Variance of Non-Homogeneous Poisson Process

Question

I have a question about this Non-Homogeneous Poisson Process with intensity function :

$f(t)=1+t$ if $t [0,2]$

$f(t)=3$ otherwise

I need to compute $Var(N(2)+N(3)|N(1))$

So

$Var(N(2)+N(3)|N(1))$=$Var(N(2)+N(3))$

=$Var(N(2))+Var(N(3))+2Cov(N(2),N(3))$

Is this the correct way to do it ?

Answer 1

Use the fact that the number of events on disjoint intervals is independent (independent increments). So \begin{align*} {\rm Var}(N(2)+N(3)|N(1)) &= {\rm Var}\big(2N(1) + 2(N(2)-N(1)) + (N(3)-N(2)) | N(1)\big) \\ &= 4{\rm Var}(N(2)-N(1)) + {\rm Var}(N(3)-N(2)). \end{align*} Then use that the distribution of $N(t)-N(s)$ is Poisson with mean $\int_s^t \lambda(s)$, where $\lambda$ is the intensity function.