Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 5

Let $R=|U|$. Uniform volume measure on the unit ball gives $\mathbb{P}(R\le r)=r^n$, $0\le r\le 1$, hence radius density $n{r}^{n-1}$. Conditional on $R=r>0$, independence leaves $W$ a standard Brownian motion. Its exit time from the radius-$r$ ball is finite: applying optional stopping to $|W_t{|}^2-nt$ stopped at that exit and then letting $t\to \infty$ gives $\mathbb{E}{\tau }_r=r^2/n$.

The probability law of the exit point is invariant under every orthogonal transformation, because both Brownian motion and the centred sphere are invariant under those transformations. The unique invariant probability on that sphere is normalized surface measure; for $n=1$ it is the equally weighted pair $\{-r,r\}$. This is also the conditional angular law of $U$ given $|U|=r$. Mixing these identical conditional laws against $n{r}^{n-1}dr$ proves the uniform ball sampling by Brownian stopping identityThe event $R=0$ has probability zero and causes no difficulty. Independence of $R$ permits its value to be included in the initial filtration, making $T$ a stopping time.

Allow $g_D$ to take the value $+\infty$. Fix a closed ball $\overline{B(z,r)}\subset D$, independently choose a radius $R$ with density $n{s}^{n-1}/r^n$ on $(0,r)$, and let $S$ be the first time the Brownian motion from $z$ reaches that radius. Conditional on $R=s$, the Strong Markov property at $S$ decomposes the remaining Brownian exit time from $D$. Since $S<T_D$ and the conditional exit time has mean $s^2/n$, Tonelli's theorem and part (a) give the Brownian exit-time ball averaging identityThis identity is valid in the extended nonnegative reals; no finiteness was assumed to derive it.

If $g_D(z)<\infty$, the integral over $B(z,r)$ is finite. For any $y\in B(z,r)$ choose $\rho >0$ with $\overline{B(y,\rho )}\subset B(z,r)$. The same identity at $y$ has a finite right side because its integral is over a subset of the integrable ball. Thus $g_D(y)<\infty$. The set $F=\{z\in D:g_D(z)<\infty \}$ is therefore open relative to $D$.

It is also relatively closed. If $z_j\in F$ converges to $y\in D$, choose $r>0$ with $\overline{B(y,3r)}\subset D$. For large $j$, $|z_j-y|<r$, and $\overline{B(z_j,2r)}\subset D$. The preceding argument shows that every point in $B(z_j,2r)$, including $y$, belongs to $F$. Hence $F$ is closed in $D$. Since $F$ is nonempty and $D$ is connected, $g_D(y)<\infty$ for every $y\in D$. This proves finiteness propagation of Brownian mean exit times without assuming beforehand that $g_D$ solves a smooth Poisson equation.

Start at $x=(x_1,\dots ,x_n)$ with all $x_i>0$, and let ${\tau }_i$ be the first zero of the $i$th coordinate. The orthant exit time is $T={\min }_i{\tau }_i$. The coordinate processes are independent, soHere the Brownian reflection principle gives ${\mathbb{P}}_a(\tau \le t)=2\mathbb{P}(B_t\le -a)$; $\Phi$ is the standard normal distribution function. Since $\Phi (z)-1/2=z/\sqrt{2\pi }+o(z)$ as $z\to 0$,The tail integral formula for expectation, ${\mathbb{E}}_{x}T={\int }_0^{\infty }{\mathbb{P}}_x(T>t)dt$, has a bounded integrand on $(0,1)$ and converges at infinity exactly when $n/2>1$. Therefore the Brownian exit-time expectation in an orthant givesThe divergence for $n=2$ is logarithmic; almost-sure finiteness of the exit time does not imply a finite expectation.