# 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!

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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

Since the stochastic integral process $$Y(t)=\int_{0}^{t}X_{s}\,\text{d}W_{s}\ , \quad t \geq 0$$ is a martingale and $\tau_{b}$ is an (almost-surely) bounded stopping time (adapted to the same filtration as process $Y$) then by the Optional Stopping Theorem the expected value of the random variable $Y(\tau_{b})$ is equal to that of the random variable $Y(0)$, which is zero.