Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.1. Distributions of Two Random Variables 95

Example 2.1.8.LetX 1 andX 2 have the pdf

f(x 1 ,x 2 )=

{

8 x 1 x 2 0 <x 1 <x 2 < 1

0elsewhere.

Figure 2.1.3 shows the space for (X 1 ,X 2 ). Then

E(X 1 X 22 )=

∫∞

−∞

∫∞

−∞

x 1 x^22 f(x 1 ,x 2 )dx 1 dx 2.

To compute the integration, as shown by the arrow on Figure 2.1.3, we fixx 2 and
then integratex 1 from 0 tox 2 .Wethenintegrateoutx 2 from 0 to 1. Hence,

∫∞

−∞

∫∞

−∞

x 1 x^22 f(x 1 ,x 2 )=

∫ 1

0

[∫x 2

0

8 x^21 x^32 dx 1

]

dx 2 =

∫ 1

0

8

3 x

6

2 dx^2 =

8

21.

In addition,

E(X 2 )=

∫ 1

0

[∫x 2

0

x 2 (8x 1 x 2 )dx 1

]

dx 2 =^45.

SinceX 2 has the pdff 2 (x 2 )=4x^32 , 0 <x 2 <1, zero elsewhere, the latter expecta-
tion can also be found by

E(X 2 )=

∫ 1

0

x 2 (4x^32 )dx 2 =^45.

Using Theorem 2.1.1,

E(7X 1 X 22 +5X 2 )=7E(X 1 X 22 )+5E(X 2 )

=(7)( 218 ) + (5)(^45 )=^203.

Example 2.1.9.Continuing with Example 2.1.8, suppose the random variableY
is defined byY =X 1 /X 2. We determineE(Y) in two ways. The first way is by
definition; i.e., find the distribution ofY and then determine its expectation. The
cdf ofY,for0<y≤1, is

FY(y)=P(Y≤y)=P(X 1 ≤yX 2 )=

∫ 1

0

[∫yx 2

0

8 x 1 x 2 dx 1

]

dx 2

=

∫ 1

0

4 y^2 x^32 dx 2 =y^2.

Hence, the pdf ofYis

fY(y)=FY′(y)=

{

2 y 0 <y< 1

0elsewhere,

which leads to

E(Y)=

∫ 1

0

y(2y)dy=

2

3

.