# Calculating distribution of Poisson process at time t when a future value is known

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?

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?

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 $$p$$.
• 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? Commented Nov 13, 2022 at 3:04