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{equation} \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} \end{equation} 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{equation} \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} \end{equation} 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?

Thanks for your help!


It seems that the Levy measure of p is Lebesgue measure. However, a Levy measure ν should satisfy ∫ℝx2∧1ν(dx)<∞. Why is the compensator Q has the form dt⊗dz ?

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?


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