5.1.1.1 Discrete Random Variables
Let us fir st dispose of the case when X is a discrete random variable, since it
requires only simple point-to-point mapping. Suppose that the possible values
taken by X can be enumerated as x 1 ,x 2 ,.... Equation (5.2) shows that the
corresponding possible values of Y may be enumerated as y 1 g(x 1 ), y 2
g(x 2 ),.... Let the pmf of X be given by
The pmf of y is simply determined as
Ex ample 5. 1. Problem: the pmf of a random variable X is given as
Determine the pmf of Y if Y is related to X by Y 2 X 1.
Answer: the corresponding values of Y are: g( 1) 2( 1) 1 1;
g(0) 1; g(1) 3; and g(2) 5. Hence, the pmf of Y is given by
Ex ample 5. 2. Problem: for the same X as given in Example 5.1, determine the
pmf of Y if Y 2 X^2 1.
Answer: in this case, the corresponding values of Y are: g( 1) 2( 1)^2
1 3; g(0) 1; g(1) 3; and g(2) 9, resulting in
Functions of Random Variables 121
pX
xipi; i 1 ; 2 ;...:
5 : 3
pY
yipYg
xipi; i 1 ; 2 ;...:
5 : 4
pX
x
1
2
; forx
1 ;
1
4
; forx 0 ;
1
8
; forx 1 ;
1
8 ; forx^2 ;
8
>>>
>>
>>
<
>>>
>>
>>
:
pY
y
1
2 ; fory
^1 ;
1
4
; fory 1 ;
1
8 ; fory^3 ;
1
8
; fory5.
8
>>
>>
>>
><
>>
>>
>>
>:
pY
y
1
4 ; fory^1 ;
5
8
^1
2
^1
8
; fory 3 ;
1
8
; fory 9 :