Question: Consider a Poisson process of rate λ. Let N be the number of arrivals in (0, t], and let M be the number of arrivals in [0, t + s]. Find the expectation of NM.
Attempt to Answer: The expectation is given by the following expression: $$\sum_{n=0}^\infty\sum_{m=0}^\infty = nmP(N=n,M=m)$$
We now find P(N = n, M = m).
Let N' be the number of arrivals in (t, t + s]. Then M = N + N'. Also, N~Poi(λt), and N'~Poi(λs), and N and N' are independent. So,
$$P(N = n, M = m)=P(N = n,N+N'=m)=P(N=n,N'=m-n)=P(N=n)P(N'=m-n)=\frac{(λt)^ne^{-λt}}{n!}*\frac{(λs)^{m-n}e^{-λs}}{(m-n)!}$$ For $0\le n \le m$. Otherwise, P(N = n, M = m) = 0.
So our expectation becomes: $$\sum_{n=0}^\infty\sum_{m=n}^\infty nm*\frac{(λt)^ne^{-λt}}{n!}*\frac{(λs)^{m-n}e^{-λs}}{(m-n)!}$$
I'm not sure how to simplify this further. Any ideas?
You are on the right track but it's much easier to just use linearity and independence like: $$ E(NM) = E(N(N+N')) = E(N^2)+E(N)E(N').$$ $N$ and $N'$ are (independent) Poissons with means $\lambda t$ and $\lambda s,$ so these expectations can be computed: $$ E(N^2) = Var(N) + E(N)^2 = \lambda t+ (\lambda t)^2 \\ E(N) = \lambda t\\E(N') = \lambda s.$$
EDIT
I guess the hard thing here is computing $E(N^2)$ where $N$ is poisson $\lambda$ (and thus the variance). We have $$ E(N^2) = \sum_{k=0}^\infty k^2 e^{-\lambda} \frac{\lambda^k}{k!}\\ = e^{-\lambda}\sum_{n=0}^\infty (n+1)\frac{\lambda^{n+1}}{n!} \\= \lambda e^{-\lambda} \frac{d}{d\lambda}\sum_{n=0}^\infty \frac{\lambda^{n+1}}{n!}\\=\lambda e^{-\lambda} \frac{d}{d\lambda} \lambda e^\lambda\\=\lambda e^{-\lambda} (e^\lambda + \lambda e^\lambda) \\=\lambda+\lambda^2$$ One could also get this from differentiating the moment generating function, which is relatively easy to compute.