I would like to ask a question about Brownian motion:
Let $B$ be a standard brownian motion. How to show that $\mathbb P( \max\limits_{0 \leq s \leq t} B(s) \in (a,b) )$ decreases exponentially in t with a, b fixed?
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$\begingroup$ $\max_{s\leq t}B(s)$ equals $\vert B(t) \vert$ in law. Is this enough to prove the result ? $\endgroup$– ChivalCommented Jul 27, 2015 at 21:46
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$\begingroup$ Apparently, yes. We can estimate the integral over (a,b) of the |Y| where Y is of the normal law N(0,t) with t large. $\endgroup$– mrvuiveCommented Jul 28, 2015 at 13:16
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1 Answer
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The precise answer depends on the initial condition, let's assume you start the Brownian motion (with diffusion constant $D$) at time $t=0$ in some point $x_0\in(a,b)$, then you ask for the probability $$P(t)=\mathbb P[ \max\limits_{0 \leq s \leq t} B(s) \in (a,b) ]$$ that the particle has not crossed the point $x=b$ for all times up to time $t$. The point $x=b$ functions as an absorbing boundary for the Brownian motion. The survival probability is given by $$P(t)={\rm Erf}\,\left(\frac{b-x_0}{2\sqrt{Dt}}\right)$$ For large $t$ this decays as as $1/\sqrt t$, $$P(t)=\frac{b-x_0}{\sqrt{\pi Dt}}+{\rm order}(1/t)$$ so much slower than exponentially.
answered Jul 19, 2015 at 15:19
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$\begingroup$ Thanks a lot for your answer. I assumed that $B(0)=0$ (a standard BM). The answer seems correct but quit surprising to me at the first sight because I thought that the integral in $\mathbb R$ w.r.t. time $t$ of $P(t)$ is finite. $\endgroup$– mrvuiveCommented Jul 22, 2015 at 11:32