Digital Signal Processing Reference
In-Depth Information
Exercise 4.13
The normal random variable
X
has a zero mean value and a variance
of
s
2
. Find the mean value of the random variable
Y
Y ¼
2
þ
3
X
2
3
X
3
:
(4.198)
Answer
From (
4.198
), the mean value of
Y
is given as:
Y ¼
2
þ
3
X
2
3
X
3
:
(4.199)
A mean squared value of the random variable
X
is equal to the variance
s
2
.
Additionally, the third moment is equal to zero because the density function is
symmetrical around zero.
Therefore, from (
4.199
) we have:
Y ¼
2
þ
3
s
2
:
(4.200)
Exercise 4.14
The random variable
X
is a normal random variable
N
(2, 4) with the
probability 0.6, and it is a normal random variable
N
(
2, 9) with the probability
0.4. Find the PDF of the random variable
X
.
Answer
The PDF of the random variable
X
is given as:
2
2
e
ð
x
2
Þ
e
ð
x
þ
2
Þ
1
2
1
3
8
18
f
X
ðxÞ¼
0
:
p
þ
0
:
p
:
6
4
(4.201)
2
p
2
p
Let us verify that (
4.201
) is a PDF. To this end, the condition (
2.83
) must be
satisfied:
2
e
ð
x
2
Þ
1
1
1
2
8
f
X
ðxÞ
d
x ¼
0
:
6
p
d
x
2
p
1
1
2
e
ð
x
þ
2
Þ
1
1
3
18
þ
0
:
4
p
d
x ¼
0
:
6
1
þ
0
:
4
1
¼
1
;
2
p
1
2
2
e
ð
x
2
Þ
e
ð
x
þ
2
Þ
1
1
1
2
1
3
p
8
p
18
d
x ¼
1
d
x ¼
1
:
(4.202)
;
2
p
2
p
1
1
Exercise 4.15
The random variables
X
and
Y
are independent. Their corresponding
density functions are:
2
e
ð
x
3
Þ
1
4
1
=
3
for
1
x
2
32
f
X
ðxÞ¼
p
;
f
Y
ðyÞ¼
(4.203)
0
otherwise
:
2
p
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