Independence of conditional hitting distribution and hitting time

Question

Suppose given a d-dimensional Brownian motion $B_t$ starting from the origin and a centered ball with radius 1. Define T as the first hitting time of the sphere (boundary of the ball). How can one prove that T and $B_T$ are independent?

Answer 1

The short, and somewhat heuristic, answer is that rotational invariance implies that given $T$ the distribution of $B_T$ is uniform on the sphere $S^{d-1}$. Since the conditional distribution does not depend on $T$, this is independence of the two variables.

In more detail, and with more rigor, let $\Omega$ denote the sample space of Brownian paths, chosen so that all paths are continuous. For $B \in \Omega$ and a rotation $R \in SO(d)$ let $R\omega$ denote the path $$ [R B](t)= R(B_t).$$ More generally, given an event $E \subset \Omega$ and $R \in SO(d)$, let $RE=\lbrace B \ : \ R^{-1} B \in E \rbrace.$ Wiener measure is rotation invariant, so $\Pr (R E)=\Pr(E)$.

Now let $M\subset S^{d-1}$ and $J\subset [0,\infty)$ be Borel sets and let $E_M$ and $F_J$ be the events $\lbrace B_T \in M\rbrace $ and $\lbrace T \in J \rbrace$ respectively. To prove independence of $B_T$ and $T$ we must show $$ \Pr (E_M \cap F_J ) = \Pr (E_M) \Pr (F_J) \quad \quad (\star)$$ for all such $M$ and $J$.

Let $R\in SO(d)$. Then $R E_M = E_{RM}$. Since the exit time $$ T(B)= \inf \lbrace t \ : \ |B(t)|\ge 1 \rbrace ,$$ we see that $T(R B)=T(B)$ and thus $R F_J = F_J$ for $R \in SO(d)$. Thus $$ \Pr (E_M \cap F_J ) = \Pr(E_{RM} \cap F_J),$$ so the measure $M \mapsto \Pr (E_M \cap F_J)$ is rotation invariant. Since it has total mass $\Pr(F_J)$, we conclude (e.g., from the uniqueness of Haar measure on $SO(d)$) that $$\Pr (E_M \cap F_J) = |M| \Pr (F_J),$$ where $|\cdot|$ is normalized Lebesgue measure on $S^{d-1}.$ Taking $J=[0,\infty)$ (for which $\Pr(F_J)=1$), we see that $\Pr(E_M)=|M|$ and $(\star)$ follows.