Interarrival times of nonhomogeneous Poisson process

2.3k Views Asked by At

It is well known that interarrival times of homogeneous Poisson process are independent and exponentially distributed.

But how about interarrival times of nonhomogeneous Poisson process:
- are they still independent random variables?
- what is their joint distribution?

Could you reccommend a textbook related to that question?

1

There are 1 best solutions below

2
On BEST ANSWER

Let $\mu$ be the reference measure of the Poisson process. By definition, this means that the number of arrivals falling into any $n$ mutually disjoint sets $B_1,\ldots,B_n$ has the same distribution as independent Poisson random variables with means $\mu(B_1),\ldots,\mu(B_n)$.

Specializing to the case of a Poisson process on $[0,\infty)$ (so that the interarrival times are well-defined) we see that if $T$ denotes the first arrival, then $$\mathbb P(T>t)=\mathbb P(0\text{ arrivals in }[0,t])=e^{-\mu([0,t])},$$ where the last equality uses the probability of a Poisson variable with mean $\mu([0,t])$ being zero.

You can see from this that if $\mu$ is not a constant multiple of Lebesgue measure, then $T$ is no longer an exponential random variable.

The interarrival times are no longer independent in general, since if $T'$ denotes the next interarrival time after $T$ then $$ \mathbb P(T'>t\mid T)=e^{-\mu([T,T+t])}. $$ Whenever $\mu$ is not a constant multiple of Lebesgue measure, this conditional probability depends on $T$, and thus $(T,T')$ are not independent.

The joint distribution of $(T,T')$ satisfies $$ \mathbb P(T'>t,T>s)=\mathbb E[e^{-\mu([T,T+t])};T>s], $$ you can not go too much farther without knowing more about the reference measure $\mu$.

Kallenberg's textbook is a good reference for Poisson process questions.