EE464 Homework #12 Solutions
Problem #1
Note that for a Poisson random variable with parameter
λ
we have
M
X
(
t
) =
exp(
λ
(
e
t

1)). Since the X’s are iid we have
M
S
(
t
) = exp(80
λ
(
e
t

1)) and
so we see that S is Poisson with parameter 80
×
.
25 = 20. We now plot this
theoretical pdf vs the approximation given in eqn 7.30. where the mean and
variance are both 20. Note how closely these plots agree.
Problem #2
(a) We first note that
¯
M
n
= max(
¯
M
n

1
, X
n
) for
n >
1 and that
¯
M
1
=
X
1
.
Since this system has memory 1 we conclude that it is indeed a Markov chain.
To show this rigorously we write
P
(
¯
M
n
=
m
n

¯
M
n

1
=
m
n

1
, ...,
¯
M
1
=
m
1
) =
P
(max(
¯
M
n

1
, X
n
) =
m
n

¯
M
n

1
=
m
n

1
, ...,
¯
M
1
=
m
1
) =
P
(max(
m
n

1
, X
n
) =
m
n

¯
M
n

1
=
m
n

1
, ...,
¯
M
1
=
m
1
) =
P
(max(
m
n

1
, X
n
) =
m
n
) =
P
(
¯
M
n
=
m
n

¯
M
n

1
=
m
n

1
), where we have used the fact that the
X
n
’s are inde
pendent of the past
¯
M
n
’s, since the previous
¯
M
n
’s are functions of only the
1
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previous
X
n
’s and the sequence is iid.
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 Spring '06
 Caire
 Probability theory, Markov chain, Andrey Markov, Xn

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