Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

3.2. The Poisson Distribution 171

The Poisson distribution satisfies the following important additive property.

Theorem 3.2.1.SupposeX 1 ,...,Xnare independent random variables and sup-
poseXihas a Poisson distribution with parameterλi.ThenY =

∑n

i=1Xihas a

Poisson distribution with parameter

∑n

i=1λi.

Proof:We obtain the result by determining the mgf ofY, which by Theorem 2.6.1
is given by

MY(t)=E

(

etY

)

=

∏n

i=1

eλi(e

t−1)

=e

Pn

i=1λi(et−1).

By the uniqueness of mgfs, we conclude thatY has a Poisson distribution with

parameter

∑n

i=1λi.

Example 3.2.3(Example 3.2.2, Continued).Suppose, as in Example 3.2.2, that
a bail of wire consists of 3000 feet. Based on the information in the example, we
expect three blemishes in a bail of wire, and the probability of five or more blemishes
is 0.1847. Suppose in a sampling plan, three bails of wire are selected at random and
we compute the mean number of blemishes in the wire. Now suppose we want to
determine the probability that the mean of the three observations has five or more
blemishes. LetXibe the number of blemishes in theith bail of wire fori=1, 2 ,3.
ThenXihas a Poisson distribution with parameter 3. The mean ofX 1 ,X 2 ,andX 3
isX=3−^1

∑ 3

i=1Xi, which can also be expressed asY/3, whereY=

∑ 3
i=1Xi.By
the last theorem, because the bails are independent of one another,Yhas a Poisson
distribution with parameter

∑ 3

i=13 = 9. Hence, the desired probability is

P(X≥5) =P(Y≥15) = 1−ppois(14,9) = 0. 0415.

Hence, while it is not too odd that a bail has five or more blemishes (probability
is 0.1847), it is unusual (probability is 0.0415) that three independent bails of wire
average five or more blemishes.

EXERCISES
3.2.1.If the random variableXhas a Poisson distribution such thatP(X=1)=
P(X= 2), findP(X=4).

3.2.2. The mgf of a random variableXise4(e
t−1)

. Show thatP(μ− 2 σ<X<
μ+2σ)=0.931.
3.2.3.In a lengthy manuscript, it is discovered that only 13.5 percent of the pages
contain no typing errors. If we assume that the number of errors per page is a
random variable with a Poisson distribution, find the percentage of pages that have
exactly one error.
3.2.4.Let the pmfp(x) be positive on and only on the nonnegative integers. Given
thatp(x)=(4/x)p(x−1),x=1, 2 , 3 ,..., find the formula forp(x).
Hint: Note thatp(1) = 4p(0),p(2) = (4^2 /2!)p(0), and so on. That is, find each
p(x)intermsofp(0) and then determinep(0) from

1=p(0) +p(1) +p(2) +···.