Independence of Brownian motion at hitting time from that hitting time Let $B_t$ be a Brownian motion for a given probability space and $T:=\inf \lbrace t\geq 0 : \vert B_t \vert = 1 \rbrace$.
Is the process at this time, $B_T$, independent of the hitting time $T$? If so, how can one show this?
 A: This result holds less obviously for Brownian motion with constant drift, not just $0$ drift. It is critical that the starting point is centered on the interval and it fails otherwise.
Stern, F. An Independence in Brownian Motion with Constant Drift. The Annals of Probability, Vol. 5 (1977), 571-572.
This holds for biased random walks because reflecting the paths to one boundary point gives paths to the other boundary with a constant magnification of probability. Taking the limit shows that the same is true for Brownian motion with constant drift. 
A: In a word, "symmetry".  (I presume you mean to have $B_0=0$.)  The law of such a Brownian motion is invariant under orthogonal transformations, and the stopping time $T$ is pointwise invariant under such transformations.  Therefore the law of $B_T$ is likewise invariant...  This argument is valid in all dimensions.
More interesting is L. Pitt's converse, asserting that if the exit time from a bounded domain $D$ (for a Brownian motion started at $0\in D$) is independent of the exit place, then $D$ is essentially a ball centered at $0$. See [Annals of Probability, vol. 17 (1989), pp. 1651–1657].
