Math-1687 · mathematics · Probability

Math-1687

Probabilistic McMillan theorem in higher dimensions

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Statement

Definitions and mathematical statement

Let $d\ge3$ and let $X=(X_t)_{t\ge0}$ be standard Brownian motion in $\mathbb R^d$, started at $0$. Thus its coordinates are independent real Brownian motions with variance $t$ at time $t$. Let $D\subseteq\mathbb R^d$ be open with $0\in D$, and define the exit time $\tau=\inf\{t>0:X_t\notin D\}$.

On the event $\tau<\infty$, continuity gives $X_\tau\in\partial D$. For $t<\tau$, put $u_t=(X_t-X_\tau)/|X_t-X_\tau|$, a point of the unit sphere $S^{d-1}$. Let $A$ consist of all limits of sequences $u_{t_n}$ with $t_n\uparrow\tau$. Equivalently, \[ A=\bigcap_{0<s<\tau}\overline{\{u_t:s<t<\tau\}}\subseteq S^{d-1}. \] Must $A$, almost surely on finite exit, equal either the whole sphere or a closed hemisphere $\{u\in S^{d-1}:u\cdot v\ge0\}$ for some unit vector $v$?

The hemisphere and the choice between the two alternatives may depend on the sample path. The assertion is universal over domains, without a smoothness or connected-boundary assumption. The finite-exit qualification is necessary because an arbitrary unbounded domain need not be exited, and then $X_\tau$ is not defined. For bounded domains it is automatic. The question concerns the entire set of limiting approach directions, rather than the existence of a single tangent direction or a limiting normal vector.

Short English statement

At a finite Brownian exit from any higher-dimensional domain, are the limiting approach directions always a sphere or a closed hemisphere?

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