# 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?

• What is your reason for asking this question? In what context did it arise? Also, reading mathoverflow.net/howtoask and implementing the suggestions from there may help attract answers... – David Roberts Dec 21 '11 at 23:28
• $B_T$ is random variable taking value 1 and -1 with probability 1/2, it is independent from $T$, you can conditionate by replacing T, by T' the first time B gets to 1, then B_T is a constant so it is independent of T'. Then taking T'' first time to -1 tell the same as T=min(T',T'') you get the result. Regards, but not an MO question more suited for mathstackexchange. – The Bridge Dec 21 '11 at 23:50
• @The Bridge: I didn't understand your argument. Where does it fail for starting point other than 0? – Ori Gurel-Gurevich Dec 22 '11 at 0:20
• This question may be easy, but I don't think the downvote is justified. – Ori Gurel-Gurevich Dec 22 '11 at 0:23
• @Ori: I haven't downvoted, but perhaps the reason for the downvote was that a baldly asked question, devoid of context (what has the OP already tried, why did they come to this problem) can seem like homework. It is, to be fair, a natural question; but I think the question would be better received if the OP acted along the lines suggested by David Roberts – Yemon Choi Dec 22 '11 at 2:27

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.