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
2
3
The Binomial No-Arbitrage Pricing Model . . . . . . . . . . . . . . . .
1
1.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Probability Theory on Coin Toss Space . . . . . . . . . . . . . . . . . . . .
7
2.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
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
X1 (H) = −X1 (T ).
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
strategy worked out in Example 1.1.1. In particular, she should short 21 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
worth $3, has $ 45 · 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 = 12 , and take the
Pn interest rate r = 4 , so that
1
p̃ = q̃ = 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,
+
v3 (s, y) = 41 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.
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
1
[p̃vn+1 (us, y + us) + q̃vn+1 (ds, y + ds)],
1+r
vn+1 (us, y + us) − vn+1 (ds, y + ds)
δn (s, y) =
.
us − ds
vn (s, y) =
(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 :
4 1
1
v2 (16, 28) =
v3 (32, 60) + v3 (8, 36) = 6.40,
5 2
2
4 1
1
v2 (4, 16) =
v3 (8, 24) + v3 (2, 18) = 1,
5 2
2
1
4 1
v3 (8, 18) + v3 (2, 12) = 0.20,
v2 (4, 10) =
5 2
2
4 1
1
v2 (1, 7) =
v3 (2, 9) + v3 (.50, 7.50) = 0.
5 2
2
4
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
1
2 v2 (16, 28)
4
5
1
2 v2 (4, 10)
+ 21 v2 (4, 16) = 2.96,
+ 21 v2 (1, 7)
= 0.08.
Finally, we may now compute
1
4 1
v1 (8, 12) + v1 (2, 6) = 1.216.
v0 (4, 4) =
5 2
2
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
u0 S0 if ω1 = H,
S1 (ω1 ) =
d0 S0 if ω1 = T,
and, for n ≥ 1, the stock price at time n + 1 is given by
un (ω1 ω2 . . . ωn )Sn (ω1 ω2 . . . ωn ) if ωn+1 = H,
Sn+1 (ω1 ω2 . . . ωn ωn+1 ) =
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
p̃0 =
1 + r 0 − d0
,
u0 − d 0
q̃0 =
and for each and and for all ω1 ω2 . . . ωn ,
u0 − 1 − r 0
,
u0 − d 0
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 )
.
q̃n (ω1 ω2 . . . ωn ) =
un (ω1 ω2 . . . ωn ) − dn (ω1 ω2 . . . ωn )
p̃n (ω1 ω2 . . . ωn ) =
In place of (1.2.16), we define for n = N − 1, N − 2, . . . , 1,
Vn (ω1 ω2 . . . ωn ) =
1 p̃n (ω1 ω2 . . . ωn )Vn+1 (ω1 ω2 . . . ωn H)
1+r
+q̃n (ω1 ω2 . . . ωn )Vn+1 (ω1 ω2 . . . ωn T ) ,
and for the the case n = 0 we adopt the definition
V0 =
1 p̃0 V1 (H) + q̃0 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
x
x . The corresponding riskneutral probabilities are
p̃n =
1 − dn
=
un − d n
q̃n =
un − 1
=
un = d n
x−10
x
− x−10
x
=
1
,
2
x+10
x −1
x+10
x−10
x −
x
=
1
.
2
1−
x+10
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 prob5
1
ability of an arbitrary sequence ω1 ω2 ω3 ω4 ω5 to be 21 = 32
.
6
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
32 . The time-zero price
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 p̃ = 12 ,
q̃ = 21 .
e 1 , ES
e 2 , and ES
e 3 . What is the average rate of growth of the
(ii) Compute ES
e
stock price under P?
(iii) Answer (i) and (ii) again under the actual probabilities p = 32 , q = 13 .
Solution.
(i) The distribution of S3 under the risk-neutral probabilities p̃ and q̃ is
32 8
2 .50
p̃3 3p̃2 q̃ 3p̃q̃ 2 q̃ 3
With p̃ = 21 , q̃ = 21 , this becomes
2 .50
32 8
.125 .375 .375 .125
(ii) By Theorem 2.4.4,
Therefore,
e
E
S3
e S2
e S1 = ES
e 0 = S0 = 4.
=E
=E
3
2
(1 + r)
(1 + r)
(1 + r)
8
2 Probability Theory on Coin Toss Space
e 1 = (1 + r)S0 = (1.25)(4) = 5,
ES
e 2 = (1 + r)2 S0 = (1.25)2 (4) = 6.25,
ES
e 3 = (1 + r)3 S0 = (1.25)3 (4) = 7.8125.
ES
In particular, we see that
e 3 = 1.25 · ES
e 2 , ES
e 2 = 1.25 · ES
e 1 , ES
e 1 = 1.25 · S0 .
ES
e is the same
Thus, the average rate of growth of the stock price under P
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
With p = 32 , q = 31 , this becomes
8
2
.50
32
.2963 .4444 .2222 .0371
To compute the average rate of growth, we reason as follows:
Sn+1
Sn+1
= S n En
= (pu + qd)Sn .
En Sn+1 = En Sn
Sn
Sn
In our case,
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
ESn+1 = E En Sn+1 = 1.5 · ESn ,
pu + qd =
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
In =
n−1
X
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.
0
be martingales under the risk(i) Let M0 , M1 , . . . , MN and M00 , M10 , . . . , MN
0
e
neutral 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 = Mn0 (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
e
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
VN
en
Vn0 = E
, n = 0, 1, . . . , N − 1.
(1 + r)N −n
Show that
V00 ,
VN0 −1
VN
V10
,...,
,
N
−1
1+r
(1 + r)
(1 + r)N
is a martingale.
(iv) Conclude that Vn = Vn0 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.
0
(i) We are given that Mn = MN
. For n between 0 and N − 1, this equality
and the martingale property imply
e n [MN ] = E
e n [M 0 ] = Mn .
