Introduction to Probability and Statistics for Engineers and Scientists

154 Chapter 5: Special Random Variables

The last equality follows since, from Equation 5.2.2,

E[Xi]=P{Xi= 1 }=

1

n

■

The Poisson distribution possesses the reproductive property that the sum of indepen-
dent Poisson random variables is also a Poisson random variable. To see this, suppose that
X 1 andX 2 are independent Poisson random variables having respective meansλ 1 andλ 2.
Then the moment generating function ofX 1 +X 2 is as follows:

E[et(X^1 +X^2 )]=E[etX^1 etX^2 ]

=E[etX^1 ]E[etX 2 ] by independence

=exp{λ 1 (et−1)}exp{λ 2 (et−1)}

=exp{(λ 1 +λ 2 )(et−1)}

Because exp{(λ 1 +λ 2 )(et−1)}is the moment generating function of a Poisson random
variable having meanλ 1 +λ 2 , we may conclude, from the fact that the moment generating
function uniquely specifies the distribution, thatX 1 +X 2 is Poisson with meanλ 1 +λ 2.

EXAMPLE 5.2f It has been established that the number of defective stereos produced daily
at a certain plant is Poisson distributed with mean 4. Over a 2-day span, what is the
probability that the number of defective stereos does not exceed 3?

SOLUTION Assuming thatX 1 , the number of defectives produced during the first day, is
independent ofX 2 , the number produced during the second day, thenX 1 +X 2 is Poisson
with mean 8. Hence,

P{X 1 +X 2 ≤ 3 }=

∑^3

i= 0

e−^8

8 i

i!

=.04238 ■

Consider now a situation in which a random number, call itN, of events will occur, and
suppose that each of these events will independently be a type 1 event with probabilitypor
a type 2 event with probability 1−p. LetN 1 andN 2 denote, respectively, the numbers of
type 1 and type 2 events that occur. (SoN =N 1 +N 2 .) IfNis Poisson distributed
with meanλ, then the joint probability mass function ofN 1 andN 2 is obtained as
follows.

P{N 1 =n,N 2 =m}=P{N 1 =n,N 2 =m,N=n+m}

=P{N 1 =n,N 2 =m|N=n+m}P{N=n+m}

=P{N 1 =n,N 2 =m|N=n+m}e−λ

λn+m

(n+m)!