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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.