# Use the martingale $M_t = \exp(\theta B_t − \theta^2t/2)$, $\theta \in \mathbb{R}$, to find $P(\tau_a < \tau_b)$

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 $$P(\tau_{b-a} < 0)$$?

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.

$$p = \frac{1-e^{\theta b}}{e^{\theta a}-e^{\theta b}} \stackrel{\theta=-2\mu}{=} \frac{1-e^{-2\mu b}}{e^{-2\mu a}-e^{-2\mu b}}.$$