Fundamentals of Probability and Statistics for Engineers

Then, by definition,

Since X 1 ,X 2 ,...,Xn are mutually independent, Equation (4.36) leads to

We thus have

which was to be proved.

In Section (4.4), we obtained moments of a sum of random variables;
Equation (4.71), coupled with the inversion formula in Equation (4.58) or
Equation (4.64), enables us to determine the distribution of a sum of random
variables from the knowledge of the distributions of Xj,j 1,2,...,n, provided
that they are mutually independent.

Ex ample 4. 16. Problem: let X 1 and X 2 be two independent random variables,
both having an exponential distribution with parameter a, and let
Determine the distribution of Y.
Answer: the characteristic function of an exponentially distributed random
variable was obtained in Example 4.15. F rom Equation (4.54), we have

According to Equation (4.71), the characteristic function of Y is simply

Hence, the den sity function of Y is, as seen from the inversion formula of
Equations (4.68),

Expectations and Moments 105

Y…t†ˆEfejtYgˆEfejt…X^1 ‡X^2 ‡‡Xn†g

ˆEfejtX^1 ejtX^2 ...ejtXng:

EfejtX^1 ejtX^2 ...ejtXngˆEfejtX^1 gEfejtX^2 g...EfejtXng:

Y…t†ˆX 1 …t†X 2 …t†...Xn…t†; … 4 : 71 †

ˆ

YˆX 1 ‡X 2.

X 1 …t†ˆX 2 …t†ˆ

a

ajt

:

Y…t†ˆX 1 …t†X 2 …t†ˆ

a^2

…ajt†^2

:

fY…y†ˆ

1

2 

Z 1

1

ejtyY…t†dt

ˆ

a^2

2 

Z 1

1

ejty

…ajt†^2

dt

ˆ

a^2 yeay; fory 0 ;

0 ; elsewhere:

… 4 : 72 †