Mn = E
N
(ii) For n between 0 and N − 1, we compute the following conditional expectation:
Vn+1
en
E
(ω1 ω2 . . . ωn )
(1 + r)n+1
Vn+1 (ω1 ω2 . . . ωn T )
Vn+1 (ω1 ω2 . . . ωn H)
+ q̃
= p̃
n+1
(1 + r)
(1 + r)n+1
Vn (ω1 ω2 . . . ωn )
=
,
(1 + r)n
where the second equality follows from (1.2.16). This is the martingal
Vn
property for (1+r)
n.
V0
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,
0
Vn+1
1
VN
e
e
e
= En
En+1
En
(1 + r)n+1
(1 + r)n+1
(1 + r)N −(n+1)
1
VN
e
e
=
En En+1
(1 + r)n
(1 + r)N −n
VN
1
e
En
=
(1 + r)n
(1 + r)N −n
0
Vn
=
.
(1 + r)n
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
r1 (H) = 41
S2 (HT ) = 8
S0 = 4
r0 = 14
S2 (T H) = 8
S1 (T ) = 2
r1 (T ) = 12
S2 (T T ) = 2
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
e
e
e H), P(T
e T ),
P(HH),
P(HT
), P(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
V2
e
V0 = E
.
(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 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
12
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.
(i) For the first toss, the up factor is u0 = 2 and the down factor is d0 = 21 .
Therefore, the risk-neutral probability of a H on the first toss is
p̃0 =
1 + 14 −
1 + r 0 − d0
=
u0 − d 0
2 − 21
1
2
=
1
,
2
and the risk-neutral probability of T on the first toss is
q̃0 =
2−1−
u0 − 1 − r 0
=
u0 − d 0
2 − 21
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
p̃1 (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
q̃1 (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
p̃1 (T ) =
13
1 + 12 − 1
1
1 + r1 (T ) − d1 (T )
=
= ,
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
q̃1 (T ) =
4−1−
u1 (T ) − 1 − r1 (T )
=
u1 (T ) − d1 (T )
4−1
1
2
=
5
.
6
The risk-neutral probabilities are
e
P(HH)
= p̃0 p̃1 (H) =
e
P(HT ) = p̃0 q̃1 (H) =
e H) = q̃0 p̃1 (T ) =
P(T
(ii) We compute
e T ) = q̃0 q̃1 (T ) =
P(T
1
2
1
2
1
2
1
2
·
·
·
·
1
2
1
2
1
6
5
6
= 41 ,
= 41 ,
=
=
1
12 ,
5
12 .
1
p̃1 (H)V2 (HH) + q̃1 (H)V2 (HT )
1 + r1 (H)
1
4 1
+
+
· (12 − 7) + · (8 − 7)
=
5 2
2
= 2.40,
1
V1 (T ) =
p̃1 (T )V2 (T H) + q̃1 (T )V2 (T T )
1 + r1 (T )
5
2 1
· (8 − 7)+ + · (2 − 7)+
=
3 6
6
= 0.111111,
1
V0 =
[p̃0 V1 (H) + q̃0 V1 (T )]
1 + r0
4 1
1
=
· 2.40 + · 0.1111
5 2
2
= 1.00444.
V1 (H) =
We can confirm this price by computing according to the risk-neutral
pricing formula in part (i) of the exercise:
14
2 Probability Theory on Coin Toss Space
V2
(1 + r0 )(1 + r1 )
V2 (HH)
V2 (HT )
e
e
=
· P(HH)
+
· P(HT
)
(1 + r0 )(1 + r1 (H))
(1 + r0 )(1 + r1 (H))
V2 (T H)
V2 (T T )
e H) +
e T)
+
· P(T
· P(T
(1 + r0 )(1 + r1 (T ))
(1 + r0 )(1 + r1 (T ))
(8 − 7)+
1
1
(12 − 7)+
+
·
·
=
1
1
1
1
(1 + 4 )(1 + 4 ) 4 (1 + 4 )(1 + 4 ) 4
e
V0 = E
(2 − 7)+
(8 − 7)+
1
5
·
1
1 · 12 +
(1 + 4 )(1 + 2 )
(1 + 41 )(1 + 12 ) 12
= 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 = E
, 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
en
Pn = E
, 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
en
, n = 0, 1, . . . , N − 1.
Fn = E
(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 ;
Case III: SN < K. Then CN = 0 and PN = K − SN = −FN .
In all three cases, we see that CN = FN + PN .
(ii)
FN + P N
CN
e
= En
(1 + r)N −n
(1 + r)N −n
FN
PN
e
e
= En
+ En
= Fn + Pn .
(1 + r)N −n
(1 + r)N −n
en
Cn = E
(iii)
SN − K
FN
e
=E
(1 + r)N
(1 + r)N
SN
K
K
e
e
.
=E
−E
= S0 −
N
N
(1 + r)
(1 + r)
(1 + r)N
e
F0 = E
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
N
SN + (1 + r) (F0 − S0 ) = SN + (1 + r)
N
K
−
(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.,
!
N
1 X
Sn ,
VN = f
N + 1 n=0
where the function f is determined by the contractual details of the option.
Pn
(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
depends
and whereas Sn and Yn depend only on the first n tosses, SSn+1
n
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
e n [hn+1 (Sn+1 , Yn+1 )] = E
e n hn+1 Sn · Sn+1 , Yn + Sn · Sn+1
E
Sn
Sn
= hn (Sn , Yn ),
where
e n+1 s · Sn+1 , y + s · Sn+1
hn (s, y) = Eh
Sn
Sn
= p̃hn+1 (su, y + su) + q̃hn+1 (sd, y + sd).
e n [hn+1 (Sn+1 , Yn+1 )] can be written as a function of (Sn , Yn ),
Because E
the two-dimensional process (Sn , Yn ), n = 0, 1, . . . , N , is a Markov process.
(ii) We have the final condition VN (s, y) = f Ny+1 . For n = N − 1, . . . , 1, 0,
we have from the risk-neutral pricing formula (2.4.12) and (i) above that
Vn =
where
1 e
1 e
En Vn+1 =
En vn+1 (Sn+1 , Yn+1 ) = vn (Sn , Yn ),
1+r
1+r
vn (s, y) =
1 p̃vn+1 (su, y + su) + q̃vn+1 (sd, y + sd) .
