I know that Poisson process is the probability of occurring an event in an interval, on the other hand, I know that the probability of the time of occurrences is uniform and the probability of waiting time is exponential distribution, but how is it even possible when the time of occurrence is uniform, the time between events is exponential?

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    $\begingroup$ I suspect you may mean "Poisson process" rather than "Poisson distribution". It leads to exponentially distributed waiting times because it is memoryless $\endgroup$ – Henry Dec 19 '16 at 21:27
  • $\begingroup$ Thanks @Henry, I'm bit confused, I read coalescent model in population genetics, in that model we generate a sample from Poisson distribution and the times of events in that Poisson are chosen uniformly, on an equivalent approach, the distance between events can be generated of an exponential distribution and I cannot connect this two approach, for more information in this link slides number 8 and 11. $\endgroup$ – user137927 Dec 19 '16 at 21:39
  • $\begingroup$ Following Henry's comment, one way to define the Poisson process is to define its interarrivals to be iid and exponentially distributed with rate $\lambda$. It can be proven that the number of arrivals in a given time $T$ has a Poisson distribution with mean $\lambda T$. Also, it can be proven that, conditioned on the number of arrivals being $k$, they are independently and uniformly located over the interval $[0,T]$. This conditioning changes the interarrival time distributions since, given there are $k$ arrivals, we know their interarrival times are constrained to sum to less than $T$. $\endgroup$ – Michael Dec 19 '16 at 21:45
  • $\begingroup$ You can reverse the procedure: Given an interval $[0,T]$, you can generate a random integer $K$ that is Poisson distributed with mean $\lambda T$, i.e., $P[K=k]=e^{-\lambda T}\frac{(\lambda T)^k}{k!}$. Given that $K=k$, independently and uniformly place the $k$ arrivals over the interval $[0,T]$. The result looks like a Poisson process of rate $\lambda$ viewed over the finite interval $[0,T]$. $\endgroup$ – Michael Dec 19 '16 at 21:47
  • $\begingroup$ Thanks @Michael you helped me a lot. $\endgroup$ – user137927 Dec 19 '16 at 21:49

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