Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

from which we obtain

M′X(t)=λeteλ(e

t−1)

,

M′′X(t)=(λ^2 e^2 t+λet)eλ(e

t−1)

.

Thus, the mean and variance of the Poisson distribution are given by

E[X]=MX′(0) =λ and V[X]=MX′′(0)−[M′X(0)]^2 =λ.

The Poisson approximation to the binomial distribution

Earlier we derived the Poisson distribution as the limit of the binomial distribution

whenn→∞andp→0 in such a way thatnp=λremains finite, whereλis the

mean of the Poisson distribution. It is not surprising, therefore, that the Poisson

distribution is a very good approximation to the binomial distribution for large

n(≥50, say) and smallp(≤ 0 .1, say). Moreover, it is easier to calculate as it

involves fewer factorials.

In a large batch of light bulbs, the probability that a bulb is defective is0.5%.Fora

sample of 200 bulbs taken at random, find the approximate probabilities that 0 , 1 and 2 of

the bulbs respectively are defective.

Let the random variableX= number of defective bulbs in a sample. This is distributed
asX∼Bin(200, 0.005), implying thatλ=np=1.0. Sincenis large andpsmall, we may
approximate the distribution asX∼Po(1), giving

Pr(X=x)≈e−^1

1 x

x!

,

from which we find Pr(X=0)≈ 0 .37, Pr(X=1)≈ 0 .37, Pr(X=2)≈ 0 .18. For comparison,
it may be noted that the exact values calculated from the binomial distribution are identical
to those found here to two decimal places.

Multiple Poisson distributions

Mirroring our discussion of multiple binomial distributions in subsection 30.8.1,

let us supposeXandYare twoindependentrandom variables, both of which

are described by Poisson distributions with (in general) different means, so that

X∼Po(λ 1 )andY∼Po(λ 2 ). Now consider the random variableZ=X+Y.We

may calculate the probability distribution ofZdirectly using (30.60), but we may

derive the result much more easily by using the moment generating function (or

indeed the probability or cumulant generating functions).

SinceXandYare independent RVs, the MGF forZis simply the product of

the individual MGFs forXandY. Thus, from (30.104),

MZ(t)=MX(t)MY(t)=eλ^1 (e

t−1)

eλ^2 (e

t−1)

=e(λ^1 +λ^2 )(e

t−1)

,

which we recognise as the MGF ofZ∼Po(λ 1 +λ 2 ). HenceZis also Poisson

distributed and has meanλ 1 +λ 2. Unfortunately, no such simple result holds for

thedifferenceZ=X−Yof two independent Poisson variates. A closed-form