Suppose given a ddimensional 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?

$\begingroup$ I took the liberty of fixing some of your spelling. It might help you to get a helpful or favourable response if you give some explanation of why you want to know the answer (research? study?) and give some indication of what level of student or researcher you are. $\endgroup$ – Yemon Choi Jul 18 '10 at 10:46

$\begingroup$ For instance, while it's been years since I studied the appropriate courses, I think this question is standard homework. But I could be misremembering $\endgroup$ – Yemon Choi Jul 18 '10 at 10:48

$\begingroup$ I am a fresh ph.d student. I asked the question for it does not seem so obvious to me. Is there a probabilistic proof of the assertion? Thanks by the way for your comment $\endgroup$ – cyan Jul 18 '10 at 11:02

3$\begingroup$ at least do you see why this is true: if I tell you where the Brownian motion exit the sphere, can you say anything interesting about the value of T ? Of course not, and this is a consequence of the symmetry of the problem. $\endgroup$ – Alekk Jul 18 '10 at 17:30

1$\begingroup$ btw, this is theorem (39.6) of "Diffusions, Markov processes, and martingales, Volume 2" by Rogers and Williams  books.google.com.vn/… $\endgroup$ – Alekk Jul 19 '10 at 1:00
The short, and somewhat heuristic, answer is that rotational invariance implies that given $T$ the distribution of $B_T$ is uniform on the sphere $S^{d1}$. 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 $M\subset S^{d1}$ 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 $S^{d1}.$ Taking $J=[0,\infty)$ (for which $\Pr(F_J)=1$), we see that $\Pr(E_M)=M$ and $(\star)$ follows.


1$\begingroup$ The first 3 lines do a great job explaining it! $\endgroup$ – Victor Protsak Jul 20 '10 at 7:10