The Poisson process

1 Chapter 3 The Poisson processThe next part of the course deals with some fundamental models of events occurring randomly incontinuous time. Many modelling applications involve events ( arrivals ) happening one by one,with randominterarrivaltimes between them. A general process of this type is arenewal process ,named because we can think of each new arrival as a renewal in which the system is reset . Inthis chapter we focus on a special renewal process called thePoisson process . Here are some typicalsituations where renewal processes might companies often model customers claims using renewal ideas. Inthis case the interarrival distribution is a crucial element of thecalculation of what insurancepremium to processes.

2 Many devices can be described ascountersin that they attempt to recordthe occurrence of successive signal pulses impinging on some instrument. For example Geigercounters for recording ionization events, or scintillation counters for recording passage of asubatomic times at which successive cars pass a monitoring station on a long single-lane road can be modelled as a renewal process . Much more generally, any sort of traffic can fit a similar model, such as data packets arriving at a server across anetwork of congestion can be answered using renewal theory and the relatedtheory large department store needs to know how much stock of a particularitem to hold, and a schedule for replenishment. The pattern of demands can often be modelledas a renewal any of these or other similar situations in which events occur randomly in time at someuniform average rate, an assumption of total randomness leads to the Poisson process as a Definition of the Poisson processWe describe the situation by thecounting processN(t),t >0, which counts the number of eventsthat have occurred between time 0 and timet.

3 Our model has a single parameter, >0, which is theaverage arrival rate per unit time. Before defining the model formally, we make some preliminarycalculations based on the following three natural assumptions:26 The probability of an event occurring in a short interval of time [t, t+h] is h+o(h) ash 0. The probability of two or more events occurring in interval [t, t+h] iso(h) ash 0. The numbers of events occurring in disjoint time intervals are the interval [0, t] intonequal pieces of lengtht/n(for largen). By the third assumption,the numbers of arrivals in each of these intervals are independent. Roughly speaking, since it isvery unlikely (for largen) that more than one event occurs in any small interval, the total numberof events in allnintervals is approximately binomially distributed, ,P(N(t) =k) (nk)( t/n)k(1 t/n)n k,which converges to the Poisson probability e t( t)k/k!

4 Asn . (The fact that binomial (n, p)is approximately Poisson (np) whennis large andp c/nis one of the oldest results in probabilitytheory, with its origin in 1711 work of de Moivre: see also MM304.) Thus the number of eventsoccurring in an interval of lengthtis Poisson distributed with mean t. This motivates the followingmore precise Poisson process withrate(orintensity) >0is a continuous-time stochasticprocess(N(t), t 0)taking values inZ+such that:(i)N(0) = 0;(ii) The pathst7 N(t)are right-continuous;(iii) For any0 t1< t2< < tkthe incrementsN(tn+1) N(tn),1 n k 1, areindependent Poisson random variables with means (tn+1 tn).Note that properties (i) and (iii) imply thatN(t) is Poisson with mean arrive at a shop as a Poisson process with rate 20 per hour.

5 In a half-hourperiod, calculate the probability that 4 customers arrive in the first 15 minutes and 6 in the next15 minutes. The two 1/4-hour periods are disjoint, so the arrivals in each are independent Poissonrandom variables with mean 20 1/4 = 5. So we gete 5544! e 5566!= e 105104!6! in the Seine are caught at rate >0. LetN(t) be the number of fishcaught up to timet. How long do we have to wait until we catch the first fish? LetN(0) = 0, andconsiderX1= inf{t 0 :N(t) 1},the first time thatN(t) reaches 1. SinceN(t) is non-decreasing,P(X1> t) =P(N(t) = 0) = e t,using the Poisson distribution. SoP(X1 t) = 1 e t, ,X1is anexponentialrandom variable with parameter .This is an important property of the Poisson process . Before saying more on this, we recallsome facts about continuous random Revision of continuous random variablesAs in the case of discrete random variables, recall that thedistribution functionfor a randomvariableXis given byF(x) :=P(X x).

