2.1. Distributions of Two Random Variables 93

(0,1). The reader should sketch this region on the space of (X 1 ,X 2 ). Fixingx 1 and
integrating with respect tox 2 ,wehave

P(X 1 +X 2 ≤1) =

∫ 1

0

[∫ 1 −x 1

0

(x 1 +x 2 )dx 2

]

dx 1

=

∫ 1

0

[

x 1 (1−x 1 )+

(1−x 1 )^2

2

]

dx 1

=

∫ 1

0

(

1

2

−

1

2

x^21

)

dx 1 =

1

3

.

This latter probability is the volume under the surfacef(x 1 ,x 2 )=x 1 +x 2 above
the set{(x 1 ,x 2 ):0<x 1 ,x 1 +x 2 ≤ 1 }.

Example 2.1.7(Example 2.1.3, Continued).Recall that the random variablesX
andYof Example 2.1.3 were the lifetimes of two batteries installed in an electrical
component. The joint pdf of (X, Y) is sketched in Figure 2.1.1. Its space is the
positive quadrant ofR^2 so there are no constraints involving bothxandy.Using
the change-in-variablew=y^2 ,themarginalpdfofXis

fX(x)=

∫∞

0

4 xye−(x

(^2) +y (^2) )
dy=2xe−x
2
∫∞
0
e−wdw=2xe−x
2
,
forx>0. By the symmetry ofxandyin the model, the pdf ofY is the same as
that ofX. To determine the median lifetime,θ, of these batteries, we need to solve
1
2

∫θ
0
2 xe−x
2
dx=1−e−θ
2
,
where again we have made use of the change-in-variablesz=x^2 .Solvingthis
equation, we obtainθ=
√
log 2≈ 0 .8326. So 50% of the batteries have lifetimes
exceeding 0.83 units.

2.1.2 Expectation............................

The concept of expectation extends in a straightforward manner. Let (X 1 ,X 2 )bea
random vector and letY=g(X 1 ,X 2 ) for some real-valued function; i.e.,g:R^2 →R.
ThenYis a random variable and we could determine its expectation by obtaining
the distribution ofY. But Theorem 1.8.1 is true for random vectors also. Note the
proof we gave for this theorem involved the discrete case, and Exercise 2.1.12 shows
its extension to the random vector case.
Suppose (X 1 ,X 2 ) is of the continuous type. ThenE(Y)existsif
∫∞

−∞

∫∞

−∞

|g(x 1 ,x 2 )|fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2 <∞.

Then

E(Y)=

∫∞

−∞

∫∞

−∞

g(x 1 ,x 2 )fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2. (2.1.10)