Expectation stopped Brownian motion with drift
Hints:
- 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).
- By the optional stopping theorem, $$\mathbb{E}(M_{\tau \wedge t}) = \mathbb{E}(M_0) = 1, \qquad t \geq 0.$$
- Show that $|M_{t \wedge \tau}| \leq e^{\alpha a}$. Deduce from the dominated convergence theorem that $$\mathbb{E}(M_{\tau}) = 1.$$
- Since $(X_t)_{t \geq 0}$ has continuous sample paths, we have $X_{\tau}=a$. Hence, $$M_{\tau} = e^{\alpha a} \exp \left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right).$$
- 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.