Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

t+∆tis given by

Px(t+∆t)=Px(t)(1−λ∆t)+Px− 1 (t)λ∆t.

Rearranging the equation, dividing through by ∆tand letting ∆t→0, we obtain

the differential recurrence equation

dPx(t)

dt

=λPx− 1 (t)−λPx(t). (30.101)

Forx= 0 (i.e. no calls received), however, (30.101) simplifies to

dP 0 (t)

dt

=−λP 0 (t),

which may be integrated to giveP 0 (t)=P 0 (0)e−λt. But since the probabilityP 0 (0)

of receiving no calls in a zero time interval must equal unity, we haveP 0 (t)=e−λt.

This expression forP 0 (t) may then be substituted back into (30.101) withx=1

to obtain a differential equation forP 1 (t) that has the solutionP 1 (t)=λte−λt.

We may repeat this process to obtain expressions forP 2 (t),P 3 (t),...,Px(t), and we

find

Px(t)=

(λt)x

x!

e−λt. (30.102)

By settingt= 1 in (30.102), we again obtain the Poisson distribution (30.100) for

obtaining exactlyxcalls in a unit time interval.

If a discrete random variable is described by a Poisson distribution of meanλ

then we writeX∼Po(λ). As it must be, the sum of the probabilities is unity:

∑∞

x=0

Pr(X=x)=e−λ

∑∞

x=0

λx

x!

=e−λeλ=1.

From (30.100) we may also derive thePoisson recurrence formula,

Pr(X=x+1)=

λ

x+1

Pr(X=x)forx=0, 1 , 2 ,...,

(30.103)

which enables successive probabilities to be calculated easily once one is known.

A person receives on average one e-mail message per half-hour interval. Assuming that

the e-mails are received randomly in time, find the probabilities that in any particular hour

0 , 1 , 2 , 3 , 4 , 5 messages are received.

LetX= number of e-mails received per hour. Clearly the mean number of e-mails per
hour is two, and soXfollows a Poisson distribution withλ=2,i.e.

Pr(X=x)=

2 x

x!

e−^2.

Thus Pr(X=0)=e−^2 =0.135, Pr(X=1)=2e−^2 =0.271, Pr(X=2)=2^2 e−^2 /2! = 0.271,
Pr(X=3)=2^3 e−^2 /3! = 0.180, Pr(X=4)=2^4 e−^2 /4! = 0.090, Pr(X=5)=2^5 e−^2 /5! =
0 .036. These results may also be calculated using the recurrence formula (30.103).