Calculate Probability of Bachier model hitting upper barrier

Question

Let $B_t$ be a standard brownian motion and $\mu, \sigma \in \mathbb{R}$ with $\sigma \neq 0$. Now let $X_t = \sigma B_t + \mu t$ be the Bachier model with votality $\sigma$ and drift $\mu$. Further let $a,b > 0$ and define the stopping time $\tau = \text{inf} \{t \geq 0 : X_t = a$ or $X_t = -b\}$.
As part of an exercise I need to prove that: \begin{equation} \mathbb{P}(X_\tau=a) = \frac{\text{exp}(-\frac{2\mu}{{\sigma}^2})-1}{\text{exp}(-\frac{2\mu}{{\sigma}^2}(a+b))-1} \end{equation} I am not sure how to tackle this problem. We did prove in our lecture that if we define the function $g: \mathbb{R} \to \mathbb{R}$ as: \begin{equation} g(x) = \frac{\text{exp}(-\frac{2\mu }{{\sigma}^2}x)-\text{exp}(\frac{2\mu }{{\sigma}^2}b)}{\text{exp}(-\frac{2\mu}{{\sigma}^2}a)-\text{exp}(\frac{2\mu }{{\sigma}^2}b)} \end{equation} then $g(X_t)$ is a martingale satisfying $g(a) = 1$ and $g(-b) = 0$, but I don't see how this helps. All this tells me is \begin{equation} \mathbb{P}(X_\tau=a) = \mathbb{P}(g(X_{\tau})=1) \end{equation} Any help or hints would be appreciated.

Answer 1

As you note, $g(X_t)$ is a martingale; therefore (and more useful) the stopped process $Y_t:=g(X_{t\wedge \tau})$ is a bounded martingale, with $Y_0=g(0)$. Therefore $$ g(0)=E[Y_0] = E[Y_\tau]= g(a)P[X_\tau = a]+g(-b)P[X_\tau=-b]. $$ (This requires the previously garnered knowledge (or the stipulation) that $P[\tau<\infty] = 1$.)