Chapter 1 Poisson Processes - NYU Courant

1 Chapter 1 Poisson The Basic Poisson ProcessThe Poisson Process is basically a counting processs. A Poisson Process onthe interval [0, ) counts the number of times some primitive event hasoccurred during the time interval [0, t]. The following assumptions are madeabout the Process N(t).(i). The distribution ofN(t+h) N(t) is the same for eachh >0, isindependent oft.(ii). The random variablesN(t j) N(tj) are mutually independent if theintervals [tj, t j] are nonoverlapping.(iii).N(0) = 0,N(t) is integer valued, right continuous and nondecreasingint, with Probability 1.(iv).P[N(t+h) N(t) 2 ] =P[N(h) 2 ] =o(h) ash the above assumptions, the processN( )has the fol-lowing additional properties.(1). With probability1,N(t)is a step function that increases in steps ofsize1.]

2 (2). There exists a number 0such that the distribution ofN(t+s) N(s)has a Poisson distribution with parameter 1. Poisson Processes (3). The gaps 1, 2, between successive jumps are independent identicallydistributed random variables with the exponential distributionP{ j x}={exp[ x] forx 01forx 0( ) us divide the interval [0, T] intonequal parts and compute theexpected number of intervals withN((k+1)Tn) N(kTn) 2. This expectedvalue is equal tonP[N(Tn) 2]=n . o(1n) =o(1) asn there by proving property (1).Because of right continuity we haveP[N(t) 1] 0 ast 0proving that the distribution ofN(t) is infinitesimal ast 0. By theindependence of the increments over disjoint intervals,N(t) is approximatelythe sum of [nt] independent copies ofN(1n).E{exp[ N(t)]}= limn E{exp[ N([nt]n)]}= limn [E{exp[ N(1n)]}][nt]( )= exp[ tg( )]( )whereg( ) = limn n[1 [E{exp[ N(1n)]}]]= limn n[E{1 exp[ N(1n)]}]= limn n[(1 e )P[N(1n) = 1] +o(1n)]= (1 e )( )where = limn n P[N(1n) = 1].}

3 ( ) THE BASIC Poisson PROCESS3 The limit in ( ) must necessarily exist because the limit in ( ) clearlyexists. The positivity ofE{exp[ N(t)]}and the identity ( ) guaranteesthe finiteness of the limit in ( ). The formulas ( ) and ( ) togetheridentify the distribution ofN(t) as the Poisson distribution with parameter t, thereby proving property (2).Finally, we turn to the proof of prpoerty (3). First let us prove that 1has the right [ 1> x] =P[N(x) = 0 ] =e xbecause of the Poisson distribution. We will prove 1is a regenerative timein the sense thatN( 1+t) N( 1) =N( 1+t) 1 is again a Poisson Processindependent of 1. This will prove that 2will have the same distribution as 1and will be independent of it. Repeating the step and induction onnwillcomplete the proof. Let be a stopping time that takes a countable set ofvalues{vj}.

4 Since the set =vjis measurable with respect to {N(t) :t vj}, the processN(t+vj) N(vj) is independent of the set =vj, and isconditionally again a Poisson Process. Therefore for such a ,N( +t) N( )is again a Poisson process independent of . Finally, 1is a stopping timeand for anyk, (k)=[k 1]+1kis a stopping time that takes only a countablenumber of values. ThereforeN( (k)+t) N( (k)) is a Poisson Process withparameter that is independent of (k). We letk . Because (k) 1,and the process is right continuous we are we have proved is that ifPis a measure on the space ofnondecreasing right continuous functions on [0, ) satisfying the propertieslisted in the assumptions of Theorem , thenP=P is determined by asingle parameter and the additional properties (1)-(3) of Theorem willbe valid for it.]

5 The processP is referred to as the Poisson process withparameter or rate . proof of property (3) of Theorem Let 0 bearbitrary. The processesM (t) = exp[ N(t) t(e 1)]are martingales with repect to ( ,Ft, P) and Doob s stopping theorem pro-vides the relationE{M ( +t)|F }=M ( ) it into a proof that 1is 1. Poisson that for any Poisson process with parameter N(t) and [N(t) t]2 tare martingales with respect to ( ,Ft, P), whereFt= {N(s) : 0 s t} distribution of 1+ + k+1is a Gamma distributionwith density kk!e xxk 1. Why does it look like the Poissson probability forkjumps? Compound Poisson thatX1, X2 , Xn is a sequence of independent identically dis-tributed random variables with a common distribution having partial sumsSn=X1+X2+ +Xn. We define a procesX(t) byX(t) =SN(t).

