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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Community, E. Joseph, user26857, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.


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 $t$. $$P(S_1\le x\ |\ N_t=1)={P(S_1\le x,N_t=1)\over P(N_t=1)}\\={P(N_x= 1,N_t-N_x=0)\over P(N_t=1)}\\={P(N_x= 1)P(N_t-N_x=0)\over P(N_t=1)}\\={P(N_x= 1)P(N_{t-x}=0)\over P(N_t=1)}\\={\alpha xe^{-\alpha x}\cdot e^{-\alpha (t-x)}\over \alpha te^{-\alpha t}} ={x\over t} $$

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.