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.