Proving that $N(\cdot \cap A)$ is a poisson random measure

Question

If $N$ is a simple poisson random measure on $E$ with mean measure $\mu$ ($\mu$ is diffuse), then I want to show that for a fixed $A \subseteq E$, that $N(\cdot \cap A)$ is also a poisson random measure.

I was wondering if the following argument by cases is sufficient. For an arbitrary $B \in$Borel$(E)$:

Case 1: If $A,B$ are disjoint, $ A \cap B = \emptyset$.

Case 2: If $A \subseteq B$, then $A \cap B = A$ and $N (A \cap B ) = N(A)$ is a simple PRM($\mu$) on E. Identical reasoning holds for when $B \subseteq A$

Case 3: If $A \cap B \neq \emptyset$ and one doesn't contain the other, then $A \cap B = A - B$ which is just a member of Borel$(E)$. So then N is a PRM($\mu$) on E

Having exhausted the cases, $N(\cdot \cap A)$ is also a simple PRM$(\mu)$ on E

The only doubt I have is that I have not really used the fact that $\mu$ is diffuse, it seems that this result would hold even if it wasn't the case?

Answer 1

If $\{N_B\}_{B\in\mathcal A}$ is a Poisson random measure with intensity measure $\mu$, and $A$ is an element of $\mathcal A$ then $\{M_B\}_{B\in\mathcal A}$ prescribed by $M_B=N_{A\cap B}$ can be shown to be a Poisson random measure with intensity $\nu$ where $\nu$ is prescribed by $\nu(B)=\mu(A\cap B)$.

To prove that you only need to prove the defining properties:

• (i) $M_B=N_{A\cap B}$ is a Poisson random variable with rate $\mu(A\cap B)=\nu(B)$
• (ii) If sets $B_1,\dots,B_n$ don't intersect then also the sets $A\cap B_1,\dots,A\cap B_n$ don't intersect and consequently the random variables $N_{A\cap B_1}=M_{B_1},\cdots,N_{A\cap B_n}=M_{B_n}$ are independent.
• (iii) for every $\omega\in\Omega$ the map $A\cap B\mapsto N_{A\cap B}(\omega)=M_B(\omega)$ is a measure on $(E,\mathcal A)$.

That $\mu$ is diffuse implies that $\nu$ is diffuse, but plays no role in the proof.

You are saying that $N(.\cap B)$ is (also) a PRM($\mu$). That is not true. It is a PRM($\nu$).