1+r
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:
e
(i0 ) P
1
Z
> 0 = 1;
e 1 = 1;
(ii0 ) E
Z
(iii0 ) for any random variable Y ,
e
EY = E
1
·Y .
Z
e to E in the same way Z
In other words, Z1 facilitates the switch from E
e
facilitates the switch from E to E.
Solution
e
(i0 ) Because P(ω) > 0 and P(ω)
> 0 for every ω ∈ Ω, the ratio
1
P(ω)
=
e
Z(ω)
P(ω)
0
is defined and positive for every ω ∈ Ω.
(ii ) We compute
e1 =
E
Z
X
ω∈Ω
X P(ω)
X
1 e
e
P(ω)
=
P(ω) =
P(ω) = 1.
e
Z(ω)
ω∈Ω P(ω)
ω∈Ω
20
3 State Prices
(iii0 ) We compute
X
X
P(ω)
1
e
e
·Y =
Y (ω)P(ω)
=
E
Y (ω)P(ω) = EY.
e
Z
ω∈Ω P(ω)
ω∈Ω
Exercise 3.3. Using the stock price model of Figure 3.1.1 and the actual
probabilities p = 32 , 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) = 32 S3 (HHH) + 12 S3 (HHT ) =
M2 (T H) =
2
3 S3 (HT H)
2
3 S3 (T HH)
M2 (T T ) =
2
3 S3 (T HH)
M2 (HT ) =
=
+
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.
We next compute M1 from the formula M1 = E1 [S3 ]:
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.
M1 (H) =
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
!!
M2 (HH) = 24
aa
!
!!
aa
M1 (H) = 18
Z
Z
M0 = 13.50
Z
Z
M2 (HT ) = M2 (T H) = 6
M1 (T ) = 4.50
Z
Z
!!
!!
!
21
S3 (HHH) = 32
aa S3 (HHT ) = S3 (HT H)
= S3 (T HH) = 8
Z
Z S (HT T ) = S3 (T HT )
!! 3
= S3 (T T H) = 2
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:
E1 [M2 ](H) = 32 M2 (HH) + 31 M2 (HT ) =
E1 [M2 ](T ) =
M0 =
1
2 M2 (T H)
2
3 M1 (H)
+
+
1
2 M2 (T T )
1
3 M1 (T )
=
=
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 .
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-Nikodým derivative Z(HH), Z(HT ), Z(T H) and
e with respect to P
Z(T T ) of P
(ii) The Radon-Nikodým derivative process Z0 , Z1 , Z2 satisfies Z2 = Z. Compute 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
1 e
1 e
1 e
5
e
P(HH)
= , P(HT
) = , P(T
H) =
, P(T T ) =
.
4
4
12
12
Therefore, the Radon-Nikodým derivative is
Z(HH) =
Z(T H) =
e
1 9
9
P(HH)
=
·
=
,
P(HH)
4 4
16
Z(HT ) =
e H)
P(T
1 9
3
=
· = ,
P(T H)
12 2
8
Z(T T ) =
e
P(HT
)
1 9
9
=
·
= ,
P(HT )
4 2
8
e T)
P(T
5 9
15
=
· =
,
P(T T )
12 1
4
(ii)
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 )
2
3
· 2 9
8 9+
3
= ,
2
Z0 = E0 [Z1 ]
=
1
9
+
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 2
3
2 1
= ·
+ ·
+
+
4
9 9
2
9 9
= 1.
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} =
4
9
2
4
+
9
9
P{ω2 = T given that ω1 = H} =
2
9
4
2
+
9
9
P{ω2 = H given that ω1 = T } =
Therefore,
e 2 = T given that ω1 = T } =
P{ω
2
9
2
1
+
9
9
1
9
2
1
9+9
2
,
3
1
= ,
3
2
= ,
3
1
= .
3
=
24
3 State Prices
1
E1 [Z2 V2 ](H)
Z1 (H) 1 + r1 (H)
−1
3 5
=
Z2 (HH)V2 (HH)P{ω2 = H given that ω1 = H}
·
4 4
V1 (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
E1 [Z2 V2 ](T )
V1 (T ) =
Z1 (T ) 1 + r1 (T )
−1
3 3
·
Z2 (T H)V2 (T H)P{ω2 = H given that ω1 = T }
=
2 2
+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
Z 2 V2
V0 = E
(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 ))
−1
−1
−1
5 5
9
9
3
4
5 5
2
5 3
2
=
·
·5· +
·
·1· +
·
·1·
4 4
16
9
4 4
8
9
4 2
8
9
−1
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 = X
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 ).
When U (x) = ln x, U 0 (x) =
Therefore,
1
x
3.7 Solutions to Selected Exercises
25
and the inverse function of U 0 is I(y) =
1
y.
XN =
1
λζN
We must choose λ to satisfy (3.3.26), which in this case takes the form
1
X0 = E ζN I(λζN ) = .
λ
Substituting this into the previous equation, we obtain
XN =
Because
Xn
(1+r)n
X0
.
ζN
e we have
is a martingale under the risk-neutral measure P,
1
1
Xn
X0
XN
Z N XN
en
=
=
=
E
E
En [ζN XN ] =
.
n
(1 + r)n
(1 + r)N
Zn
(1 + r)N
Zn
Zn
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,
λ
∗
Z ,
(3.6.1)
XN = I
(1 + r)N
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 0 (x) − y, and this is zero if and only if U 0 (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
λZ
λZ
λZ
λZXN
≤
U
I
I
−
.
U (XN ) −
(1 + r)N
(1 + r)N
(1 + r)N
(1 + r)N
Taking expectations under P and using the fact that Z is the Radone with respect to P, we obtain
Nikodým derivative of P
XN
(1 + r)N
Z
λZ
λZ
− λE
.
