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Consider a Poisson process. Given that a single arrival occurred in a given interval [0,t], why is the resulting distribution for the arrival time uniform?

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closed as off-topic by Did, user223391, E. Joseph, user26857, Watson Nov 5 '16 at 15:51

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Let the arrival times be $\sim \exp(\alpha).$ Denote $S_1$ to be the time of the first arrival,and $N_t $ to be the number of arrivals upto time $t$. $$P(S_1\le x\ |\ N_t=1)={P(S_1\le x,N_t=1)\over P(N_t=1)}\\={P(N_x= 1,N_t-N_x=0)\over P(N_t=1)}\\={P(N_x= 1)P(N_t-N_x=0)\over P(N_t=1)}\\={P(N_x= 1)P(N_{t-x}=0)\over P(N_t=1)}\\={\alpha xe^{-\alpha x}\cdot e^{-\alpha (t-x)}\over \alpha te^{-\alpha t}} ={x\over t} $$

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  • $\begingroup$ Shouldn't the final expression evaluate to x/t? $\endgroup$ – John Smith Nov 5 '16 at 10:48
  • $\begingroup$ @JohnSmith I make too many typos! Thanks! $\endgroup$ – Qwerty Nov 5 '16 at 10:54
  • $\begingroup$ How did you arrive at the expressions for $P(N_x= 1)$ and $P(N_{t-x}=0)$ in the last line? $\endgroup$ – John Smith Nov 5 '16 at 12:14
  • $\begingroup$ @JohnSmith In a Poission process with rate $\alpha$ , it is a well known fact that by the way it is defined, $N_t\sim \text{Poission} (\alpha t)$ $\endgroup$ – Qwerty Nov 5 '16 at 12:17

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