Showing that $P(T_{-a}<T_b)=\frac{e^{2\mu b}}{e^{2\mu(a+b)}-1}$

Question

Let $X(t)= \mu t +W(t)$ be brownian motion with drift. And define the stopping time,

$$T_a = \{\inf{t\geq:X(t)=a}\} $$

I want to stop an appropriate martingale in order to show that,

$$P(T_{-a}<T_b)=\frac{e^{2\mu b}-1}{e^{2\mu(a+b)}-1} $$

with $a>0,b>0$

Answer 1

Hint

You need to find a function $f:\mathbb R\to \mathbb R$ such that $f(X_t)$ is a martingale. Assuming you can do that, then letting $T=\min(T_{-a},T_b)$, you would have on the one hand that $$ E[f(X_T)]=P(T_{-a}<T_{b})\cdot f(-a)+P(T_{-a}>T_b)\cdot f(b) $$ but on the other hand that $$E[f(X_T)]=f(X_0)=f(0).$$ Combined, these two equations woud prove $$ P(T_{-a}<T_b)=\frac{f(b)-f(0)}{f(b)-f(-a)} $$ which would then hopefully simplify to the expression you want.

Now, apply Itô's formula to find $df(X_t)$. If you assert that the $dt$ part of the result is equal to zero, you get a differential equation that $f$ must satisfy, allowing you to solve for $f$.