Solutions to Stochastic Calculus for Finance I (Steven Shreve)

Dr. Guowei Zhao

Department of Mathematics and Statistics

McMaster University
Hamilton,ON L8S 4K1
September 9, 2013
Contents
1 Chapter 1 2
1.1 Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Exercise 1.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Chapter 2 4
2.1 Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.3 Exercise 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.4 Exercise 2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.5 Exercise 2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.6 Exercise 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 Chapter 3 7
3.1 Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 Exercise 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.3 Exercise 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.4 Exercise 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.5 Exercise 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4 Chapter 4 11
4.1 Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.2 Exercise 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4.3 Exercise 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5 Chapter 5 16
5.1 Exercise 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
5.2 Exercise 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
5.3 Exercise 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Email: zhaog22@mcmaster.ca
1
Introduction
This solution manual will be updated anytime, and is NOT intended for any business use. The author
suggests this manual as a reference book to the metioned book by Steven Shreve. Also anyone involved
in any mathematical nance courses shall not copy the solutions in this book directly. This is ideally
for self-check purpose.
If anyone nd any mistakes or misprints in this book, please inform me. Thanks.
1 Chapter 1
1.1 Exercise 1.3
p =
1 +
1
4

1
2
2
1
2
=
1
2
, q =
1
2
Then by the neutral pricing formula (1.1.10) of Text,
V
0
=
1
1 +
1
4
[ pV
1
(H) + qV
1
(T)]
=
4
5
[ pS
1
(H) + qS
1
(T)]
=
4
5
_
1
2
8 +
1
2
2
_
=4
Therefore V
1
= S
1
implies V
0
= S
0
.
1.2 Exercise 1.4
The case of n = 0 is given by the denition. Suppose its true for n, now for n + 1:
X
n+1
(
1
. . .
n
T) =
n
(
1
. . .
n
)dS
n
(
1
. . .
n
) +(1+r)(X
n
(
1
. . .
n
)
n
(
1
. . .
n
)S
n
(
1
. . .
n
))
To simplify the notation, we suppress (
1
. . .
n
) and rewrite it as
X
n+1
(T) =
n
dS
n
+ (1 +r)(X
n

n
S
n
)
=(1 +r)X
n
+
n
S
n
(d (1 +r))
=(1 +r)V
n
+ (d (1 +r))
V
n+1
(H) V
n+1
(T)
u d
= pV
n+1
(H) + qV
n+1
(T) ( pV
n+1
(H) + qV
n+1
(T))
=V
n+1
(T)
This completes the proof.
1.3 Exercise 1.8
1. For the three-period model of the stock price
2
S
0
= 4
S
1
(T) = 2
S
2
(TT) = 1
S
3
(TTT) = 0.5
S
3
(TTH) = 2
S
2
(TH) = 4
S
3
(THT) = 2
S
3
(THH) = 8
S
1
(H) = 8
S
2
(HT) = 4
S
3
(HTT) = 2
S
3
(HTH) = 8
S
2
(HH) = 8
S
3
(HHT) = 8
S
3
(HHH) = 32
The correponsding pairs of (s, y) are:
(4, 4)
(2, 6)
(1, 7)
(0.5, 7.5)
(2, 9)
(4, 10)
(2, 12)
(8, 18)
(8, 12)
(4, 16)
(2, 18)
(8, 24)
(16, 28)
(8, 36)
(32, 60)
At time three, there are 8 possible pairs for (S
3
, Y
3
), namely
(32, 60), (8, 36), (8, 24), (2, 18), (8, 18), (2, 12), (2, 9), (0.5, 7.5)
Since the payo of the option at time three is given by v
3
(s, y) = (
1
4
y 4)
+
, then
v
3
(32, 60) = 11, v
3
(8, 36) = 5, v
3
(8, 24) = 2, v
3
(2, 18) = 0.5
v
3
(8, 18) = 0.5, v
3
(2, 12) = 0, v
3
(2, 9) = 0, v
3
(0.5, 7.5) = 0
Then the algorithm of Theorem 1.2.2 of Text can be written as
v
n
(s, y) =
1
1 +
1
4
_
1
2
v
n+1
(2s, y + 2s) +
1
2
v
n+1
(
s
2
, y +
s
2
)
_
=
2
5
_
v
n+1
(2s, y + 2s) +v
n+1
(
s
2
, y +
s
2
)
_
(1.1)
3
2. By (1.1), we continue the computation as
v
2
(16, 28) =
2
5
[v
3
(32, 60) +v
3
(8, 36)] = 6.4
v
2
(4, 16) =
2
5
[v
3
(8, 24) +v
3
(2, 18)] = 1
v
2
(4, 10) =
2
5
[v
3
(8, 18) +v
3
(2, 12)] = 0.2
v
2
(1, 7) =
2
5
[v
3
(2, 9) +v
3
(0.5, 7.5)] = 0
Then we compute
v
1
(8, 12) =
2
5
[v
2
(16, 28) +v
2
(4, 16)] = 2.96
v
1
(2, 6) =
2
5
[v
2
(4, 10) +v
2
(1, 7)] = 0.08
Eventually
v
0
(4, 4) =
2
5
[v
1
(8, 12) +v
1
(2, 6)] = 1.216
3. At each time n = 0, 1, 2, the number of shares of stock that shoudl be held by replicating portfolio
is

