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I observe a Poisson process for a finite time interval. Let $X_i$ denote the time of the $i^{th}$ arrival. I start my timer as soon as I get my first event so that $X_0:=0$ and store the time-stamps $0=:X_0<X_1<X_2<\ldots <X_{N(T)}<T$ for a random number of arrivals $N(T)$ in the time interval $[0,T)$.

Let $Y_k = X_{k}-X_{k-1}$, $1\leq k \leq N(T)$, be the sequence of interarrival times. How do I find the distribution of the mean interarrival time: $m(T):=\frac{1}{N(T)}\sum_{i=1}^{N(T)} Y_i = \frac{X_{N(T)}}{N(T)}$?

(If $N(T)=0$, I define $m(T)=\infty.$)

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  • $\begingroup$ There is not a nice closed form for the distribution of $X_{N(T)}$. See here: stats.stackexchange.com/questions/273902/… $\endgroup$ – Math1000 May 10 '18 at 2:35
  • $\begingroup$ Thanks @Math1000. I was a bit confused if the method in that stats.SE link works here because I recall somewhat vaguely a result that says that conditioned on the number of events, the Poisson arrival times look like a uniform distribution. $\endgroup$ – Atul Ingle May 10 '18 at 13:46
  • $\begingroup$ Yes - conditioned on $\{N(t) = n\}$, the joint distribution of the order statistics of the $n$ arrival times is the same as the joint distribution of $n$ $\mathsf{Unif}(0,t)$ random variables. But that is the arrival times, not the interarrival times. $\endgroup$ – Math1000 May 10 '18 at 14:52

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