I have the following problem. Let $\pi$ be a measure on $(\mathbb{R}, B(\mathbb{R}))$ such that $\pi(\{0\})=0$ and a poisson random measure $N$ over $(\mathbb{R}^+ \times \mathbb{R},B(\mathbb{R}^+) \times B(\mathbb{R}))$ with mean $\nu(ds,dx) = ds \pi(dx)$. Given a $B \in B(\mathbb{R})$ with $0 < \pi(B) < \infty$, prove that, for $t \geq 0$
\begin{equation}
X_t = \int_{[0,t)}\int_B x N(ds,dx)
\end{equation}
is a compound Poisson process with parameter $\lambda = \pi(B)$ and jump distribution $\frac{\pi(dx)}{\pi(B)}\mid_{x\in B}$
I have no idea how to proceed. My main ideas are to use a characterization of the compound poisson process through its Laplace transform.
Best Answer
Let $K_t=[0,t)\times B$. Then the following holds
\begin{equation} X_t = \int_{[0,t)}\int_B x N(ds,dx)= \sum_{(x_i, y_i)\in supp M\cap K_t}x_i. \end{equation}
The properties can be shown from that pretty easily. E.g. for $0<w<s<t$ is $X_t-X_s$ independent of $X_s-X_w$, because:
$X_t-X_s=\sum_{(x_i, y_i)\in supp M\cap (K_t-K_s)}x_i$ where I mean $K_t-K_s=[s,t)\times B$. Obviously is $X_t-X_s$ independent of $X_s-X_w$, because $M\cap K_t-K_s$ is independent of $M\cap K_s-K_w$ from the definition of PPP.