Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.2. Transformations: Bivariate Random Variables 101

A random variable of interest isY 1 =X 1 +X 2 ; i.e., the total number of reported
cases of A and B flu during a week. By Theorem 2.1.1, we knowE(Y 1 )=μ 1 +μ 2 ;
however, we wish to determine the distribution ofY 1. Ifweusethechangeof
variable technique, we need to define a second random variableY 2. BecauseY 2 is
of no interest to us, let us choose it in such a way that we have a simple one-to-one
transformation. For this example, we takeY 2 =X 2 .Theny 1 =x 1 +x 2 andy 2 =x 2
represent a one-to-one transformation that mapsSonto

T ={(y 1 ,y 2 ):y 2 =0, 1 ,...,y 1 and y 1 =0, 1 , 2 , ...}.
Note that if (y 1 ,y 2 )∈T,then0≤y 2 ≤y 1. The inverse functions are given by
x 1 =y 1 −y 2 andx 2 =y 2 .ThusthejointpmfofY 1 andY 2 is

pY 1 ,Y 2 (y 1 ,y 2 )=

μy 11 −y^2 μy 22 e−μ^1 −μ^2

(y 1 −y 2 )!y 2!

, (y 1 ,y 2 )∈T,

and is zero elsewhere. Consequently, the marginal pmf ofY 1 is given by

pY 1 (y 1 )=

∑y^1

y 2 =0

pY 1 ,Y 2 (y 1 ,y 2 )

=

e−μ^1 −μ^2

y 1!

∑y^1

y 2 =0

y 1!

(y 1 −y 2 )!y 2!

μ 1 y^1 −y^2 μy 22

=

(μ 1 +μ 2 )y^1 e−μ^1 −μ^2

y 1!

,y 1 =0, 1 , 2 ,...,

and is zero elsewhere, where the third equality follows from the binomial expansion.

For the continuous case we begin with an example that illustrates the cdf tech-
nique.

Example 2.2.2.Consider an experiment in which a person chooses at random
apoint(X 1 ,X 2 ) from the unit squareS={(x 1 ,x 2 ):0<x 1 < 1 , 0 <x 2 < 1 }.
Suppose that our interest is not inX 1 or inX 2 but inZ=X 1 +X 2. Once a suitable
probability model has been adopted, we shall see how to find the pdf ofZ.Tobe
specific, let the nature of the random experiment be such that it is reasonable to
assumethat the distribution of probability over the unit square is uniform. Then
the pdf ofX 1 andX 2 may be written

fX 1 ,X 2 (x 1 ,x 2 )=

{

10 <x 1 < 1 , 0 <x 2 < 1

0elsewhere,

(2.2.1)

and this describes the probability model. Now let the cdf ofZbe denoted by

FZ(z)=P(X 1 +X 2 ≤z). Then

FZ(z)=

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

0 z< 0

∫z

0

∫z−x 1

0 dx^2 dx^1 =

z^2

2 0 ≤z<^1

1 −

∫ 1

z− 1

∫ 1

z−x 1 dx^2 dx^1 =1−

(2−z)^2

2 1 ≤z<^2

12 ≤z.