Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.2. Transformations: Bivariate Random Variables 107

In addition to the change-of-variable and cdf techniques for finding distributions
of functions of random variables, there is another method, called themoment
generating function (mgf) technique, which works well for linear functions of
random variables. In Subsection 2.1.2, we pointed out that ifY=g(X 1 ,X 2 ), then
E(Y), if it exists, could be found by

E(Y)=

∫∞

−∞

∫∞

−∞

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

in the continuous case, with summations replacing integrals in the discrete case.
Certainly, that functiong(X 1 ,X 2 )couldbeexp{tu(X 1 ,X 2 )}, so that in reality
we would be finding the mgf of the functionZ=u(X 1 ,X 2 ). If we could then
recognize this mgf as belonging to a certain distribution, thenZwould have that
distribution. We give two illustrations that demonstrate the power of this technique
by reconsidering Examples 2.2.1 and 2.2.4.
Example 2.2.6(Continuation of Example 2.2.1).HereX 1 andX 2 have the joint
pmf

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

{

μx 11 μx 22 e−μ^1 e−μ^2

x 1 !x 2! x^1 =0,^1 ,^2 ,^3 ,..., x^2 =0,^1 ,^2 ,^3 ,...

0elsewhere,

whereμ 1 andμ 2 are fixed positive real numbers. LetY=X 1 +X 2 and consider

E(etY)=

∑∞

x 1 =0

∑∞

x 2 =0

et(x^1 +x^2 )pX 1 ,X 2 (x 1 ,x 2 )

=

∑∞

x 1 =0

etx^1

μx^1 e−μ^1

x 1!

∑∞

x 2 =0

etx^2

μx^2 e−μ^2

x 2!

=

[

e−μ^1

∑∞

x 1 =0

(etμ 1 )x^1

x 1!

][

e−μ^2

∑∞

x 2 =0

(etμ 2 )x^2

x 2!

]

=

[

eμ^1 (e

t−1)][

eμ^2 (e

t−1)]

= e(μ^1 +μ^2 )(e

t−1)

.

Notice that the factors in the brackets in the next-to-last equality are the mgfs of

X 1 andX 2 , respectively. Hence, the mgf ofY isthesameasthatofX 1 exceptμ 1

has been replaced byμ 1 +μ 2. Therefore, by the uniqueness of mgfs, the pmf ofY

must be

pY(y)=e−(μ^1 +μ^2 )

(μ 1 +μ 2 )y

y!

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

which is the same pmf that was obtained in Example 2.2.1.

Example 2.2.7(Continuation of Example 2.2.4).HereX 1 andX 2 have the joint

pdf

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

{ 1

4 exp

(

−x^1 + 2 x^2

)

0 <x 1 <∞, 0 <x 2 <∞

0elsewhere.