I
≤ EU I
(1 + r)N
(1 + r)N
(1 + r)N
e
EU (XN ) − λE
From (3.3.19) and (3.3.26), we have
XN
Z
λZ
e
E
= X0 = E
I
.
(1 + r)N
(1 + r)N
(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
interest rate be r = 41 so the risk-neutral probabilities are p̃ = q̃ = 12 .
(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+r p̃
=
1
1+r q̃
= 0,
= V3P (T HH) = (4 − 8)+ = 0,
= V3P (T T H) = (4 − 2)+ = 2,
= 3.50.
= 25 , the value of the put at time two is
28
4 American Derivative Securities
+ 2 P
2
, V3 (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}
V2P (HH) = max
4 − S2 (HH)
= 0.80,
+ 2
2
V2P (T T ) = max 4 − S2 (T T ) , V3P (T T H) + V3P (T T T )
5
5
2
2
= max (4 − 1)+ , · 2 + · 3.50
5
5
= max{3, 2.20}
= 3.
At time one the value of the put is
+ 2
2
V1P (H) = max 4 − S1 (H) , V2P (HH) + V2P (HT )
5
5
2
2
= max (4 − 8)+ , · 0 + · 0.80
5
5
= max{0, 0.32}
= 0.32,
+ 2 P
2 P
P
V1 (T ) = max 4 − S1 (T ) , V2 (T H) + V2 (T T )
5
5
2
2
= max (4 − 2)+ , · 0.80 + · 3
5
5
= max{2, 1.52}
= 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+r p̃
=
1
1+r q̃
= 28,
= V3C (T HH) = (8 − 4)+ = 4,
= V3C (T T H) = (2 − 4)+ = 0,
= 0.
= 25 , the value of the call at time two is
+ 2 C
2
, V3 (HHH) + V3C (HHT )
5
5
2
2
= max (16 − 4)+ , · 28 + · 4
5
5
= max{12, 12.8}
V2C (HH) = max
S2 (HH) − 4
= 12.8,
+ 2
2
V2C (HT ) = max S2 (HT ) − 4 , V3C (HT H) + V3C (HT T )
5
5
2
2
= max 4 − 4)+ , · 4 + · 0
5
5
= max{0, 1.60}
= 1.60,
+ 2
2
V2C (T H) = max S2 (T H) − 4 , 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}
= 0.
At time one the value of the call is
30
4 American Derivative Securities
+ 2 C
2
, V2 (HH) + V2C (HT )
5
5
2
2
= max (8 − 4)+ , · 12.8 + · 1.60
5
5
= max{4, 5.76} = 5.76,
+ 2 C
2 C
C
V1 (T ) = max S1 (T ) − 4 , V2 (T H) + V2 (T T )
5
5
2
2
= max (2 − 4)+ , · 1.60 + · 0
5
5
= max{0, 0.64} = 0.64.
V1C (H) = max
S1 (H) − 4
The value of the call at time zero is
2 C
+ 2 C
C
V0 = max (S0 − 4) , V1 (H) + V1 (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
p̃ = 1+r
q̃ = 25 , the value of the straddle at time two is
2 S
2 S
S
V2 (HH) = max S2 (HH) − 4 , V3 (HHH) + V3 (HHT )
5
5
2
2
= max |16 − 4|, · 28 + · 4
5
5
= max{12, 12.8}
= 12.8,
2
+ 2
V2S (HT ) = max S2 (HT ) − 4 , V3S (HT H) + V3S (HT T )
5
5
2
2
= max 4 − 4)+ , · 4 + · 2
5
5
= max{0, 2.40}
= 2.40,
4.9 Solutions to Selected Exercises
+ 2 S
2
, V3 (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.
V2S (T H) = max
S2 (T H) − 4
31
One can verify in every case that V2S = V2P + V2C . At time one the value
of the straddle is
2 S
2 S
S
V1 (H) = max S1 (H) − 4 , V2 (HH) + V2 (HT )
5
5
2
2
= max |8 − 4|, · 12.8 + · 2.40
5
5
= max{4, 6.08}
= 6.08,
2 S
2 S
S
V1 (T ) = max S1 (T ) − 4 , V2 (T H) + V2 (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 S
2 S
S
V0 = max |S0 − 4|, V1 (H) + V1 (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
+ 2 P
2 P
P
V1 (T ) = max 4 − S1 (T ) , V2 (T H) + V2 (T T )
5
5
2
2
= max (4 − 2)+ , · 0.80 + · 3
5
5
= max{2, 1.52}
= 2,
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
2 S
2 S
S
V1 (T ) = max S1 (T ) − 4 , V2 (T H) + V2 (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 = 14 so the risk-neutral probabilities are p̃ = q̃ = 12 . Find
the time-zero price and optimal exercise policy (optimal stopping time) for
the path-dependent American derivative security whose intrinsic value at each
+
Pn
1
time n, n = 0, 1, 2, 3, is 4 − n+1
. This intrinsic value is a put on
j=0 Sj
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 =
G1 (H) =
G1 (T ) =
(4 − S0 )
4−
4−
+
S0 +S1 (H)
2
S0 +S1 (T )
2
=
+
+
=
+
2 (HH)
G2 (HH) = 4 − S0 +S1 (H)+S
3
+
2 (HT )
G2 (HT ) = 4 − S0 +S1 (H)+S
3
+
2 (T H)
G2 (T H) = 4 − S0 +S1 (T )+S
3
+
G2 (T T ) = 4 − S0 +S1 (T 3)+S2 (T T )
=
=
=
=
=
(4 − 4)+
+
4 − 4+8
2
+
4 − 4+2
2
+
4 − 4+8+16
3
4 − 4+8+4
3
+
4 − 4+2+4
3
r − 4+2+1
3
= 0,
= 0,
= 1,
= 0,
= 0,
= 0.6667,
= 1.6667.