n
(s) =
v
n+1
(2s, y + 2s) +v
n+1
(
s
2
, y +
s
2
)
2s
s
2
2 Chapter 2
2.1 Exercise 2.3
By Jensens Inequality
E
n
[(M
n+1
)] (E
n
[M
n+1
]) = (M
n
)
The latter is by the martingale property of M
n
.
2.2 Exercise 2.4
1.
E
n
(M
n+1
)(
1
. . .
n
) =
1
2
M
n+1
(
1
. . .
n
H) +
1
2
M
n+1
(
1
. . .
n
T)
=
1
2
(M
n
(
1
. . .
n
) + 1) +
1
2
(M
n
(
1
. . .
n
) 1)
= M
n
(
1
. . .
n
)
so M
n
is martingalge.
2.
E
n
(S
n+1
)(
1
. . .
n
) =
1
2
S
n+1
(
1
. . .
n
H) +
1
2
S
n+1
(
1
. . .
n
T)
=
1
2
S
n
(
1
. . .
n
)e
1

2
e

+e

+
1
2
S
n
(
1
. . .
n
)e
(1)

2
e

+e

= S
n
(
1
. . .
n
)
so S
n
is martingalge.
4
2.3 Exercise 2.6
We suppress the arguments (
1
. . .
n
) for simplier notations.
E
n
[I
n+1
]
=E
n
_
_
n

j=0

j
(M
j+1
M
j
)
_
_
=E
n
_
_
n1

j=0

j
(M
j+1
M
j
)
_
_
+E
n
[
j
(M
n+1
M
n
)]
=
n1

j=0

j
(M
j+1
M
j
) +
n
E
n
[M
n+1
]
n
M
n
=I
n
+
n
M
n

n
M
n
=I
n
Therefore I
0
,I
1
,I
2
,. . . ,I
N
is a martingale.
2.4 Exercise 2.7
We construct the following binomial tree.
S
0
= 4
S
1
(T) = 2
S
2
(TT) = 1
S
3
(TTT) = 0
S
3
(TTH) = 2
S
2
(TH) = 3
S
3
(THT) = 0
S
3
(THH) = 6
S
1
(H) = 6
S
2
(HT) = 3
S
3
(HTT) = 2
S
3
(HTH) = 4
S
2
(HH) = 9
S
3
(HHT) = 6
S
3
(HHH) = 12
Here if the current result is H, then the next step would be 3 with equal probabilities; if the
current result is T, then then next step will be 1 with equal probabilities. Then it is very clear that
for any n
E
n
[X
n+1
] = X
n
which means this process is martingale.
Now let f(x) = x
2
. We investigate
E
2
[f(M
3
)](HT) =
1
2
4
2
+
1
2
2
2
= 10
E
2
[f(M
3
)](TH) =
1
2
6
2
+
1
2
0
2
= 18
5
Note that M
2
(HT) = M
2
(TH) = 3, so there cannot be a function g such that
_
E
2
[f(M
3
)](HT) = g(M
2
(HT))
E
2
[f(M
3
)](TH) = g(M
2
(TH))
i.e.,
_
10 = g(3)
18 = g(3)
Therefore this process is not Markov.
2.5 Exercise 2.9
1.

