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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1$\begingroup$ 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... $\endgroup$– David Roberts ♦Commented Dec 21, 2011 at 23:28
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$\begingroup$ $B_T$ 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. $\endgroup$– The BridgeCommented Dec 21, 2011 at 23:50
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$\begingroup$ @The Bridge: I didn't understand your argument. Where does it fail for starting point other than 0? $\endgroup$– Ori Gurel-GurevichCommented Dec 22, 2011 at 0:20
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$\begingroup$ This question may be easy, but I don't think the downvote is justified. $\endgroup$– Ori Gurel-GurevichCommented Dec 22, 2011 at 0:23
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$\begingroup$ @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 $\endgroup$– Yemon ChoiCommented Dec 22, 2011 at 2:27
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2 Answers
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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].
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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.
answered Dec 22, 2011 at 5:59