Expected Exit Times from Bounded Intervals

The Exit-Time Question

Knowing which endpoint Brownian motion reaches first does not say how long the path typically takes to get there. For an interval with two finite endpoints, the mean exit time can be found without deriving the full distribution of the exit time. The key is to stop the martingale given by squared Brownian motion minus time. Because stopping times can be unbounded, the argument must first control the stopped process at bounded times and establish that the exit time has finite expectation.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying a standard Brownian motion $(B_t)_{t\ge 0}$, with $B_0=0$, adapted to its Brownian filtration $(\mathcal{F}_t)_{t\ge0}$. Fix finite real numbers $a<b$ and a starting point $x$ satisfying $a<x<b$. The Brownian motion started at $x$ is $X_t=x+B_t$. Define its exit time by

\tau=\inf\{t\ge0:X_t\notin(a,b)\}.

Continuity of Brownian paths means that, if $\tau<\infty$, then $X_\tau$ is one of the two endpoints $a$ and $b$. The time $\tau$ is a stopping time for the Brownian filtration. We use the standard bounded-time optional stopping theorem: if $M_t$ is a martingale and $\sigma$ is a bounded stopping time, then $\mathbb{E}[M_\sigma]=\mathbb{E}[M_0]$, provided the stopped variables are integrable. The martingales used below are $B_t$ and $B_t^2-t$.

Theorem: If $a<x<b$, then the exit time $\tau$ is almost surely finite, has finite expectation, and $$ \mathbb{E}[\tau]=(x-a)(b-x). $$ Moreover, $\mathbb{P}(X_\tau=b)=\frac{x-a}{b-a}$ and $\mathbb{P}(X_\tau=a)=\frac{b-x}{b-a}$.

Why Bounded-Time Stopping Comes First

We first stop at $\tau\wedge t=\min(\tau,t)$, which is a bounded stopping time for every fixed $t\ge0$. Until exit, $B_s=X_s-x$ lies between $a-x$ and $b-x$; at exit it equals one of those endpoints. Thus $B_{\tau\wedge t}$ is bounded in absolute value by $\max(x-a,b-x)$. In particular, the stopped variables are integrable. Applying bounded-time optional stopping to $B_s^2-s$ gives

\mathbb{E}\!\left[B_{\tau\wedge t}^{\,2}\right] =\mathbb{E}[\tau\wedge t].

The left side is at most $\max((x-a)^2,(b-x)^2)$, uniformly in $t$. Since $\tau\wedge t$ increases to $\tau$ as $t$ increases, monotone convergence now shows that $\mathbb{E}[\tau]\le\max((x-a)^2,(b-x)^2)<\infty$. A nonnegative random variable that equals infinity with positive probability has infinite expectation, so $\tau<\infty$ almost surely. This establishes finiteness rather than assuming it in an optional-stopping calculation.

Proof. For each fixed $t$, optional stopping applied to $B_s^2-s$ at $\tau\wedge t$ is valid because $\tau\wedge t\le t$ and $B_{\tau\wedge t}$ is bounded. It gives the displayed identity. Also, path continuity and the definition of $\tau$ imply $|B_{\tau\wedge t}|\le\max(x-a,b-x)$. Therefore $\mathbb{E}[\tau\wedge t]\le\max((x-a)^2,(b-x)^2)$. Taking $t$ increasingly large and using monotone convergence gives the same finite upper bound for $\mathbb{E}[\tau]$, which implies $\tau<\infty$ almost surely. $\square$

We can now pass from bounded stopping times to the exit time itself. As $t$ increases, $B_{\tau\wedge t}$ converges almost surely to $B_\tau$, and the variables remain bounded by the same constant. Bounded convergence gives convergence of their expectations and of their squared expectations. Meanwhile, monotone convergence gives $\mathbb{E}[\tau\wedge t]\to\mathbb{E}[\tau]$. Applying optional stopping to both martingales at $\tau\wedge t$ consequently yields

\mathbb{E}[B_\tau]=0, \qquad \mathbb{E}[B_\tau^2]=\mathbb{E}[\tau].

Since $X_\tau\in\{a,b\}$, the first equality determines the exit probabilities. Writing $p=\mathbb{P}(X_\tau=b)$, we have $B_\tau=b-x$ on an upper exit and $B_\tau=a-x$ on a lower exit. Hence

0=p(b-x)+(1-p)(a-x), \qquad p=\frac{x-a}{b-a}.

