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I have a Poisson process with parameter $\lambda$. I stop if $k$ events happen during the last unit of time. What is the expected time until I stop?

For example, I get an email on average every 30 minutes, how often (= how long do I have to wait in expectation) does it happen that at least 5 email come within 5 minutes long window?

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  • $\begingroup$ Look up Poisson Distribution and Gamma distribution $\endgroup$ Oct 25 '19 at 15:54
  • $\begingroup$ I have but it didn't help. I understand that the waiting time for 5 emails has a gamma distribution but I require that the 5 emails come within a 5-minute window of time. How does this fit into the picture? $\endgroup$ Oct 26 '19 at 11:00
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Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda >0$. Let $k>0$ be a positive integer and $T>0$ a positive time. Let $\tau=\inf\{t>T: N(t) - N(t-T) = k\}$. Then $$ \mathbb P(\tau>T) = \int_T^\infty T*e^{-\lambda T} \frac{(\lambda T)^k}{k!}\ \mathsf d T = \frac{\Gamma[k+2, \lambda T]}{\lambda ^2k!} = \frac{\int_{\lambda T}^\infty t^{k+1}e^{-t}\ \mathsf dt}{\lambda^2k!}, $$ so $$ \mathbb E[\tau] = \int_0^\infty \mathbb P(\tau>T)\ \mathsf dT = \frac1{\lambda^2 k!}\int_0^\infty\int_{\lambda T}^\infty t^{k+2}e^{-t}\ \mathsf dt = \frac1{\lambda^3}(k+3)(k+2)(k+1).$$

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    $\begingroup$ If I understand correctly, the integral is the probability that, starting at $j$-th trial, the next $k-1$ trials will be withing (at most) one unit of time. I do not see, however, how to get the expected waiting time from this. Am I missing something? $\endgroup$ Jan 10 '20 at 16:50
  • $\begingroup$ I have only now noticed that you have updated the answer after my comment. Thank you very much! There is one part that I don't get and that is the first equality. Why does it hold? Also, shouldn't some of the T's be t (not capitalized)? $\endgroup$ Apr 12 '20 at 8:28

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