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I am currently confused with the moment of non-homogeneous compound Poisson process and a Brownian Motion. I know that generally Poisson Process and Brownian Motion are independent if they are adapted to the same filtration. But what if the intensity of the Poisson Process and the Brownian Motion are correlated?

For example, we have a CIR process $dX(t)=\kappa_1\Big(\theta_1-X(t)\Big)dt+\sigma_1\sqrt{X(t)}dW_1(t)$, and a non-homogeneous compound Poisson Process $J(t)=\int_0^t\int_{\mathbb{R}^+}Q\mu(dx,ds)$, with a stochastic intensity $\lambda$ satisfies $d\lambda(t)=\kappa_2\Big(\theta_2-\lambda(t)\Big)dt+\sigma_2\sqrt{\lambda(t)}dW_2(t)$ (another CIR process), and its jump size $Q$ has a normal distribution $N(\mu_Q, \sigma_Q^2)$, the correlation coefficient between $W_1(t)$ and $W_2(t)$ is $\rho$, but the jump size is independent of them. BTW, $\mu(dx,ds)$ is a Poisson random measure, $\tilde{\mu}(dx,ds)$ is a compensated Poisson measure.

Question is, what exactly are the mixed moment of $X(t)$ and $J(t)$, i.e. $\mathbb{E}[X(t)J(t)]$, and $\mathbb{E}[X(t)\int_0^t\int_{\mathbb{R}^+}Q\tilde{\mu}(dx,ds)]$?

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  • $\begingroup$ $X$ is exponentially affine. If $J$, too, is exponentially affine then you will be able to compute the joint characteristic function of $J$ and $X$ and you are in business. $\endgroup$ – batman Aug 5 '17 at 12:22
  • $\begingroup$ @batman :thanks for your comment, $J(t)$ here I don't think it is exponentially affine, but its intensity is exponentially affine. Someone in Mathflowsn gave me some hints, and my current confusion is just how to calculate higher joint moment of it, since finding the joint law of higher moments of $X(t)$ and $J(t)$ seems impossible. Here is the link in Mathflows link $\endgroup$ – random demon Aug 7 '17 at 18:31
  • $\begingroup$ you are right, J is not exponentially affine. Then I suppose conditioning on cumulative arrival intensity is the right way to proceed, the same way that one conditions on realised quadratic variation in stochastic volatility models. $\endgroup$ – batman Aug 8 '17 at 8:58
  • $\begingroup$ @batman : actually I think conditioning on the sigma-algebra generated by $W_1$ $W_2$ is the right way to proceed for the mixed moment $\mathbb{E}(X(t)J(t))$, since the jump size $Q$ is independent from these two Brownian Motion. $\endgroup$ – random demon Aug 8 '17 at 13:42

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