# Poisson Process: Expectation of product

Assume that calls arrive at a call centre according to a Poisson arrival process with a rate of 15 calls per hour. For $0 ≤ s < t$, let $N(s,t)$ denote the number of calls which arrive between time $s$ and $t$ where time is measured in hours.

What is $E(N(1,4)N(2,6))$?

I am confused by this because I know $E(XY) = E(X) E(Y)$ if X and Y are independent but in this case the periods of time overlap and are not independent.

$$N(1,4)=N(1,2)+N(2,4)\qquad\text{ and }\qquad N(2,6)=N(2,4)+N(4,6).$$
\begin{eqnarray*} E(N(1,4)N(2,6)) &=& E([N(1,2)+N(2,4)]\;N(2,6))) \\ &=& E(N(1,2)N(2,6) + N(2,4)\;[N(2,4)+N(4,6)]) \\ &=& E(N(1,2)N(2,6) + N(2,4)^2 + N(2,4)N(4,6)) \\ &=& E(N(1,2))E(N(2,6)) + E(N(2,4)^2) + E(N(2,4))E(N(4,6)) \\ &=& \lambda\cdot 4\lambda + (2\lambda)^2 + 2\lambda + 2\lambda\cdot 2\lambda \\ &&\qquad\qquad\qquad\qquad\text{since $E(N(s,t)^2)=((t-s)\lambda)^2+(t-s)\lambda$} \\ &=& 12\lambda^2 + 2\lambda \\ &=& 2730 \quad\text{with $\lambda=15$}. \end{eqnarray*}