Fundamentals of Probability and Statistics for Engineers

(John Hannent) #1

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


ˆ ˆ

pX…xi†ˆpi; iˆ 1 ; 2 ;...: … 5 : 3 †

pY…yi†ˆpY‰g…xi†Šˆpi; iˆ 1 ; 2 ;...: … 5 : 4 †

pX…x†ˆ

1

2

; forxˆ
1 ;
1
4
; forxˆ 0 ;
1
8
; forxˆ 1 ;
1
8 ; forxˆ^2 ;

8

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pY…y†ˆ

1

2 ; foryˆ
^1 ;
1
4
; foryˆ 1 ;
1
8 ; foryˆ^3 ;
1
8
; foryˆ5.

8

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ˆ‡


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pY…y†ˆ

1

4 ; foryˆ^1 ;
5
8

ˆ^1

2

‡^1

8



; foryˆ 3 ;
1
8
; foryˆ 9 :

8

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