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Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,7.1.2 7.4.1 7.4.2 7.5.1 7.6.1 8.1.1 6.5.1 6.5.2 and 6.5.3
Problem 7.1.2
(a) Since
Y
=
X
1
+(

X
2
)
, Theorem 7.1 says that the expected value of the difference is
E
[
Y
] =
E
[
X
1
]+
E
[

X
2
] =
E
[
X
]

E
[
X
] =
0
(b) By Theorem 7.2, the variance of the difference is
Var
[
Y
] =
Var
[
X
1
]+
Var
[

X
2
] =
2Var
[
X
]
Problem 7.4.1
K
i
has PMF
P
K
(
k
) =
1

p
k
=
0
p
k
=
1
0
otherwise
(a) The MGF of
K
is
φ
K
(
s
) =
E
£
e
sK
¤
=
1

p
+
pe
s
(b) By Theorem 7.10,
M
=
K
1
+
K
2
+
...
+
K
n
has MGF
φ
M
(
s
) = [
φ
K
(
s
)]
n
= [
1

p
+
pe
s
]
n
(c) Although we could just use the fact that the expectation of the sum equals the sum of the ex
pectations, the problem asks us to find the moments using
φ
M
(
s
)
. In this case,
E
[
M
] =
d
φ
M
(
s
)
ds
¯
¯
¯
¯
s
=
0
=
n
(
1

p
+
pe
s
)
n

1
pe
s
¯
¯
s
=
0
=
np
(d) The second moment of
M
can be found via
E
£
M
2
¤
=
d
φ
M
(
s
)
ds
¯
¯
¯
¯
s
=
0
=
np
(
(
n

1
)(
1

p
+
pe
s
)
pe
2
s
+(
1

p
+
pe
s
)
n

1
e
s
)¯
¯
s
=
0
=
np
[(
n

1
)
p
+
1
]
The variance of
M
is
Var
[
M
] =
E
£
M
2
¤

(
E
[
M
])
2
=
np
(
1

p
) =
n
Var
[
K
]
1
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 Spring '09
 MarshallSylvan

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