EE 5375/7375 Random Processes
October 7, 2003
Homework #5 Solutions
Problem 1. textbook problem 6.2
A discretetime random process
X
n
is defined as follows.
A fair coin is tossed.
If the outcome is heads,
X
n
= 1 for all
n
; if the outcome is tails,
X
n
=

1 for all
n
. (a) Sketch some sample paths of the process. (b)
Find the PMF for
X
n
. (c) Find the joint PMF for
X
n
and
X
n
+
k
. (d) Find the mean and autocovariance
functions of
X
n
.
(a) If the outcome is heads, then
X
n
= 1
,
1
,
1
, . . .
. If the outcome is tails, then
X
n
=

1
,

1
,

1
, . . .
(b) Since a head or tail is equally likely, it is equally likely that the process will be +1 or 1:
P
(
X
n
= 1) =
P
(
X
n
=

1) = 1
/
2
(c) In either case of heads or tails,
X
n
and
X
n
+
k
will have the same value:
P
(
X
n
= 1
, X
n
+
k
= 1) =
P
(
X
n
=

1
, X
n
+
k
=

1) =
1
2
In neither case will
X
n
and
X
n
+
k
have different values:
P
(
X
n
= 1
, X
n
+
k
=

1) =
P
(
X
n
=

1
, X
n
+
k
= 1) = 0
(d) Since
X
n
is 1 or 1 with equal likelihood, the mean is
E
(
X
n
) = 1
·
(
1
2
)

1
·
(
1
2
) = 0. The autocovariance
function is
C
(
n, n
+
k
) =
E
(
X
n
X
n
+
k
)

E
(
X
n
)
E
(
X
n
+
k
) =
E
(
X
n
X
n
+
k
)
= 1
·
P
(
X
n
= 1
, X
n
+
k
= 1) + (

1)(

1)
·
P
(
X
n
=

1
, X
n
+
k
=

1)
=
1
2
+
1
2
= 1
Problem 2. textbook problem 6.21
Let
S
n
denote a binomial counting process. (a) Show that
P
(
S
n
=
j, S
m
=
i
) =
P
(
S
n
=
j
)
P
(
S
m
=
i
). (b)
Find
P
(
S
n
=
j

S
m
=
i
) where
n > m
. (c) Show that
P
(
S
n
=
j

S
m
=
i, S
l
=
k
) =
P
(
S
n
=
j

S
m
=
i
) where
n > m > l
.
(a) We know that
S
n
has independent increments that follow a binomial distribution. That is, for
n > m
,
the increment
S
n

S
m
is the sum of
n

m
Bernoulli trials, which has the same distribution as
S
n

m
. Hence,
P
(
S
n
=
j, S
m
=
i
) =
P
(
S
n

S
m
=
j

i, S
m
=
i
)
=
P
(
S
n

m
=
j

i
)
P
(
S
m
=
i
)
=
P
(
S
n
=
j
)
P
(
S
m
=
i
)
The last step follows because
S
n

m
is the sum of
n

m
Bernoulli trials whereas
S
n
is the sum of
n
Bernoulli
trials.
(b) Given
S
m
=
i
, the event
S
n
=
j
is the same as the increment
S
n

S
m
=
j

i
. This increment is the
sum of
n

m
Bernoulli trials, so it is a binomial:
P
(
S
n
=
j

S
m
=
i
) =
P
(
S
n

S
m
=
j

i
) =
n

m
j

i
p
j

i
(1

p
)
n

m

(
j

i
)
1
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(c) We have to use the property of independent increments.
P
(
S
n
=
j

S
m
=
i, S
l
=
k
) =
P
(
S
n
=
j, S
m
=
i, S
l
=
k
)
P
(
S
m
=
i, S
l
=
k
)
=
P
(
S
n

S
m
=
j

i, S
m

S
l
=
i

k, S
l
=
k
)
P
(
S
m

S
l
=
i

k, S
l
=
k
)
=
P
(
S
n

S
m
=
j

i
)
P
(
S
m

S
l
=
i

k
)
P
(
S
l
=
k
)
P
(
S
m

S
l
=
i

k
)
P
(
S
l
=
k
)
=
P
(
S
n

S
m
=
j

i
)
=
P
(
S
n
=
j

S
m
=
i
)
Problem 3. textbook problem 6.23
Consider the following moving average processes:
Y
n
=
1
2
(
X
n
+
X
n

1
)
, X
0
= 0
Z
n
=
2
3
X
n
+
1
3
X
n

1
, X
0
= 0
(a) Flip a coin 10 times to obtain a realization of a Bernoulli random process
X
n
.
Find the resulting
realizations of
Y
n
and
Z
n
.
(b) Repeat part (a) with
X
n
given by the random step process introduced in
Example 6.12 (i.e.,
D
n
= 2
I
n

1 where
I
n
is the Bernoulli random process. At any time, the process takes
value +1 with probability
p
and value 1 with probability 1

p
). (c) Find the mean, variance,and covariance
of
Y
n
and
Z
n
if
X
n
is a Bernoulli random process. Are the sample means of
Y
n
and
Z
n
in part (a) close to
their respective means? (d) Repeat part (c) if
X
n
is the random step process.
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 Fall '08
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 Normal Distribution, Probability theory, Sn, Xn, Poisson PMF

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