THE POISSON PROCESS

G A Vignaux

Abstract

These notes examine the Poisson Process which models ``random'' arrivals to Markovian queue models. The properties of the process, such as the distribution of time between events, are studied. The Erlang which appears as the distribution of time between one event and the k-th next event is described.

This is Revision: 2.1 .

Contents

1  Introduction
2  The Poisson Process
    2.1  Merging (or adding) two Poisson streams
    2.2  Partitioning two streams
    2.3  The probability that NO events occur in an interval T
    2.4  The probability of n events in an interval T.
    2.5  How long to go before the next event?
    2.6  How long between events?
    2.7  The distribution of time to the k-th next event
3  The Erlang Distribution, Ek
    3.1  The Erlang distribution as sum of exponentials

References

[(Hillier and Lieberman)(1990)]
Hillier, F. S. and Lieberman, G. J. (1990). Introduction to Operations Research. McGraw-Hill, 5th edition.

[(Taha)(1992)]
Taha, H. A. (1992). Operations Research: An Introduction. Macmillan, 5th edition.

[(Winston)(1994)]
Winston, W. L. (1994). Operations Research, Applications and Algorithms. Duxbury Press, 3rd edition.

1  Introduction

The Poisson Process is the model we use in Queuing theory for ``Random arrivals''. It is in many ways the simplest model we could use for arrivals into a queue system because it assumes that the probability of an arrival in a small interval of time depends only on the size of the interval, not on any history of the process to that time.

The main reference for these notes is [(Hillier and Lieberman)(1990)]. Other OR books, such as [(Taha)(1992)] and [(Winston)(1994)] also explain the model. Many other books on statistics and queue models deal with this important stochastic process.

2  The Poisson Process

figs/quel040.gif

Small time interval h. Prob(event in interval h) = lh

figs/quel041.gif

2.1  Merging (or adding) two Poisson streams

figs/quel048.gif

Prob{event in stream 1 is in h } = l1 h
Prob{event in stream 2 is in h } = l2 h

figs/quel047.gif

Prob{some event is in h}
=
Prob{stream 1 event} +
Prob{stream 2 event}
=
l1 h + l2 h
=
(l1 + l2)h

Merged stream acts like a Poisson process with rate
l = l1 + l2

2.2  Partitioning two streams

figs/quel049.gif

Whenever an event (arrival) occurs we decide to divert it into A or B stream according to fixed probabilities, pA and pB.

Prob{A event in h}
=
prob{event in h} × Prob{choosing A stream}
=
lpA h

The A stream acts like a Poisson stream with a rate of events = pAl

2.3  The probability that NO events occur in an interval T

figs/quel044.gif

P{0 events in interval [t, t+h] } = 1 - lh where lh << 1

put
p0(t)
=
P{0  events in [0, t] }
p0(t+h)
=
P{ 0  events in [0, t+h]}

Since probability of an event in h is independent of any others:


p0(t+h)
=
p0(t).(1 - lh)
p0(t+h) - p0(t)
=
- lh p0(t)
p0(t+h) - p0(t)
h
=
- lp0(t)

and as h ® 0
dp0(t)
dt
=
- lp0(t)
dp0(t)
p0(t)
=
- ldt

integrate
lnp0(t)
=
const - lt,
p0(t)
=
Ae-lt

But note that if t ® 0, p0(t) ® 1 ( certain to have 0 events!)


p0(t) = e-lt

2.4   The probability of n events in an interval T.

figs/quel042.gif


pn(T)
=
Prob { n  events in   [0, T] }
p0(T)
=
Pr { 0 events in [0, T] } = e- lT
p1(T)
=
Pr{ 0 ev. in [0, T]}
×Pr{1 ev. in [t, t+h]}
×Pr{0 ev. in [t+h, T] }
        (int over all t from 0 to T), h ® 0
=
ó
õ
T

0 
e- lt l dt e- l(T - t)
=
le- lT ó
õ
T

0 
 dt
p1(T)
=
lT e- lT

The probability of 2 events in [0, t] is
p2(T)
=
ó
õ
Prob{1 event in [0, t] }
×(l dt)
×Prob{0 events in [t, T-t] }
        (int over all t from 0 to T), h ® 0
=
ó
õ
T

0 
(lt e- lt)(l dt) e- l(T-t)
=
l2 e- lT ó
õ
T

0 
t dt
p2(T)
=
(lT)2
2
e- lT

This leads us to the Poisson distribution


pn(T) = (lT)n
n!
e- lT


E(n) = lT     Var(n) = lT

figs/quel004.gif

2.5  How long to go before the next event?

figs/quel043.gif

Prob{next event after 0 occurs in [t, t+h] }
=
Prob{0 events in [0, t] } ×Prob {1 event in [t, t+h] }
=
e - lt.lh

thus p.d.f.


f(t) = le- lt

figs/quel020.gif

NOTE

2.6  How long between events?

figs/quel046.gif

Since no memory the zero point can be an event.

Thus time between events is also exponential, l.
f(t) = le- lt

the average time between events = 1/l.

2.7  The distribution of time to the k-th next event

figs/quel045.gif

Prob{kth next event is in [t, t+h] }


=
Prob{k-1 events in [0, t] } (lh)
=
(lt)k - 1
(k-1)!
e- lt lh
Thus p.d.f. is


fk(t) = l(lt)k-1
(k-1)!
e-lt

This is an Erlang distribution with mean = k /l and variance = k/l2. It has shape parameter of k.

3  The Erlang Distribution, Ek

The Erlang Distribution is important in queueing theory and simulation.

Named after A K Erlang, a Danish telephone engineer, considered to be the founder of queueing theory. He wrote papers in 1917-1918.

The Ek is a continuous rv with density function
f(t) = (mk)k
(k-1)!
tk-1e-kmt

A special type of Gamma distribution with an integer shape parameter. 2 parameters control its distribution (the scale parameter, m and its shape parameter, k). Mean is 1/m and its variance 1/km2.

figs/quel028.gif

figs/quel027.gif

The Erlang distribution gets narrower as k increases. As k tends to infinity the variance tends to zero. Thus the Erlang can also represent the ``distribution'' of a constant - a deterministic value.

An intermediate value of k gives a shape between exponential and constant.

3.1  The Erlang distribution as sum of exponentials

The Erlang can be thought of as the convolution (that is the sum of) k independent exponential random variables (each with the same mean).

To get a mean of 1/m these k exponentials must each have mean 1/km.

The variance of each exponential is 1/k2m2. Thus variance of their sum is k/k2m2 = 1/km2.

It is an important distribution in queueing theory because

  1. we can simulate an Erlang rv , Ek, by summing k exponential rvs,
  2. we can consider it as if it was a sequence of phases each of exponential type. Thus we can use the Markovian property and get some theoretical results with the Erlang. This is what Erlang did.


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On 10 Jun 2002, 14:27.