Fundamentals of Probability and Statistics for Engineers

Equation (5.68) now applies and we have

where

On substituting Equations (5.75) and (5.76) into Equation (5.74), we have

We note that the result can again be expressed as the product of fY 1 (y 1 )and
fY 2 (y 2 ), with

Thus random variables Y 1 and Y 2 are again independent in this case where Y 1
has the so -called Raleigh distribution and Y 2 is Cauchy distributed. We remark
that the factor (1/ ) is assigned to fY 2 (y 2 ) to make the area under each pdf
equal to 1.

Ex ample 5. 20. Let us determine the pdf of Y considered in Example 5.11 by
using the formulae developed in this section. The transformation is

152 Fundamentals of Probability and Statistics for Engineers

fY 1 Y 2 …y 1 ;y 2 †ˆfX‰g^11 …y†ŠjJ 1 j‡fX‰g^21 …y†ŠjJ 2 j; … 5 : 74 †

fX…x†ˆfX 1 …x 1 †fX 2 …x 2 †ˆ

1

2 

exp

x^21 ‡x^22

2



; … 5 : 75 †

J 1 ˆJ 2 ˆ

qg^111

qy 1

qg^111

qy 2

qg^121

qy 1

qg^121

qy 2

ˆ

y 1

1 ‡y^22

: … 5 : 76 †

fY 1 Y 2 …y 1 ;y 2 †ˆ

y 1

1 ‡y^22



2

2 

exp

y^21 y^22 ‡y^21

2 … 1 ‡y^22 †



;

ˆ

y 1

… 1 ‡y^22 †

exp

y^21

2



; fory 1  0 ;and 

1 <y 2 < 1 ;

ˆ 0 ; elsewhere: … 5 : 77 †

fY 1 …y 1 †ˆ

y 1 exp

y^21

2



; fory 1  0 ;

0 ; elsewhere;

8

><

>:

fY 2 …y 2 †ˆ

1

… 1 ‡y^22 †

; for 

1 <y 2 < 1 :



YˆX 1 X 2 : … 5 : 78 †