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HMWK 7 Solutions
1.
X
n
=
Y
1
×
Y
2
×···×
Y
n
where
X
0
= 1 and the
Y
i
are iid with
P
(
Y
= 0
.
5) =
P
(
Y
=
1) =
P
(
Y
= 1
.
5) = 1
/
3. Argue that
X
n
converges and ﬁnd the limit.
SOLUTION:
Since
E
(
Y
) = 1,
X
n
is a nonnegative martingale, so it must con
verge wp1 to a rv
X
such that
E
(
X
)
<
∞
by the martingale convergence theorem.
Because
R
n
= ln(
X
n
) = Σ
n
i
=1
ln(
Y
n
) is a random walk with iid increments ln(
Y
n
),
we conclude by SLLN that lim
n
R
n
n
=
E
[ln(
Y
)] =
1
3
ln
3
4
<
0. So the random walk
has negative drift; lim
n
R
n
=
∞
, wp1. Thus
X
n
=
e
R
n
→
0, wp1.
2. Consider the Gambler’s ruin Markov chain
{
X
n
}
on
{
0
,
1
,
2
,
3
,
4
}
(
N
= 4) with
transition matrix
P
=
1
0
0
0
0
1
/
2
0
1
/
2
0
0
0
1
/
2
0
1
/
2
0
0
0
1
/
2
0
1
/
2
0
0
0
0
1
Suppose that
X
0
= 1. Does
X
n
converge wp1? Find the limiting rv
X
if so. Repeat
for each initial condition
X
0
=
i, i
= 0
,
1
,
2
,
3
,
4.
SOLUTION:
First we can easily check that
{
X
n
}
is a nonnegative martingale
for each initial state
X
0
=
i, i
∈ {
0
,
1
,
2
,
3
,
4
}
. This is because
X
n
is the simple
symmetric random walk (already a known MG) away from the boundaries (e.g.,
when
X
0
=
i
∈ {
1
,
2
,
3
}
), and at a boundary (e.g.
X
0
= 0 or
X
0
= 4) it remains
constant. Thus
X
n
converges wp1 to a rv
X
with
E
(
X
)
<
∞
. From the gambler’s
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This note was uploaded on 10/20/2010 for the course IEOR 4106 taught by Professor Whitt during the Fall '08 term at Columbia.
 Fall '08
 Whitt
 Operations Research

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