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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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  • $\begingroup$ What is the intensity measure in $x$ ? I guess your Poisson measure is $ \mu \sim P( \nu \otimes \lambda) $ where $ (\lambda_s)_s $ is your CIR process and $ \nu $ is to be defined. Since you integrate on $ x \in \mathbb{R}_+ $ it has to be integrable and cannot be the Lebesgue measure. $\endgroup$ – Synia Jul 31 '17 at 14:36
  • $\begingroup$ @Synia :Sorry for the ambiguity. If I understand you correctly, here I mean the intensity is $\lambda$, and $\lambda$ follows CIR process at the same time. While I take the other part of the Poisson random measure as a simple counting process. Should I ease the restriction of $x$ to $x\in\mathbb{R}$, so the jump size could be negative. $\endgroup$ – random demon Jul 31 '17 at 15:17
  • $\begingroup$ I was just asking what was this other part that I call $ \nu $. Seemingly, this is the counting measure (but hopefully, on a compact set, since you want $ \nu(\mathbb{R}_+) < \infty $). $ \mathbb{R} $ or $ \mathbb{R}_+ $ is not a problem. $\endgroup$ – Synia Jul 31 '17 at 15:21
  • $\begingroup$ @Synia : thanks for your comments. First time ask questions and I am happy to make it clearer. $\endgroup$ – random demon Jul 31 '17 at 15:37
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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.

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  • $\begingroup$ Thx, I'm currently using your first route to tackle it. Yeh, the $\sqrt{X\lambda}$ might be a big problem, I've tried some methods. By further assuming that $\sqrt{X(t)}$ and $\sqrt{\lambda(t)}$ follow simply Ornstein-Uhlenbeck in the same way like: $d\sqrt{X(t)}=-\frac{\kappa_1}{2}\sqrt{X(t)}dt+\frac{1}{2}\sigma_1dW_1(t)$, seemingly I can get $\mathbb{E}(X\lambda)$ solved by solving first order ODE. BTW, if I want to calculate $\mathbb{E}[X(t)J(t)^2]$ or $\mathbb{E}[X(t)^2J(t)^2]$, can I do it in a similar way? $\endgroup$ – random demon Jul 31 '17 at 18:13
  • $\begingroup$ If $X$ (or $ \lambda $) follow a CIR dynamic, this is unlikely that $ \sqrt{X} $ will follow an OU dynamic. This is then a different problem. In this case, indeed, you can solve the first order equation. Concerning the computation of all the moments, if you are brave enough :), you can try to compute the joint Laplace transform of $ (X_t, \int_{ \mathbb{R}+\times [0, t] } d\mu )_t $ using the Campbell formula (see wikipedia if you don't know it). $\endgroup$ – Synia Jul 31 '17 at 21:21
  • $\begingroup$ (This is basically the functional version of the Laplace transform of a Poisson-distributed random variable : $ \mathbb{E}( e^{ \langle f, \mu \rangle }) = e^{ \langle e^f - 1, I \rangle } $ where $I$ is the intensity measure). Nevertheless, supposing that $\nu(\mathbb{R}_+) = 1$, the fact that you have this $ Q $ will give you something like $ \mathbb{E}( e^{- x X_t - y \mu_t }) = \mathbb{E}(e^{ -x X_t + \lambda_t^2 (e^{-y} - 1)^2/2 } ) $ so you need in fact the law of $(X_t, \lambda_t^2)$. For OU in place of CIR, this is relatively easy. The moments are then given by differentiating in 0. $\endgroup$ – Synia Jul 31 '17 at 21:23
  • $\begingroup$ By the way, if you think the question is solved, you can vote for it :) $\endgroup$ – Synia Jul 31 '17 at 21:24
  • $\begingroup$ Thx, I accepted it. It seems that getting the law of $(X_t, \lambda_t^2)$ is impossible. IDK whether it is helpful to use It\^o's formula, $X(t)Y(t)=X(0)Y(0)+\int X(t-)dY(t)^c+\int Y(t-)dX(t)^c+[X^c,Y^c]_{[0, t]}+\sum \Delta X(t)Y(t)$. Write $J(t)=\int_0^tQdN(s)$, $\Delta J(t)=J(t)-J(t-)=QdN(t)$ it gives that $X(t)J(t)^2=X(0)J(0)^2+\int J(t-)^2dX(t)^c+\sum\Delta X(t)J(t)^2$, from which I have $X(t)J(t)^2=X(0)J(0)^2+\int\Big(J(t)-QdN(t)\Big)^2dX(t)^c+\int X(t)(2Q-Q^2)dN(t)$, is this a right path and should I continue to take expectations on both sides? $\endgroup$ – random demon Aug 1 '17 at 16:32

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