Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.3. Conditional Distributions and Expectations 113

zero elsewhere. The conditional pdf ofX 2 ,givenX 1 =x 1 ,is

f 2 | 1 (x 2 |x 1 )=

6 x 2

3 x^21

=

2 x 2

x^21

, 0 <x 2 <x 1 ,

zero elsewhere, where 0<x 1 <1. The conditional mean ofX 2 ,givenX 1 =x 1 ,is

E(X 2 |x 1 )=

∫x 1

0

x 2

(

2 x 2

x^21

)

dx 2 =

2

3

x 1 , 0 <x 1 < 1.

NowE(X 2 |X 1 )=2X 1 /3 is a random variable, sayY.ThecdfofY=2X 1 /3is

G(y)=P(Y≤y)=P

(

X 1 ≤

3 y

2

)

, 0 ≤y<

2

3

.

From the p dff 1 (x 1 ), we have

G(y)=

∫ 3 y/ 2

0

3 x^21 dx 1 =

27 y^3

8

, 0 ≤y<

2

3

.

Of course,G(y)=0ify<0, andG(y)=1if^23 <y. The pdf, mean, and variance
ofY=2X 1 /3are

g(y)=

81 y^2

8

, 0 ≤y<

2

3

,

zero elsewhere,

E(Y)=

∫ 2 / 3

0

y

(

81 y^2

8

)

dy=

1

2

,

and

Var(Y)=

∫ 2 / 3

0

y^2

(

81 y^2

8

)

dy−

1

4

=

1

60

.

Since the marginal pdf ofX 2 is

f 2 (x 2 )=

∫ 1

x 2

6 x 2 dx 1 =6x 2 (1−x 2 ), 0 <x 2 < 1 ,

zero elsewhere, it is easy to show thatE(X 2 )=^12 and Var(X 2 )= 201. That is, here

E(Y)=E[E(X 2 |X 1 )] =E(X 2 )

and
Var(Y)=Var[E(X 2 |X 1 )]≤Var(X 2 ).

Example 2.3.2 is excellent, as it provides us with the opportunity to apply many
of these new definitions as well as review the cdf technique for finding the distri-
bution of a function of a random variable, namelyY =2X 1 /3. Moreover, the two
observations at the end of this example are no accident because they are true in
general.