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It is obvious that given that in the interval $(0,t)$ the number of arrivals following Poisson process is $N(t)=n$, these $n$ arrivals are independently and uniformly distributed in the interval. It is also clear that the interarrival time of the events following poisson process follows the exponential distribution. Here is my point. I think If $n$ arrivals follows uniform distribution, the interarrivail time of them should follow uniform distribution as well. Am I doing wrong? Anybody have an Idea, pls explain it to me.

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    $\begingroup$ "If n arrivals follows uniform distribution, the interarrivail time of them should follow uniform distribution as well" Frankly wrong. Did you try to check the case n=2? None of the three intervals is uniform... $\endgroup$ – Did Oct 8 '15 at 9:01
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The joint distribution of inter-arrival times is uniform on $\{x_1+\dots + x_n<t\}$ conditionally on the fact that $N(t) = n$. Unconditionally, they are independent and have exponential distribution. There is no contradiction; you can try to verify this fact yourself, it's not that hard.

Moreover, as @Did wrote, while the joint (conditional) distribution of the inter-arrival times is uniform, the individual (conditional) distribution of each inter-arrival time is not.

The reason of this non-uniformity comes out of the fact that, conditionally on $N(t) = n$, the inter-arrival times are no longer independent. If they were, then, naturally, you would be able to deduce the uniformity of marginal distributions from that of their joint distribution.

Another way to put this. The conditional distribution of the arrival times is uniform on $[0,t]^n$, so indeed they can be understood as iid uniform random variables on $[0,t]$. However, if you are interested in the inter-arrival times, then you should first arrange the arrival times in the increasing order. And at this point you already lose independence of individual components.

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  • $\begingroup$ What is the reason of voting the answer down? Can the downvoter explain? $\endgroup$ – zhoraster Oct 8 '15 at 11:41
  • $\begingroup$ No idea. +1 from me as soon as I am not out of votes anymore. $\endgroup$ – Did Oct 12 '15 at 20:31
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Following @zhoraster's answer, the interarrival time distribution of n arrivals uniform in $[0,T]$ is equal the 1st arrival time distribution by symmetry - the marginal distribution of time between any 2 consecutive arrivals is an integral of the uniform joint distribution over the same volume. Therefore, the interarrival time cdf equals the 1st arrival time cdf $F(t;n)=F_\mathrm{Beta}(\frac{t}{T};1,n)$.

We can recover the exponential cdf from $\mathrm{lim}_{T\rightarrow\infty} F(t;\lambda T)$. Alternatively, $\frac{\sum_{i=1}^\infty f_{Poi}(i; \lambda T) F(t;i)}{\sum_{i=1}^\infty f_{Poi}(i; \lambda T)}$ gives the $[0,T]$-truncated exponential cdf as the 1st arrival time cdf conditional on 1st arrival being before $T$; $\sum_{i=1}^\infty f_{Poi}(i; \lambda T) F(t;i) + f_{Poi}(0; \lambda T) F_T(t;\lambda)$, with $F_T(t;\lambda) = \frac{F_{Exp}(t;\lambda)-F_{Exp}(T;\lambda)}{1-F_{Exp}(T;\lambda)}H(t-T)$ being the $[T,\infty]$-truncated exponential cdf, gives the exponential unconditional interarrival time cdf.

Conversely, if n arrivals in [0,T] have exponential interarrival times, the conditional pdf of the i-th arrival time is the product of the pmf of n arrivals conditional on the i-th arrival time and the Erlang pdf of the i-th arrival time, divided by the Poisson pmf of n arrivals:

$$ f_i(t;n)=\frac{f_{Poi}(n-i;\lambda (T-t))}{f_{Poi}(n;\lambda T)}f_{Erlang}(t;i,\lambda) $$

The uniform pdf is recovered as the conditional arrival time pdf from mean $f_i(t;n)$.

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