STA3007
Applied probability (0809)
Solution to Assignment 2
Chapter 2
Problem 5.1
Let
X, Y, Z
denote
X
n
+2
, X
n
+1
,
(
X
0
,
· · ·
, X
n
)
T
, then, according to the given formula
E[E(
X

Y, Z
)

Z
] = E[
X

Z
]
,
we have
E[
X
n
+2

X
0
,
· · ·
, X
n
] = E[E(
X
n
+2

X
0
,
· · ·
, X
n
+1
)

X
0
,
· · ·
, X
n
]
.
By the definition of martingale, we have
E(
X
n
+2

X
0
,
· · ·
, X
n
+1
) =
X
n
+1
.
Hence
E[
X
n
+2

X
0
,
· · ·
, X
n
] = E[
X
n
+1

X
0
,
· · ·
, X
n
] =
X
n
.
By using induction, it is easy to generalize the above formula to
E[
X
n
+
m

X
0
,
· · ·
, X
n
] =
X
n
,
for any m
>
0
.
Problem 5.3
Since
ε
1
,
· · ·
, ε
n
are identically and independently distributed as exponential distribu
tion with mean 1, thus
E(

X
n

) = E(
X
n
) =
n
i
=1
∞
0
2 exp(

ε
i
)
·
exp(

ε
i
)
dε
i
= 1
<
∞
.
Observe
S
n
+1
=
S
n
+
ε
n
+1
, we have
X
n
+1
= 2
n
+1
exp(

S
n
+1
) = 2
·
2
n
exp(

S
n

ε
n
+1
) =
X
n
·
2 exp(

ε
n
+1
)
,
hence,
E(
X
n
+1

X
0
,
· · ·
, X
n
)
=
E(
X
n
·
2 exp(

ε
n
+1
)

X
n
)
=
X
n
E(2 exp(

ε
n
+1
))
=
X
n
∞
0
2 exp(

2
ε
n
+1
)
dε
n
+1
=
X
n
1
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Therefore,
X
n
is a martingale.
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 Spring '11
 afdsa
 Probability theory, Xn, n. λ

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