Brownian exit time ($T_D$)

For a domain $D$, the Brownian exit time is $T_D=\inf \{t\ge 0:B_t\notin D\}$. It is a stopping time, and it has finite expectation when $D$ is bounded.

For $d\ge 3$ and $|x|>\epsilon$, stop the harmonic function $u(x)=|x{|}^{2-d}$ at the two boundaries of $\{\epsilon <|x|<R\}$. Bounded optional stopping gives the probability of hitting the inner sphere first as $(u(x)-u(R))/(u(\epsilon )-u(R))$. Letting $R\to \infty$ gives the displayed formula; a finite hitting path is contained in some finite-radius ball.

For $d\ge 3$, fixed $x\ne 0$ and $t>0$, apply optional stopping to $u(B_{t\wedge T_{\epsilon }})$ with $u(z)=|z{|}^{2-d}$. The resulting identity is ${\epsilon }^{2-d}{\mathbb{P}}_x(T_{\epsilon }\le t)=u(x)-\mathbb{E}u(B_t)+\mathbb{E}[u(B_t);T_{\epsilon }\le t]$. The last term tends to zero by integrability of $u(B_t)$ and the vanishing total hitting probability. The Newtonian potential of the Brownian heat kernel identifies the remaining difference.

For independent coordinates of standard Brownian motion and $x_i>0$, the Brownian reflection principle gives ${\mathbb{P}}_x(T>t)={\prod }_i[2\Phi (x_i/\sqrt{t})-1]$. This is asymptotic to $(2/\pi {)}^{n/2}{\prod }_i{x}_i{t}^{-n/2}$. The tail integral formula for expectation therefore proves the stated criterion, with logarithmic divergence in dimension two.

For standard Brownian motion started at zero, choose an independent radius $R$ with $\mathbb{P}(R\le s)=(s/r{)}^n$, $0\le s\le r$, and stop on first reaching that radius. Rotational invariance of Brownian motion makes the conditional exit position uniform on each sphere. Mixing with radius density $n{s}^{n-1}/r^n$ gives the uniform probability distribution on the ball. In dimension one, the conditional sign has probabilities $1/2,1/2$. The stopped point samples volume measure even though each conditional exit law is supported on a boundary.

Let $g_D(x)={\mathbb{E}}_x{T}_D$, possibly infinite, and suppose the closed ball $\overline{B(x,r)}$ lies in the open set $D$. A Strong Markov property decomposition at an independent random-radius exit time, using uniform ball sampling by Brownian stopping, gives the displayed identity in the nonnegative extended reals. Indeed the expected exit time from a radius-$s$ ball started at its centre is $s^2/n$, and averaging this against $n{s}^{n-1}/r^n$ gives $r^2/(n+2)$. The identity gives local integrability whenever $g_D(x)$ is finite.

For a connected open set $D$, finiteness of the Brownian exit time expectation at one point implies finiteness everywhere. The Brownian exit-time ball averaging identity gives integrability on a ball centred at any finite point. Every point of that ball has a smaller centred ball contained in it; the same identity then gives a finite value there. The set of finite points is open. It is also relatively closed: near a limit point in $D$, choose a fixed ball radius smaller than a quarter of its distance to $D^c$ and centre it at a nearby finite point. The limit point lies in that ball and hence is finite. Connectedness completes the proof.

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.

For a suitable twice differentiable function $u$ and an integrable stopping time $T$,

It is the Dynkin's formula with the Brownian motion generator $\frac{1}{2}\Delta$.