Introduction to Probability and Statistics for Engineers and Scientists

120 Chapter 4:Random Variables and Expectation

Hence, since it was shown in Example 4.4a thatE[X]=^72 , we obtain from Equation
4.6.1 that

Var(X)=E[X^2 ]−(E[X])^2

=^916 −

( 7

2

) 2

=^3512 ■

EXAMPLE 4.6b Variance of an Indicator Random Variable.If, for some eventA,

I=

{

1 if eventAoccurs

0 if eventAdoes not occur

then

Var(I)=E[I^2 ]−(E[I])^2

=E[I]−(E[I])^2 sinceI^2 =I(as 1^2 =1 and 0^2 =0)

=E[I](1−E[I])

=P(A)[ 1 −P(A)] sinceE[I]=P(A) from Example 4.4b ■

A useful identity concerning variances is that for any constantsaandb,

Var(aX+b)=a^2 Var(X) (4.6.2)

To prove Equation 4.6.2, letμ=E[X]and recall thatE[aX+b]=aμ+b. Thus, by
the definition of variance, we have

Var(aX+b)=E[(aX+b−E[aX+b])^2 ]

=E[(aX+b−aμ−b)^2 ]

=E[(aX−aμ)^2 ]

=E[a^2 (X−μ)^2 ]

=a^2 E[(X−μ)^2 ]

=aaVar(X)

Specifying particular values foraandbin Equation 4.6.2 leads to some interesting
corollaries. For instance, by settinga=0 in Equation 4.6.2 we obtain that

Var(b)= 0

That is, the variance of a constant is 0. (Is this intuitive?) Similarly, by settinga=1we
obtain

Var(X+b)=Var(X)