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Stochastic Signals and Systems
Homework 3 Book Solutions
Problem Solutions
: Yates and Goodman,3.1.1 3.2.1 3.4.1 3.6.7 5.2.1 5.5.4 5.9.3 5.10.1 and
5.10.4
Problem 3.1.1
In this problem, it is helpful to label points with nonzero probability on the
X
,
Y
plane:
y
x
P
X
,
Y
(
x
,
y
)
•
c
•
3
c
•
2
c
•
6
c
•
4
c
•
12
c
0
1
2
3
4
0
1
2
3
4
(a) We must choose
c
so the PMF sums to one:
∑
x
=
1
,
2
,
4
∑
y
=
1
,
3
P
X
,
Y
(
x
,
y
) =
c
∑
x
=
1
,
2
,
4
x
∑
y
=
1
,
3
y
=
c
[
1
(
1
+
3
)+
2
(
1
+
3
)+
4
(
1
+
3
)] =
28
c
Thus
c
=
1
/
28.
(b) The event
{
Y
<
X
}
has probability
P
[
Y
<
X
] =
∑
x
=
1
,
2
,
4
∑
y
<
x
P
X
,
Y
(
x
,
y
) =
1
(
0
)+
2
(
1
)+
4
(
1
+
3
)
28
=
18
28
(c) The event
{
Y
>
X
}
has probability
P
[
Y
>
X
] =
∑
x
=
1
,
2
,
4
∑
y
>
x
P
X
,
Y
(
x
,
y
) =
1
(
3
)+
2
(
3
)+
4
(
0
)
28
=
9
28
(d) There are two ways to solve this part. The direct way is to calculate
P
[
Y
=
X
] =
∑
x
=
1
,
2
,
4
∑
y
=
x
P
X
,
Y
(
x
,
y
) =
1
(
1
)+
2
(
0
)
28
=
1
28
The indirect way is to use the previous results and the observation that
P
[
Y
=
X
] =
1

P
[
Y
<
X
]

P
[
Y
>
X
] = (
1

18
/
28

9
/
28
) =
1
/
28
(e)
P
[
Y
=
3
] =
∑
x
=
1
,
2
,
4
P
X
,
Y
(
x
,
3
) =
(
1
)(
3
)+(
2
)(
3
)+(
4
)(
3
)
28
=
21
28
=
3
4
1
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View Full DocumentProblem 3.2.1
On the
X
,
Y
plane, the joint PMF
P
X
,
Y
(
x
,
y
)
is
y
x
P
X
,
Y
(
x
,
y
)
•
c
•
3
c
•
2
c
•
6
c
•
4
c
•
12
c
0
1
2
3
4
0
1
2
3
4
By choosing
c
=
1
/
28, the PMF sums to one.
(a) The marginal PMFs of
X
and
Y
are
P
X
(
x
) =
∑
y
=
1
,
3
P
X
,
Y
(
x
,
y
) =
4
/
28
x
=
1
8
/
28
x
=
2
16
/
28
x
=
4
0
otherwise
P
Y
(
y
) =
∑
x
=
1
,
2
,
4
P
X
,
Y
(
x
,
y
) =
7
/
28
y
=
1
21
/
28
y
=
3
0
otherwise
(b) The expected values of
X
and
Y
are
E
[
X
] =
∑
x
=
1
,
2
,
4
xP
X
(
x
) = (
4
/
28
)+
2
(
8
/
28
)+
4
(
16
/
28
) =
3
E
[
Y
] =
∑
y
=
1
,
3
yP
Y
(
y
) =
7
/
28
+
3
(
21
/
28
) =
5
/
2
(c) The second moments are
E
£
X
2
¤
=
∑
x
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 Spring '08
 Ferry

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