For independent X 1 and X 2 ,
The pdf of Y is thus given by, on differentiating Equation (5.49) with respect
to y,
and, on substituting Equation (5.48) into Equation (5.50), it takes the form
Again, it is easy to check that
Example 5.13.To show that it is elementary to obtain solutions to problems
discussed in this section when X 1 ,X 2 ,..., and Xn are discrete, consider again
Y X 1 /X 2 given that X 1 and X 2 are discrete and their joi nt probability mass
function (jpmf) pX 1 X 2 (x 1 ,x 2 ) is tabulated in Table 5.1. In this case, the pmf of Y
is easily determined by assignment of probabilities pX 1 X 2 (x 1 ,x 2 ) to the corres-
ponding values of y x 1 /x 2. Thus, we obtain:
142 Fundamentals of Probability and Statistics for Engineers
FY
y
Z 1
0
Zx 2 y
0
fX 1
x 1 fX 2
x 2 dx 1 dx 2 ;
Z 1
0
FX 1
x 2 yfX 2
x 2 dx 2 ; fory> 0 ;
0 ; elsewhere:
5 : 49
fY
y
Z 1
0
x 2 fX 1
x 2 yfX 2
x 2 dx 2 ; fory> 0 ;
0 ; elsewhere;
8
><
>:
5 : 50
fY
y
Z 1
0
x 2 e^ x^2 ye^ x^2 dx 2
1
1 y^2
; fory> 0 ;
0 ; elsewhere:
8
><
>:
5 : 51
Z 1
0
fY
ydy 1 :
pY
y
0 : 5 ; fory
1
2
;
0 : 24 0 : 04 0 : 28 ; fory 1 ;
0 : 04 ; fory
3
2
;
0 : 06 ; fory 2 ;
0 : 12 ; fory 3 :