Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background: Let $C$ be the space of continuous function on $[0,T]$, $f, \sigma \in C$ bounded with $\sigma^2 \geq \varepsilon > 0$ and let $X=(X_t)_{t\in [0,T]}$ be a diffusion process of infinitesimal generator $A^X$ given by $A^Xg(x) = f(x)g'(x)+\frac{\sigma^2(x)}{2}g''(x)$ for all $g$ in the domain $D$. Introduce, for all $\omega\in C$, the functional $\tau_t(\omega)=\inf\{s\geq 0: \int_0^s \sigma^2(\omega_r)dr>t\},\ t\geq 0$ and then define a new process: $Y_t:=X_{\tau_t(X)}$, $t\geq 0$.

Thanks to a result of Volkonskii we know that $Y$ is also a diffusion process of domain $D$ and infinitesimal generator given by: $$A^Yg(x)=\frac{1}{\sigma^2(x)}A^Xg(x),\quad g\in D.$$

Question: Consider now the solution of the SDE: $$\bar X_0=0,\ d\bar X_t=\bar f_n(t,\bar X)dt+\bar\sigma_n(t,\bar X) dB_t, \ t\in[0,T],$$ where $B$ is a standard Brownian motion and $\bar f_n,\bar \sigma_n$ are approximations of $f,\sigma$, piecewise constant, in the following sense: $$\bar{f}_n(t,\omega)=\sum_{i=1}^n f\big(\omega(t_{i-1})\big)\mathbb{I}_{(t_{i-1},t_i]}(t), \ \bar{\sigma}_n(t,\omega)=\sum_{i=1}^n\sigma\big(\omega(t_{i-1})\big)\mathbb{I}_{(t_{i-1},t_i]}(t),\ t_i=T \frac{i}{n}.$$ Remark that, in particular, $\bar{f}_n$ and $\bar{\sigma}_n$ are not functions of $X_t$ only but of the whole trajectory. Define a new process $\bar{Y}_t:=\bar X_{\bar \tau_t(\bar X)}$, where $\bar\tau_t(\omega)=\inf\{s\geq 0: \int_0^s \bar\sigma^2(r,\omega)dr>t\}$.

Is there any result relating the generator of $\bar{Y}$ to that of $\bar{X}$, similar to the one above?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.