At time three, the intrinsic value G3 agrees with the option value V3 . In other
words,
V3 (HHH) = G3 (HHH)
+
S0 + S1 (H) + S2 (HH) + S3 (HHH)
= 4−
4
+
4 + 8 + 16 + 32
= 4−
4
= 0,
V3 (HHT ) = G3 (HHT )
+
S0 + S1 (H) + S2 (HH) + S3 (HHT )
= 4−
4
+
4 + 8 + 16 + 8
= 4−
4
= 0,
V3 (HT H) = G3 (HT H)
+
S0 + S1 (H) + S2 (HT ) + S3 (HT H)
= 4−
4
+
4+8+4+8
= 4−
4
= 0,
V3 (HT T ) = G3 (HT T )
+
S0 + S1 (H) + S2 (HT ) + S3 (HT T )
= 4−
4
+
4+8+4+2
= 4−
4
= 0,
34
4 American Derivative Securities
V3 (T HH) = G3 (T HH)
+
S0 + S1 (T ) + S2 (T H) + S3 (T HH)
= 4−
4
+
4+2+4+8
= 4−
4
= 0,
V3 (T HT ) = G3 (T HT )
+
S0 + S1 (T ) + S2 (T H) + S3 (T HT )
= 4−
4
+
4+2+4+2
= 4−
4
= 1,
V3 (T T H) = G3 (T T H)
+
S0 + S1 (T ) + S2 (T T ) + S3 (T T H)
= 4−
4
+
4+2+1+2
= 4−
4
= 1.75,
V3 (T T T ) = G3 (T T T )
+
S0 + S1 (T ) + S2 (T T ) + S3 (T T T )
= 4−
4
+
4 + 2 + 1 + 0.50
= 4−
4
= 2.125.
We use the algorithm of Theorem 4.4.3, noting that
p̃
1+r
=
q̃
1+r
= 52 , 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
+
Pn
1
defined in the exercise, it does not matter what
Gn = 4 − n+1
j=1 Sj
Pn
1
exercise rule we choose in this case. If the payoff were 4 − n+1
S
j=1 j , 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
τ are
11). For each stopping time τ in the latter set,
compute E I{τ ≤2} 54 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.
τ
4
Gτ1 = G0 = 1,
E I{τ1 ≤2}
5
τ
1 16
1 4
4
E I{τ17 ≤2}
Gτ17 = · G2 (HT ) + · G1 (T )
5
4 26
2 5
4
2
=
· 1 + · 3 = 1.36,
5
25
τ
1 16
1 16
1 16
4
Gτ18 = · G2 (HT ) + · G2 (T H) + · G2 (T T )
E I{τ18 ≤2}
5
4 26
4 25
4 25
4
4
4
·1+
·1+
· 4 = 0.96,
=
25
25
25
τ
1 16
1 16
4
Gτ19 = · G2 (HT ) + · G2 (T H)
E I{τ19 ≤2}
5
4 26
4 25
4
4
=
·1+
· 1 = 0.32,
25
25
τ
1 16
4
1 16
E I{τ20 ≤2}
Gτ20 = · G2 (HT ) + · G2 (T T )
5
4 26
4 25
4
4
=
·1+
· 4 = 0.80,
25
τ
25
4
1 16
Gτ21 = · G2 (HT )
E I{τ21 ≤2}
5
4 26
4
=
· 1 = 0.16,
25
τ
1 4
4
Gτ22 = · G1 (T )
E I{τ22 ≤2}
5
2 5
2
= · 3 = 1.20,
τ
5
4
1 16
1 16
E I{τ23 ≤2}
Gτ23 = + · G2 (T H) + · G2 (T T )
5
4 25
4 25
4
4
=
·1+
· 4 = 0.80,
25
25
38
4 American Derivative Securities
τ
1 16
4
Gτ24 = + · G2 (T H)
E I{τ24 ≤2}
5
4 25
4
· 1 = 0.16,
=
τ
25
4
1 16
E I{τ25 ≤2}
Gτ25 = + · G2 (T T )
5
4 25
4
= + · 4 = 0.64,
25
τ
4
E I{τ26 ≤2}
Gτ26 = 0.
5
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
Sn+1
K
e
e
e
En [Yn+1 ] = En
− En
(1 + r)n+1
(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
e N ∧τ ≤ EY
e N whenever τ is a stopping
4.3.3 (Optional Sampling—Part II), EY
time. If τ is a stopping time satisfying τ (ω) ≤ N for every sequence of coin
e τ ≤ EY
e N , or equivalently,
tosses ω, this becomes EY
1
1
e
e
E
Gτ ≤ E
GN .
(1 + r)τ
(1 + r)N
Therefore,
V0 =
max
τ ∈S0 ,τ ≤N
e
E
1
1
e
Gτ ≤ E
GN .
(1 + r)τ
(1 + r)N
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
V0 =
max
τ ∈S0 ,τ ≤N
e
E
1
1
e
=
E
.
G
G
τ
N
(1 + r)τ
(1 + r)N
39
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 12 < p < 1 so that
0 < q < 21 . 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
1
f (σ)
n
is a martingale.
(iii) Show that for σ > 0,
e
−σ
= E I{τ1 <∞}
1
f (σ)
τ 1 .
Conclude that P{τ1 < ∞} = 1.
(iv) Compute Eατ1 for α ∈ (0, 1).
(v) Compute Eτ1 .
Solution.
(i) The function f (σ) satisfies f (0) = 1 and f 0 (σ) = peσ −qe−σ , f 00 (σ) = f (σ).
Since f is always positive, f 00 is always positive and f is convex. Under
the assumption p > 1/2 > q, we have f 0 (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
e n [Sn+1 ] =
E
=
1
f (σ)
1
f (σ)
n+1
n+1
h
i
e n eσ(Mn +Xn+1 )
E
eσMn E eσXn+1
n+1
1
eσMn peσ + qe−σ
f (σ)
n
1
σMn
=e
= Sn .
f (σ)
=
(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 e
σMn∧τ1
n→∞
1
f (σ)
n∧τ1
= I{τ1 <∞} e
σ
1
f (σ)
τ
.
