Applied Statistics and Probability for Engineers

5-3

The summation in this last expression is the binomial expansion of so

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 Xcontinuous and the transformation is one to one.

fY 11 y 12 

e^1 ^1 ^22 1  1  22 y^1

y 1!

, y 1 0, 1, p

1  1  22 y^1 ,



e^1 ^1 ^22

y 1!

(^) a
y 1
y 2  0
y 1!  11 y^1 y^22  2 y^2
1 y 1 y 22! y 2!

e^1 ^1 ^22
y 1!
(^) a
y 1
y 2  0
a
y 1
y 2 b^ ^1
1 y 1 y (^22) 
2
y 2
Suppose that Xis a continuousrandom variable with probability distribution fX(x).
The function Y h(X) is a one-to-one transformation between the values of Yand X
so that the equation y h(x) can be uniquely solved for xin terms of y. Let this
solution be x u(y). The probability distribution of Yis
(S5-3)
where (y) is called the Jacobianof the transformation and the absolute value
of Jis used.
Ju¿
fY 1 y 2 fX 3 u 1 y 240 J 0
Equation S5-3 is shown as follows. Let the function y h(x) be an increasing function of x.
Now
If we change the variable of integration from xto yby using x u(y), we obtain dx u(y) dy
and then
Since the integral gives the probability that Yafor all values of acontained in the feasible
set of values for y, must be the probability density of Y. Therefore, the proba-
bility distribution of Yis
If the function y h(x) is a decreasing function of x, a similar argument holds.
fY 1 y 2 fX 3 u 1 y 24 u¿ 1 y 2 fX 3 u 1 y 24 J
fX 3 u 1 y 24 u¿ 1 y 2

P 1 Ya 2  

a

   

fX 3 u 1 y 24 u¿ 1 y 2 dy

 

u 1 a 2

   

fX 1 x 2 dx

P 1 Ya 2 P 3 Xu 1 a 24

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