Normal Random Variable - Random Signals and Processes Primer with MATLAB - page 269

Fig. 4.24 The characteristic
of the transformation block

PfY ¼ 0 g¼PfX> 1 : 5 g¼ 1 PfY ¼ 1 g: (4.175)

From ( 4.30 ), we have

1
2

1 : 5

2

PfX 1 : 5 g¼F X ð 1 : 5 Þ¼ 1 þ erf p ¼ 0 : 9332 : (4.176)

From ( 4.174 )to( 4.176 ), we have:

PfY ¼ 1 g¼ 0 : 9332 ; PfY ¼ 0 g¼ 0 : 0668 : (4.177)

From ( 2.419 ), we have:
f Y ðoÞ¼ X
i

e joy i PðY ¼ y i Þ¼ e jo 1 PfY ¼ 1 gþ e jo 0 PfY ¼ 0 g: (4.178)

Using ( 4.174 )-( 4.178 ), we arrive at:

9332 e jo f Y ðoÞ¼ 0 : þ 0 : 0668 : (4.179)

Exercise 4.9 Find the characteristic function of the random variable

Y ¼ X 2 ; (4.180)

where X is the normal random variable X ¼ N (1,4).

Answer The random variable Y is the linear transformation of the normal random
variable X and thus itself is a normal random variable with the parameters,

EfYg¼EfXg 2 ¼ 1 2 ¼ 1 : (4.181)

The variance of the random variable Y is:

s 2

Y ¼ s 2

X ¼ 4

: (4.182)

Using ( 4.100 ), the characteristic function is:

o 2 s 2
Y

¼ e jo 2 o 2 f Y ðoÞ¼ e jom Y : (4.183)

2