EE 5375/7375 Random Processes
September 30, 2003
Homework #4 solutions
Problem 1. textbook problem 6.4
A discretetime random process is deﬁned by
X
n
=
s
n
for
n
≥
0 where
s
is selected at random from the
interval (0,1). (a) Sketch a sample path of the process. (b) Find the PDF of
X
n
. (c) Find the joint PDF of
X
n
and
X
n
+1
. (d) Find the mean and autocovariance functions of
X
n
.
(a) The sample path will decrease geometrically by
X
0
= 1,
X
1
=
s
,
X
2
=
x
2
, and so on, as shown in Fig. 1.
(b) The PDF of
X
n
at a speciﬁc
n
is
P
(
X
n
≤
x
) =
P
(
s
n
≤
x
) =
P
(
s
≤
x
1
/n
) =
x
1
/n
The last step follows because
s
is uniformly distributed in (0,1).
(c) By deﬁnition, the joint PDF is
P
(
X
n
≤
x, X
n
+1
≤
y
) =
P
(
s
n
≤
x, s
n
+1
≤
y
)
=
P
(
s
≤
x
1
n
, x
≤
y
1
n
+1
)
=
P
(
s
≤
min(
x
1
n
, y
1
n
+1
))
= min(
x
1
n
, y
1
n
+1
)
We use the minimum of
x
1
n
and
y
1
n
+1
because the joint event that
s
is less than both of them means that
s
must be less than the smaller of them. Again, the last step follows because
s
is uniformly distributed.
(d) The mean of
X
n
depends on knowledge of the pdf of
s
which is
f
s
(
s
) = 1 over the interval (0,1):
E
(
X
n
) =
E
(
s
n
) =
Z
1
0
s
n
f
s
(
s
)
ds
=
Z
1
0
s
n
ds
=
1
n
+ 1
The autocovariance function for lag
k
is deﬁned as
C
X
(
n, n
+
k
) =
E
(
X
n
X
n
+
k
)

E
(
X
n
)
E
(
X
n
+
k
)
=
E
(
s
n
s
n
+
k
)

1
n
+ 1
1
n
+
k
+ 1
=
E
(
s
2
n
+
k
)

1
(
n
+ 1)(
n
+
k
+ 1)
=
Z
1
0
s
2
n
+
k
ds

1
(
n
+ 1)(
n
+
k
+ 1)
=
1
2
n
+
k
+ 1

1
(
n
+ 1)(
n
+
k
+ 1)
Problem 2. textbook problem 6.11
Find an expression for
E
(

X
t
2

X
t
1

2
) in terms of autocorrelation function.
The absolute value does not matter because it is squared.
E
(

X
t
2

X
t
1

2
) =
E
((
X
t
2

X
t
1
)
2
)
=
E
(
X
2
t
2

2
X
t
1
X
t
2
+
X
2
t
1
)
=
E
(
X
2
t
2
)

2
E
(
X
t
1
X
t
2
) +
E
(
X
2
t
1
)
=
R
X
(
t
2
, t
2
)

2
R
X
(
t
1
, t
2
) +
R
X
(
t
1
, t
1
)
1
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View Full Documentwhere
R
X
(
t, s
) is the autocorrelation function.
Problem 3. textbook problem 6.20
Let
Y
n
be the process that results when individual 1’s in a Bernoulli process are erased with probability
α
.
Find the PMF of
S
0
n
, the counting process for
Y
n
. Does
Y
n
have independent and stationary increments?
Let
I
n
be a Bernoulli process where a 1 occurs with probability
p
and a 0 occurs with probability 1

p
.
Whenever a 1 occurs, it is not erased with probability (1

α
) in the derived process
Y
n
. The probability of a
1 in
Y
n
is therefore (1

α
)
p
. The counting process
S
0
n
gives the number of ones in the ﬁrst
n
Bernoulli trials
where each trial has probability (1

α
)
p
of success. Therefore
S
0
n
is binomial with parameters (
n,
(1

α
)
p
)
and has PMF
P
(
S
0
n
=
x
) =
±
n
x
²
((1

α
)
p
)
x
(1

(1

α
)
p
)
n

x
Y
n
has independent and stationary increments because it is a Bernoulii process (sequence of independent
Bernoulli random variables).
Problem 4. textbook problem 6.24
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 LORENZELLI

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