P(HH) =
1 +
1
4
1
12
8
1
=
1
2

P(HT) =
1
2

P(TH) =
1 +
1
2
1
8
2
1
=
1
6

P(TT) =
5
6
2. Since V
2
(HH) = 5,V
2
(HT) = V
2
(TH) = 1,V
2
(TT) = 0,
V
1
(H) =
4
5
(
1
2
V
2
(HH) +
1
2
V
2
(HT)) = 2.4
V
1
(T) =
2
3
(
1
6
V
2
(TH) +
5
6
V
2
(TT)) =
1
9
0.1111
V
1
(T) =
4
5
(
1
2
V
1
(H) +
1
2
V
1
(T)) 1.0044
3.

0
=
V
1
(H) V
1
(T)
S
1
(H) S
1
(T)
=
2.4 0.1111
8 2
0.3815
To see it more clearly, we compute the value of the portfolio at time one
X
1
(H) =
0
S
1
(H) + (1 +r)(V
0

0
S
0
) = 0.3815 8 + (1 +
1
4
)(1 0.3815 4) = 2.3945 2.40 = V
1
(H)
X
1
(T) =
0
S
1
(T) + (1 +r)(V
0

0
S
0
) = 0.3815 2 + (1 +
1
4
)(1 0.3815 4) = 0.1055 0.1111 = V
1
(T)
4.

1
(H) =
V
2
(HH) V
2
(HT)
S
2
(HH) S
2
(HT)
=
5 1
12 8
= 1
2.6 Exercise 2.13
1. Let Z = S
n+1
/S
n
, which only depends on the (n + 1)th coin toss, then
S
n+1
=S
n

S
n+1
S
n
Y
n+1
=Y
n
+S
n+1
= Y
n
+S
n
Z
We need to compute for arbitrary choice of function h,
E
n
[h(S
n+1
, Y
n+1
)] = E
n
[h(S
n
Z, Y
n
+S
n
Z)]
6
Now we dene
g(s, y) = E[h(sZ, y +sZ)] = ph(su, y +su) +qh(sd, y +sd)
Then we have
E
n
[h(S
n+1
, Y
n+1
)] = g(S
n
, Y
n
)
which means (S
n
, Y
n
) is a Markov process.
Similarly for the case under the risk-neutral probability.
2. First for n = N,
V
N
(S
N
, Y
N
) = V
N
= f
_
1
N + 1
N

n=0
S
n
_
= f
_
Y
N
N + 1
_
Replacing by dummy variables, we have
V
N
(s, y) = f
_
y
N + 1
_
Then for n = N 1, N 2, . . . , 0, we have
v
n
(S
n
, Y
n
)
=

E
n
_
v
N
(S
N
, Y
N
)
(1 +r)
Nn
_
=
1
r + 1

E
n
[V
n+1
]
=
1
r + 1

E
n
[v
n+1
(S
n+1
, Y
n+1
)]
=
1
r + 1

E
n
_
v
n+1
(S
n

S
n+1
S
n
, Y
n
+S
n

S
n+1
S
n
)
_
Replacing by dummy variables, we have
v
n
(s, y) =
1
r + 1
[ pv
n+1
(us, y +us) + qv
n+1
(ds, y +ds)]
3 Chapter 3
3.1 Exercise 3.2
1.

P() =

P() =

Z()P() = EZ = 1
2.

EY =

Y ()

P() =

Y ()Z()P() = E[ZY ]
3. Since P(Z 0) = 1, we can apply Holders inequality as:
0

P(A) =

P() =

A
Z()P()
_

A
P()
_

A
Z()
_
= P(A)
_

A
Z()
_
= 0
therefore P(A) = 0 implies

P(A) = 0.
7
4. Under the assumption P(Z > 0) = 1,
0 P(A) =

A
P() =

A
1
Z()

P()
_

P()
_

A
1
Z()
_
=

P(A)
_

A
1
Z()
_
= 0
therefore P(A) = 0.
5.
P(A) = 1 P(A
c
) = 0

P(A
c
) = 0

P(A) = 1
which means the equivalent probability measures also agree with which events have probability
1.
6. We consider the following example in which the given two probability measures are not equivalent:
S
1
(H) = 2
S
0
= 1
S
1
(T) = 0.5
with P(H) = 1/2 = P(T) and

