1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.