# Stochastic integral with Poisson random measure

The following is what I read in paper and I am confused by some parts.

We consider a one-dimensional Itô semimartingale $$X$$ which is defined on some probability space $$(\Omega,\mathcal F,\{\mathcal F_t\},P)$$ and can be represented \begin{aligned} X_t=&X_0+\int_0^tb_sds+\int_0^t\sigma_sdW_s+\int_0^t\int_{\mathbb R}\delta(s,z)1_{\{|\delta(s,z)|\leq1\}}(p-q)(ds,dz)\\ &+\int_0^t\int_{\mathbb R}\delta(s,z)1_{\{|\delta(s,z)|>1\}}p(ds,dz), \end{aligned} where $$W$$ is a standard Brownian motion and $$p$$ is a Poisson random measure on $$\mathbb R^+\times \mathbb R$$ with compensator $$q(dt,dz)=dt\otimes dz$$. Assume that for some $$r\in[0,2]$$, $$|\delta(\omega,t,z)|^r\wedge1\leq J(z)$$ where $$J$$ is a Lebesgue-integrable function on $$\mathbb R$$. If $$r=1$$, we can rewrite the above equation (up to modifying $$b_s$$) as $$X_t=X_0+\int_0^tb_sds+\int_0^t\sigma_sdW_s+\int_0^t\int_{\mathbb R}\delta(s,z)p(ds,dz).$$ My questions are as follows:

1. It seems that the Levy measure of $$p$$ is Lebesgue measure. However, a Levy measure $$\nu$$ should satisfy $$\int_\mathbb Rx^2\wedge1\nu(dx)<\infty$$. Why is the compensator $$Q$$ has the form $$dt\otimes dz$$ ?

2. If the Levy measure is the Lebesgue measure, for the case $$r=1$$, \begin{aligned} &E\int_0^t\int_{\mathbb R}|\delta(s,z)|1_{\{|\delta(s,z)|\leq1\}}p(ds,dz)\\ =&E\int_0^t\int_{\mathbb R}|\delta(s,z)|1_{\{|\delta(s,z)|\leq1\}}ds\otimes dz\\ \leq&E\int_0^t\int_{\mathbb R}|J(z)|ds\otimes dz\\ <&\infty. \end{aligned} Therefore, we can decompose $$\int_0^t\int_{\mathbb R}\delta(s,z)1_{\{|\delta(s,z)|\leq1\}}(p-q)(ds,dz)$$ into two parts and combine $$\int_0^t\int_{\mathbb R}\delta(s,z)1_{\{|\delta(s,z)|\leq1\}}p(ds,dz)$$ with $$\int_0^t\int_{\mathbb R}\delta(s,z)1_{\{|\delta(s,z)|>1\}}p(ds,dz)$$ to get $$\int_0^t\int_{\mathbb R}\delta(s,z)p(ds,dz)$$. Since $$\int_\mathbb R|\delta(s,z)|1_{\{|\delta(s,z)|\leq1\}}dz\leq\int_\mathbb R|J(z)|dz<\infty,$$ we can define a new drift $$b_s'=b_s-\int_\mathbb R\delta(s,z)1_{\{|\delta(s,z)|\leq1\}}dz$$ which is still locally bounded. Is my computation of the transform correct?

I don't think the Levy measure of $$p$$ is Lebesgue measure. I think it is dirac measure at 1, i.e. $$\delta_1$$. This is because all jumps are of size 1 (hence the subscript) and the intensity is a constant 1 (coefficient in front of $$\delta_1$$. I think you might be confusing Levy measure with the intensity of a Poisson measure. Since the compensator has form $$q(dt,dz)=dt\otimes dz$$, the process $$p$$ is a random Poisson measure with intensity 1 on $$\mathbb{R}^+ \times \mathbb{R}$$. That might clear up some confusion?