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?
Best Answer
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 , $E(\exp(-\frac{\tau_{b}\theta^{2}}{2})\mathbf{1}_{\tau_{b}<\tau_{-a}})=\frac{\sinh(a\theta)}{\sinh(\theta(b+a))}$