P(H) = 1,

P(T) = 0. Then
Z(H) =

P(H)
P(H)
= 2, Z(T) =

P(T)
P(T)
= 0
which gives P(Z 0) = 1. However

P(T) = 0 = 1/2 = P(T), so they are not equivalent.
3.2 Exercise 3.3
For n = 3,
M
3
= E
3
[S
3
] = S
3
For n = 2,
M
2
(HH) =E
2
[S
3
](HH) =
2
3
32 +
1
3
8 = 24
M
2
(HT) =E
2
[S
3
](HT) =
2
3
8 +
1
3
2 = 6
M
2
(TH) =E
2
[S
3
](HH) =
2
3
8 +
1
3
2 = 6
M
2
(TT) =E
2
[S
3
](HH) =
2
3
2 +
1
3

1
2
= 1.5
For n = 1,
M
1
(H) =E
1
[S
3
](H) = E
1
[E
2
[S
3
]](H) =
2
3
24 +
1
3
6 = 18
M
1
(T) =E
1
[S
3
](T) = E
1
[E
2
[S
3
]](T) =
2
3
6 +
1
3
1.5 = 4.5
8
For n = 0,
M
0
= E
0
[S
3
](H) = E
0
[E
1
[S
3
]](H) =
2
3
18 +
1
3
4.5 = 13.5
Therefore the binomial tree of M
n
is:
M
3
(HHH) = 24
M
2
(HH) = 24
M
1
(H) = 18 M
3
(HHT) = M
3
(HTH) = M
3
(THH) = 8
M
0
= 13.5 M
2
(HT) = M
2
(TH) = 6
M
1
(T) = 4.5 M
3
(HTT) = M
3
(TTH) = M
3
(THT) = 2
M
2
(TT) = 1.5
M
3
(TTT) = 0.5
2
3
1
3
Note that the above computation of M
n
is also the vercation of the martingale property by
iteration property of the conditional expectation, which is for n = 0, 1, 2:
E
n
[M
n+1
] = E
n
[E
n+1
[S
3
]] = E
n
[S
3
] = M
n
3.3 Exercise 3.4
1.

3
(HHH) =
Z
3
(HHH)
(1 +r)
3
=
27
64
(1 +
1
4
)
3
=
27
125

3
(HHT) =
3
(HTH) =
3
(THH) =
Z
3
(HHH)
(1 +r)
3
=
27
32
(1 +
1
4
)
3
=
54
125

3
(HTT) =
3
(HTT) =
3
(TTH) =
Z
3
(HTT)
(1 +r)
3
=
27
16
(1 +
1
4
)
3
=
108
125

3
(TTT) =
Z
3
(TTT)
(1 +r)
3
=
27
8
(1 +
1
4
)
3
=
216
125
9
2.
V
0
=E[V
3
] =

V
3
()
3
()P()
=11
27
125

8
27
+ 5
54
125

4
27
+ 2
54
125

4
27
+ 0.5
108
125

2
27
+
+ 0.5
54
125

4
27
+ 0
108
125

2
27
+ 0
108
125

2
27
+ 0
216
125

1
27
=
4104
125 27
=1.216
which is the same as Exercise (1.3).
3.

2
(HT) =
2
(TH) =
Z
2
(HT)
(1 +
1
4
)
2
=
18
25
4.
V
2
(HT) =
1

2
(HT)
=
_
2
3

3
(HTH)V
3
(HTH) +
1
3

3
(HTT)V
3
(HTT)
_
=
25
18

_
2
3

54
125
2 +
1
3

108
125
0.5
_
=
25
18

5 54
3 125
= 1
V
2
(TH) =
1

2
(TH)
=
_
2
3

3
(THH)V
3
(THH) +
1
3

3
(THT)V
3
(THT)
_
=
25
18

_
2
3

54
125
0.5 +
1
3

108
125
0
_
=
25
18

54
3 125
= 0.2
Note that V
2
(HT) = v
2
(4, 16) = v
2
(4, 10) = V
2
(TH).
3.4 Exercise 3.6
When the utility function U(x) = ln x, U

(x) = 1/x, then the equation (3.3.24) of Textbook yields

1
X
N
=
Z
(1 +r)
N
X
N
=
(1 +r)
N
Z
Then (3.3.29) of Text becomes
E
_
Z
(1 +r)
N

(1 +r)
N
Z
_
= X
0
which implies = 1/X
0
. Therefore
X
N
=
(1 +r)
N
Z
=
(1 +r)
N
Z/X
0
=
X
0
(1 +r)
N
Z
=
X
0

