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Let $(B_t)$ denote the standard Brownian motion on the interval $[0,1]$. For a given confidence level $\alpha \in (0,1)$ a confidence band on $[0,1]$ is any function $u$ with the property that $$ P(\omega; |B_t(\omega)| < u(t), \quad \forall t\in [0,1])=\alpha. $$ In other words, the probability that a path of the Brownian motion stays within a confidence band is $\alpha$. Additionally the boundary hitting position for those paths leaving the band must be uniformly distributed on $[0,1]$. This condition can be stated using the stopping time $$\tau(\omega) = \inf [ t \in [0,1], |B_t(\omega)|=u(t) ]. $$ Then $\tau $ is the time of the first hitting, and one asks that $\tau$ is uniformly distributed on $[0,1]$ conditionally on the event that $\tau$ is finite.

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The context of the problem is a rather boring one, so will not state it here. The problem itself seem to be non-trivial and interesting.

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    What, exactly, is your question?2010-10-30
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    The question is, how does $u$ look like? Does it exists for all values of alpha? If yes, is it unique?2010-10-30
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    @wergrew: I edited your question. Please check the result suits you.2011-03-20
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    My guess is that $u$ exists, is unique, and is continuous. For $\alpha=1$ it should satisfy $u(t)\sim\sqrt{2t\log\log(1/t)}$ as $t\to0$ (because of the law of the iterated logarithm) and $u(1)=0$. Then the solution for arbitrary $\alpha$ should follow from rescaling. $u_\alpha(t)=u_1(\alpha t)/\sqrt{\alpha}$.2011-03-20
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    My comment above is taking $\alpha$ to be the probability that te barrier is hit (rather than the probability that it is not hit).2011-03-20
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    @George Any proof for these guesses?2011-03-20
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    @Didier: No, but there are things you can say. If it is continuous then I don't think it is too hard to show that $u_1$ must satisfy the properties mentioned. Also, there must be a unique solution in discrete time and, if you can prove some bound on the modulus of continuity of $u$, then you would be able to obtain the continuous time solution as a limit of the discrete time case. It does seem approachable, and I think you should get a unique solution.2011-03-20
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    ...well, unique almost everywhere. You can always increase $u$ on a zero measure set without changing the hitting distribution.2011-03-21
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    Sorry guys! I missed that interesting discussion in March. I particularly like the discretization idea. The method could give us the idea how such a curve might look like. Thanks!2011-11-30

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