Let $B$ be a Brownian motion. For all $y\in \Bbb{R}_+^\star$ we define $\tau_y:=\inf\{t\geq 0: B_t\geq y\}$. Now let us fix $a,b>0$ and define $\tau_{a,b}:=\tau_{-a}\wedge \tau_b$.
Now my first question arises here, what is $\tau_{-a}$? Because since $-a<0$ we cannot say $\tau_{-a}=\inf\{t\geq 0: B_t\geq -a\}$ since this would only be defined for $y>0$.
This question is really relevant since I cannot solve the following:
I want to compute $\Bbb{E}\left(\exp\left(-\frac{\theta^2}{2} \tau_b\right) \Bbb{1}_{\tau_b<\tau_{-a}}\right)=\frac{\sinh(\theta a)}{\sinh(\theta(a+b))}$ where $\theta\in \Bbb{R}$.
The hint was to use that from class we know that $X_t:=\sinh(\theta(B_t+a))\exp\left(-\frac{\theta^2}{2} t\right) $ is a martingale.
My idea was then to apply the optional stopping theorem since I know that $\tau_{a,b}$ is a stopping time. Then I would get that $$\Bbb{E}(X_{\tau_{a,b}})=\Bbb{E}(X_0)=\sinh(\theta a)$$ Now on the other hand I would need to compute $\Bbb{E}(X_{\tau_{a,b}})=\Bbb{E}\left(X_{\tau_{-a}}\Bbb{1}_{\tau_{-a}<\tau_b}\right)+\Bbb{E}\left(X_{\tau_{b}}\Bbb{1}_{\tau_{b}<\tau_{-a}}\right)$ . Now Since I don't know that $\tau_{-a}$ exactly is I also don't know what $X_{\tau_{-a}}$ is therefore I'm stuck here. Can someone help me further?
You have guessed all steps correctly but you are having trouble to finish the proof. Well, contrary to my comment, I am here again posting an answer to your question :) . To answer your first question, $\tau_{-a}$ is the first time the Brownian motion hits the level $-a$ which is equivalent to saying $\tau_{-a}=\inf\{t\geq 0: B_{t}\leq -a\}$ . You should remember that $\lim\inf B_{t}=-\infty$ and $\lim\sup B_{t}=\infty$ . And not only that , the hitting time of any real number $\tau_{r}$ is finite almost surely . In fact, due to recurrence, the Brownian motion will hit $r$ infinitely often.
First notice that $T=\tau_{-a}\wedge\tau_{b}$ is such that $X_{T\wedge t}$ is uniformly bounded in $L^{\infty}$ (i.e. the supremum of this Martingale is a finite fixed real number) where $X_t =\sinh(\theta(B_t+a))\exp\left(-\frac{\theta^2}{2} t\right) $ for $\theta \in \mathbb{R}$ . This is just because $|B_{t\wedge T}|\leq \max(-a,b)$ . And thus $X_{t\wedge T}$ is uniformly integrable.
So you can apply the optional stopping theorem to the Martingale $M_{t\wedge T}$ to conclude that $E(X_{T})=E(X_{0})=\sinh(a\theta)$
As you again correctly thought ,
$E(X_T)=E\left(X_{\tau_{-a}}\mathbf{1}_{\tau_{-a}<\tau_b}\right)+E\left(X_{\tau_{b}}\mathbf{1}_{\tau_{b}<\tau_{-a}}\right)$
But notice that $X_{\tau_{-a}}=\sinh(\theta(B_{\tau_{-a}+a}))\exp(-\frac{t\theta^{2}}{2})=\sinh((-a+a)\theta)\exp(-\frac{t\theta^{2}}{2})=0$
So $E\left(X_{\tau_{b}}\mathbf{1}_{\tau_{b}<\tau_{-a}}\right)=E(X_{T})=\sinh(a\theta)$
And $$E(X_{\tau_{b}}\mathbf{1}_{\tau_{b}<\tau_{-a}})=E\bigg((\sinh(\theta(b+a))\exp(-\frac{\tau_{b}\theta^{2}}{2})\mathbf{1}_{\tau_{b}<\tau_{-a}})\bigg)=\\\sinh(\theta(b+a))E(\exp(-\frac{\tau_{b}\theta^{2}}{2})\mathbf{1}_{\tau_{b}<\tau_{-a}})$$
Thus you have , $E(\exp(-\frac{\tau_{b}\theta^{2}}{2})\mathbf{1}_{\tau_{b}<\tau_{-a}})=\frac{\sinh(a\theta)}{\sinh(\theta(b+a))}$