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5-5

The joint probability distribution of X 1 and X 2 is

f 1x , x 2 4e 21x 1 x 2 2 , x 0, x 0
1
2
2
1
X 1 X 2
because X 1 and X 2 are independent. Let Y 1 h 1 (X 1 , X 2 ) X X and Y 2 h 2 (X 1 , X 2 )
1
2
1 2
X 1 X 2 . The inverse solutions of y 1 x x and y 2 x 1 x 2 are x y y 11 y 2 and
1
1
1
2
2
x y 11 y 2 , and it follows that
2
1
x 1 1 d, x 1 y 1
2
y 1 y c 11 y 2 2 y 2 c 11 y 2 d
1
1
x 2 1 d, x 2 1
2
y 1 y c 11 y 2 2 y 2 c 11 y 2 d
1
1
Therefore
y 2 y 1
11 y 2 2 11 y 2
1
1
J ∞ ∞ y 2
y 2 1 11 y 2 2
1
11 y 2 2 11 y 2
1
1
and from Equation S5-4 the joint probability distribution of Y 1 and Y 2 is
f 1y , y 2 f 3u 1y , y 2, u 1 y , y 24 0 J 0
2
1
1
1
1
2
2
2
Y 1 Y 2 X 1 X 2
y 2
4e 23y 1 y 2
11 y 1 2 y 2
11 y 1 24 ` 2 `
11 y 2
1
2
4e 2y 2 y 11 y 2 2
1
for y 1
0, y 2
0. We need to find the distribution of Y X X . This is the marginal prob-
1
2
1
ability distribution of Y , or
1
1
f 1 y 2 f 1y , y 2 dy
Y 1 Y 1 Y 2 1 2 2
0
2
4e 2y 2 3y 2
11 y 1 2 4 dy 2
0
1
2 , y
0
1
11 y 2
1
An important application of Equation S5-4 is to obtain the distribution of the sum of two
independent random variables X and X . Let Y X X and let Y X . The inverse so-
2
2
1
2
2
1
1
lutions are x y y and x y . Therefore,
2
2
2
1
1
x
x 1 1
1 1
y 1 y 2
x 2 x 2
0 1
y 1 y 2