# Expected waiting time for $k$ arrivals at most $t$ apart in a poisson process

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?

• Look up Poisson Distribution and Gamma distribution Oct 25 '19 at 15:54
• 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? Oct 26 '19 at 11:00

## 1 Answer

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 $$\mathbb E[\tau] = \int_0^\infty \mathbb P(\tau>T)\ \mathsf dT = \frac1{\lambda^2 k!}\int_0^\infty\int_{\lambda T}^\infty t^{k+2}e^{-t}\ \mathsf dt = \frac1{\lambda^3}(k+3)(k+2)(k+1).$$

• 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? Jan 10 '20 at 16:50
• 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)? Apr 12 '20 at 8:28