Random Variable - Random Signals and Processes Primer with MATLAB - page 74

The corresponding probability from ( 1.13 ) is:

Pfy 0 i g ¼ X

N

Pfx j g: (2.194)

j¼ 1

The following example illustrates this concept.

Example 2.6.8 The discrete random variable X has the following values:

x 1 ¼ 1 ; x 2 ¼ 0 ; and x 3 ¼ 1 ; (2.195)

with the corresponding probabilities:

Pfx 1 g ¼ 0 : ; Pfx 2 g ¼ 0 : ; Pfx 3 g ¼ 0 : :

3 1 6 (2.196)

Find the PDF and distribution for the random variable:

Y ¼ X 2 þ 1 : (2.197)

Solution From ( 2.195 ) and ( 2.197 ), it follows:

y 1 ¼ x 1 2 þ 1 ¼ 2 ;

y 2 ¼ x 2 2

þ 1 ¼ 1 ;

y 3 ¼ x 3 2 þ 1 ¼ 2 : (2.198)

From ( 2.198 ), we can see that the random variable Y takes only the value
y 1 0 ¼ 2, if X is equal to either x 1 or x 3 , and the value y 2 0 ¼ 1 if X is equal to x 2 .
Consequently, we have:

y 1 0 ¼ y 1 ¼ y 3 ¼ 2 (2.199)

and

P y 1 0

f g ¼ Pfy 3 g ¼ Pfy 1 g ¼ Pf 2 g ¼ Pfx 1 gþPfx 3 g ¼ 0 : 9 : (2.200)

Similarly, from ( 2.198 ), we have:

y 2 0 ¼ y 2 ¼ 1 (2.201)

and

P y 2 0 f g ¼ Pfy 2 g ¼ Pf 1 g ¼ Pfx 2 g ¼ 0 : 1 : (2.202)