64 Probability and DistributionsProof:For the continuous case, existence follows from the hypothesis, the triangle
inequality, and the linearity of the integral; i.e.,
∫∞−∞|k 1 g 1 (x)+k 2 g 2 (x)|fX(x)dx ≤|k 1 |∫∞−∞|g 1 (x)|fX(x)dx+|k 2 |∫∞−∞|g 2 (x)|fX(x)dx <∞.The result (1.8.7) follows similarly using the linearity of the integral. The proof for
the discrete case follows likewise using the linearity of sums.
The following examples illustrate these theorems.Example 1.8.6.LetXhave the pdf
f(x)={
2(1−x)0<x< 1
0elsewhere.Then
E(X)=∫∞−∞xf(x)dx =∫ 10(x)2(1−x)dx=1
3
,E(X^2 )=∫∞−∞x^2 f(x)dx =∫ 10(x^2 )2(1−x)dx=1
6,and, of course,
E(6X+3X^2 )=6(
1
3)
+3(
1
6)
=5
2.Example 1.8.7.LetXhave the pmfp(x)={ x
6 x=1,^2 ,^3
0elsewhere.
ThenE(6X^3 +X)=6E(X^3 )+E(X)=6∑^3x=1x^3 p(x)+∑^3x=1xp(x)=301
3.Example 1.8.8.Let us divide, at random, a horizontal line segment of length 5
into two parts. IfXis the length of the left-hand part, it is reasonable to assume
thatXhas the pdf
f(x)={ 1
5 0 <x<^5
0elsewhere.
The expected value of the length ofXisE(X)=^52 and the expected value of the
length 5−xisE(5−x)=^52. But the expected value of the product of the two
lengthsisequalto
E[X(5−X)] =∫ 5
0 x(5−x)(1
5 )dx=25
6
=(5
2 )(^2).
That is, in general, the expected value of a product is not equal to the product of
the expected values.