Letting n → ∞ in (5.8.1), we obtain
τ1 1
σ
,
1 = E I{τ1 <∞} e
f (σ)
or equivalently,
e
−σ
= E I{τ1 <∞}
1
f (σ)
τ 1 .
(5.8.2)
This equation holds for all positive σ. We let σ ↓ 0 to obtain the formula
1 = EI{τ1 <∞} = P{τ1 < ∞}.
(iv) We now introduce α ∈ (0, 1) and solve the equation α =
This equation can be written as
αpeσ + αqe−σ = 1,
which may be rewritten as
αq e−σ
2
and the quadratic formula gives
− e−σ + αp = 0,
1
f (σ)
for e−σ .
5.8 Solutions to Selected Exercises
e
−σ
=
1±
p
1 − 4α2 pq
.
2αq
43
(5.8.3)
We want σ to be positive, so we need e−σ to be less than 1. Hence we
take the negative sign, obtaining
p
1 − 1 − 4α2 pq
−σ
e =
.
2αq
Substituting this into (5.8.2), we obtain the formula
p
1 − 1 − 4α2 pq
τ1
Eα =
.
2αq
(5.8.4)
(v) Differentiating (5.8.4) with respect to α leads to
p
1 − 1 − 4α2 pq
τ −1
p
Eτ α
=
,
2α2 q 1 − 4α2 pq
and letting α ↑ 1, we obtain
p
√
1 − (1 − 2q)2
1
1
1 − (1 − 2q)
1 − 1 − 4pq
p
√
=
=
=
.
=
Eτ1 =
2
2q(1 − 2q)
1 − 2q
p−q
2q 1 − 4pq
2q (1 − 2q)
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, reaches
the level 2. According to Theorem 5.2.3,
!2
√
1 − 1 − α2
τ2
Eα =
for all α ∈ (0, 1).
α
Using the power series (5.2.21), we may write the right-hand side as
!2
√
√
2 1 − 1 − α2
1 − 1 − α2
= ·
−1
α
α
α
= −1 +
=
=
∞ 2j−2
X
(2j − 2)!
α
2
j!(j
− 1)!
j=1
∞ 2j−2
X
(2j − 2)!
α
2
j!(j
− 1)!
j=2
∞ 2k
X
α
k=1
2
(2k)!
.
(k + 1)!k!
(i) Use the above power series to determine P{τ2 = 2k}, k = 1, 2, . . . .
44
5 Random Walk
(ii) Use the reflection principle to determine P{τ2 = 2k}, k = 1, 2, . . . .
Solution
(i) Since the random walk can reach the level 2 only on even-numbered steps,
P{τ2 = 2k − 1} = 0 for k = 1, 2, . . . . Therefore,
Eατ2 =
∞
X
k=1
α2k P{τ2 = 2k} =
∞ 2k
X
α
k=1
2
(2k)!
.
(k + 1)!k!
Equating coefficients, we conclude that
2k
(2k)!
1
P{τ = 2k} =
, k = 1, 2, . . . .
(k + 1)!k! 2
(ii) For k = 1, 2, . . . , the number of paths that reach or exceed level 2 by time
2k is equal to the number of paths that are at level 2 at time 2k plus the
number of paths that exceed level 2 at time 2k plus the number of paths
that reach the level 2 before time 2k but are below level 2 at time 2k. A
path is of the last type if and only if its reflected path exceeds level 2 at
time 2k. Thus, the number of paths that reach or exeed level 2 by time 2k
is equal to the number of paths that are at level 2 at time 2k plus twice
the number of paths that exeed level 2 at time 2k. For the symmetric
random walk, every path has the same probability, and hence
P{τ2 ≤ 2k} = P{M2k = 2} + 2P{M2k ≥ 4}
= P{M2k = 2} + P{M2k ≥ 4} + P{M2k ≤ −4}
= 1 − P{M2k = 0} − P{M2k = −2}
2k
2k
1
(2k)!
(2k)! 1
−
.
= 1−
k!k! 2
(k + 1)!(k − 1)! 2
Consequently,
P{τ2 = 2k}
= P{τ2 ≤ 2k} − P{τ2 ≤ 2k − 2}
2k−2 (2k − 2)!
(2k − 2)!
1
+
=
2
(k − 1)!(k − 1)! k!(k − 2)!
2k 1
(2k)!
(2k)!
+
−
2
k!k!
(k + 1)!(k − 1)!
2k 1
4(2k − 2)!
(2k)!
−
=
2
(k − 1)!(k − 1)!
k!k!
2k 1
4(2k − 2)!
(2k)!
−
+
2
k!(k − 2)!
(k + 1)!(k − 1)!
5.8 Solutions to Selected Exercises
45
2k 1
2k · 2k(2k − 2)! − 2k(2k − 1)(2k − 2)!
=
2
k!k!
2k (2k + 2)(2k − 2)(2k − 2)! − 2k(2k − 1)(2k − 2)!
1
+
2
(k + 1)!(k − 1)!
2k 2
2
4k (2k − 2)! − (4k − 2k)(2k − 2)!
1
=
2
k!k!
2k 2
1
(4k − 4)(2k − 2)! − (4k 2 − 2k))(2k − 2)!
+
2
(k + 1)!(k − 1)!
2k 1
2k(2k − 2)! 2(k − 2)(2k − 2)!
=
+
2
k!k!
(k + 1)!(k − 1)!
2k
1
=
2
2k
1
=
2
2k
1
=
2
(2k − 2)! (2k(k + 1) + 2(k − 2)k
(k + 1)!k!
(2k − 2)!
2k(2k − 1)
(k + 1)!k!
(2k)!
.
(k + 1)!k!
