Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,3.1.2 3.3.2 3.4.2 3.5.2 3.6.2 3.6.3 and 3.7.3
Problem 3.1.2
On the
X
,
Y
plane, the joint PMF is
y
x
P
X
,
Y
(
x
,
y
)
•
3
c
•
2
c
•
c
•
c
•
c
•
c
•
2
c
•
3
c
1
2
1
(a) To find
c
, we sum the PMF over all possible values of
X
and
Y
. We choose
c
so the sum equals
one.
∑
x
∑
y
P
X
,
Y
(
x
,
y
) =
∑
x
=

2
,
0
,
2
∑
y
=

1
,
0
,
1
c

x
+
y

=
6
c
+
2
c
+
6
c
=
14
c
Thus
c
=
1
/
14.
(b)
P
[
Y
<
X
] =
P
X
,
Y
(
0
,

1
)+
P
X
,
Y
(
2
,

1
)+
P
X
,
Y
(
2
,
0
)+
P
X
,
Y
(
2
,
1
)
=
c
+
c
+
2
c
+
3
c
=
7
c
=
1
/
2
(c)
P
[
Y
>
X
] =
P
X
,
Y
(

2
,

1
)+
P
X
,
Y
(

2
,
0
)+
P
X
,
Y
(

2
,
1
)+
P
X
,
Y
(
0
,
1
)
=
3
c
+
2
c
+
c
+
c
=
7
c
=
1
/
2
(d) From the sketch of
P
X
,
Y
(
x
,
y
)
given above,
P
[
X
=
Y
] =
0.
(e)
P
[
X
<
1
] =
P
X
,
Y
(

2
,

1
)+
P
X
,
Y
(

2
,
0
)+
P
X
,
Y
(

2
,
1
)+
P
X
,
Y
(
0
,

1
)+
P
X
,
Y
(
0
,
1
)
=
8
c
=
8
/
14
1
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Problem 3.3.2
On the
X
,
Y
plane, the joint PMF is
y
x
P
X
,
Y
(
x
,
y
)
•
3
c
•
2
c
•
c
•
c
•
c
•
c
•
2
c
•
3
c
1
2
1
(a) To find
c
, we sum the PMF over all possible values of
X
and
Y
. We choose
c
so the sum equals
one.
∑
x
∑
y
P
X
,
Y
(
x
,
y
) =
∑
x
=

2
,
0
,
2
∑
y
=

1
,
0
,
1
c

x
+
y

=
6
c
+
2
c
+
6
c
=
14
c
Thus
c
=
1
/
14.
(b)
P
[
Y
<
X
] =
P
X
,
Y
(
0
,

1
)+
P
X
,
Y
(
2
,

1
)+
P
X
,
Y
(
2
,
0
)+
P
X
,
Y
(
2
,
1
)
=
c
+
c
+
2
c
+
3
c
=
7
c
=
1
/
2
(c)
P
[
Y
>
X
] =
P
X
,
Y
(

2
,

1
)+
P
X
,
Y
(

2
,
0
)+
P
X
,
Y
(

2
,
1
)+
P
X
,
Y
(
0
,
1
)
=
3
c
+
2
c
+
c
+
c
=
7
c
=
1
/
2
(d) From the sketch of
P
X
,
Y
(
x
,
y
)
given above,
P
[
X
=
Y
] =
0.
(e)
P
[
X
<
1
] =
P
X
,
Y
(

2
,

1
)+
P
X
,
Y
(

2
,
0
)+
P
X
,
Y
(

2
,
1
)+
P
X
,
Y
(
0
,

1
)+
P
X
,
Y
(
0
,
1
)
=
8
c
=
8
/
14
Problem 3.4.2
In Problem 3.2.2, we found that the joint PMF of
X
and
Y
was
y
x
P
X
,
Y
(
x
,
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 Spring '09
 MarshallSylvan
 Variance, Probability theory, Conditional expectation, pX,Y

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