Page 215 - Applied Statistics And Probability For Engineers

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5-3

e 1 1 2 2 y 1 y 1 ! 1 y 1 y 2 2 y 2 e 1 1 2 2 y 1 y 1
2
1

a a a b 1 y 1 y 2 2 y 2

1
2
y ! y 2 0 1 y y 2! y ! y ! y 2 0 y 2
1
1
1
2
2
The summation in this last expression is the binomial expansion of 1 2 , so
y 1
2
1
e 1 1 2 2 1 2 y 1
1
2
f 1y 2 , y 0, 1, p
y !
Y 1 1 1
1
We recognize this as a Poisson distribution with parameter 1 2 . Therefore, we have shown
that the sum of two independent Poisson random variables with parameters 1 and 2 has a
Poisson distribution with parameter 1 2 .
We now consider the situation where the random variables are continuous. Let Y h(X),
with X continuous and the transformation is one to one.
Suppose that X is a continuous random variable with probability distribution f (x).
X
The function Y h(X) is a one-to-one transformation between the values of Y and X
so that the equation y h(x) can be uniquely solved for x in terms of y. Let this
solution be x u(y). The probability distribution of Y is
f 1 y2 f 3u1y24 0 J 0 (S5-3)
X
Y
where J u¿ (y) is called the Jacobian of the transformation and the absolute value
of J is used.

Equation S5-3 is shown as follows. Let the function y h(x) be an increasing function of x.
Now

P1Y a2 P3X u1a24
u1a2

f 1x2 dx
X

If we change the variable of integration from x to y by using x u(y), we obtain dx u (y) dy
and then
a

P1Y a2 f 3u1 y24u¿1y2 dy
X

Since the integral gives the probability that Y a for all values of a contained in the feasible
3u1y24u¿1 y2 must be the probability density of Y. Therefore, the proba-
set of values for y, f X
bility distribution of Y is
f 1y2 f 3u1 y24u¿1 y2 f 3u1 y24J
X
Y X
If the function y h(x) is a decreasing function of x, a similar argument holds.