# Poisson integral and discontinuous martingale (Ito-Levy formula)

Consider compounded Poisson process $P$ given by $P_t = \int_0 ^t \int _{\mathbb R}z~ N(dr,dz)$ where $N$ is a Poisson random measure of intensity $dt \otimes \nu$ and $\nu$ is a Levy measure.

Why $P$ is well defined if $\mathbb R$ is not lower bounded ?

Now consider a process $M$ defined as $M_t =(P_t- m_t)^2 - \int_0 ^t \int _{\mathbb R}z^2 ~ \nu(dz) ~ dt$ where $m_t = \mathbb E [ P_t]$

How to show $M$ is a $\mathcal F^P$- martingale ?

I've tried to apply Ito - Levy formula but I'm doing something wrong since at the end it does not some up giving a martingale.

Could someone help with that showing this calculation or show me another approach ? Thanks in advance.

We can write

$$P_t = \int_0^t \int_{|z|<1} z \, N(dr,dz) + \int_0^t \int_{|z| \geq 1} z \, N(dr,dz).\tag{1}$$

Since $P$ is a compound Poisson process, its Lévy measure $\nu$ is finite, i.e. $\nu(\mathbb{R})<\infty$. This implies that

$$\mathbb{E} \left( \int_0^t \int_{|z|<1} z \, N(dr,dz) \right) = \int_0^t \int_{|z|<1} z \nu(dz) \, dr \leq t \nu(\mathbb{R})< \infty$$

and therefore the first addend in $(1)$ is well-defined.

If $\int z^2 \, \nu(dz)< \infty$, then in particular $m_t<\infty$ and

$$P_t-m_t = \int_0^t \int_{\mathbb{R}} z \tilde{N}(dr,dz)$$

where $\tilde{N}$ denotes the compensated Poisson measure. Use the fact that this process has independent increments in order to show that $(M_t)_{t \geq 0}$ is a martingale. Hint:

$$(P_t-m_t)^2 = \bigg[(P_t-P_s-(m_t-m_s))+(P_s-m_s)\bigg]^2$$