Random Variable - Random Signals and Processes Primer with MATLAB - page 82

The random variable Y has only two values:

y 1 ¼ 1 and y 2 ¼ 4 : (2.234)

The value y 1 occurs if X ¼ x 1 or X ¼ x 2 . Therefore, according to ( 1.13 ), the
corresponding probability is:

Pfy 1 g¼Pfx 1 gþPfx 2 g¼ 1 = 4 þ 1 = 4 ¼ 1 = 2 : (2.235)

Similarly, we find:

Pfy 2 g¼Pfx 3 gþPfx 4 g¼ 1 = 4 þ 1 = 4 ¼ 1 = 2 : (2.236)

From ( 2.231 ) and ( 2.234 )-( 2.236 ), we have:

g¼EfYg¼ X

2

EfX 2 y i PfY ¼y i g¼ 1 1 = 2 þ 4 1 = 2 ¼ 5 = 2 : (2.237)

i¼ 1

This is the same result as was obtained in ( 2.232 ).
A more general expression ( 2.230 ) is obtained considering the infinite value for N :

EfgðXÞg ¼ 1
i¼1

gðx i ÞPfX ¼ x i g: (2.238)

The following condition must be satisfied in order to show that the expected
value ( 2.238 ) exists

1

j gðx i Þ jPfX ¼ x i g<1: (2.239)

i¼1

2.7.2.2 Properties

We now review some properties of the mean value for discrete random variable:

P.1 The mean value of the constant is the constant itself.

Consider the expected value of the constant b . In this case, there is no random
variable but only one discrete value b with the probability P { b } ¼ 1. From ( 2.220 ),
it follows:

Efbg¼b 1 ¼ b ; b ¼ const :

(2.240)