Brownian exit from an interval

For one-dimensional Brownian motion started at zero and $T={\tau }_b\wedge {\tau }_{-a}$ with $a,b>0$,Optional stopping of $B_t$ and $B_t^2-t$ proves the two identities.

For standard Brownian motion started at zero and exit time $T$ from $(-b,a)$, $a,b>0$, stop the two exponential Brownian martingales with parameters $\lambda$ and $-\lambda$. Their stopped values are bounded by $e^{|\lambda |\max (a,b)}$. The optional stopping theorem and dominated convergence theorem give $e^{\lambda a}p+e^{-\lambda b}q=1$ and $e^{-\lambda a}p+e^{\lambda b}q=1$, where $p,q$ are the discounted upper and lower exit probabilities. Solving gives the displayed formula and $q=\sinh (\lambda a)/\sinh (\lambda (a+b))$. ThusAt $\lambda =0$, the first ratio has its removable value $b/(a+b)$, and the second equals one. Finiteness of $T$ follows from a uniformly positive one-unit-time chance to leave any fixed bounded interval.

Successive exits from intervals of radius $a$ around the current position embed a simple symmetric random walk into Brownian motion. The Strong Markov property makes the durations independent with common law $a^2{\sigma }_1$; the displacements divided by $a$ are independent symmetric signs. The displacement sequence is independent across exits; no independence between a given displacement and its duration is needed for these conclusions.

The position is the sum of $k$ independent symmetric signs. Its characteristic function after division by $\sqrt{k}$ is $\cos (u/\sqrt{k}{)}^k\to e^{-u^2/2}$. The Lévy continuity theorem gives the normal distribution limit. Scaling the same skeleton to radius $2^{-n}$ and using Brownian exit-skeleton clock convergence identifies its other limit as $B_{\mathbb{E}{\sigma }_1}$, forcing $\mathbb{E}{\sigma }_1=1$.

Unit-radius Brownian exit time has finite variance, since independent unit-time increments give an exponential upper tail. The Brownian exit-time skeleton clock after $2^{2n}$ radius-$2^{-n}$ exits has mean $\mathbb{E}{\sigma }_1$ and variance $2^{-2n}\mathrm{Var}({\sigma }_1)$. Chebyshev's inequality and the Borel-Cantelli first lemma yield the displayed pathwise convergence. Independence between different rows indexed by $n$ is unnecessary.

For standard Brownian motion started at zero and first exit ${\tau }_x$ from $(-x,x)$, the Laplace transform is $1/\cosh (x\sqrt{2\lambda })$ for $\lambda >0$. The stochastic process $e^{-\lambda t}\cosh (\sqrt{2\lambda }{B}_t)$ is the average of two copies of the Exponential martingale for Brownian motion. Its values stopped at $t\wedge {\tau }_x$ are bounded, so bounded-time application of the optional stopping theorem followed by dominated convergence evaluates the transform without assuming uniform integrability of an unstopped exponential martingale on the infinite horizon.

For standard Brownian motion started at zero, let ${\tau }_x$ be its first exit from $(-x,x)$, with $x>0$. Bounded-time stopping of $B_t^2-t$ first proves integrability of the exit time and then gives its mean $x^2$. Stopping the quartic Hermite polynomial martingale gives a uniform bound on $\mathbb{E}(t\wedge {\tau }_x{)}^2$, proving second-moment integrability before passing to the limit. The resulting second moment is $5x^4/3$ and the variance is $2x^4/3$.

The exit time conditioned on leaving through the upper endpoint satisfiesOptional stopping of the cubic martingale $B_t^3-3t{B}_t$, together with the lower moments of the interval exit, gives the formula.