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STAT 870 Lecture 14

Properties of Poisson Processes

1)
If tex2html_wrap_inline178 and tex2html_wrap_inline180 are independent Poisson processes with rates tex2html_wrap_inline182 and tex2html_wrap_inline184 , respectively, then tex2html_wrap_inline186 is a Poisson processes with rate tex2html_wrap_inline188 .

2)
Suppose N is a Poisson process with rate tex2html_wrap_inline192 . Suppose each point is marked with a label, say one of tex2html_wrap_inline194 , independently of all other occurrences. Suppose tex2html_wrap_inline196 is the probability that a given point receives label tex2html_wrap_inline198 . Let tex2html_wrap_inline200 count the points with label i (so that tex2html_wrap_inline204 ). Then tex2html_wrap_inline206 are independent Poisson processes with rates tex2html_wrap_inline208 .

3)
Suppose tex2html_wrap_inline210 independent rvs, each uniformly distributed on [0,T]. Suppose M is a Poisson tex2html_wrap_inline216 random variable independent of the U's. Let

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Then N is a Poisson process on [0,T] with rate tex2html_wrap_inline192 .

4)
Suppose N is a Poisson process with rate tex2html_wrap_inline192 . Let tex2html_wrap_inline232 be the times at which points arrive Given N(T)=n, tex2html_wrap_inline236 have the same distribution as the order statistics of a sample of size n from the uniform distribution on [0,T].

5)
Given tex2html_wrap_inline242 , tex2html_wrap_inline236 have the same distribution as the order statistics of a sample of size n from the uniform distribution on [0,T].

Indications of some proofs:

1) tex2html_wrap_inline206 independent Poisson processes rates tex2html_wrap_inline252 , tex2html_wrap_inline254 . Let tex2html_wrap_inline256 be the event of 2 or more points in N in the time interval (t,t+h], tex2html_wrap_inline262 , the event of exactly one point in N in the time interval (t,t+h].

Let tex2html_wrap_inline268 and tex2html_wrap_inline270 be the corresponding events for tex2html_wrap_inline200 .

Let tex2html_wrap_inline274 denote the history of the processes up to time t; we condition on tex2html_wrap_inline274 . Technically, tex2html_wrap_inline274 is the tex2html_wrap_inline282 -field generated by

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We are given:

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and

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Note that

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Since

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and

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we have checked one of the two infinitesimal conditions for a Poisson process.

Next let tex2html_wrap_inline294 be the event of no points in N in the time interval (t,t+h] and tex2html_wrap_inline300 the same for tex2html_wrap_inline200 . Then

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shows

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Hence N is a Poisson process with rate tex2html_wrap_inline306 .

2) The infinitesimal approach used for 1 can do part of this. See text for rest. Events defined as in 1): The event tex2html_wrap_inline270 that there is one point in tex2html_wrap_inline200 in (t,t+h] is the event, tex2html_wrap_inline262 that there is exactly one point in any of the r processes together with a subset of tex2html_wrap_inline256 where there are two or more points in N in (t,t+h] but exactly one is labeled i. Since tex2html_wrap_inline326

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Similarly, tex2html_wrap_inline268 is a subset of tex2html_wrap_inline256 so

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This shows each tex2html_wrap_inline200 is Poisson with rate tex2html_wrap_inline336 . To get independence requires more work; see the homework for an easier algebraic method.

3) Fix s;SPMlt;t. Let N(s,t) be the number of points in (s,t]. Given N=n the conditional distribution of N(s,t) is Binomial(n,p) with p=(s-t)/T. So

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4): Fix tex2html_wrap_inline352 for tex2html_wrap_inline354 such that

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Given N(T)=n we compute the probability of the event

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Intersection of A,1 N(T)=n is ( tex2html_wrap_inline366 ):

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whose probability is

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So

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Divide by tex2html_wrap_inline370 and let all tex2html_wrap_inline372 go to 0 to get joint density of tex2html_wrap_inline236 is

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which is the density of order statistics from a Uniform[0,T] sample of size n.

5) Replace the event tex2html_wrap_inline382 with tex2html_wrap_inline384 . With A as before we want

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Note that B is independent of tex2html_wrap_inline390 and that we have already found the limit

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We are left to compute the limit of

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The denominator is

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Thus

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This gives the conditional density of tex2html_wrap_inline236 given tex2html_wrap_inline382 as in 4).

Inhomogeneous Poisson Processes

The idea of hazard rate can be used to extend the notion of Poisson Process. Suppose tex2html_wrap_inline402 is a function of t. Suppose N is a counting process such that

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and

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Then N has independent increments and N(t+s)-N(t) has a Poisson distribution with mean

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If we put

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then mean of N(t+s)-N(T) is tex2html_wrap_inline422 .

Jargon: tex2html_wrap_inline192 is the intensity or instantaneous intensity and tex2html_wrap_inline426 the cumulative intensity.

Can use the model with tex2html_wrap_inline426 any non-decreasing right continuous function, possibly without a derivative. This allows ties.

Space Time Poisson Processes

Suppose at each time tex2html_wrap_inline430 of a Poisson Process, N, we have rv tex2html_wrap_inline434 with the tex2html_wrap_inline434 iid and independent of the Poisson process. Let M be the counting process on tex2html_wrap_inline440 (where tex2html_wrap_inline442 is the range space of the Ys) defined by

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Then M is an inhomogeneous Poisson process with mean function tex2html_wrap_inline450 a measure extending

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This means that each M(A) has a Poisson distribution with mean tex2html_wrap_inline456 and if tex2html_wrap_inline458 are disjoint then tex2html_wrap_inline460 are independent. The proof in general is a monotone class argument. The first step is: if tex2html_wrap_inline462 are disjoint intervals and tex2html_wrap_inline464 disjoint subsets of tex2html_wrap_inline442 then the rs rvs tex2html_wrap_inline470 are independent Poisson random variables. See the homework for proof of a special case.


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Richard Lockhart
Tuesday October 31 11:03:24 PST 2000