Chapter 9 Poisson processes - Yale University

1 Page 1 Chapter 9 Poisson processesThe Binomial distribution and the geometric distribution describe the behavior of tworandom variables derived from the random mechanism that I have called coin tossing . Thenamecoin tossingdescribes the whole mechanism; the namesBinomialandgeometricreferto particular aspects of that mechanism. If we increase the tossing rate to m tosses per sec-ond and decrease the probability of heads to a small p, while keeping the expected numberof heads per second fixed at Dmp, the number of heads in atsecond interval will haveapproximately a ;p/distribution, which is close to the Poisson . t/. Also, the num-bers of heads tossed during disjoint time intervals will still be independent random the limit, asm!1, we get an idealization called aPoisson process. Poisson process< > Poisson process with rate on[0;1/is a random mechanism that gener-ates points strung out along[0;1/in such a way that(i) the number of points landing in any subinterval of lengthtis a random variable witha Poisson .]]

2 T/distribution(ii) the numbers of points landing in disjoint (= non-overlapping) intervals are indepen-dent random variables. The double use of the name Poisson is unfortunate. Much confusion would be avoidedif we all agreed to refer to the mechanism as idealized-very-fast-coin-tossing , or somesuch. Then the Poisson distribution would have the same relationship to idealized-very-fast-coin-tossing as the Binomial distribution has to coin-tossing. Obversely, we could createmore confusion by renaming coin tossing as the binomial process . Neither suggestion islikely to be adopted, so you should just get used to having two closely related objects withthe name bother about Poisson processes ? When we pass to the idealized mechanism ofpoints generated in continuous time, several awkward artifacts of discrete-time coin tossingdisappear. The Examples and Exercises in this Chapter will illustrate the simplifications.

3 < > the distribution of the time to thekth point in a Poisson process on [0;1/with rate .Solution:Denote the time to thekth point byTk. It has a continuous distribution, whichis specified by a density function. Fort>0 and small >0,Pft Tk<tC gDPfexactlyk 1 points in [0;t/, exactly one point in [t;tC /gCsmaller order termsThe smaller order terms contribute probability less thanPf2 or more points in [t;tC /gDPfPoisson. / 2gDe . /22!C:::Statistics 241: 28 October 1997c David PollardChapter 9 Poisson processesPage 2By the independence property (ii) for Poisson processes , the main term factorizes asPfexactlyk 1 points in [0;t/gPfexactly one point in [t;tC /gDe t. t/k 1/!e t. /11!De t ktk 1 .k 1/!Csmaller order termsThat is, the distribution ofTkhas densitye t ktk 1/!fort>0: Gamma function and gamma densityIt is easier to remember the form of the density forTkif one rescales, using an argumentYou should try thiscalculation at to the one for theN.]]]]]]

4 ; 2/distribution in Chapter 7, to show that Tkhas a distri-bution with density< >e ttk 1/!fort>0:This density is called generally, for each >0, the densitye tt 10. /fort>0is called thegamma. /density. The scaling constant,0. /, which ensures that the den- gamma. /densitysity integrates to one, is given by0. /DZ10e xx 1dxfor each >0:The function0. /is called thegamma function. Don t confuse the gamma density with gamma functionthe gamma function.< > waiting timeTkfrom Example< >has expected valueETkDZ10te t ktk 1/!dtD1 Z10e 1/!dxputtingxD t, (cf. distribution of Tk)Dk (Use integration by parts.)Does it make sense to you thatETkshould decrease as increases?More generally, for >0,0. C1/DZ10e xx dxD e xx 10C Z10e xx 1dxD 0. /In particular, 1 1 1/.k 2 2/D:::Statistics 241: 28 October 1997c David PollardChapter 9 Poisson processesPage 1/.k 2/.k 3/:::.2/.1 1/! xdxD1. Compare with the fact that the in< >integrates to one.

5 Exponential distributionSpecializing the the casekD1 we get the densitye tfort>0;which is called the (standard)exponential distribution. The time to the first point in exponential distributionthe Poisson process has density e tfort>0;an exponential distribution with expected value 1= . Don t confuse the exponential densitywith the exponential the parallels between the negative binomial distribution (in discrete time) and thegamma distribution (in continuous time). Each distribution corresponds to the waiting timeto thekth occurrence of something, for various values ofk. Just as (see Problem Sheet 4)the negative binomial can be written as a sum of independent random variables, each with ageometric distribution, so can the written as a sum ofkindependent randomcts. time$discrete timegamma$neg. binomialexponential$geometricFor counts: Poisson $Binomialvariables, each with an exponential distribution.

