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?



Independence of Brownian motion at hitting time from that hitting time
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].