Exercise 5.7. (Hedging a short position in the perpetual American
put). Suppose you have sold the perpetual American put of Section 5.4 and
are hedging the short position in this put. Suppose at the current time the
stock price is s and the value of your hedging portfolio is v(s). Your hedge is
to first consume the amount
1 s
4 1
v(2s) + v
(5.7.3)
c(s) = v(s) −
5 2
2 2
and then take a position
v(2s) − v
δ(s) =
2s − 2s
s
2
(5.7.4)
in the stock. (See Theorem 4.2.2 of Chapter 4. The processes Cn and ∆n in
that theorem are obtained by replacing the dummy variable s by the stock
price Sn in (5.7.3) and (5.7.4); i.e., Cn = c(Sn ) and ∆n = δ(Sn ).) If you hedge
this way, then regardless of whether the stock goes up or down on the next
step, the value of your hedging portfolio should agree with the value of the
perpetual 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 that
the 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 value
of the perpetual American put at that time).
Solution
(i) Throughout the following computations, we use (5.4.6):
(
4 − 2j , if j ≤ 1,
j
4
v(2 ) =
,
if j ≥ 1.
2j
For j ≤ 0, we have
1
4 1
j+1
j−1
v(2 ) + v(2 )
c(2 ) = v(2 ) −
5 2
2
2
4 − 2j+1 + 4 − 2j−1
= 4 − 2j −
5
2
j
= 4−2 −
8 − 5 · 2j−1
5
4
= .
5
j
j
For j = 1, we have
1
4 1
v(4) + v(1)
c(2) = v(2) −
5 2
2
2
= 2 − [1 + 3]
5
2
= .
5
Finally, for j ≥ 2,
4 1
1
j+1
j−1
c(2 ) = v(2 ) −
v(2 ) + v(2 )
5 2
2
4
2
4
4
= j −
+ j−1
2
5 2j+1
2
2
16
4
4
+ j+1
= j −
2
5 2j+1
2
= 0.
j
(ii) For j ≤ 0, we have
j
4 − 2j+1 − 4 − 2j−1
v(2j+1 ) − v(2j−1 )
=
= −1.
δ(2 ) =
2j+1 − 2j−1
2j+1 − 2j−1
j
5.8 Solutions to Selected Exercises
47
For j = 1, we have
δ(2) =
1−3
2
v(4) − v(1)
=
=− .
4−1
4−1
3
Finally, for j ≥ 2,
v(2j+1 ) − v(2j−1 )
2j+1 − 2j−1
4
4
− j−1
j+1
2
2
= j+1
2
− 2j−1
4 − 16
1
= j+1 ·
2
(4 − 1)2j−1
4
= − 2j .
2
δ(2j ) =
(iii) If the stock price is 2j , where j ≤ 0, then we have a portfolio valued
at v(2j ) = 4 − 2j . We consume c(2j ) = 45 and short one share of stock
(δ(2j ) = −1). This creates a cash position of
4 − 2j −
4
16
+ 2j =
,
5
5
which is invested in the money market. When we go to the next period,
the money market investment grows to 45 · 16
5 = 4. If the stock goes up to
2j+1 , our short stock position has value −2j+1 , so our portfolio is valued
at 4 − 2j+1 , which is the option value in this case. If the stock goes down
to 2j−1 , our portfolio is valued at 4 − 2j−1 , which is the option value also
in this case.
When the stock price is 2j with j ≤ 0, the owner of the option should
exercise, 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 cover
our short position. But when she exercises, we must pay her 4, so we are
maintaining a cash position of 4. At the beginning of any period in which
she fails to exercise, we get to consume the present value of the interest
that 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. We
consume c(2) = 25 and short 32 of a share of stock (δ(2) = − 32 ). This
creates a cash position of
2−
44
2 2
+ ·2=
,
5 3
15
which is invested in the money market. When we go to the next period,
11
the money market investment grows to 45 · 44
15 = 3 . If the stock goes up
48
5 Random Walk
to 4, the short stock position has value − 32 · 4 = − 83 , so the portfolio is
8
valued at 11
3 − 3 = 1, which is the option value in this case. If the stock
goes down to 1, the short stock position has value − 32 · 1 = − 23 , so the
2
portfolio is valued at 11
3 − 3 = 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 valued
at 24j . We consume c(2j ) = 0 and short 242j shares of stock (δ(2j ) = − 242j ).
This creates a cash position of
4
8
4
+ 2j · 2j = j ,
2j
2
2
which is invested in the money market. When we go to the next period,
the money market investment grows to
5 8
10
· j = j.
4 2
2
If the stock goes up to 2j+1 , the short stock position has value
−
8
4
· 2j+1 = − j ,
22j
2
so the portfolio is valued at
10
8
2
4
− j = j = j+1 ,
j
2
2
2
2
which is the option value in this case. If the stock goes down to 2j−1 , the
short stock position has value
−
2
4
· 2j−1 = − j ,
22j
2
so the portfolio is valued at
10
2
8
4
− j = j = j−1 ,
j
2
2
2
2
which is the option value also in this case.
Exercise 5.9. (Provided by Irene Villegas.) Here is a method for solving
equation (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 not
optimal to exercise the put, so the maximum in (5.4.13) will be given by
the second term,
1 s
2 s
4 1
2
v(2s) + v
.
= v(2s) + v
5 2
2 2
5
5 2
5.8 Solutions to Selected Exercises
49
We thus seek solutions to the equation
v(s) =
2 s
2
.
v(2s) + v
5
5 2
(5.7.5)
All such solutions are of the form sp for some constant p, or linear combinations of functions of this form. Substitute sp into (5.7.5), obtain a
quadratic equation for 2p , and solve to obtain 2p = 2 or 2p = 21 . This
leads to the values p = 1 and p = −1, i.e., v1 (s) = s and v2 (s) = 1s are
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.,
B
(5.7.6)
v(s) = As + .
s
For large values of s, the value of the perpetual American put must be
given by (5.7.6). It remains to evaluate A and B. Using the second boundary 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 some
constant B still to be determined. For small values of s, the value of the
put is its intrinsic value 4 − s. We must choose B so these two functions
coincide at some point, i.e., we must find a value for B so that, for some
s > 0,
B
fB (s) =
− (4 − s)
s
equals zero. Show that, when B > 4, this function does not take the value
0 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 equation
fB (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 the
perpetual American put exercises the first time the stock price is sB or
smaller. Then the discounted risk-neutral expected payoff of the put is
vB (S0 ), where vB (s) is given by the formula
4 − s, if s ≤ sB ,
(5.7.7)
vB (s) = B
if s ≥ sB .
s,
Which values of B and sB give the owner the largest option value?