6 The times between points in a Poisson pro-cess are independent, exponentially distributed, random gamma distribution turns up in a few unexpected places.< > a standard normal distribution, with density .t/Dexp. t2=2/=p2 for 1<t<1, show thatZ2=2 has a gamma(1/2) :WriteYforZ2=2. It has a continuous distribution concentrated on the posi-tive half ;1/.Fory>0, and >0 small,Pfy<Y<yC gDPf2y<Z2<2yC2 gDPfp2y<Z<p2yC2 or p2yC2 <Z< p2ygNotice the two contributions; the square function is not one-to-one. Students who memorizeand blindly apply transformation formulae quite often overlook such multiple gives a good approximation to the length of the short interval fromp2ytop2yC2 . Temporarily Thenp2yC2 / =p2yThe interval from p2yC2 to p2yhas the same length. Using the approximationPfx<Z<xC g .x/for small >0;deduce thatPfy<Y<yC g p2y .p2y/C p2y . p2y/D2 p2y1p2 exp p2y 2=2 D p y 1=2e yThat is,Yhas the distribution with density1p y 1=2e yfory>0:Statistics 241: 28 October 1997c David PollardChapter 9 Poisson processesPage 4 Compare with the ,y1 1=2e >0:The distribution ofZ2=2 is gamma (1/2), as : From the fact that the density must integrate to 1, we get a 1e ydyDp Actually, you could arrive at the same conclusion by making the change of variableyDx2=2 in the integral which is effectively what we have done in finding the density for therandom variableZ2=2.

7 The Poisson process is often used to model the arrivals of customers in a waiting line,or the arrival of telephone calls at an exchange. The underlying idea is that of a large pop-ulation of potential customers, each of whom acts independently of all the others. The nextExample will derive probabilities related to waiting times for Poisson processes of part of the calculations we will need to find probabilities by conditioning on the valuesof a random variable with a continuous distribution. As before, the trick is first to conditionon a discretized approximation to the the variable, and then pass to a densityf. /, and letXbe another random variable. If 0, T<tC /Break the whole range forTinto small intervals. Rule E4 for expectations givesEXD1 XjD T<.jC1/ /Pfj T<.jC1/ gApproximate the last probability / . Temporarily ,wethen get1 XjD / as an approximation Think of the sum as an approximation toR1 tends to zero, the errors of approximation to both the expectation and the integral tendto zero, leaving (in the limit) each random variableX:As a special case, whenXis replaced by the indicator function of an event, we get< > each eventA;Rule E4 for expectations strikes again!

8 < > an office receives two different types of inquiry: persons who walkin off the street, and persons who call by telephone. Suppose the two types of arrival aredescribed by independent Poisson processes , with rate wfor the walk-ins, and rate cforthe callers. What is the distribution of the number of telephone calls received before the firstwalk-in customer?WriteTfor the arrival time of the first walk-in, and letNbe the number of calls in[0;T/. The timeThas a continuous distribution, with the exponential we wtfort>0:We need to calculatePfNDigforiD0;1;2;:::. Invoke formula< >, 241: 28 October 1997c David PollardChapter 9 Poisson processesPage 5 The conditional distribution ofNis affected by the walk-in process only insofar as that pro-cess determines the length of the time interval over whichNcounts. GivenTDt, the ran-dom variableNhas a Poisson . ct/conditonal distribution. ThusPfNDigDZ10e ct.]

9 Ct/ii! we wtdtD w ici!Z10 x cC w ie xdx cC wputtingxD. cC w/tD w cC w c cC w i1i!Z10xie xdxThe 1=i! and the last integral cancel. (Compare ) Writingpfor w=. cC w/we p/iforiD0;1;2;:::Compare with the The random variableNhas the distribution ofthe number of tails tossed before the first head, for independent tosses of a coin that landsheads with a nice clean result couldn t happen just by accident. Maybe we don t need all thecalculus to arrive at the distribution forN. In fact, the properties of the Poisson distributionand Problem show what is going on, as I will now the process of all inquiries, both walk-ins and calls. In an interval of lengtht,the total number of inquiries is the sum of a Poisson . wt/distributed random variable andan independent Poisson . ct/distributed random variable; the total has a Poisson . wtC ct/distribution. Both walk-ins and calls contribute independent counts to disjoint intervals; thetotal counts for disjoint intervals are independent random variables.

10 It follows that the pro-cess of all arrivals is a Poisson process with rate wC consider an interval of lengthtin which there areXwalk-ins andYcalls. FromProblem , given thatXCYDn, the conditional distribution ofXis ;p/, wherepD wt wtC ctD w wC cThat is,Xhas the conditional distribution that would be generated by the following mecha-nism:(1) Generate inquiries as a Poisson process with rate wC c.(2) For each inquiry, toss a coin that lands heads with probabilitypD w=. wC c/.Fora head, declare the arrival to be a walk-in, for a tail declare it to be a formal proof that this two-step mechanism does generate a pair of independent Pois-son processes , with rates wand c, would involve:(10) Prove independence between disjoint intervals. (Easy)(20) If step 2 generatesXwalk-ins andYcalls in an interval of lengtht, show thatPfXDi;YDjgDPfXDigPfYDjgX Poisson . wt/andY Poisson . ct/You should be able to write out the necessary conditioning argument for two-step mechanism explains the appearance of the geometric distribution in theproblem posed at the start of the Example.