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Let $X=\{X_t\}_{t\geq0}$ be an $n$-dimensional Markov process, defined by the SDE

$$dX_t = \mu(t, X_t) \, dt + \sigma(t,X_t) \, dB_t+\beta(t-,X_{t-}) \, dN_t,$$

where $\mu, \sigma$ and $\beta$ are vector functions, $B_t$ is an $n$-dimensional Brownian motion and $N_t$ is an $n$-dimensional counting process, whose conditional intensity is given by $\lambda(t-, X_{t-})$, where $\lambda$ is a continuous function.

Does anyone know the explicit formula for the infinitesimal generator of $X_t$?

As always, help is greatly appreciated.

Edit: I should maybe mention that the result for the one-dimensional case is

$$ \mathcal{A}f = \frac{\partial f}{\partial t}(t,x) + \mu(t,x)\frac{\partial f}{\partial x}(t,x) + \frac{1}{2}\sigma^{2}(t,x)\frac{\partial^{2} f}{\partial x^{2}}(t,x) + f_{\beta}(t,x)\lambda(t,x)\text{,}$$

where

$$f_{\beta}(t,x) = f(t,x+\beta(t,x)) - f(t,x)\text{.}$$

The only thing that I really have no clue about is what the fourth term in the previous equation looks like in the $n$-dimensional case.

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1 Answer 1

Perhaps the following will be of use. In "Levy Processes and Stochastic Calculus" by David Applebaum, Theorem 6.7.4, the following is argued: Consider a SDE of the form

$$ dY_t = b(Y_{t-})dt + \sigma(Y_{t-})dW_t + \int_{|x|<c} F(Y_{t-},x) \tilde{N}(dt,dx) + \int_{|x|\ge c} G(Y_{t-},x) N(dt,dx),$$

where $W$ is a $n$-dimensional Brownian motion, $N$ a $n$-dimensional Poisson random measure with intensity measure $\nu$ and compensator $\tilde{N}$, independent of $W$, and coefficients satisfying appropriate regularity conditions. Then this SDE has a unique solution, which is a Feller process with generator $A$ whose value on $C_c^2(\mathbb{R}^n)$ is

$$ Af(y) = \sum_{i=1}^n b_i(y)\frac{\partial}{\partial y_i}f(y) + \frac{1}{2}\sum_{i=1}^n \sum_{j=1}^n (\sigma(y)\sigma(y)^t)_{ij}\frac{\partial^2}{\partial y_i\partial y_j}f(y)\\ +\int_{|x|\le c} f(y+F(y,x))-f(y) - F_i(y,x)\frac{\partial}{\partial x_i}f(y) d\nu(x)\\ +\int_{|x|\ge c}f(y+G(y,x)) - f(y) d \nu(x). $$

The only difference with your set-up appears to be that you have a nondeterministic intensity measure $\nu$.

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True...I'll think about this - thank you very much, though! –  Stojan Jovanović May 24 '13 at 13:33
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