0
(v) For s < sB , the derivative of vB (s) is vB
(s) = −1. For s > sB , this
B
0
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
0
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
2
(2p ) −
5 p
· 2 + 1 = 0,
2
a quadratic equation in 2p . The solution to this equation is
!
r
1
5
25
1 5 3
p
2 =
.
±
−4 =
±
2 2
4
2 2 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 to
zero:
B
− 2 + 1 = 0.
s
√
This results in the critical point sc = B. We note that the second derivative, 2B
s3 , is positive on (0, ∞), so the function is convex, and hence f B
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, ∞).
(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 lies
below the curve 4s (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 value
y = 4 − s at the point (2,2) in Figure 5.8.1 that y = 4s and y = 4 − s have
the same derivative at s = 2. Indeed,
d 4
ds s
s=2
=−
4
s2
s=2
= −1,
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
Fig. 5.8.1. The curve y =
3
B
s
y=
4
for B = 3 and B = 4.
3
s
s
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,
e n [c1 X + c2 Y ](ω 1 . . . ω n )
E
X
c1 X(ω 1 . . . ω n ω n+1 . . . ω N ) + c2 Y (ω 1 . . . ω n ω n+1 . . . ω N )
=
ω n+1 ,...,ω N
= c1
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
X
X(ω 1 . . . ω n ω n+1 . . . ω N )
ω n+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
X
Y (ω 1 . . . ω n ω n+1 . . . ω N )
+c2
ω n+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
e n [X](ω 1 . . . ω n ) + c2 E
e n [Y ](ω 1 . . . ω n ).
= c1 E
(ii) Taking out what is known. If X depends only on the first n coin
tosses, then
54
6 Interest-Rate-Dependent Assets
e n [XY ](ω 1 . . . ω n )
E
X
X(ω 1 . . . ω n )Y (ω 1 . . . ω n ω n+1 . . . ω N )
=
ω n+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
X
Y (ω 1 . . . ω n ω n+1 . . . ω N )
= X(ω 1 . . . ω n )
ω n+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
e n [Y ](ω 1 . . . ω n )
= X(ω 1 . . . ω n )E
e m [X] de(iii) Iterated conditioning. If 0 ≤ n ≤ m ≤ N , then because E
pends only on the first m coin tosses and
X
e n+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N
P{ω
ω m+1 ,...,ω N
|ω1 = ω 1 , . . . , ωn = ω n }
e n+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n },
= P{ω
we have
en E
e m [X] (ω 1 . . . ω n )
E
X
e m [X](ω 1 . . . ω m )
E
=
ω n+1 ,...,ω N
=
e n+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N
×P{ω
|ω1 = ω 1 , . . . , ωn = ω n }
X
X
e m [X](ω 1 . . . ω m )
E
ω n+1 ,...,ω m
=
X
ω n+1 ,...ω m
=
ω m+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N
×P{ω
|ω1 = ω 1 , . . . , ωn = ω n }
e m [X](ω 1 . . . ω m )
E
e n+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
X
X
X(ω 1 . . . ω m ω m+1 . . . ω N )
ω n+1 ,...,ω m ω m+1 ,...,ω N
e m+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m }
×P{ω
e n+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n },
×P{ω
where we have used the definition
6.9 Solutions to Selected Exercises
55
e m [X](ω 1 . . . ω m )
E
X
X(ω 1 . . . ω m ω m+1 . . . ω N )
=
ω m+1 ,...,ω N
e m+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m }
×P{ω
in the last step. Using the fact that
e m+1 = ω m+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωm = ω m }
P{ω
e n+1 = ω n+1 , . . . , ωm = ω m |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
e n+1 = ω n+1 , . . . , ωm = ω m , ωm+1 = ω m+1 , . . . , ωN = ω N
= P{ω
|ω1 = ω 1 , . . . , ωn = ω n }
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n },
= P{ω
e m [X] (ω 1 . . . ω n )
en E
we may write the last term in the above formula for E
as
e m [X] (ω 1 . . . ω n )
en E
E
X
X
X(ω 1 . . . ω m ω m+1 . . . ω N )
=
ω n+1 ,...,ω m ω m+1 ,...,ω N
=
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
X
X(ω 1 . . . ω N )
ω n+1 ,...,ω N
e n+1 = ω n+1 , . . . , ωN = ω N |ω1 = ω 1 , . . . , ωn = ω n }
×P{ω
e n [X](ω 1 . . . ω n ).
=E
(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
e
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
e k [Dk+1 Xk+1 ] = E
e k [Dk+1 Sk+1 ] − Sn E
e k [Dk+1 Bk+1,m ] .
E
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 =
e k+1 [Dm ] and then in the form Dk Bk,m = E
e k [Dm ], we may rewrite this as
E
e k+1 [Dm ]
ek E
e k [Dk+1 Xk+1 ] = Dk Sk − Sn E
E
Bn,m
Sn e
= 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
hedge for a short position in the caplet paying (R2 − 31 )+ 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 13 at time two in the event ω1 = H, ω2 = H, then one can simply
invest this 31 in the money market in order to have the 23 required to pay off
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
at time zero and invest in the money market
(ii) Show how to begin with 21
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,
1 e
E1 [D2 V2 ](H)
D1 (H)
e 2 = H|ω1 = H}D2 (HH)V2 (HH)
= P{ω
e 2 = T |ω1 = H}D2 (HT )V2 (HT )
+P{ω
V1 (H) =
2 6 1 1 6
4
· · + · ·0 =
,
3 7 3 3 7
21
1 e
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
It is straight-forward to verify that this works. Set X0 = V0 =
2
21
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
58
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.