Problem:
Let $B_t$ be a Brownian motion with $B_0 = 0$, and for $\mu > 0$, let $X_t = B_t + \mu t$ (a
drifted Brownian motion). For any $a \in \mathbb{R}$, let $\tau_a = \inf\{t > 0; X_t = a\}$. Fix $a < 0 < b$. Use the martingale $M_t = \exp(\theta B_t − \theta^2t/2)$, $\theta \in \mathbb{R}$, to find $P(\tau_a < \tau_b)$.
Idea:
I'm not sure how to approach this with the drifted Brownian motion. Maybe there is some theorem I should be looking at?
I know that if $a < x < b$ then $P_x(\tau_a < \tau_b) = \frac{b − x}{b − a}$, with $\tau_a, \tau_b$ bounded stopping times. (in this situation $x=0$)
But how can we utilize this given $M_t$? Maybe look at $P(\tau_{b-a} < 0)$?
Best Answer
If we choose $\theta=-2\mu$, then $$M_t = \exp(\theta X_t).\tag{1}$$
Since $(M_t)_{t \geq 0}$ is a martingale, it follows from the optional stopping theorem that
$$\mathbb{E}(M_{t \wedge \tau})=\mathbb{E}(M_0)=1$$
for $\tau:=\min\{\tau_a,\tau_b\}$, i.e.
$$\mathbb{E}\exp(\theta X_{t \wedge \tau}) = 1.$$
By the continuity of the sample paths, we have $|X_{t \wedge \tau}| \leq \max\{|a|,|b|\}$ for all $t \geq 0$ and $X_{t \wedge \tau} \to X_{\tau}$ almost surely. Hence, by dominated convergence,
$$\mathbb{E}\exp(\theta X_{\tau})=1.$$
Moreover, the continuity of the sample paths yields that $X_{\tau}$ can only take the values $a$ and $b$; more precisely $X_{\tau}=a$ on $\{\tau_a<\tau_b\}$ and $X_{\tau}=b$ on $\{\tau_b<\tau_a\}$. Hence,
$$e^{\theta a} \mathbb{P}(\tau_a<\tau_b) + e^{\theta b} \mathbb{P}(\tau_b<\tau_a)=1. \tag{2}$$
On the other hand, $$\mathbb{P}(\tau_a<\tau_b) + \mathbb{P}(\tau_b<\tau_a)=1. \tag{3}$$
If we set $p:= \mathbb{P}(\tau_a<\tau_b)$, then
$$e^{\theta a} p + e^{\theta b} (1-p)=1,$$
i.e.
$$p = \frac{1-e^{\theta b}}{e^{\theta a}-e^{\theta b}} \stackrel{\theta=-2\mu}{=} \frac{1-e^{-2\mu b}}{e^{-2\mu a}-e^{-2\mu b}}.$$