88 Multivariate Distributions

In the next example, we use the general fact that double integrals can be ex-
pressed as iterated univariate integrals. Thus double integrations can be carried
out using iterated univariate integrations. This is discussed in some detail with
examples in Section 4.2 of the accompanying resourceMathematical Comments.^2
The aid of a simple sketch of the region of integration is valuable in setting up the
upper and lower limits of integration for each of the iterated integrals.

Example 2.1.3.Suppose an electrical component has two batteries. LetXandY
denote the lifetimes in standard units of the respective batteries. Assume that the
pdfof(X, Y)is

f(x, y)=

{

4 xye−(x

(^2) +y (^2) )
x> 0 ,y > 0
0elsewhere.
The surfacez =f(x, y) is sketched in Figure 2.1.1 where the grid squares are
0 .1by0.1. From the figure, the pdf peaks at about (x, y)=(0. 7 , 0 .7). Solving
the equations∂f/∂x=0and∂f/∂y= 0 simultaneously shows that actually the
maximum off(x, y) occurs at (x, y)=(
√
2 / 2 ,
√
2 /2). The batteries are more likely
to die in regions near the peak. The surface tapers to 0 asxandyget large in any
direction. for instance, the probability that both batteries survive beyond
√
2 / 2
units is given by
P
(
X>
√
2
2
,Y >
√
2
2
)

∫∞
√
2 / 2
∫∞
√
2 / 2
4 xye−(x
(^2) +y (^2) )
dxdy

∫∞
√ 2 / 22 xe
−x^2
[∫
∞
√ 2 / 22 ye
−y^2 dy
]
dx

∫∞
1 / 2
e−z
[∫
∞
1 / 2
e−wdw
]
dz=
(
e−^1 /^2
) 2
≈ 0. 3679 ,
where we made use of the change-in-variablesz=x^2 andw=y^2. In contrast to the
last example, consider the regionsA={(x, y):|x−(1/2)|< 0. 3 ,|y−(1/2)|< 0. 3 }
andB={(x, y):|x− 2 |< 0. 3 ,|y− 2 |< 0. 3 }. The reader should locate these regions
on Figure 2.1.1. The areas ofAandBare the same, but it is clear from the figure
thatP[(X, Y)∈A] is much larger thanP[(X, Y)∈B]. Exercise 2.1.6 confirms this
by showing thatP[(X, Y)∈A]=0.1879 whileP[(X, Y)∈B]=0.0026.
For a continuous random vector (X 1 ,X 2 ), thesupportof (X 1 ,X 2 ) contains all
points (x 1 ,x 2 )forwhichf(x 1 ,x 2 )>0. We denote the support of a random vector
byS. As in the univariate case,S⊂D.
As in the last two examples, we extend the definition of a pdffX 1 ,X 2 (x 1 ,x 2 )
overR^2 by using zero elsewhere. We do this consistently so that tedious, repetitious
references to the spaceDcan be avoided. Once this is done, we replace
∫∫
D
fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2 by
∫∞
−∞
∫∞
−∞
f(x 1 ,x 2 )dx 1 dx 2.
(^2) Downloadable at the site listed in the Preface.