6 Each time the Poisson ProcessN(t) jumps by 1, the processX(t) jumps bya random amount which has distribution . The jumps at different instancesare independent. Such a processX(t) inherits the independent incrementsproperty fromN(t). For any collection of nonoverlapping intervals [tj, t j] theincrementsX(t j) X(tj) are independent random variables. The distributionof any incrementX(t+s) X(s) is that ofXN(t), and is calculated to bethe distribution ofSnwherenis random and has a Poisson distribution withparameter {exp[i y X(t)]}= je t( t)jj![ (y)]j=e te t (y)=e t[ (y) 1]= exp[ t (ei y x 1)d (x)]Inother wordsX(t) has an infinitely divisible distribution with a Levy mea-sure given by t . If we denote byM= thenE{exp[i y X(t)]}= exp[t (ei y x 1)dM(x)] INFINITE NUMBER OF SMALL :Assuming that has two moments, show that are constantsAandBsuch thatX(t) Atand [X(t) At]2 Btare martingales.

7 ComputeAandBin terms of and . [A= x dM= x d ;B= x2dM= x2d ] Infinite number of small as a Poisson process cannot have an infinite number of jumps in afinite interval, once we start considering compound Poissonprocesses we canin principle sum an infinite number of small jumps so that we still have afinite answer. For example supposeXk(t) is a compound Poisson Processthat corresponds to k k= {Xk(t)}=t x dMk(x)andV ar[Xk(t)] =t x2dMk(x)Let us takeXk(t) to be mutually independent Processes with independentincrements and try to sum upX(t) = kXk(t)for eacht. If the sum exists then it is a process with independent fact we can do a little better. We may center these Processes with suitableconstants{ak}, and writeX(t) = k[Xk(t) akt]to help in convergence. We shall assume that k x21 +x2dMk(x)< which amounts to two conditions kMk[|x| 1]< 6 Chapter 1.

8 Poisson PROCESSESand k |x| 1x2dMk(x)< We can always decompose anyMasM1+M2and this will result in a de-composition of the processX(t) =X1(t) +X2(t), two mutually independentprocesses with independent increments corresponding toM1andM2respec-tively. It is better for us to decompose eachMkasM(1)k+M(2)kcorrespondingto jumps of size|x| 1 and|x|> kM(2)ksums to a finite measure and offers no difficulty. The processX(2)(t) = kX(2)k(t)exists very nicely becauseP{X(2)k(t)6= 0} 1 e tM(2)k[|x|>1 ] tM(2)k[|x|>1]and kP{X(2)k(t)6= 0}< Borel-Cantelli does it! As for a discussion of kX(1)k(t), it is now clearthat we can assume thatM(1)k[|x|>1 ] = 0 for everyk. If we takeak=E{Xk(1)}= x dM(1)k(x),E{[X(1)k(t) akt]2}= x2dM(2)k(x)and by the two-series theorem k[X(1)k(t) akt]converges almost surely.

9 A simple application of Doob s inequality yieldsin fact almost sure uniform convergence in finite time intervals. If we nowreassemble the pieces we see INFINITE NUMBER OF SMALL {ei y X(t)}= exp[t |x| 1(ei y x 1 i y x)dM(x) + |x|>1(ei y x 1)dM(x)]which is essentially the same as Levy-Khintchine representation for infinitelydivisible distributions except for the Gaussian term that is missing. That isthe continuous part, that cannot be built from jumps and needs the BrownianMotion or Wiener 1. Poisson PROCESSESC hapter 2 Continuous time Jump Markov we have a Markov Chain{Xn}on a state spaceX, with transition probabil-ities (x , dy), and a Poisson ProcessN(t) with intensity , we can combinethe two to define a continuous time Markov processx(t) withXas statespace by the formulax(t) =XN(t)The transition probabilities of this Markov process are given byPx{x(t) A}=P{x(t) A|x(0) =x}=p(t , x , A)= k=1e t( t)kk!

10 (k)(x , A)Because of the Markov property of{Xn}and the independence of theincrements ofN(t) it is not difficult to see thatP{x(t) A|Fs}=P{XN(t) N(s) A|X0=XN(s)}=P{XN(t s) A|X0=x(s)}=p(t s , x(s), A)thereby proving the Markov property ofx(t).910 Chapter 2. CONTINUOUS TIME MARKOV PROCESSESIn general, if we want to define a Markov process on a state spaceX, weneed the transition probabilitiesp(t , x , A) defined fort >0, x XandA B, a - field of measurable subsets ofX. It is natural to definep(0, x , A) = A(x). For eacht >0,p(t , , ) will be a transition probability fromX Xand the collection will have to satisfy the Chapman-Kolmogorov equationsp(t+s , x , A) = Xp(s , y , A)p(t , x , dy)Given such a collection, for each starting pointx X, we can define a con-sistent family of finite dimensional distributions on the space of trjectories{x( ) : [0, ) X}.]