Let {$X(t);t\geq 0$} be a Poisson process with parameter $\lambda=2$.
Find $E[X(s)|X(t)=n]$, where $0<s<t$.
I really am unsure how to solve this problem. I know $E[X(s)]=\lambda s$. I think $E[X(t)=n]=\frac{n}{\lambda}$ since $X(t)$ is a Poisson process (conditioned on n?).
I think this becomes $\frac{E[X(s),X(t)=n]}{E[X(t)=n]}=\frac{sn}{\frac{n}{\lambda}}=s\lambda$ but this does not account for the condition $0<s<t$ and probably isn't the correct application of conditional expected value. I'm pretty sure this needs to be an integral over the interval $(0,t)$ but I'm unsure how to set it up.
Any suggestions are appreciated.
$E[X(s)| X(t) = n]$ is the sum of $x P[X(s)=x | X(t)=n]$ for every value $x = X(s)$ with non-zero conditional probability. Since $0<s<t$, the number of occurrences $X(s)$ will be between 0 and $X(t) =n$, so $$ E[X(s)| X(t) = n] = \sum_{x = 0}^{n}{x P[X(s) = x|X(t) = n]} $$ Now the problem is to determine $P[X(s) = x|X(t) = n]$. Using Bayes' theorem, $$P[X(s) = x|X(t) = n] = P[X(t) = n|X(s) = x] \frac{P[X(s) = x]}{P[X(t) = n]}$$ Notice $P[X(s) = x] = (\lambda s)^x e^{-(\lambda s)}/x!$ and $P[X(t) = n] = (\lambda t)^n e^{-(\lambda t)}/n!$. Also, remember that the number of occurrences in Poisson processes are independent for disjoint time intervals, so even when we are given that $X(s) = x$, the occurrences between time $s$ and $t$ will still be Poisson distributed with same mean, therefore $$P[X(t) = n|X(s) = x] = P[X(t-s) = n-x] = (\lambda (t-s))^{n-x} \frac{e^{-\lambda (t-s)}}{(n-x)!}$$ After substituting the probabilities and making some simplifications, we get $$\implies P[X(s) = x|X(t) = n] = \frac{n!}{x!(n-x)!}\frac{s^x(t-s)^{n-x}}{t^x} = {n \choose x}\left(\frac{s}{t}\right)^x \left(1-\frac{s}{t}\right)^{n-x}$$
It is possible to go back and evaluate the first summation, but that's not really necessary because the preceding probability follows a binomial distribution $B(n,s/t)$, so the expectation is already known to be $n \frac{s}{t}$.