Solution to Homework 1
1. What is the distribution of
B
s
+
B
t
,
s
≤
t
?
Solution
.
E
(
B
s
+
B
t
) = 0,
var
(
B
s
+
B
t
) =
var
(
B
s
)+
var
(
B
t
)+2
E
(
B
s
B
t
) =
s
+
t
+2
s
=
3
s
+
t
. So
B
s
+
B
t
∼
N
(0
,
3
s
+
t
).
2. Compute
E
[
B
t
1
B
t
2
B
t
3
] for
t
1
< t
2
< t
3
.
Solution
. Let with 0 =
t
0
≤
t
1
≤
t
2
≤
t
3
, Δ
t
i
=
t
i

t
i

1
and Δ
B
t
i
=
B
t
i

B
t
i

1
.
So Δ
B
t
i
’s are independent
N
(0
,
Δ
t
i
), and
E
[
B
t
1
(
B
t
2
+ Δ
B
t
2
)
B
t
3
] =
E
[Δ
B
t
1
(Δ
B
t
1
+ Δ
B
t
2
)(Δ
B
t
1
+ Δ
B
t
2
+ Δ
B
t
3
)] = 0
.
3. Calculate
P
(
B
(2)
≤
0) and
P
(
B
(1)
≤
0
, B
(2)
≤
0).
Solution
.
P
(
B
(2)
≤
0) = 1
/
2 as
B
(2)
∼
N
(0
,
2).
Next, Let
X
=
B
(1) and
Y
=
B
(2)

B
(1). Then
X, Y
are i.i.d.
N
(0
,
1). Hence,
P
(
B
(1)
≤
0
, B
(2)
≤
0) =
P
(
X
≤
0
, X
+
Y
≤
0)
=
Z
∞
x
=
∞
P
(
X
≤
0
, X
+
Y
≤
0

X
=
x
)
d
Φ(
x
) =
Z
0
x
=
∞
P
(
Y
≤ 
x

X
=
x
)
d
Φ(
x
)
=
Z
0
x
=
∞
Φ(

x
)
d
Φ(
x
) =
Z
0
x
=
∞
[1

Φ(
x
)]
d
Φ(
x
)
=
Z
1
/
2
t
=0
[1

t
]
dt
= (
t

t
2
/
2)

1
/
2
t
=0
= 3
/
8
.
4. Recall that a stochastic process
{
X
t
, t
≥
0
}
is said to be
strictly stationary
if (
X
t
1
, ..., X
t
n
)
has the same joint distribution as (
X
t
1
+
a
, ..., X
t
n
+
a
) for all
n
,
a
,
t
1
, ..., t
n
.
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 Spring '11
 D
 Brownian Motion, Stochastic process, Stationary process, Xt, BTI

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