# For $X_{t}=\exp\left\{\left(\mu-r-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t}\right\}$, do we have $\mathbb{E}[\int_{0}^{\tau_{b}}X_{s}dW_{s}]=0$?

Let $X_{t}$ denote the solution to the SDE: $$dX_{t}=(\mu -r)X_t dt+\sigma X_t d W_{t}, \ X_{0}=1$$

i.e. $X_{t}$ is the process: $$X_{t}:=\exp\left\{\left(\mu-r-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t}\right\}$$

(here we assume $\mu <r$ and $\sigma >0$, and $W$ is a standard 1-d Brownian motion). Fix $0<b<1$ and let $\tau_{b}$ denote the hitting time of the level b: $$\tau_{b}:=\inf\{t \geq0, \ X_{t}=b\}.$$ Do we have $\mathbb{E}\left(\int\limits_{0}^{\tau_{b}}X_{s}dW_{s}\right)=0$?

Thank you!

-
Just a remark if $Y_t=\int_0^tX_sdW_s$ a uniformly integrable martingale then I think you are done by optimal stopping theorem, is it true ? –  TheBridge Sep 18 '12 at 20:22