Conditional expectation of Poisson Process given two counts

Question

Let ${N(t),t\ge 0}$ be a Poisson Process with rate $\lambda=2.$ How do I go about calculating $E[N(6)\mid N(4)=4,N(10)=12]$?

Answer 1

$$N(6) - N(4) \sim\operatorname{Poisson}(\lambda(6-4)) = \operatorname{Poisson}(2\lambda).$$

\begin{align} & \operatorname E[N(6)\mid N(4)=4,N(10)=12] \\[10pt] = {} & 4 + \operatorname E(N(6)-N(4)\mid N(10) - N(4) = 8) \end{align}

So a question is: What is the conditional distribution of $N(6)-N(4)$ given that $N(10)-N(4)=8\text{?}$

\begin{align} & \Pr(N(6)-N(4) = n \mid N(10)-N(4) = 8) \\[10pt] = {} & \frac{\Pr(N(6) - N(4) = n\ \&\ N(10)-N(4) = 8)}{\Pr(N(10)- N(4) =8) } \tag 1 \\[8pt] = {} & \frac{\Pr(N(6) - N(4) = n\ \&\ N(10)-N(6) = 8-n)}{\Pr(N(10)- N(4) =8)} \tag 2 \end{align} The reason the probabilities in the numerators in lines $(1)$ and $(2)$ are equal is that the actual events whose probabilities are taken are equal. On line $(2),$ you can exploit the fact that the two events with "and" between them are independent.

After that, when you simplify the fraction, everything depending on $\lambda$ will cancel out and you will get a binomial distribution with parameters $8$ and $1/3,$ the latter number coming from the fact that the length of the interval $[4,6]$ is $1/3$ the length of the interval $[4,10].$