Fundamentals of Probability and Statistics for Engineers

where the integration limits are determined from the requirements y x 2 >0,
and x 2 > 0. The result is

Let us note that this problem has also been solved in Example 4.16, by means of
characteristic functions. It is to be stressed again that the method of character-
istic functions is another powerful technique for dealing with su ms of independ-
ent random variables. In fact, when the number of random variables involved
in a sum is large, the method of characteristic function is preferred since there is
no need to consider only two at a time as required by Equation (5.56).

Ex ample 5. 17. Problem: the random variables X 1 and X 2 are independent
and uniformly distributed in intervals 0 x 1 1, and 0 x 2 2. Determine
the pdf of Y X 1 X 2.

Answer: the convolution of fX 1 (x 1 )1,0x 1 1, and f (^) X 2 (x 2 )1/2,
0 x 2 2, results in
In the above, the limits of the integrals are determined from the requirements
0 yx 2 1, and 0 x 2 2. The shape of fY (y) is that of a trapezoid, as
shown in Figure 5.21.

5.3 m Functions of n Random Variables

We now consider the general transformation given by Equation (5.1), that is,

Functions of Random Variables 147

fY…y†ˆ

a^2 ye^ ay; fory 0 ;

0 ; elsewhere:



… 5 : 59 †



ˆ‡

ˆˆ



fY…y†ˆ

Z 1

1

fX 1 …y x 2 †fX 2 …x 2 †dx 2 ;

ˆ

Zy

0

… 1 †

1

2



dx 2 ˆ

y

2

; for 0<y 1 ;

ˆ

Zy

y 

1

… 1 †

1

2



dx 2 ˆ

1

2

; for 1<y 2 ;

ˆ

Z 2

y 

1

… 1 †

1

2



dx 2 ˆ

3

 y

2

; for 2<y 3 ;

ˆ 0 ; elsewhere:





Yjˆgj…X 1 ;...;Xn†; jˆ 1 ; 2 ;...;m;mn: … 5 : 60 †