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Let $\left \{ B (t) \right \}$ be a standard Brownian motion, and let $T_a$ be the hitting time for that motion. We know that for $b < 0 < a$, the probability that $B (t)$ hits $a$ before $b$ is given by $$P( B(t) \text{ hits } a \text{ before } b ) = \frac{|b|}{a + |b|}.$$ Say $X (t)$ is a Brownian motion with drift coefficient $\mu$. Is it possible to find a similar formula? In particular, if $X (t) = \mu t + \sigma B (t)$, then can we find a closed form expression for $P ( X (t) \text{ hits } a \text{ before } b )$?

The progress I've been able to make is below. $$P( X(t) \text{ hits } a \text{ before } b ) = P \left( B(t) \text{ hits } \frac{a - \mu t}{\sigma} \text{ before } \frac{b - \mu t}{\sigma} \right)$$ $$ = \frac{ \left| \frac{b - \mu t}{\sigma} \right| }{ \frac{a - \mu t}{\sigma} + \left| \frac{b - \mu t}{\sigma} \right| }$$ My intuition is that this is not how you solve the problem. Any help would be appreciated.

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One way to get the "probability to hit $a$ before $b$" formula is to use the fact that Brownian motion is a martingale, by applying optional stopping at the time $T=T_a\wedge T_b$: $$ \eqalign{ 0=E(B_0)=E(B_T)&=a\cdot P(T_a<T_b)+b\cdot P(T_a<T_b)\cr &=a\cdot P(T_a<T_b)+b\cdot [1-P(T_a<T_b)].\cr } $$ For Brownian motion with drift $\mu$, the same line of reasoning works, but you need to use the martingale $e^{-2\mu B(t)}$.

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