Martingale expectation value I want to show that for $(B_t)_t$ being the Brownian motion and a stopping time $\tau:= \text{inf}_{t \ge 0} \{B_t= a+bt\}$ where $a,b>0$ we have that the expectation value $E(e^{-\lambda  \tau}, \tau< \infty) = e^{-a(b+ \sqrt{b^2 +2 \lambda})}$ for $\lambda >0.$
There is also a hint saying I should use the martingale $M(t)=e^{c B(t)- \frac{c^2t}{2}}$ where $c \ge 0.$
I guess this can be derived somehow from the optional stopping theorem, but I don't see how all this fits together. So I would start with 
$1=E(M(0))=E(M(\tau))$ but I don't see how I can evaluate the right hand-side so that I get the desired expression. Also I am not sure that the conditions of the optional stopping theorem are satisfied.
 A: Not so sure it is still useful after 2 years, anyway 
1) first you need to compute the Laplace transform of the hitting time of llevel $a$ for a standard Brownian Motion, say $B_t$, under a measure $P$. 
Define $\tilde{\tau}:= \inf \{t:\ B_t\geq a \}$. In order to do so, we use the hint you proposed, that is
$$E_P(e^{\theta B_t- \frac{\theta^2t}{2}})=1,$$
which entails that, using Doob's Optional Stopping Theorem, since $B_{\tilde{\tau}}=a$,
$$E_P(e^{\theta a- \frac{\theta^2\tilde{\tau}}{2}})=1,$$
or equivalently:
$$e^{-\theta a}= E_P( e^{\frac{\theta^2\tilde{\tau}}{2}}).$$
Now define $\lambda:=\frac{\theta^2}{2}  $, so that $\theta = \sqrt{2 \lambda}$.
Replacing this quantities in the above equation, it follows that
$$ E_P( e^{-\lambda\tilde{\tau}})= e^{\sqrt{2 \lambda}}.$$
2) Now we use Girsanov's Theorem. In particular there exists a measure $Q_t$ for which $\frac{dQ_t}{dP_t} = e^{ \mu B_t+ \frac{\mu^2}{2} t}$ and $ \tilde{B}_t =B_t - \mu t $ is a Brownian motion under $Q_t$.
We can apply the result in step 1) for $\tilde{B}_t$, in particular:
$$E_P( e^{-\lambda\tilde{\tau}})= E_Q( e^{-\lambda \tau} e^{ \mu B_\tau+ \frac{\mu^2}{2} \tau} ) = e^{\sqrt{2 \lambda}}.$$
Therefore
$$E_Q(e^{-\tau(-\lambda+\frac{\mu^2}{2} )})= e^{-a(\mu+\sqrt{2\lambda})}.$$
Define $\hat{\lambda}:=-\lambda+\frac{\mu^2}{2}, $ so $\lambda = \tilde{\lambda}+\frac{\mu^2}{2}$. Replacing in the above equation this quantity you get the conclusion.
