STAT 870 Lecture 14
Properties of Poisson Processes
Then N is a Poisson process on [0,T] with rate
.
Indications of some proofs:
1)
independent Poisson processes rates
,
. Let
be the event of 2 or more points
in N in the time interval (t,t+h],
, the event of exactly
one point in N in the time interval (t,t+h].
Let
and
be the corresponding events for
.
Let
denote the history of the processes up to time t; we
condition on
. Technically,
is the
-field generated by
We are given:
and
Note that
Since
and
we have checked one of the two infinitesimal conditions for a Poisson process.
Next let
be the event of no points in
N in the time interval (t,t+h] and
the same
for
.
Then
shows
Hence N is a Poisson process with rate
.
2) The infinitesimal approach used for 1 can do part of this.
See text for rest. Events defined as in 1):
The event
that there is one point in
in (t,t+h] is the event,
that there is exactly one point in any of the r processes together
with a subset of
where there are two or more points in N in
(t,t+h] but exactly one is labeled i. Since
Similarly,
is a subset of
so
This shows each
is Poisson with rate
. 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
4): Fix
for
such that
Given N(T)=n we compute the probability of the event
Intersection of A,1 N(T)=n is (
):
whose probability is
So
Divide by
and let all
go to 0 to
get joint density of
is
which is the density of order statistics from a Uniform[0,T] sample of size n.
5) Replace the event
with
. With
A as before we want
Note that B is independent of
and that we have
already found the limit
We are left to compute the limit of
The denominator is
Thus
This gives the conditional density of
given
as in 4).
Inhomogeneous Poisson Processes
The idea of hazard rate can be used to extend the notion of Poisson
Process. Suppose
is a function of t. Suppose
N is a counting process such that
and
Then N has independent increments and N(t+s)-N(t) has a Poisson distribution with mean
If we put
then mean of N(t+s)-N(T) is
.
Jargon:
is the intensity or instantaneous intensity
and
the cumulative intensity.
Can use the model with
any non-decreasing right continuous
function, possibly without a derivative. This allows ties.
Space Time Poisson Processes
Suppose at each time
of a Poisson Process, N, we have rv
with the
iid and independent of the Poisson process. Let M be the counting
process on
(where
is the range space of the Ys)
defined by
Then M is an inhomogeneous Poisson process with mean function
a measure extending
This means that each M(A) has a Poisson distribution with mean
and
if
are disjoint then
are independent.
The proof in general is a monotone class argument. The first step is:
if
are disjoint intervals and
disjoint
subsets of
then the rs rvs
are independent
Poisson random variables. See the homework for proof of a special case.