Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

170 Some Special Distributions

for all real values oft.Since

M′(t)=eλ(e

t−1)

(λet)

and

M′′(t)=eλ(e

t−1)

λetλet+eλ(e

t−1)
λet
then
μ=M′(0) =λ
and
σ^2 =M′′(0)−μ^2 =λ^2 +λ−λ^2 =λ.
That is, a Poisson distribution hasμ=σ^2 =λ>0.
IfXhas a Poisson distribution with parameterλ,thenP(X=k) is computed
by the R commanddpois(k,lambda)and the cumulative probabilityP(X≤k)is
calculated byppois(k,lambda).

Example 3.2.1. LetXbe the number of automobile accidents at a busy inter-
section per week. Suppose thatX has a Poisson distribution withλ=2. Then
the expected number of accidents per week is 2 and the standard deviation of the
number of accidents is

√

2. The probability of at least one accident in a week is

P(X≥1) = 1−P(X=0)=1−e−^2 =1−dpois(0,2) = 0. 8647

and the probability that there are between 3 and 8 (inclusive) accidents is

P(3≤X≤8) =P(X≤8)−P(X≤2) = ppois(8,2)−ppois(2,2) = 0. 3231.

Suppose we want to determine the probability that there are exactly 16 accidents
in a 4 week period. By Remark 3.2.1, the number of accidents over a 4 week period
has a Poisson distribution with parameter 2×4 = 8. So the desired probability is
dpois(16,8) = 0.0045. The following R code computes a spiked plot of the pmf
ofXover{ 0 , 1 ,..., 7 }, a subset of the range ofX.

rng=0:7; y=dpois(rng,2); plot(y~rng,type="h",ylab="pmf",xlab="Rng");

points(y~rng,pch=16,cex=2)

Example 3.2.2.Let the probability of exactly one blemish in 1 foot of wire be
about 10001 and let the probability of two or more blemishes in that length be, for
all practical purposes, zero. Let the random variableXbe the number of blemishes
in 3000 feet of wire. If we assume the independence of the number of blemishes in
nonoverlapping intervals, then by Remark 3.2.1 the postulates of the Poisson process
are approximated, withλ= 10001 andt= 3000. ThusXhas an approximate Poisson
distribution with mean 3000( 10001 ) = 3. For example, the probability that there are
five or more blemishes in 3000 feet of wire is

P(X≥5) =

∑∞

k=5

3 ke−^3

k!

=1−ppois(4,3) = 0. 1847.