I was wondering if someone could give me some hints on the following question? Thanks a lot!
Consider the Ito process $$dX_t=\mu(X_t) dt+\sigma dB_t, t\ge0, X_0=x(0) \text{ a.s}., \sigma>0$$ and suppose that for all $\tau\ge 0$ $$\mu(X_\tau)=\begin{cases}\bar\mu, &\text{if } X_\tau\ge x^c\\\underline \mu,&\text{if }X_\tau<x^c\end{cases}$$ where $\bar\mu, \underline \mu\in\mathbb R$ and $x^c\in\mathbb R$ are fixed. Then what is $\mathbb E(X_t|X_s=x)$ ($t\ge s\ge 0$)? (I guess whether $x\ge x^c$ or $x<x^c$ should make a difference.)
As far as I'm aware, there's not much you can show except $\mathbb{E}[X_t|X_s]=X_s+\int_s^t\mathbb{E}[\mu(X_r)]dr=\underline{\mu}\lambda(A_{x_c,s,t})+\overline{\mu}\lambda(A_{x_c,s,t}^c)$ where $A_{x_c,s,t}=\{r\in[s,t]:X_r\leq x_c\}$ and $\lambda$ is Lebesgue measure. $A_{x_c,s,t}$ is a random set that depends on $X(s)$, but not in a way that I think you could easily get a handle on.