N
Then we treat it as the payo of a derivative security. Then by (3.2.6) of Text,
X
n
= V
n
=
1

n
E
n
[
N
V
N
] =
1

n
E
n
[
N
X
N
] =
1

n
E
n
[
N
X
0

N
] =
X
0

n
10
3.5 Exercise 3.7
When the utility function U(x) =
1
p
x
p
, U

(x) = x
p1
, then the equation (3.3.24) of Textbook yields
X
p1
N
=
Z
(1 +r)
N
X
N
=
_
Z
(1 +r)
N
_ 1
p1
Then (3.3.29) of Text becomes
E
_
Z
(1 +r)
N

_
Z
(1 +r)
N
_ 1
p1
_
= X
0

1
p1
E
_
Z
1+
1
p1
(1 +r)
N(1+
1
p1
)
_
= X
0
which implies

1
p1
=
X
0
(1 +r)
p/(p1)
E[Z
p/(p1)
]
Therefore
X
N
=
_
Z
(1 +r)
N
_ 1
p1
=
_
Z
(1 +r)
N
_ 1
p1

X
0
(1 +r)
p/(p1)
E[Z
p
p1
]
=
X
0
Z
1/(p1)
E[Z
p
p1
]
(1 +r)
N(
p
p1

1
p1
)
=
X
0
(1 +r)
N
Z
1
p1
E[Z
p
p1
]
4 Chapter 4
4.1 Exercise 4.1
1. For the American put, g(s) = 4s, (or (4s)
+
), then we follow the American algorithm (4.2.5)
and (4.2.6) of Text as:
v
3
(32) =max{4 32, 0} = 0
v
3
(8) =max{4 8, 0} = 0
v
3
(2) =max{4 2, 0} = 2
v
3
(0.5) =max{4 0.5, 0} = 3.5
v
2
(16) =max{4 16,
2
5
(v
3
(32) +v
3
(8))} = max{12, 0} = 0
v
2
(4) =max{4 4,
2
5
(v
3
(8) +v
3
(2))} = max{0, 0.8} = 0.8
v
2
(1) =max{4 1,
2
5
(v
3
(2) +v
3
(0.5))} = max{3, 2.2} = 3
v
1
(8) =max{4 8,
2
5
(v
2
(16) +v
2
(4))} = max{4, 0.32} = 0.32
v
1
(2) =max{4 2,
2
5
(v
2
(4) +v
2
(1))} = max{2, 1.52} = 2
v
0
(4) =max{4 4,
2
5
(v
1
(8) +v
1
(2))} = max{0, 0.928} = 0.928
So V
P
0
= 0.928.
11
2. For the American call, g(s) = s 4, (or (s 4)
+
), then we follow the American algorithm (4.2.5)
and (4.2.6) of Text as:
v
3
(32) =max{32 4, 0} = 28
v
3
(8) =max{8 4, 0} = 4
v
3
(2) =max{2 4, 0} = 0
v
3
(0.5) =max{0.5 4, 0} = 0
v
2
(16) =max{16 4,
2
5
(v
3
(32) +v
3
(8))} = max{12, 12.8} = 12.8
v
2
(4) =max{4 4,
2
5
(v
3
(8) +v
3
(2))} = max{0, 1.6} = 1.6
v
2
(1) =max{1 4,
2
5
(v
3
(2) +v
3
(0.5))} = max{3, 0} = 0
v
1
(8) =max{8 4,
2
5
(v
2
(16) +v
2
(4))} = max{4, 5.76} = 5.76
v
1
(2) =max{2 4,
2
5
(v
2
(4) +v
2
(1))} = max{2, 0.64} = 0.64
v
0
(4) =max{4 4,
2
5
(v
1
(8) +v
1
(2))} = max{0, 2.56} = 2.56
So V
C
0
= 2.56.
3. For the straddle option, g(s) = (4 s)
+
+ (s 4)
+
= |s 4|. Then we follow the American
algorithm (4.2.5) and (4.2.6) of Text as:
v
3
(32) =max{|32 4| , 0} = 28
v
3
(8) =max{|8 4| , 0} = 4
v
3
(2) =max{|4 2| , 0} = 2
v
3
(0.5) =max{|4 0.5| , 0} = 3.5
v
2
(16) =max{|16 4| ,
2
5
(v
3
(32) +v
3
(8))} = max{12, 12.8} = 12.8
v
2
(4) =max{|4 4| ,
2
5
(v
3
(8) +v
3
(2))} = max{0, 2.4} = 2.4
v
2
(1) =max{|1 4| ,
2
5
(v
3
(2) +v
3
(0.5))} = max{3, 2.2} = 3
v
1
(8) =max{|8 4| ,
2
5
(v
2
(16) +v
2
(4))} = max{4, 6.08} = 6.08
v
1
(2) =max{|2 4| ,
2
5
(v
2
(4) +v
2
(1))} = max{2, 2.16} = 2.16
v
0
(4) =max{|4 4| ,
2
5
(v
1
(8) +v
1
(2))} = max{0, 3.2964} = 3.296
So V
P
0
= 3.296.
4. Note that V
P
0
= 3.296 < 3.488 = 0.928 + 2.56 = V
P
0
+V
C
0
.
Inspried by the above computation, especially the three v
1
(2) values, 2.16 < 2 + 0.64, we can
conclude the reason why V
P
0
< V
P
0
+V
C
0
. For the American call, it is never optimal to exercise
the option before time three. For the American put, it is optimal to make the early exercise at
time one at the node v
1
(2). However, for the straddle, the optimal policy is not to exercise at
time one at the node v
1
(2). As the result, the expected return of the straddle will be strickly
less than the summation of two separate American call and put options which have their own
dierent optimal exercise time.
12
4.2 Exercise 4.3
For the three-period model of the stock price
S
0
= 4
S
1
(T) = 2
S
2
(TT) = 1
S
3
(TTT) = 0.5
S
3
(TTH) = 2
S
2
(TH) = 4
S
3
(THT) = 2
S
3
(THH) = 8
S
1
(H) = 8
S
2
(HT) = 4
S
3
(HTT) = 2
S
3
(HTH) = 8
S
2
(HH) = 8
S
3
(HHT) = 8
S
3
(HHH) = 32
Denote Y
n
=

