Brownian Exit Time

Let $B_t$ be a standard Brownian motion started at $0$ , and let $\tau$ be the first time it reaches either $+1$ or $-1$ . What is the mean of $\tau$ ?

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#A martingale built from the path

For standard Brownian motion the process $B_t^2 - t$ is a martingale, because the variance of $B_t$ grows exactly like $t$ . Started at $0$ , this process begins at the value $0$ .

#Stop at the barrier

At the exit time $\tau$ the path sits on a barrier, so $B_\tau^2 = 1$ no matter which side it hits. Optional stopping keeps the martingale's mean pinned at its starting value,

\E[B_\tau^2 - \tau] = 0 \quad\Longrightarrow\quad \E[\tau] = \E[B_\tau^2] = 1. \tag{1}

Because B_t squared minus t is a martingale and the path squares to 1 the moment it is absorbed, optional stopping turns the expected hitting time into the expected squared position, which is 1.

#Read it off

\E[\tau] = 1. \tag{2}

More generally, with barriers at $-a$ and $b$ the same martingale gives an expected exit time of $ab$ , the product of the two distances. Symmetric barriers at $\pm 1$ make that product simply $1$ .