Multidimensional Random Variables - Random Signals and Processes Primer with MATLAB - page 166

where the denominator in ( 3.52 ) is:

F X 2 ðx 22 ÞF X 2 ðx 21 Þ¼Pfx 21 < X 2 x 22 g 6¼ 0 : (3.53)

From ( 3.27 ) and ( 3.52 ), the corresponding conditional density is:

Ð

x 22

f X 1 X 2 ðx 1 ; x 2 Þ d x 2

x 21

f X 1 x 1 jx 21 < X 2 x 22 ð Þ ¼

f X 1 X 2 ðx 1 x 2 Þ d x 1 d x 2 :

(3.54)

Ð Ð 1

x 22

x 21 1

Example 3.2.7 Consider a random variable X 1 with the density function:

l e lx 1

for x 1 > 0

f X 1 ðx 1 Þ¼

(3.55)

0 otherwise

and the conditional density

x 1 e x 1 x 2

for x 1 > 0 ; x 2 > 0

f X 2 x 2 jx 1 ð Þ¼ : (3.56)

0 otherwise

Find the conditional density f X 1 ðx 1 jx 2 Þ .
Solution From ( 3.51 ), ( 3.55 ), and ( 3.56 ), we get:

f X 1 X 2 ðx 1 ; x 2 Þ¼f X 2 x 2 jx 1 ð Þf X 1 ðx 1 Þ: (3.57)

Placing ( 3.55 ) and ( 3.56 ) into ( 3.57 ), we arrive at:

lx 1 e x 1 ðlþx 2 Þ

for x 1 > 0 ; x 2 > 0

f X 1 X 2 ðx 1 ; x 2 Þ¼ : (3.58)

0 otherwise

From here, using ( 3.42b ), we find:

1 1

lx 1 e x 1 ðlþx 2 Þ d x 1 : f X 2 ðx 2 Þ¼ f X 1 X 2 ðx 1 ; x 2 Þ d x 1 ¼ (3.59)

0 0

Using integral 1 from Appendix A , we obtain:

l
ðlþx 2 Þ

for x 2 > 0

f X 2 ðx 2 Þ¼

2

: (3.60)

0 otherwise