ACTL2003&5103 Tutorial 12 Solutions
1. Ross (8th edition): Chapter 19, Q25
(a)
E
[
Y
]
=
E
Z
1
0
tdB
t
¶
=
E
B
1

Z
1
0
B
t
dt
¶
=
E
[
B
1
]

Z
1
0
E
[
B
t
]
dt
= 0
and the variance is
V ar
[
Y
]
=
V ar
Z
1
0
tdB
t
¶
=
Z
1
0
t
2
dt
=
1
3
.
(b)
E
[
Z
]
=
E
Z
1
0
t
2
dB
t
¶
=
E
B
1

Z
1
0
2
B
t
tdt
¶
=
E
[
B
1
]

Z
1
0
2
E
[
B
t
]
tdt
= 0
and the variance is
V ar
[
Z
]
=
V ar
Z
1
0
t
2
dB
t
¶
=
Z
1
0
t
4
dt
=
1
5
.
1
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2.
Z
2
.
5
1
.
5
f
(
t
)
dB
t
=
Z
2
1
.
5
3
dB
t
+
Z
2
.
5
2
4
dB
t
= 3(
B
2

B
1
.
5
) + 4(
B
2
.
5

B
2
)
Since we know that 3(
B
2

B
1
.
5
) + 4(
B
2
.
5

B
2
) is the sum of two inde
pendent normal distributions
N
(0
,
4
.
5) and
N
(0
,
8) respectively, then
Z
2
.
5
1
.
5
f
(
t
)
dB
t
is
N
(0
,
13
.
5)
3.
(a) First, note that we can rewrite the stochastic process
Y
t
as
Y
t
= log (
X
t
) = log (
X
0
) +
μt
+
σB
t
.
Let
F
(
x, t
) = log(
X
0
) +
μt
+
σx
. We apply Ito’s lemma on
Y
t
=
F
(
B
t
, t
)
.
Since
∂
∂x
F
(
x, t
) =
σ
,
∂
2
∂x
2
F
(
x, t
) = 0 and
∂
∂t
F
(
x, t
) =
μ
, We have
dY
t
=
[
∂
∂x
F
(
x, t
)]
x
=
B
t
dB
t
+ [
∂
∂t
F
(
x, t
)]
x
=
B
t
dt
+
1
2
σ
2
t
[
∂
2
∂x
2
F
(
x, t
)]
x
=
B
t
dt
=
σdB
t
+
μdt
or
=
μdt
+
σdB
t
.
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 Three '11
 kim
 Normal Distribution, Brownian Motion, dt, Yt, dRT

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