n
j=0
S
j
as Exercise 1.8. Then the correponsding pairs of (s, y) are:
(4, 4)
(2, 6)
(1, 7)
(0.5, 7.5)
(2, 9)
(4, 10)
(2, 12)
(8, 18)
(8, 12)
(4, 16)
(2, 18)
(8, 24)
(16, 28)
(8, 36)
(32, 60)
At time three, there are 8 possible pairs for (S
3
, Y
3
), namely
(32, 60), (8, 36), (8, 24), (2, 18), (8, 18), (2, 12), (2, 9), (0.5, 7.5)
13
Since the payo of the option at time three is given by v
n
(s, y) = (4
y
n+1
))
+
, then
v
3
(32, 60) =
_
4
1
4
60
_
+
= 0
v
3
(8, 36) =
_
4
1
4
36
_
+
= 0
v
3
(8, 24) =
_
4
1
4
24
_
+
= 0
v
3
(2, 18) =
_
4
1
4
18
_
+
= 0
v
3
(8, 18) =
_
4
1
4
18
_
+
= 0
v
3
(2, 12) =
_
4
1
4
12
_
+
= 1
v
3
(2, 9) =
_
4
1
4
9
_
+
= 0.75
v
3
(0.5, 7.5) =
_
4
1
4
7.5
_
+
= 2.125
Then we continue Theorem 4.4.3 of Text as:
v
2
(16, 28) =max{(4
1
3
28)
+
,
2
5
(0 + 0)} = 0
v
2
(4, 16) =max{(4
1
3
16)
+
,
2
5
(0 + 0)} = 0
v
2
(4, 10) =max{(4
1
3
10)
+
,
2
5
(0 + 1)} =
2
3
v
2
(1, 7) =max{(4
1
3
7)
+
,
2
5
(1.75 + 2.125)} =
5
3
v
1
(8, 12) =max{(4
1
2
12)
+
,
2
5
(0 + 0)} = 0
v
1
(2, 6) =max{(4
1
2
6)
+
,
2
5
(
2
3
+
5
3
)} = 1
v
0
(4, 4) =max{(4
1
1
4)
+
,
2
5
(0 + 1)} = 0.4
Now we note that v
2
(4, 10), v
2
(1, 7) and v
1
(2, 6) are the nodes where V
n
= G
n
. Then by Denition
4.3.1 or (4.4.21) of Text, we can conclude that the optimal exercising policy/stopping time as:
(HHH) = (HHT) = (HTH) = (HTT) =
(THH) = (THT) = (TTH) = (TTT) = 1
4.3 Exercise 4.5
By the property of the stopping times, the cases such as (TT) = 1 but (TH) = 2, or (HH) = 1
but (HT) = 1, are not allowed. Therefore, all the 26 stopping times in S
0
are:
14
(HH) (HT) (TH) (TT)
0 0 0 0
1 1 1 1
1 1 2 2
1 1 2
1 1 2
1 1
2 2 1 1
2 2 2 2
2 2 2
2 2 2
2 2
2 1 1
2 2 2
2 2
2 2
2
2 1 1
2 2 2
2 2
2 2
2
1 1
2 2
2
2

