Page 164 - Fundamentals of Probability and Statistics for Engineers

Functions of Random Variables 147
where the integration limits are determined from the requirements y
x 2 >0,
and x 2 > 0. The result is

2
ay
a ye ; for y 0;
f y†ˆ 5:59†
Y
0; elsewhere:
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 sums 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).

Example 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 ˆ X 1 ‡ X 2 .
Answer: the convolution of f (x 1 ) ˆ 1, 0 1, and f (x 2 ) ˆ 1/2,
X 1 x 1 X 2
0 x 2 2, results in
Z 1
f y†ˆ f X 1 y
x 2 †f X 2 x 2 †dx 2 ;
Y

1
1
Z y y
ˆ 1† dx 2 ˆ ; for 0 < y 1;
0 2 2
1
Z y 1
ˆ 1† dx 2 ˆ ; for 1 < y 2;
y
1 2 2
1
Z 2 3
y
ˆ 1† dx 2 ˆ ; for 2 < y 3;
y
1 2 2
ˆ 0; elsewhere:

In the above, the limits of the integrals are determined from the requirements
0 y
x 2 1, and 0 x 2 2. The shape of f (y) is that of a trapezoid, as
Y
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,

Y j ˆ g X 1 ; .. . ; X n †; j ˆ 1; 2; ... ; m; m n: 5:60†
j

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