6.262 Discrete Stochastic Processes, Spring 2010
Problem Set 2 — Solutions
due: Friday, February 19, 2010
Problem 1
a We have
P(
X
n
= 1) =
1
n
and
P(
X
n
= 0) = 1

1
n
.
Equivalently, looking at the cumulative distribution function,
P(
X
n
≤
a
) =
0
a <
0
1

1
n
0
≤
a <
1
1
1
≤
a
Consider the random variable
Y
distributed as P(
Y
= 0) = 1. The corresponding CDF is
given by
P(
Y
≤
a
) =
±
0
a <
0
1 0
≤
a
Then, clearly,
lim
n
→∞
P(
X
n
≤
a
) = P(
Y
≤
a
)
for all
a
∈
R
.
Therefore,
X
n
→
Y
in distribution.
Now let
Z
n
=

X
n

Y
n

and note that
P(
Z
n
= 1) =
1
n
and
P(
Z
n
= 0) = 1

1
n
.
Thus, E(

X
n

Y

2
) = E(
Z
2
n
) =
1
n
, and therefore lim
n
→∞
E(

X
n

Y

2
) = 0, from which we
conclude that
X
n
converges to
Y
in mean square.
b We now have
P(
X
n
≤
a
) =
0
a <
0
1

1
n
0
≤
a < n
1
n
≤
a
Letting
Y
be as above, we again obtain
lim
n
→∞
P(
X
n
≤
a
) = P(
Y
≤
a
)
for all
a
∈
R
.
However, letting
Z
n
=

X
n

Y
n

, we obtain
P(
Z
n
=
n
) =
1
n
and
P(
Z
n
= 0) = 1

1
n
,
and therefore E(

X
n

Y

2
) = E(
Z
2
n
) =
n
2
n
=
n
. It follows that lim
n
→∞
E(

X
n

Y

2
)
6
= 0, from
which we conclude that
X
n
does not converge to
Y
in mean square.
1
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View Full Documentc Recall that in the doubleorquarter game, P(
X
k
= 2) = P(
X
i
= 1
/
4) = 1
/
2 and
W
n
=
X
1
X
2
...X
n
, assuming an initial bet of $1. First we claim that
W
n
converges in distribution
to
Y
, where
Y
is a random variable distributed as P(
Y
= 0) = 1. Here, it suﬃces to show
that lim
n
→∞
P
(
W
n
≤
a
) = 1 for all
a >
0. (Why are we not concerned about the case where
a <
0? How about
a
= 0? Think carefully.) Taking complements, it is equivalent to show
that lim
n
→∞
P
(
W
n
> a
) = 0 for all
a >
0.
Fix
a >
0 and let
k
be some positive integer such that
a >
(1
/
2)
k
. Then P(
W
n
> a
)
≤
P(
W
n
≥
(1
/
2)
k
). Now let
n
be even and large enough so that 1
≤
k < n/
2. Consider
n
coin
tosses and note that in order to have
W
n
>
(1
/
2)
k
, the number of heads must at least equal
to the number of tails, that is,
{
W
n
≥
(1
/
2)
k
} ⊂ {
H
n
≥
n/
2
}
,
(1)
where
H
n
denotes the number of heads in
n
tosses. One way to see this is to show the
contrapositive, that is, taking complements, we wish to show that
{
W
n
<
(1
/
2)
k
} ⊃ {
H
n
< n/
2
}
.
(2)
This is easy: since
W
n
= 2
H
n
(1
/
4)
n

H
n
and 1
≤
k < n/
2, we have that
H
n
< n/
2
=
⇒
W
n
<
2
n/
2
(1
/
4)
n

n/
2
= (1
/
2)
n/
2
<
(1
/
2)
k
,
which is exactly what we need to show to prove (2).
Taking probabilities in (1) yields that
P(
W
n
≥
(1
/
2)
k
)
≤
P(
H
n
≥
n/
2)
.
Now by Chebyshev inequality,
P(
H
n
≥
n/
2)
≤
V ar
(
H
n
)
(
n/
2)
2
=
n/
4
(
n/
2)
2
.
The righthand side of the above equation gets arbitrarily small as
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 Spring '08
 Moon,J
 Poisson Distribution, Probability theory, td, Poisson process, Lévy process

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