Brownian Motion Exit From an Interval
Let $B_t$ be a standard Brownian motion starting at $0$. Fix $a, b > 0$ and define the first exit time
$$\tau = \inf\{t \geq 0 : B_t \notin (-b,\, a)\}$$
so $\tau$ is the first time the process leaves the interval $(-b, a)$.
- Compute $P(B_\tau = a)$ -- the probability that the process exits through the upper barrier.
- Compute $E[\tau]$ -- the expected time to exit.
- Interpret part (1) as the fair odds for a one-touch wager with asymmetric barriers. If a bookmaker offers a bet that pays $\$1$ when the process first hits $a$ (and nothing if it hits $-b$ first), what is the fair price?
Hints
- Brownian motion is a martingale. If you can apply the Optional Stopping Theorem at the exit time $\tau$, you can read off the hitting probability directly from $E[B_\tau] = 0$.
- For the expected exit time, you need a second martingale. Consider $B_t^2 - t$ -- verify it has zero drift using Ito's lemma, then apply OST again.
- Write $E[B_\tau^2] = p \cdot a^2 + (1-p) \cdot b^2$ where $p = b/(a+b)$ from part (1), and use $E[B_\tau^2] = E[\tau]$ to get the answer in one line.
Worked Solution
How to Think About It: You have a Brownian motion trapped between two walls at $-b$ and $a$. It will eventually escape through one of them. The question is: which wall does it hit, with what probability, and how long does it take on average? The tool here is the Optional Stopping Theorem (OST) -- Brownian motion is a martingale, so if you can apply OST to the exit time, you get the hitting probability for free. For the expected exit time, you need a second martingale.
Quick Estimate: Suppose $a = 3$ and $b = 1$. The starting point $0$ is much closer to the lower barrier $-b = -1$ than to the upper barrier $a = 3$. So intuitively, the process should hit $-1$ more often than it hits $3$. The formula gives $P(B_\tau = a) = b/(a+b) = 1/4$, which makes sense -- three-quarters of the time you exit downward through the closer barrier. For expected exit time: $E[\tau] = ab = 3$. If instead $a = b = 1$, symmetry gives $P = 1/2$ and $E[\tau] = 1$.
Approach: Apply OST to two different martingales: first $B_t$ itself (to get the hitting probability), then $B_t^2 - t$ (to get the expected exit time).
Formal Solution:
Part (1): Hitting probability
Since $B_t$ is a martingale and $\tau$ is a stopping time with $E[\tau] < \infty$, the OST gives
$$E[B_\tau] = E[B_0] = 0$$
At time $\tau$, the process is at either $a$ or $-b$. Let $p = P(B_\tau = a)$. Then
$$p \cdot a + (1-p)(-b) = 0$$
Solving:
$$pa + pb - b = 0 \implies p = \frac{b}{a+b}$$
This is a "gamblers ruin" result: the hitting probability depends only on the ratio of the distances to the two barriers, not on the volatility or any other parameter.
Part (2): Expected exit time
The process $M_t = B_t^2 - t$ is also a martingale (verify by Ito's lemma: $dM_t = 2B_t\,dB_t$, which has zero drift). Applying OST:
$$E[M_\tau] = E[M_0] = 0$$
$$E[B_\tau^2] - E[\tau] = 0$$
$$E[\tau] = E[B_\tau^2] = p \cdot a^2 + (1-p) \cdot b^2$$
Substituting $p = b/(a+b)$:
$$E[\tau] = \frac{b}{a+b} \cdot a^2 + \frac{a}{a+b} \cdot b^2 = \frac{a^2 b + a b^2}{a+b} = \frac{ab(a+b)}{a+b} = ab$$
Part (3): One-touch wager interpretation
The contract pays $\$1$ if $B_t$ hits $a$ before $-b$. Under risk-neutral pricing (Brownian motion is already a martingale), the fair price equals the probability of payout:
$$\text{Fair price} = P(B_\tau = a) = \frac{b}{a+b}$$
Notice the intuition: the fair price is determined entirely by the ratio of the lower barrier distance to the total barrier span. A symmetric barrier ($a = b$) gives a fair price of $1/2$. As the lower barrier gets closer ($b \to 0$), the contract becomes worthless (the process almost surely exits downward first). As $b \to \infty$, the price approaches $1$ (the lower barrier is so far away the process will hit $a$ first with near certainty). A bookmaker offering this bet at any price above $b/(a+b)$ is giving you an edge.
Answer:
$$P(B_\tau = a) = \frac{b}{a+b}, \qquad E[\tau] = ab, \qquad \text{Fair price} = \frac{b}{a+b}$$
Intuition
This problem is the continuous-time analog of the gambler's ruin. A gambler starts with $b$ dollars and plays a fair game until they either go broke or reach a total of $a + b$ dollars. The probability of reaching the target is $b/(a+b)$ -- exactly the same formula, because the discrete random walk converges to Brownian motion in the limit. The hitting probability depends only on relative distances to the barriers, not on the volatility or the time scale. This is a deep consequence of the scale invariance of Brownian motion.
The result $E[\tau] = ab$ is worth memorizing -- it shows up constantly in barrier option pricing, mean-reversion trading, and interview problems. The key technique is the "two-martingale trick": use $B_t$ for probabilities and $B_t^2 - t$ for expectations of stopping times. Any time you see a Brownian motion with absorbing barriers, these two applications of OST should be your first move. In practice, this framework extends to pricing double-barrier options and computing expected holding times for mean-reverting positions between profit targets and stop losses.