Page 14 - A First Course In Stochastic Models

THE POISSON PROCESS 5

same exponential distribution as the time elapsed between epoch 0 and the epoch
of the first arrival after epoch 0. Next mimic the proof of Theorem 1.1.1.
To illustrate the foregoing, we give the following example.

Example 1.1.1 A taxi problem

Group taxis are waiting for passengers at the central railway station. Passengers for
those taxis arrive according to a Poisson process with an average of 20 passengers
per hour. A taxi departs as soon as four passengers have been collected or ten
minutes have expired since the first passenger got in the taxi.

(a) Suppose you get in the taxi as first passenger. What is the probability that you
have to wait ten minutes until the departure of the taxi?
(b) Suppose you got in the taxi as first passenger and you have already been waiting
for five minutes. In the meantime two other passengers got in the taxi. What
is the probability that you will have to wait another five minutes until the taxi
departs?

To answer these questions, we take the minute as time unit so that the arrival
rate λ = 1/3. By Theorem 1.1.1 the answer to question (a) is given by

P {less than 3 passengers arrive in (0, 10)}
2 k
−10/3 (10/3)
= e = 0.3528.
k!
k=0
The answer to question (b) follows from the memoryless property stated in Theo-
rem 1.1.2 and is given by
−5/3
P {γ 5 > 5} = e = 0.1889.
In view of the lack of memory of the Poisson process, it will be intuitively clear
that the Poisson process has the following properties:

(A) Independent increments: the numbers of arrivals occurring in disjoint intervals
of time are independent.

(B) Stationary increments: the number of arrivals occurring in a given time interval
depends only on the length of the interval.

A formal proof of these properties will not be given here; see Exercise 1.8. To
give the infinitesimal-transition rate representation of the Poisson process, we use

h 2 h 3
−h
1 − e = h − + − · · · = h + o(h) as h → 0.
2! 3!