I'm working on a counting process and trying to prove that this counting process is a Poisson point process (non homogeneous). I have the 3 conditions :
- $N(t)>0$
- $N(t)\in \mathbb N,\forall t \in \mathbb R$
- If $s \leq t$ then $N(s) \leq N(t)$
I now need to have these 3 conditions :
a) N($0$)=$0$
b) $\{N(t) , t>0\}$ has independent increments
c) $N(t)\sim\mathcal{P}(\lambda(t))$
I have N($0$)=$0$ and feel like b) and c) will always be true because for every counting process $N(t)$, I could find a function $\lambda(t)$ that satisfies $N(t)\sim\mathcal{P}(\lambda(t))$. so my question is :
Do I have to show that my counting process is a non homogeneous poisson process or is it trivial ?
The statement $N(t)\sim\mathcal{P}(\lambda(t))$ is actually much stronger than you think. If such a function $\lambda(t)$ exists, then it is entirely determined by $\Pr[N(t)=0] = e^{-\lambda(t)}$. So $\Pr[N(t)=1]$ and so forth should be determined by this probability.
To give a particular example, $\Pr[N(t)=1] = \lambda(t) e^{-\lambda(t)}$. So a Poisson process must satisfy $$\Pr[N(t)=1] = \Pr[N(t)=0] \cdot \ln \frac1{\Pr[N(t)=0]}.$$This will certainly not be true of any counting process.
Usually, the property of independent increments is not checked in conjunction with checking $N(t) \sim \mathcal P(\lambda(t))$, but in conjunction with checking two limit conditions:
This shows that the condition is important, because these two limit conditions alone are much weaker than a Poisson distribution. It is trickier to come up with an example where $N(t) \sim \mathcal P(\lambda(t))$, but the increments are not independent; however, I suspect that there are some.
For a very specific example, consider the deterministic counting process $N(t) = \lfloor t\rfloor$. This satisfies the three conditions you have, but is certainly not any kind of inhomogeneous Poisson process!