Brownian Exit Probability

Key idea. Brownian motion is a martingale, and it is bounded up to the exit time, so optional stopping applies.

First we show that $\tau<\infty$ almost surely. The process $W_t^2-t$ is a martingale, so optional stopping at the bounded time $\tau\wedge t$ gives

Letting $t\to\infty$, monotone convergence gives $\mathbb{E}[\tau]\le 4$. Hence $\tau<\infty$ almost surely.

Next, $W_{\tau\wedge t}$ is a martingale with $|W_{\tau\wedge t}|\le 2$, so $\mathbb{E}[W_{\tau\wedge t}]=0$ for every $t$. As $t\to\infty$, $W_{\tau\wedge t}\to W_\tau$ almost surely. Bounded convergence then gives $\mathbb{E}[W_\tau]=0$.

Write $p=\mathbb{P}[W_\tau=-2]$. Since $W_\tau\in\{-2,1\}$,

so $3p=1$ and

Alternative. This uses symmetry and the strong Markov property.

• Starting from $0$, the path hits $\pm1$ first with probability $\tfrac12$ each.
• If it hits $-1$, then from there it reaches $-2$ before $0$ with probability $\tfrac12$.
• If it returns to $0$ instead, the problem restarts.

Hence $p=\tfrac12\big(\tfrac12+\tfrac12p\big)$, which gives $p=\tfrac13$.

Generalisation. For $a,b>0$, the same argument with the levels $-a$ and $b$ gives

Optional stopping applied to $W_t^2-t$ also gives $\mathbb{E}[\tau]=ab$. Here $\mathbb{E}[\tau]=2$.