Let $(B_t)$ a standard Brownian motion, $\mu \in \mathbb{R}$, $a \in \mathbb{R_+}$ and ${\tilde{B}}_t = B_t + \mu t$. Now we define the stopping time $\tau_a = \inf(t\geq0,{\tilde{B}}_t\geq0) $.
I have to show that $$E[\exp(\lambda {\tilde{B}}_{t\land \tau_a} - (\frac{\lambda^2}{2}+\lambda\mu)(t\land \tau_a)] = 1 $$
I simply write that $\exp(\lambda {\tilde{B}}_{t} - (\frac{\lambda^2}{2}+\lambda\mu)t) = \exp(\lambda B_t - \frac{\lambda^2}{2}t)$ which is a continuous martingale and then, $\tau_a$ being a stopping time, $\exp(\lambda B_{t\land \tau_a} - \frac{\lambda^2}{2}{t\land \tau_a})$ is a martingale too with $$E[\exp(\lambda B_{t\land \tau_a} - \frac{\lambda^2}{2}({t\land \tau_a})) = E[\exp(\lambda B_{0\land \tau_a} - \frac{\lambda^2}{2}({0\land \tau_a})) = 1$$
This should gives us the answer but in the correction, we use the fact that $t\land \tau_a$ is a bounded stopping time for a fixed $t$, and then we can use the Stopping theorem to the martingale $\exp(\lambda B_{t} - \frac{\lambda^2}{2}{t})$ and get the result.
I understand this point but I wonder if my answer is correct. It seems more direct than the correction, more "natural", did I miss something?