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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?

tern
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1 Answers1

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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.

pre-kidney
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    Thanks a lot for the comprehensive answer.
    It is interesting that process has independent increments, but interarrival times are dependent.
    – tern Aug 12 '19 at 09:16
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    Good answer. Could use please explain the notation $E[e^{−μ([T,T+t])};T>s]$? Especially ";", which I guess is not to indicate conditional on. – TOMILO87 Dec 17 '19 at 19:45