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?
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Sign up to join this communityLet $B_t$ 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?
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].
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.
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.