2
3· 1 +
7
6
(4
21+
2
3· 4
7
)= 0 = V2(HT ).
58 6 Interest-Rate-Dependent Assets
We do not use the maturity two bond because B2,2(HH) = B2,2(HT ), andthis bond therefore provides no hedge against the second coin toss.