Random Variable - Random Signals and Processes Primer with MATLAB - page 134

Exercise E.2.14 The random variable X has the following PDF:

0 : 5 ðx þ 1 Þ 1 x 1 ;

f X ðxÞ ¼ (2.501)

0 otherwise :

as shown in Fig. 2.68 . Find the PDF of the random variable Y where

Y ¼ 1 X 2 : (2.502)

Answer From ( 2.502 ), it follows:
x 1 ; 2 ¼

p

1 y (2.503)

and

j ¼ 2

p

d y
d x

¼ 2 x

j 1 y : (2.504)

The PDF is obtained using ( 2.170 ), ( 2.501 ), ( 2.503 ), and ( 2.504 ):

x 1 ¼ x 2 ¼

f X ð x Þ
d d x

f X ð x Þ
d d x

f Y ðyÞ ¼ þ

p p

1 y 1 y

5

p 1 y þ 1

5

p 1 y þ 1

0 : ð Þ 0 : ð Þ

¼

2

p þ

2

p : (2.505)

1 y 1 y

From here, we have:

8
<

1
2

p 1 y

0 y 1 ;

f Y ðyÞ ¼ (2.506)

:

0 otherwise :

Fig. 2.68 PDF of the input
variable X