6 IfFis differentiable, then its derivativeF =fexists, and is called theprobability density functionofX. If this is the case,Xis acontinuousrandom variable. Note thatF(x) = x f(y)dy,and, more generally,P(a X b) = baf(y) (Expectation).For a random variableXwith densityf, defineE(X) = Rxf(x) exponential distribution with parameter >0 has densityf(x) = e xforx 0,f(x) = 0 forx <0. Moreover, ifXis exponential with parameter ,E(X) = 0x e xdx=1 .Theorem (Properties of expectation).are the same as for the discrete case (linearity, mono-tonicity, triangle inequality). IfXhas a density, we have the following version of the law of theunconscious statistician, ifh:X( ) [0, ),E(h(X)) = Rh(x)f(x) (Independence).A family of continuous random variablesXi,i Iis independentifP j J{Xj xj} = j JP(Xj xj),for allj J,xj Rand all finiteJ Properties of the Poisson (t) be the number of radioactive disintegrations detected by aGeiger counter up to timet.]

7 Then, as long astis small compared to the half-life of the substance,(N(t), t 0) can be modelled as a Poisson process with rate . As above, the timeX1until thefirst decay event is exponential with parameter . In fact, the time between the first and secondevents isalsoexponential with parameter , and independent ofX1. We state this property morecarefully below, after some more (Jump times and interarrival times).LetS0= 0and letSnbe the time of thentharrival. More formally, we can defineSniteratively bySn= inf{t > Sn 1:N(t)> N(Sn 1)},(n 1);soS1, S2, S3, ..are the times at whichN(t) jumps (increases by1). SetX1=S1and forn 2defineXn=Sn Sn say thatXnis thenthinterarrival useful relationship betweenSnandN(t) is thatN(t) nif and only ifSn t(a picture ishelpful to see this).

8 Is the probability that the 3rd arrival occurs after timet? We haveP(S3> t) =P(N(t)<3) =P(N(t) = 0,1 or 2) = e t(1 + t+( t)22).Theorem (Interarrival times).For a Poisson process of rate , the random variables(X1, X2, X3, ..)are independent exponential random variables with parameter .We will not prove the theorem yet, but will return to this resultlater. For now we look at someimportant consequences of this for a arrive as a Poisson process of rate = 4 per hour after start waiting for a bus at 5pm, and, knowing about the exponential distribution, expect to waitfor about 1/ hours = 15 minutes for a bus. By , no bus has turned up. How much longerdo I expect to wait?We are working withX1, the time until the first arrival.

9 NowX1is exponential with parameter = 4, as we have seen. But we are told thatX1> (already waited half an hour). So we needaconditionaldistribution. We can calculate the probability of waiting a further timet >0, giventhat we have already waited , as followsP(X1> +t|X1> ) =P(X1> +t)P(X1> )=e 4( +t)e 4( )= e 4t,which is another exponential random time with parameter 4. So I am no better off now than when Istarted! The (conditional) waiting time has the same distribution as the original, and the expectedwaiting time isstill15 (more) phenomenon in the last example is general to Poisson processes. Here is a general (Memoryless property of the exponential distribution.).IfXis exponential withparameter then for alls >0,t >0,P(X > s+t|X > s) =P(X > t).

10 Idea as the previous example. Also see the for more owned by two different companies, A and B, arrive atthe same stop. Buses owned by A arrive as a Poisson process of rate 2 per hour. Buses owned byBarrive as a Poisson process of rate 1 per hour. Assuming that the two processes are independent,what is the distribution of the waiting time to the arrival of the firstbus (of either kind)?The first arrival time for type A,TA, has an exponential distribution with parameter 2, whileTB, the first arrival time for type B, is exponential with parameter 1. We are interested inT=min{TA, TB}. ThenP(T > t) =P({TA> t} {TB> t}) =P(TA> t)P(TB> t),by independence. SoP(T > t) = e 2t e t= e 3t,soThas an exponential distribution with parameter 3, andE(T) = 1 , this example shows a general (Superposition).