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I found this interesting question on the internet, but unfortunately I could not solve it.

What is probability that Brownian motion (starting at origin) has value 1 before having value -2?

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Two-thirds. $ $ – Did Jan 9 '13 at 12:58
AFAIK, the hitting of $a$ before $b$ by the Brownian motion starting from $x$ is a solution of the following equation $f''(x )= 0$ with boundary conditions $f(a) = 1$ and $f(b) = 0$. There is the unique solution, which is also given by a piecewise-affine function, and in your case $f(0) = 2/3$. – Ilya Jan 9 '13 at 13:00
@Ilya Can you please elaborate the claim about function $f$? I appreciate your help. – John Peter Jan 9 '13 at 13:03
@JohnPeter: I am not sure, where I've seen such approach - maybe Oksendal. Unfortunately, I can't provide you some guidelines how to obtain this equation. – Ilya Jan 9 '13 at 13:09
up vote 2 down vote accepted

Take $B$ to be a Brownian motion. Let

$$\tau = \inf\{t\geq 0: B_t\geq 1, B_t\leq -2\}.$$

Then $\tau$ is a stopping time. $B$ is a martingale, so $B^\tau$ - $B$ stopped at time $\tau$, is also a martingale. $B^\tau$ is uniformly integrable, since it is bounded, so we may apply the Optional Stopping Theorem to $B^\tau$ at time $\tau$ to get

$$ 0 = \mathbb{E}[B^\tau_\tau] = \mathbb{E}[B_\tau] = 1\mathbb{P}[B \text{ hits 1 before -2}]+(-2)\mathbb{P}[B \text{ hits -2 before 1}]. $$

Since precisely one of these events occurs, we may solve for $\mathbb{P}[B \text{ hits 1 before -2}]$, giving the probability as $\frac23$.

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