# Expectation stopped Brownian motion with drift

Let $\{X_t:t\geq 0\}$ be a Brownian motion with drift $\mu>0$ and define a stopping time $\tau$ by $$\tau=\inf\{t\geq 0:X_t=a\}.$$ Now I want to show that $$\mathbb{E}(e^{-\lambda\tau})=e^{(\mu-\sqrt{\mu^2+2\lambda})a}$$ for $\lambda>0$. Now as a hint I know that I need to use the martingale $M_t=e^{\alpha X_t-\alpha\mu t-\frac{1}{2}\alpha^2t}$. Obviously I need to use Doobs optional stopping theorem but I do not know how. Anyone has a suggestion?

1. Check that for fixed $\alpha>0$ the process $$M_t := \exp \left(\alpha X_t-\alpha \mu t- \frac{1}{2} \alpha^2 t \right)$$ is a martingale (with respect to the canonical filtration of the Brownian motion).
2. By the optional stopping theorem, $$\mathbb{E}(M_{\tau \wedge t}) = \mathbb{E}(M_0) = 1, \qquad t \geq 0.$$
3. Show that $|M_{t \wedge \tau}| \leq e^{\alpha a}$. Deduce from the dominated convergence theorem that $$\mathbb{E}(M_{\tau}) = 1.$$
4. Since $(X_t)_{t \geq 0}$ has continuous sample paths, we have $X_{\tau}=a$. Hence, $$M_{\tau} = e^{\alpha a} \exp \left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right).$$
5. It follows from step 3 and 4 that $$\mathbb{E} \exp\left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right) = e^{-\alpha a}.$$ Setting $\lambda := \mu \alpha + \frac{1}{2} \alpha^2$ proves the assertion.
• Thanks for the help, it helped me a lot! I have a follow-up question though: I want to prove that with this drift $a>0$, that $\tau<\infty$ with probability one by taking the limit $\lambda \to 0$, but I can't figure out how $a>0$ plays a big role in proving this using equation from step 5? – higuys Jan 28 '17 at 15:26
• @higuys I suppose you mean with drift $\mu>0$ (and not drift $a>0$)....? (Just as a side remark: If you find the answer useful, you can upvote it by clicking on the up arrow next to it.) – saz Jan 28 '17 at 15:34
• @higuys I used in step 3 that $\tau<\infty$ almost surely. Note that if $\tau(\omega)=\infty$ for some $\omega$, then $$M_{t \wedge \tau}(\omega) = M_{t}(\omega) \xrightarrow[]{t \to \infty} 0,$$ this follows from the fact that $X_t(\omega) < a$ for all $t$ (as $\tau(\omega)=\infty$) and $\mu>0$. Hence, by the dominated convergence theorem, $$\mathbb{E}(M_{\tau} 1_{\{\tau<\infty\}}+ 0 \cdot 1_{\{\tau=\infty\}})=1.$$ Therefore, we get $$\mathbb{E}(e^{-\lambda \tau} 1_{\{\tau<\infty\}}) = e^{(\mu-\sqrt{\mu^2+2\lambda})a}.$$ Now you can let $\lambda \to 0$ to conclude $\mathbb{P}(\tau<\infty)=1$. – saz Jan 28 '17 at 17:55
• @DaneelOlivaw 1.What do you mean by "couldn't we just state that [...]"? How do you justify that $\mathbb{E}(M_{t \wedge \tau}) = \mathbb{E}(M_{0 \wedge \tau})$? The optional sampling theorem shows that $(M_{t \wedge \tau})_t$ is a martingale and so the assertion follows; I fail to see how to prove this without the optional sampling theorem. 2. Typically you have to assume continuity of the sample paths. What exactly do you mean by "$X_{\tau}$ is not random variable"? $X_{\tau}=a$ is a random variable for sure (well, it's constant, but nevertheless it's a random variable). – saz Jun 30 '17 at 10:48
• @DaneelOlivaw Well, yes, there are several names for this result (optional sampling/optional stopping/...). In my answer I use the fact that $(M_t)_t$ martingale implies that $(M_{t \wedge \tau})_t$ is a martingale, and that's it. – saz Jun 30 '17 at 11:24