Expectation stopped Brownian motion with drift

Question

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?

Answer 1

Hints:

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.