Introduction to Probability and Statistics for Engineers and Scientists

4.5Properties of the Expected Value 113

By differentiatingFY(a), we obtain the density ofY,

fY(a)=

1

3

a−2/3,0≤a< 1

Hence,

E[X^3 ]=E[Y]=

∫∞

−∞

afY(a)da

=

∫ 1

0

a

1

3

a−2/3da

=

1

3

∫ 1

0

a1/3da

=

1

3

3

4

a4/3|^10

=

1

4

■

While the foregoing procedure will, in theory, always enable us to compute the expec-
tation of any function ofXfrom a knowledge of the distribution ofX, there is an easier
way of doing this. Suppose, for instance, that we wanted to compute the expected value
ofg(X). Sinceg(X) takes on the valueg(X) whenX=x, it seems intuitive thatE[g(X)]
should be a weighted average of the possible valuesg(X) with, for a givenx, the weight
given tog(x) being equal to the probability (or probability density in the continuous case)
thatXwill equalx. Indeed, the foregoing can be shown to be true and we thus have the
following proposition.

PROPOSITION 4.5.1 EXPECTATION OF A FUNCTION OF A RANDOM VARIABLE
(a)IfXis a discrete random variable with probability mass functionp(x), then for any
real-valued functiong,

E[g(X)]=

∑

x

g(x)p(x)

(b)IfXis a continuous random variable with probability density functionf(x), then

for any real-valued functiong,

E[g(X)]=

∫∞

−∞

g(x)f(x)dx