Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.3. Conditional Distributions and Expectations 109

2.2.6.SupposeX 1 andX 2 have the joint pdffX 1 ,X 2 (x 1 ,x 2 )=e−(x^1 +x^2 ), 0 <xi<
∞,i=1,2, zero elsewhere.

(a)Use formula (2.2.5) to find the pdf ofY 1 =X 1 +X 2.

(b)Find the mgf ofY 1.

2.2.7.Use the formula (2.2.5) to find the pdf ofY 1 =X 1 +X 2 ,whereX 1 andX 2

have the joint pdffX 1 ,X 2 (x 1 ,x 2 )=2e−(x^1 +x^2 ), 0 <x 1 <x 2 <∞, zero elsewhere.

2.2.8.SupposeX 1 andX 2 have the joint pdf

f(x 1 ,x 2 )=

{

e−x^1 e−x^2 x 1 > 0 ,x 2 > 0

0elsewhere.

For constantsw 1 >0andw 2 >0, letW=w 1 X 1 +w 2 X 2.

(a)Show that the pdf ofWis

fW(w)=

{ 1

w 1 −w 2 (e

−w/w (^1) −e−w/w (^2) ) w> 0
0elsewhere.
(b)Verify thatfW(w)>0forw>0.
(c)Note that the pdffW(w) has an indeterminate form whenw 1 =w 2 .Rewrite
fW(w)usinghdefined asw 1 −w 2 =h. Then use l’Hˆopital’s rule to show that
whenw 1 =w 2 ,thepdfisgivenbyfW(w)=(w/w^21 )exp{−w/w 1 }forw> 0
and zero elsewhere.

2.3 Conditional Distributions and Expectations

In Section 2.1 we introduced the joint probability distribution of a pair of random
variables. We also showed how to recover the individual (marginal) distributions
for the random variables from the joint distribution. In this section, we discuss
conditional distributions, i.e., the distribution of one of the random variables when
the other has assumed a specific value. We discuss this first for the discrete case,
which follows easily from the concept of conditional probability presented in Section
1.4.
LetX 1 andX 2 denote random variables of the discrete type, which have the joint
pmfpX 1 ,X 2 (x 1 ,x 2 ) that is positive on the support setSand is zero elsewhere. Let
pX 1 (x 1 )andpX 2 (x 2 ) denote, respectively, the marginal probability mass functions
ofX 1 andX 2 .Letx 1 be a point in the support ofX 1 ; hence,pX 1 (x 1 )>0. Using
the definition of conditional probability, we have

P(X 2 =x 2 |X 1 =x 1 )=

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

P(X 1 =x 1 )

=

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

pX 1 (x 1 )

for allx 2 in the supportSX 2 ofX 2. Define this function as

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

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

pX 1 (x 1 )

,x 2 ∈SX 2. (2.3.1)