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Suppose a Poisson process $N(t)\sim\text{Poisson}(\lambda t)$. Let $T(N)$ be the time of the last arrival before time $t$ and $T(N+1)$ be the time of the first arrival after time $t$. From simulations, I believe that $T(N+1)-T(N)$ is a Gamma$(2,\lambda)$ and I have a sense that it has to do with the fact that $T(N+1)-t\sim\exp(\lambda)$ and $t-T(N)\sim\exp(\lambda)$ due to memorylessness but I don't know how to show it.

Specifically, I think I want to show that $t-T(N)$ is exponential and then consider the sum of the (independent?) random variables $T(N+1)-t$ and $t-T(N)$. Note: I have heard this referred to as the "Hitchhiker's Paradox."

medley56
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  • What you are referring to is the stationary length bias distribution of a memoryless renewal process. In that sense, you are correct that the age and residual are both Exponential and their sum is Gamma. However, your problem is not defined well because $T(N)$ depends on some finite $t$, and for the result to hold you need to take $t\to\infty$ – QQQ Dec 10 '15 at 05:50
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    You can have a look here: http://www.columbia.edu/~ww2040/6711F12/lect1011.pdf. For the special case of exponential renewal times (Poisson Process) you will get your desired result. – QQQ Dec 10 '15 at 06:13
  • Hmm, I'm not confident that I understand the article you mention. The context of my question feels very different than everything presented there and I'm sure my claim can be proven in the framework of a simple homogeneous Poisson process. Also, I don't think the result relies on $t\to\infty$ as I obtained a very good fit to a Gamma distribution from simulation with $t=3$, $\lambda=2$. – medley56 Dec 12 '15 at 19:32
  • So your question is not well defined because it lacks initial conditions. Suppose $t$ is very small, there is a very big chance that there were no arrivals before it. What does $T(N)$ equal then? – QQQ Dec 13 '15 at 06:05

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