Conditional Expectation of Poisson Process?

Question

Fix s,t such that s < t. If we are given that $N(t)=n$ then what is $\mathbb{E}[N(s)|N(t)=n]$? Intuitively I would imagine a uniform distribution so that $\mathbb{E}[N(s)|N(t)=n]=\frac{ns}{t}$ But is this correct? And if so, how would I prove it?

Answer 1

Note that given $N(t) = n$, $N(s)$ can take only the values $0,1,...,n$, because $N$ has increasing paths. Now, for each $ 0 \leq k \leq n$ integer, $$P[N(s) = k | N(t) = n] = \frac{P(N(s) =k , N(t) = n)}{P(N(t) = n)} = P(N(s) = k) \frac{P(N(t) - N(s) = n-k)}{P(N(t) = n)}$$

where you know that $N(t) - N(s)$ is a Poisson random variable independent of $N(s)$ which is also Poisson. Thus one can calculate the above ratio.

Finally, $$ \mathbb E[N(s) | N(t) = n] = \sum_{k=0}^n k P[N(s) = k | N(t) = n] $$

so we can calculate this sum as well.

Answer 2

Your guess is right. This follows from the uniformity of the Poisson process: The distribution of the number of points in a subset $S$ of $[0,t]$ depends only on the measure of $S$, and, given the number of points in $[0,t]$, this is only compatible with the linearity of expectation if the expectation is proportional to the measure of $S$.

(If you want a more rigorous proof, this is straightforward if $\frac st$ is rational, since we can then divide $[0,t]$ into equal parts, some of which make up $[0,s]$. For $\frac st$ irrational, you might have to make a limit argument using countable additivity.)