# SDE and expectation of stopping time

Let $X_t$ be a solution to the SDE, $dX_t=X_t \,dt+X_t\,dW_t$, $X_0=x>0$ where $W_t$ is brownian motion, then the solution to this SDE is $X_t=xe^{\frac{t}{2}+W_t}$.

Let $\tau=\inf_{t>0} \{t:X_t\ge R\}$. I am not sure how to calculate the expectation of the stopping time $\mathbb{E}_x[\tau]$.

Thank you

• Do you know what the expected value of the hitting time of a given level $a$ is for standard Brownian motion (with or without drift)? – cardinal Nov 13 '11 at 22:39
• Can you solve similar hitting times problems? Which ones? – Did Nov 13 '11 at 22:52
• I am not very comfortable with hitting time actually :( – Tasha Chen Nov 13 '11 at 22:55
• I know how to do the basic ones, like first passage time – Tasha Chen Nov 13 '11 at 22:55
• @cardinal, I know how to calculate the problem you mentioned – Tasha Chen Nov 13 '11 at 23:00

Observe that $(W_t)_{t\geqslant0}$ is a martingale, hence the optional stopping theorem yields $\mathbb{E}[W_\tau]=\mathbb{E}[W_0]$. Thus, $\mathbb{E}[W_\tau]=\mathbb{E}_x[\log(X_\tau)-\log(X_0)-\frac{1}{2}\tau]=0$ and $\mathbb{E}_x[\tau]=2\log(R/x)$.