Page 169 - Fundamentals of Probability and Statistics for Engineers

152 Fundamentals of Probability and Statistics for Engineers

Equation (5.68) now applies and we have

1

1
f y 1 ; y 2 †ˆ f ‰g y†ŠjJ 1 j‡ f ‰g y†ŠjJ 2 j; 5:74†
Y 1 Y 2 X 1 X 2
where
2 2
1
x ‡ x
f x†ˆ f x 1 † f x 2 †ˆ exp 1 2 ; 5:75†
X
X 2
X 1
2 2
qg qg

1
1
11 11

qy 1 y 1

ˆ
2 : 5:76†
qy 2
J 1 ˆ J 2 ˆ
qg
1 qg
1 1 ‡ y 2
12
12

qy 1 qy 2
On substituting Equations (5.75) and (5.76) into Equation (5.74), we have
2 2 2
y 1 2 y y ‡ y 1
1 2
f y 1 ; y 2 †ˆ exp
;
Y 1 Y 2 2 2 2
1 ‡ y 2 1 ‡ y †
2 2
2
y 1
y 1
ˆ exp ; for y 1 0; and
1 < y 2 < 1;
2
1 ‡ y † 2
2
ˆ 0; elsewhere: 5:77†
(y 1 ) and
We note that the result can again be expressed as the product of f Y 1
f (y 2 ), with
Y 2
8
2

y
> 1
< y 1 exp ; for y 1 0;
f y 1 †ˆ 2
Y 1
>
0; elsewhere;
:
1
y 2 †ˆ ; for
1 < y 2 < 1:
f Y 2 2
1 ‡ y †
2
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 f (y 2 ) to make the area under each pdf

Y 2
equal to 1.

Example 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
Y ˆ X 1 X 2 : 5:78†
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