Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
I think in $f_{\beta}(t,x)$ should be also $\frac{\partial f}{dx}(t,x)\cdot \beta(t,x)$. – Leon Feb 12 at 20:38

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 $\nu$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.