The lower-exit probability is $1-p=(b-x)/(b-a)$. Substituting these probabilities into the second-moment identity gives

\begin{aligned} \mathbb{E}[\tau] &=\mathbb{E}[B_\tau^2]\\ &=\frac{x-a}{b-a}(b-x)^2 +\frac{b-x}{b-a}(a-x)^2\\ &=\frac{(x-a)(b-x)}{b-a}\big((b-x)+(x-a)\big)\\ &=(x-a)(b-x), \end{aligned}

where $(a-x)^2=(x-a)^2$ and $(b-x)+(x-a)=b-a$. This proves both the exit probabilities and the expected exit-time formula. $\square$

Worked Calculations

Worked Example: Exit from a Symmetric Interval

Let $X_t=B_t$ start at $x=0$, and let $\tau$ be the first exit from $(-2,2)$. The upper-exit probability is $(0-(-2))/(2-(-2))=2/4=1/2$; the lower-exit probability is $1-1/2=1/2$. The terminal Brownian values are $2$ and $-2$, whose squares are both $4$. Thus the stopped-square calculation gives

\mathbb{E}[\tau] =\tfrac12(2^2)+\tfrac12((-2)^2) =\tfrac12(4)+\tfrac12(4) =4.

The interval formula gives the same result: $(0-(-2))(2-0)=2\cdot2=4$. Check: the exit probabilities sum to $1/2+1/2=1$, and their weighted squared displacements sum to $4$.

Worked Example: Unequal Distances to the Endpoints

Start at $x=2$ in the interval $(0,5)$. The distances to the lower and upper endpoints are $2$ and $3$, respectively. The probability of exiting at $5$ is $2/5$, and the probability of exiting at $0$ is $3/5$. Relative to the starting point, the terminal Brownian values are $3$ on an upper exit and $-2$ on a lower exit. Therefore

\mathbb{E}[\tau] =\tfrac25(3^2)+\tfrac35((-2)^2) =\tfrac{18}{5}+\tfrac{12}{5} =\tfrac{30}{5} =6.

Direct substitution into the formula gives $(2-0)(5-2)=2\cdot3=6$. Check: $2/5+3/5=1$, and the weighted squared terminal displacements equal $6$.

Worked Example: An Off-Centre Start

Let the interval be $(-1,3)$ and the starting point be $x=0$. The upper-exit probability is $(0-(-1))/(3-(-1))=1/4$, leaving lower-exit probability $3/4$. At exit, $B_\tau=3$ on an upper exit and $B_\tau=-1$ on a lower exit. The expected exit time is

\mathbb{E}[\tau] =\tfrac14(3^2)+\tfrac34((-1)^2) =\tfrac94+\tfrac34 =\tfrac{12}{4} =3.

The formula verifies this value directly: $(0-(-1))(3-0)=1\cdot3=3$. Check: the probabilities total $1/4+3/4=1$, and their weighted squared displacements also total $3$.

Where Is the Mean Exit Time Largest?

For a fixed interval, write its length as $L=b-a$ and its midpoint as $m=(a+b)/2$. The formula can be rewritten as

(x-a)(b-x)=\frac{L^2}{4}-(x-m)^2.

Corollary: Among starting points $x\in(a,b)$, the expected exit time is largest at the midpoint and its maximum value is $(b-a)^2/4$.

Proof. Since $(x-m)^2\ge0$, the displayed identity gives $(x-a)(b-x)\le L^2/4$. Equality holds exactly when $(x-m)^2=0$, which is equivalent to $x=m$. The midpoint belongs to $(a,b)$ because $a<b$, so the bound is attained there. $\square$

This result makes the dependence on location explicit. Starting closer to either boundary reduces the mean time to exit; in the limiting boundary case the exit time is zero. The formula also scales quadratically with interval length: multiplying both distances from the starting point to the endpoints by a factor multiplies their product, and hence the expected exit time, by the square of that factor.

A Common Probability Error

A frequent mistake is to assign the upper-exit probability using the distance from the starting point to the upper endpoint. The probability is instead proportional to the distance from the starting point to the opposite, lower endpoint. For example, in $(0,4)$ starting at $x=1$, the correct upper-exit probability is $(1-0)/(4-0)=1/4$, not $3/4$. The correct weighted-square calculation is

\tfrac14(4-1)^2+\tfrac34(0-1)^2 =\tfrac94+\tfrac34 =3.

Using the incorrect probability $3/4$ for the upper exit would give $\tfrac34(3^2)+\tfrac14(1^2)=7$, which is not the expected exit time. The formula $(1-0)(4-1)=3$ confirms the correct calculation. The weights must be the endpoint probabilities derived from $\mathbb{E}[B_\tau]=0$, not an intuitive guess based on the nearer endpoint.

The method gives the mean and exit probabilities, but not the full distribution of $\tau$. In particular, the expectation alone does not describe the spread or shape of exit times. The essential discipline here is to apply optional stopping first at bounded times, prove finiteness, and only then take the limit to the exit time.

Check Your Understanding

For each question, assume Brownian motion starts strictly inside a bounded interval and use exact values.

For an interval $(-3,1)$ and starting point $x=-1$, find both endpoint exit probabilities and the expected exit time.
For an interval $(2,8)$, at which starting point is the expected exit time largest, and what is that maximum?
Explain why applying optional stopping directly at an unbounded exit time needs justification, even though the formula for the expected exit time is finite.
For the interval $(0,5)$ starting at $x=2$, calculate the expected squared terminal Brownian value from the two exit possibilities.
What changes in the exit-time formula if the starting point is exactly one of the endpoints?