Expected Time for Brownian Motion to Hit -1 or +1
Let $B_t$ be a standard Brownian motion with $B_0 = 0$, and let
$$T = \inf\{t \ge 0 : B_t = 1 \text{ or } B_t = -1\}$$
be the first time it reaches either $+1$ or $-1$. Find $E[T]$.
Generalize: if the barriers are at $+\alpha$ and $-\beta$ (with $\alpha, \beta > 0$), what is the expected exit time?
Hints
- $B_t$ itself is a martingale, but it only tells you which barrier is hit. You need a martingale that involves $t$.
- Apply Ito's lemma to $B_t^2$: $d(B_t^2) = 2B_t\,dB_t + dt$, so $B_t^2 - t$ is a martingale.
- Optional stopping at $T$ (justified because $E[T] < \infty$ and $B_{t \wedge T}$ is bounded) gives $E[B_T^2] = E[T]$, and $B_T^2 = 1$ always.
Worked Solution
How to Think About It: To extract an expected time from a stopping problem you need a martingale in which $t$ appears explicitly. For Brownian motion the canonical one is $B_t^2 - t$. At the exit time the position is known exactly ($\pm 1$), so its square is deterministic and optional stopping hands you $E[T]$ immediately.
Quick Estimate: $B_t$ has standard deviation $\sqrt{t}$; it typically first wanders a distance of 1 when $\sqrt{t} \approx 1$, i.e. $t \approx 1$. The exact answer is exactly 1.
Formal Solution:
*Step 1 -- A martingale involving time.* By Ito's lemma with $f(x) = x^2$,
$$d(B_t^2) = 2B_t\,dB_t + \tfrac{1}{2}\cdot 2\,(dB_t)^2 = 2B_t\,dB_t + dt.$$
Hence $M_t = B_t^2 - t$ satisfies $dM_t = 2B_t\,dB_t$, a driftless Ito integral, so $M_t$ is a martingale.
*Step 2 -- Optional stopping.* $T$ is a stopping time with $E[T] < \infty$ (for instance because in each unit time interval the process has probability at least $P(|N(0,1)| > 2) > 0$ of exiting regardless of its start point inside $(-1,1)$, so $T$ has a geometric tail). For the bounded stopped process $B_{t \wedge T}$, optional stopping applied to $M_{t \wedge T}$ and dominated/monotone convergence give
$$E[B_T^2] - E[T] = E[M_T] = E[M_0] = 0.$$
*Step 3 -- Evaluate.* At time $T$, $B_T \in \{-1, +1\}$, so $B_T^2 = 1$ with certainty:
$$E[T] = E[B_T^2] = 1.$$
*Step 4 -- Asymmetric barriers $+\alpha$, $-\beta$.* First use the martingale $B_t$: $E[B_T] = 0$ gives $p\alpha - (1-p)\beta = 0$ with $p = P(B_T = \alpha)$, so $p = \beta/(\alpha + \beta)$. Then
$$E[T] = E[B_T^2] = \frac{\beta}{\alpha+\beta}\alpha^2 + \frac{\alpha}{\alpha+\beta}\beta^2 = \frac{\alpha\beta(\alpha + \beta)}{\alpha + \beta} = \alpha\beta.$$
With $\alpha = \beta = 1$ this is again $1$.
Answer: $E[T] = 1$; in general the expected time to exit $(-\beta, \alpha)$ is $\alpha\beta$.
Intuition
Brownian motion spreads out like $\sqrt{t}$, so reaching distance 1 should take time of order $1^2 = 1$, and the martingale $B_t^2 - t$ makes that exact: at the exit time $B_T^2$ is identically 1, so $E[T] = 1$. With asymmetric barriers the same argument gives $E[T] = \alpha\beta$, the continuous-time gambler's-ruin duration. This "quadratic martingale" trick is the standard way to get expected holding times for a position closed at a take-profit or stop-loss level, and it is the first computation in any barrier-option or mean-reversion timing problem.