But we note that when (HH) = 2, or (HH) = 1, the option will be out of money as the intrinsic
value will be negative, so we shall not exercise in these cases. Therefore the remaining 11 possible sets
of stopping times are:
(HH) (HT) (TH) (TT)
0 0 0 0
2 1 1
2 2 2
2 2
2 2
2
1 1
2 2
2
2

15
Then we compute the V
0
= max
0

E
_
I
{2}
_
4
5
_

with the values in Fig.4.4.1 of Text. In

advance we compute the terms for the immediate later use:
_
4
5
_
1
G
1
(T) =2.4
_
4
5
_
2
G
2
(HT) =0.64
_
4
5
_
2
G
2
(TH) =0.64
_
4
5
_
2
G
2
(TT) =2.56
Then we have
(HH) (HT) (TH) (TT) V
0
0 0 0 0 1
2 1 1 1.36
2 2 2 0.96
2 2 0.32
2 2 0.80
2 0.16
1 1 1.2
2 2 0.80
2 0.16
2 0.64
0
Note that the largest value is the second row with (, 2, 1, 1), which gives 1.36 as (4.4.7) of Text.
5 Chapter 5
5.1 Exercise 5.2
1. Since
f

() = pe

qe

= e

(pe
2
q)
and also 0 < q < 1/2 < p, for any 0, pe
2
q p q > 0, so we have f

() > 0 for any

0. Then f() > f(0) = p +q = 1 for any > 0.
2. By
E
n
[S
n+1
] =E
n
_
S
n

1
f()
e
Xn+1
_
=S
n

1
f()

_
pe

+qe

_
=S
n
we can conclude that S
n
is a martingale.
3. As the symmetric random walk in Section 5.2 of Text, since S
n1
is a martingale, hence
1 = S
0
= E[S
n1
] = E
_
e
Mn
1
_
1
f()
_
n1
_
(5.1)
16
We would like to take the limit n .
First, since 0 <
1
f()
< 1,
lim
n
_
1
f()
_
n1
=
__
1
f()
_
1
, if
1
<
0
1
=
(5.2)
Second, since
0 e
Mn
1
e

(5.3)
and
lim
n
e
Mn
1
= e
M
1
= e

if
m
< (5.4)
Then by (5.2), (5.3) and (5.4), we have
lim
n
e
Mn
1
_
1
f()
_
n1
=
_
e

_
1
f()
_
1
, if
1
<
0
1
=
which is equivalent to
lim
n
e
Mn
1
_
1
f()
_
n1
= I
{1<}
e

_
1
f()
_
1
Now we apply this result to (5.1) by taking the limit n , we will have
E
_
I
{1<}
e

_
1
f()
_
1
_
= 1
Immediately,
E
_
I
{1<}
_
1
f()
_
1
_
= e

(5.5)
By taking the limit 0+, we will have E[I
{1<}
] = 1, which implies P(
1
< ) = 1.
4. For any xed number (0, 1), there is always a such that
=
1
f()
=
1
pe

+qe

which is
pe

+qe

1 =0
p +qe
2
e

=0
e

=
1
_
1 4
2
pq
2q
e

=
1
_
1 4
2
pq
2q
since > 0, e

< 1
It is also clear that this > 0.
Then immediately (5.5) yields
E [
1
] =
1
_
1 4
2
pq
2q
(5.6)
by dropping I
{1<}
due to the condition P(
1
< ) = 1.
17
5. By taking the derivative with respect to in (5.6), we have
E[
1

11
] =

E [
1
]
=

1
_
1 4
2
pq
2q
=

1
2
(1 4
2
pq)

1
2
(8pq)2q
_
1
_
1 4
2
pq
_
2q
4
2
q
2
Then by taking the limit 1, we have
E[
1
] =
8pq
2
(1 4pq)

