Introduction to Probability and Statistics for Engineers and Scientists

114 Chapter 4:Random Variables and Expectation

EXAMPLE 4.5c Applying Proposition 4.5.1 to Example 4.5a yields

E[X^2 ]= 02 (0.2)+(1^2 )(0.5)+(2^2 )(0.3)=1.7

which, of course, checks with the result derived in Example 4.5a. ■

EXAMPLE 4.5d Applying the proposition to Example 4.5b yields

E[X^3 ]=

∫ 1

0

x^3 dx (sincef(x)=1, 0<x<1)

=

1

4

■

An immediate corollary of Proposition 4.5.1 is the following.

Corollary 4.5.2

Ifaandbare constants, then

E[aX+b]=aE[X]+b

Proof

In the discrete case,

E[aX+b]=

∑

x

(ax+b)p(x)

=a

∑

x

xp(x)+b

∑

x

p(x)

=aE[X]+b

In the continuous case,

E[aX+b]=

∫∞

−∞

(ax+b)f(x)dx

=a

∫∞

−∞

xf(x)dx+b

∫∞

−∞

f(x)dx

=aE[X]+b 

If we takea=0 in Corollary 4.5.2, we see that

E[b]=b