2 Basic Options
9
b) Due to the fact that the sum of two normal random variables is a
normal variable, it is easy to show that the sum of
n
normal variables,
φ
1
√
Δt
+
· · ·
+
φ
n
√
Δt
, and the limit of the sum as
n
→ ∞
are normal
variables. Because
E [
X
(
t
)] = lim
n
→∞
E
"
n
X
i
=1
φ
i
√
Δt
#
= lim
n
→∞
n
X
i
=1
E [
φ
i
]
√
Δt
= 0
,
and
Var [
X
(
t
)] = lim
n
→∞
n
X
i
=1
Var [
φ
i
]
Δt
= lim
n
→∞
t
=
t,
X
(
t
) is a normal random variable with mean zero and variance
t
.
c) From a) we know that
dX
=
X
(
t
+
dt
)

X
(
t
) is a normal variable.
For
dX
we have
E [
dX
] = E [
X
(
t
+
dt
)]

E [
X
(
t
)] = 0
and
Var [
dX
] = E
h
(
dX
)
2
i
= E
h
(
X
(
t
+
dt
)

X
(
t
))
2
i
= E
£
X
2
(
t
+
dt
)

2
X
(
t
+
dt
)
X
(
t
) +
X
2
(
t
)
/
=
t
+
dt

2E
£
X
2
(
t
)
/

2E [
dX
·
X
(
t
)] +
t
=
dt,
so
dX
is a normal random variable with mean zero and variance
dt
.
Here we have used the fact that
X
(
t
) and
dX
are independent.
d) From
S
=
e
μt
+
σX
(
t
)
, we have ln
S
=
μt
+
σX
(
t
). Because
∂
ln
S
∂t
=
μ,
∂
ln
S
∂X
=
σ
,
∂
2
ln
S
∂X
2
= 0 and
dX
=
dX
, using Itˆ
o’s lemma, we have
d
ln
S
=
μdt
+
σdX.
4. Suppose
dS
=
a
(
S, t
)
dt
+
b
(
S, t
)
dX
,
where
dX
is a Wiener process. Let
f
be a function of
S
and
t
. Show that
df
=
∂f
∂S
dS
+
∂f
∂t
+
1
2
b
2
∂
2
f
∂S
2
¶
dt
=
b
∂f
∂S
dX
+
∂f
∂t
+
1
2
b
2
∂
2
f
∂S
2
+
a
∂f
∂S
¶
dt.
Solution
:
If we vary
S
and
t
by a small amount
dS
and
dt
, then
f
also varies by a
small amount. From the Taylor series expansion we can write
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
10
2 Basic Options
df
=
∂f
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Zhu
 Math, Variance, Probability theory, probability density function, dt, dx, dξ

Click to edit the document details