1
2
2q
_
1

1 4pq
_
4q
2
=
8pq
2
2q

1 4pq + 2q(1 4pq)

4q
2

1 4pq
=
2q 2q

1 4pq
4q
2

1 4pq
=
(1 4pq)
1/2
1
2q
5.2 Exercise 5.4
1. First note that the random walk can reach lelve 2 only on a even-numbered step. Hence we may
rewrite
E[
2
] =

k=0

2k
P(
2
= 2k)
Since we also have
E[
2
] =
_
1

1
2

_
2
=

k=0
_

2
_
2k
(2k)!
k!(k + 1)!
The above two equations are true for any (0, 1). By equating them, we must have

2k
P(
2
= 2k) =
_

2
_
2k
(2k)!
k!(k + 1)!
which implies
P(
2
= 2k) =
_
1
2
_
2k
(2k)!
k!(k + 1)!
2. By the reection principle, for any k = 1, 2, 3, . . .
P(
2
2k) =P(M
2k
= 2) + 2P(M
2k
4)
=P(M
2k
= 2) +P(M
2k
4) +P(M
2k
4)
=1 P(M
2k
= 0) P(M
2k
= 2)
Now in order for M
2k
to be 0 in the rst 2k tosses, there must be k heads and k tails. There are
_
2k
k
_
paths that have this property, and each such path has probability (1/2)
2k
. Hence
P(M
2k
= 0) =
_
1
2
_
2k
_
2k
k
_
=
_
1
2
_
2k
(2k)!
k!k!
18
Similarly,
P(M
2k
= 2) =
_
1
2
_
2k
_
2k
k 1
_
=
_
1
2
_
2k
(2k)!
(k 1)!(k + 1)!
Now for any k = 2, 3, 4, . . . ,
P(
2
= 2k)
=P(
2
2k) P(
2
2k 2)
=P(M
2k2
= 0) +P(M
2k2
= 2) P(M
2k
= 0) P(M
2k
= 2)
=
_
1
2
_
2k2
(2k 2)!
(k 1)!(k 1)!
+
_
1
2
_
2k2
(2k 2)!
(k 2)!k!

_
1
2
_
2k
(2k)!
k!k!

_
1
2
_
2k
(2k)!
(k 1)!(k + 1)!
=
_
1
2
_
2k
(2k)!
k!(k + 1)!

_
4k
2
(k + 1)
2k(2k 1)
+
4k(k 1)(k + 1)
2k(2k 1)
(k + 1) k
_
=
_
1
2
_
2k
(2k)!
k!(k + 1)!

_
8k
3
+ 4k
2
4k
2k(2k 1)
(2k + 1)
_
=
_
1
2
_
2k
(2k)!
k!(k + 1)!

_
8k
3
+ 4k
2
4k 2k(4k
2
1)
2k(2k 1)
_
=
_
1
2
_
2k
(2k)!
k!(k + 1)!

_
4k
2
2k
2k(2k 1)
_
=
_
1
2
_
2k
(2k)!
k!(k + 1)!
For k = 1, P(
2
= 2) = 1/4, which is the same as
_
1
2
_
2
(2)!
1!(1+1)!
. Therefore for any k = 1, 2, 3, . . . ,
P(
2
= 2k) =
_
1
2
_
2k
(2k)!
k!(k + 1)!
5.3 Exercise 5.5
1. For the symmetic random walk, it is easy to see that for any 0 j n,
P(M
n
= b|M
j
= k) = P(M
n
= 2k b|M
j
= k)
Then
P(M

n
m, M
n
= b) = P(M

n
m, M
n
= 2mb) = P(M
n
= 2mb) (5.7)
The latter equality is because m b, so M
n
= 2mb m implies M

n
m.
In order for M
n
to be 2mb, we need N heads and nN tails. Then the number N is determined
as
N (n N) = 2mb = N = m+
n b
2
Note that there are
_
n
N
_
paths that have this property and each such path has probability
_
1
2
_
n
.
Hence by (5.7)
P(M

n
m, M
n
= b) = P(M

n
m, M
n
= 2mb) = P(M
n
= 2mb)
=
_
1
2
_
n
n!
_
m+
nb
2
_
!
_
n m
nb
2
_
!
=
_
1
2
_
n
n!
_
m+
nb
2
_
!
_
n+b
2
m
_
!
2.
19