Poisson process with stochastic intensity correlated with a Brownian Motion 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 $\mathbb{E}[X(t)\int_0^t\int_{\mathbb{R}^+}Q\tilde{\mu}(dx,ds)]$?
 A: This is only a partial answer, as the computations can become quite involved.
We have 
$$ \mathbb{E}( X_t J_t ) = \mathbb{E}\left( X_t Q \int_{ \mathbb{R}_+ \times [0, t] } d\mu \right) = \mu_Q \mathbb{E}\left( X_t   \int_{ \mathbb{R}_+ \times [0, t] } d\mu \right). $$
by independence of $Q$ with the other processes.
By conditioning on $ \mathcal{F}_t = \sigma(W_1(s), s \leqslant t)\otimes \sigma(W_2(s), s \leqslant t) $, we have 
$$ \mathbb{E}( X_t J_t ) =  \mu_Q \nu(\mathbb{R}_+) \mathbb{E}\left( X_t \lambda_t \right). $$
The problem is now to understand the product of two CIR processes. An application of Itô formula gives $ d(X \lambda) = X d\lambda + \lambda dX + d\langle X, \lambda \rangle $ and $ \langle X, \lambda \rangle_t = \rho \sigma_1 \sigma_2  \sqrt{X(t) \lambda(t) } dt $. The problem lies in the $ (\theta_1, \theta_2) $. If these parameters are equal to $0$, you get the squared Bessel process which has a semi-group property. Here, you have correlated BM ; expressing the product $ XY = Z $ as a solution of the same SDE seems complicated with $ \theta_i \neq 0 $. 
Possible attacks: 
1/ You can have the expectation $ f(t) = \mathbb{E}( X_t ) $ (or $ \lambda_t $) by solving the ODE $ f(t) = f(0) + \kappa_1 \int_0^t (\theta_1 - f(s) ) ds $ with $ f(0) = \mathbb{E}(X_0) $ (obtained by taking the expectation of the SDE, with the martingale of expectation 0). You could try to integrate the SDE in $ Z := X\lambda $, of the form $ dZ = (\kappa_1 + \kappa_2) Z dt + \rho \sigma_1 \sigma_2 \sqrt{Z} dt + (\alpha X + \beta \lambda) dt + dM_t $ where $ M $ is the martingale term (and $\alpha, \beta = ...$). But you then need to know the quantity $ g(t) = \mathbb{E}( \sqrt{Z_t} ) $. Maybe another equation using the Itô formula for $ (x, y) \mapsto \sqrt{xy} $. I don't think this is a simple problem though. The literature has maybe some closed form using some properties of the CIR process. 
2/ If you can access the semi-group/infinitesimal generator, the theory of diffusions gives you $ \mathbb{E}_{a, b}( X_t \lambda_t) = e^{t \mathcal{L}}f(a, b) $ where $ f(x, y) = xy $ and $ (a, b) $ is the initial value of the process. You can write this generator $ \mathcal{L} $ with the help of the Itô formula (it is of the form $ \frac{\sigma_1^2}{2}x\partial^2_x + \frac{\sigma_2^2}{2}y\partial^2_y + \sigma_1\sigma_2 \rho \partial_{x, y} + ...$), but computing the exponential is complicated. The best way would be to diagonalise it, certainly with bi-variate orthogonal polynomials or special functions (I am not expecting sin and cos to be of any help here) ; since the function $f$ is a simple polynomial, this could work (try a general polynomial to see if the 2-recurrence equation can be solved). If anyone has a reference on that, this would be nice. 
The compensated case amounts to do the same computations. 
