Random Variable - Random Signals and Processes Primer with MATLAB - page 71

Solution The relations between the corresponding ranges,

y ¼ x 2

; (2.175)

does not have any real solution for y < 0. Therefore,

f Y ðyÞ¼ 0 for y < 0 : (2.176)

For y 0, ( 2.175 ) has two solutions:

p y

x 1 ¼ ;

x 2 ¼ y

(2.177)

p

:

From ( 2.175 ) and ( 2.177 ), we have:

d y

d x ðx 1 Þ¼ 2 x 1 ¼ 2 y

p

;

(2.178)

d y

d x ðx 2 Þ¼ 2 x 2 ¼ 2 y

p

:

Finally, from ( 2.170 ) and ( 2.178 ), we obtain:

8
<

f X ð p Þþ f X ð p Þ
2 y

for y 0 ;

p

f Y ðyÞ¼ (2.179)

:

0 for y < 0 :

If the density function f x ( x ) is even,

f X ðxÞ¼f X ðxÞ; (2.180)

from ( 2.179 ), we obtain:

p
y

8
<

f X y

for y 0 ;

p

f Y ðyÞ¼ (2.181)

:

0 for y < 0 :

Example 2.6.7 Consider the uniform random variable X in the interval [ 1, 5]
(as shown in Fig. 2.36a ). The random variable Y is the absolute value of random
variable X , as shown in Fig. 2.36b ,

Y ¼ j : (2.182)