Multidimensional Random Variables - Random Signals and Processes Primer with MATLAB - page 203

Similarly from ( 3.276 )to( 3.278 ), we obtain:

EfX 2 g¼ 1 P fX 2 ¼ 1 gþ 0 P fX 2 ¼ 0 gþð 1 Þ P fX 2 ¼ 1 g
¼ 2 = 15 9 = 15 ¼ 7 = 15 : (3.280)

Exercise 3.11 The discrete random variable X 1 has 0 and 1 as its discrete values,
whereas X 2 has 0, 1, and 1 as its discrete values. Find the mean value and the
variance of the random variable X ,if

X ¼ 2 X 1 þ X 2 : (3.281)

The corresponding probabilities are:

PfX 1 ¼ 0 ;X 2 ¼ 0 g¼ 0 : 2 ; PfX 1 ¼ 0 ;X 2 ¼ 1 g¼ 0 ; PfX 1 ¼ 0 ;X 2 ¼ 1 g¼ 0 : 1 ;

PfX 1 ¼ 1 ;X 2 ¼ 0 g¼ 0 : 3 ; PfX 1 ¼ 1 ;X 2 ¼ 1 g¼ 0 : 2 ; PfX 1 ¼ 1 ;X 2 ¼ 1 g¼ 0 :

(3.282)

: 2

Answer The variance of the random variable X is:

2

2

2

s 2

X ¼ X 2

X ¼ ð 2 X 1 þ X 2 Þ ð 2 X 1 þ X 2 Þ : (3.283)

From ( 3.281 ), we get the mean value of the variable X :

X ¼ 2 X 1 þ X 2 ¼ X

X

2 3

j¼ 1 ð 2 x 1 i þ x 2 j ÞPfX 1 ¼ x 1 i ; X 2 ¼ x 2 j g i¼ 1

¼ 0 : 1 þ 2 0 : 3 þð 2 þ 1 Þ 0 : 2 þð 2 þ 1 Þ 0 : 2 ¼ 0 : 1 þ 0 : 6 þ 0 : 6 þ 0 : 6 ¼ 1 :

(3.284)

: 9

Similarly, the mean squared value of X is:

¼ X

X

2 3

j¼ 1 ð 2 x 1 i þ x 2 j Þ

2

2

X 2 ¼ ð 2 X 1 þ X 2 Þ PfX 1 ¼ x 1 i ; X 2 ¼ x 2 j g

i¼ 1

¼ 0 : 1 þ 1 : 2 þ 1 : 8 þ 1 : 8 ¼ 4 : 9 : (3.285)

Placing ( 3.284 ) and ( 3.285 ) into ( 3.283 ), we arrive at:

X 2 s X ¼ X 2 9 2 ¼ 4 : 9 1 : ¼ 1 : 29 : (3.286)