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5-4

EXAMPLE S5-3 Let X be a continuous random variable with probability distribution

x
f 1x2 , 0 x 4
8
X
Find the probability distribution of Y h(X) 2X 4.
Note that y h(x) 2x 4 is an increasing function of x. The inverse solution is x
u( y) ( y 4) 2, and from this we find the Jacobian to be J u¿1 y2 dx
dy 1
2.
Therefore, from S5-3 the probability distribution of Y is

1y 42
2 1 y 4
f 1 y2 a b , 4 y 12
Y
8 2 32
and X are continuous random variables and we wish
We now consider the case where X 1 2
to find the joint probability distribution of Y h (X , X ) and Y h (X , X ) where the trans-
2
2
1
2
1
1
1
2
formation is one to one. The application of this will typically be in finding the probability dis-
tribution of Y h (X , X ), analogous to the discrete case discussed above. We will need the
1
2
1
1
following result.
Suppose that X and X are continuous random variables with joint probability distri-
1
2
bution f 1x , x 2, and let Y h (X , X ) and Y h (X , X ) define a one-to-one
2
2
1
1
2
1
1
1
2
2
X 1 X 2
transformation between the points (x , x ) and (y , y ). Let the equations y h (x ,
1
2
1
1
2
1
1
x ) and y h (x , x ) be uniquely solved for x and x in terms of y and y as x
2
1
1
2
2
2
2
1
2
1
u (y , y ) and x u (y , y ). Then the joint probability of Y and Y is
2
1
2
1
2
2
1
1
2
f 1 y , y 2 f 3u 1 y , y 2, u 1y , y 24 0 J 0 (S5-4)
Y 1 Y 2 1 2 X 1 X 2 1 1 2 2 1 2
where J is the Jacobian and is given by the following determinant:
1
x y 1 , x 1
y 2
J ` `
x 2
y 1 , x 2
y 2
and the absolute value of the determinant is used.
This result can be used to find f 1 y , y 2, the joint probability distribution of Y 1 and Y 2 . Then
2
1
Y 1 Y 2
the probability distribution of Y 1 is

f 1y 2 f 1y , y 2 dy
Y 1 1 Y 1 Y 2 1 2 2

That is, f (y 1 ) is the marginal probability distribution of Y 1 .
Y 1
EXAMPLE S5-4 Suppose that X 1 and X 2 are independent exponential random variables with f 1x 2 2e 2x 1
X 1 1
and f 1x 2 2e 2x 2 . Find the probability distribution of Y X X .
1
2
2
X 2