Digital Signal Processing Reference
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1
0.9
0.8
0.7
0.6
2 =4
0.5
m=2, s
2 =16
m= - 3, s
0.4
0.3
0.2
2 =1
m=0, s
0.1
0
-
2
-
1.5
-
1
-
0.5
0
w
0.5
1
1.5
2
Fig. 4.15 Characteristic functions (absolute values)
Example 4.5.1 Find the characteristic function of the variable Y ¼ 2 X 2, where
X ¼ N (1, 9).
Solution The random variable Y is the linear transformation of the normal random
variable X and, consequently, the random variable Y is also a normal random
variable with the parameters,
s X ¼ 9
s Y ¼ 36
m Y ¼ 2 m X 2 ¼ 0
;
;
:
(4.101)
Therefore, the characteristic function, according to ( 4.100 ), is:
f Y ðoÞ¼ e o 2 36
¼ e 18 o 2
:
(4.102)
2
The absolute values of characteristic functions of the normal random variables
X and Y are shown in Fig. 4.16 .
4.5.2 Characteristic Function of the Sum
of Independent Variables
The calculation of the characteristic function of the sum of normal variables reveals
another interesting property of normal variables, which is described below.
Consider
random variables X 1 ¼ Nðm 1 ; s 1 Þ
two independent normal
and
X 2 ¼ Nðm 2 ; s 2 Þ . Denote their sum as the variable X ,
X ¼ X 1 þ X 2 :
(4.103)
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