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Homework 4: Solutions
Chapter 2
5.5 (a) Since
M
n
+1
= 2
U
n
+1
M
n
, and
U
n
+1
is independent of
M
0
,...,M
n
we have using (1.3) that
E
(
M
n
+1

M
n
=
m
n
,...,M
0
=
m
0
)
=
m
n
E
(2
U
n
+1

M
n
=
m
n
,...,M
0
=
m
0
)
=
m
n
.
(b)

log
U
1
,

log
U
2
,...
are independent with
P
(

log
U
i
> x
) =
P
(
U
i
<
e

x
) =
e

x
, so

log
U
i
has an exponential distribution with mean
1. Using the strong law of large numbers gives (1
/n
) log
X
n
→ 
1.
(c) For any sequence of positive numbers (
X
n
), if (1
/n
) log
X
n
→ 
1
then 2
n
X
n
→
0.
5.6 (a) When
X
0
=
x
,
X
1
= Binomial(
x/N,N
), so
E
x
X
1
=
x
,
E
x
X
2
1
=
x
(1

x/N
) +
x
2
, and
E
x
(
X
1
(
N

X
1
)) =
Nx

x
+
x
2
/N

x
2
=
x
(
N

x
)(1

1
/N
). Part (a) follows from the Markov Property
(b) Since
Y
n
is a martingale,
E
x
X
n
(
N

X
n
) =
x
(
N

x
)(1

1
/N
)
n
. The
desired result now follows from the obsrvation that when 0
< X
n
< N
we have
N

1
≤
X
n
(
N

X
n
)
≤
N
2
/
4.
5.13 The pedestrian lemma, (3.3) in Chapter 1, implies that
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 Spring '10
 DURRETT

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