Fundamentals of Probability and Statistics for Engineers

As a numerical example, suppose that X 1 and X 2 are independent and

The pdf of Y is, fo llowing Equation (5.45),

In the above, the integration limits are determined from the fact that fX 1 (x 1 )
and fX 2 (x 2 ) are nonzero in intervals 0 x 1 1, and 0 x 2 2. With the
argument of fX 1 (x 1 )replacedbyy/x 2 in the integral, we have 0 y/x 2 1,
and 0 x 2 2, which are equivalent to y x 2 2. Also, range 0 y 2 for
the nonzero portion of fY (y) is determined from the fact that, since y x 1 x 2 ,
intervals 0 x 1 1, and 0 x 2 2 directly give 0 y 2.
F inally, Equation (5.46) gives

This is shown graphically in Figure 5.17. It is an easy exercise to show that

Ex ample 5. 12. Problem: let Y X 1 /X 2 where X 1 and X 2 are independent and
identically distributed according to

and similarly for X 2. Determine f (^) Y (y).
Answer: it follows from Equations (5.41) and (5.42) that
140 Fundamentals of Probability and Statistics for Engineers
fX 1 …x 1 †ˆ
2 x 1 ; for 0 x 1  1 ;
0 ; elsewhere;



fX 2 …x 2 †ˆ

2

 x 2

2

; for 0 x 2  2 ;

0 ; elsewhere:

8

<

:

fY…y†ˆ

Z 1

1

fX 1

y

x 2



fX 2 …x 2 †

1

x 2

dx^2 ;

ˆ

Z 2

y

2

y

x 2



2

 x 2

2



1

x 2



dx 2 ; for 0 y 2 ;

ˆ 0 ; elsewhere:

… 5 : 46 †

   

 

     

     

fY…y†ˆ

2 ‡y…lny 

1

 ln 2†; for 0 y 2 ;

0 ; elsewhere:



… 5 : 47 †

Z 2

0

fY…y†dyˆ 1 :

fX 1 …x 1 †ˆ

e^ x^1 ; forx 1 > 0 ;

0 ; elsewhere;



… 5 : 48 †

FY…y†ˆ

Z

…R^2 :x 1 =x 2 y†

Z

fX 1 X 2 …x 1 ;x 2 †dx 1 dx 2 :

ˆ

ˆ