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.