Let $P$ be a Poisson point process with rate $\lambda$. If it is known that $P(t) = n$, how can we retroactively derive the conditional distribution of $P(k)$, where $k=t-s$ for $s<t$?

My idea: The expected value of $P(k)$ is $n-s\cdot \mathbb{E}(f(x; \lambda)) = n-s\cdot \frac{1}{\lambda}$, where $f$ is the exponential distribution. Since we know a priori that $P(k)$ follows a Poisson distribution (whose sole parameter is identically its expected value), we conclude that $$P(k) \sim \text{Pois}(n-\frac{s}{\lambda}) = \frac{(n-\frac{s}{\lambda})^ke^{-(n-\frac{s}{\lambda})}}{k!}$$ However, this seems to contradict the assumption that $P$ had rate $\lambda$ to begin with. How should I be approaching this problem?


1 Answer 1


Given that $P(t)=n$, the conditional distribution of $P(k)$ is binomial distributed with $n$ trials and success probability $k/n$, since the $n$ events are equally likely to have occured at any time prior to time $t$. After conditioning on $P(t)$, the original rate $\lambda$ becomes irrelevant.

Another way to reach the same conclusion is to note that $$P(t) = P(s) + P(k)$$ where $P(s)$ is the number of events in the time interval $(0,s)$ and $P(k)$ is the number of events in the interval $(s,t)$. The Poisson process assumption tells us that $P(s)$ and $P(k)$ are independent with $P(s) \sim {\rm Poi}(s\lambda)$ and $P(k) \sim {\rm Poi}(k\lambda)$. It follows from basic probability that the conditional distribution of one part given the total is binomial: $$P(k) | P(t) \sim {\rm Bin}(n=P(t),p=k/(s+k)),$$ with $\lambda$ cancelling out of the expression for $p$.

  • $\begingroup$ Interesting! So it is equally likely for the $n$ successes to occur precisely at times $t, t-1, ... t-(n-1)$ as it is for them to be "uniformly" spaced among the $t$ timesteps? $\endgroup$
    – user830529
    Commented Nov 13, 2022 at 3:04
  • $\begingroup$ @user830529 Poisson processes are memoryless so, retrospectively, every possible set of event times is equally likely. $\endgroup$ Commented Nov 13, 2022 at 3:16
  • $\begingroup$ @user830529 Have I answered your question, or were you after anything else? $\endgroup$ Commented Nov 14, 2022 at 6:24
  • $\begingroup$ You have answered my question! Thank you for you help. $\endgroup$
    – user830529
    Commented Nov 15, 2022 at 7:15
  • $\begingroup$ @user830529 Glad to hear it helped you, but note the convention on StackExchange is to acknowledge answers by voting or accepting, see stats.stackexchange.com/help/someone-answers $\endgroup$ Commented Nov 15, 2022 at 7:27

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