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I'll ask this with a concrete example to be clear.

Let's say I have a Poisson process that tends to produce one event every two minutes. Then the probability of getting an event in a given minute is about 30%, and the probability of getting two events in a given minute is about 7.6%, assuming I'm applying the distribution correctly.

I'm interested in the time distribution between groups of events: How long do I have to wait to see two events within a minute? What is the general form for this?

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  • $\begingroup$ Call $(t_1,s_1)$ and $(t_2,s_2)$ the first intervals smaller than $1$ minute. Are you asking for the distribution of $s_2-s_1$? $\endgroup$
    – Did
    Jul 21, 2015 at 21:36
  • $\begingroup$ Are you asking for how long to wait until you see one event and then another event within 1 minute after the first event? $\endgroup$
    – user2566092
    Jul 21, 2015 at 21:41
  • $\begingroup$ Yes, both of those are better ways of stating the problem. In general, what is the distribution s_2 - s_1? Specifically, what is the mean of that distribution (how long to wait until I see one event then another event within a minute). $\endgroup$
    – Dan
    Jul 21, 2015 at 21:50
  • $\begingroup$ See en.m.wikipedia.org/wiki/Exponential_distribution for this $\endgroup$
    – David Quinn
    Jul 21, 2015 at 22:02
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    $\begingroup$ Call $D=s_2-s_1$ and $X=t_2-s_1$. Conditioning on $X$, one sees that $D=X$ if $X<1$ and that $D=X+D'$ if $X>1$, where $D'$ is distributed like $D$ and independent of $D$. From here, the Laplace transform is $$E(e^{-xD})=\lambda\frac{1-e^{-\lambda-x}}{\lambda+x-\lambda e^{-\lambda-x}}.$$ Inverting this to recover the density requires more work but the mean is direct (differentiate the L transform at $x=0$). $\endgroup$
    – Did
    Jul 21, 2015 at 22:11

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I have been following this Question for about a day now, and am still not exactly sure what you are asking. Here are some comments based on sequences of one-minute time intervals, for which the number of $X$ of events in one minute is Poisson with mean 1/2.

EXACTLY ONE event in such an interval: $P(X = 1) = 0.3032653.$ Geometric mean waiting time for the first interval with exactly one event: reciprocal = 3.297443. By independence, given you have just seen such an interval, this is the waiting time for the next one. Negative binomial mean waiting time for two such intervals: double = 6.594885.

AT LEAST ONE event in such an interval: $P(X \ge 1) = 1 - P(X = 0) = 0.3934693.$ Geometric mean waiting time for the first interval with at least one event: 2.541494. Negative binomial mean waiting time for two such intervals: 5.082988. Perhaps coincidentally, this is the same as in the Comment where @Did speculates $E(D) = 1/[\lambda(1-e^{-\lambda})] = 5.082988,$ for $\lambda = 1/2.$

AT LEAST TWO events in such an interval: $P(X \ge 2) = 0.09020401.$ Geometric mean waiting time for the first interval with at least two events: 11.08598.

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