Poisson process. Compute $E[N_z-N_t | N_s=3]$

Question

Let $(N_t)_{t \in[0,\infty)}$ be a Poisson process. Let $t<s<z$. Compute

$E[N_z-N_t | N_s=3]$

My idea: Using the fact that Poisson process has independent increments:

$E[N_z-N_t| N_s=3]=E[N_z-N_s+N_s-N_t | N_s=3]= E[N_z-N_s]+ E[N_s-N_t | N_s=3]$

Answer 1

As Henry pointed out $E[N_z-N_s] = E[N_{z-s}] = \lambda (z-s)$ (assuming $\lambda$ is the intensity). Also $E[N_s-N_t | N_s=3] = \frac{3(s-t)}{s}$ (since given $N_s=3$ the arrival times are uniformly distributed in $[0,s]$).

Answer 2

Yes that is the idea.

$\begin{align}\mathsf E[N_z-N_t\mid N_s=3]&=\mathsf E[N_z-N_s+N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf E[N_z-N_s\mid N_s=3]+\mathsf E[N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf E[N_z-N_s]+\mathsf E[N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf E[N_{z-s}]+\mathsf E[N_s-N_t \mid N_s=3]\end{align}$

Now you need to determine what is the conditional distributions of $N_s-N_t$ given $N_s=3$.

Binomial(3,(s-t)/s)