I encountered the following question in my research:
Let the diffusion process $\{X_t\}_{t\ge 0}$ be governed by $$d X_t=s(1-X_t)X_td B_t$$ where $X_0\in (a,b)\subseteq (0,1)$, $s>0$, and $B_t$ is the standard Brownian motion. Define the stopping time $\tau\equiv\inf\{t\ge 0\,|\,X_t\notin (a,b)\}$.
My question: Is it true that $\mathbf E[\tau]$ is decreasing in $s$? And also, is it possible to explicitly derive (or at least sharply characterize) the distribution of $\tau$ with $s$ being a parameter? If this is possible, how does $s$ shift the distribution of $\tau$?
Can anyone give me some hints on this problem? Thanks a lot!
Let $\tau(x)=E(\tau|X_0=x)$. Then $\tau(x)$ is the solution to the following boundary value problem: $$L\tau=-1\tag 1$$ $$\tau(x)=0 \text{ for } x\in\{a,b\}\tag 2$$ where $$L=\frac 1 2(s(1-x)x)^2\frac{d^2}{dx^2}$$ is the generator of the diffusion. You just need to integrate $(1)$ twice and use $(2)$ to determine constants of integration. You'll get $\tau\propto 1/s^2$- decreasing in $s$ as expected.