Let $W_t$ be the standard Wiener process, find probability $P_t(u)=P(|W(s)|\leq u,0\leq s\leq t)$.

Asked Apr 06 '24 at 11:43

Active Apr 06 '24 at 12:03

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For an one-dimensional standard Wiener process $W_t$, find $P_t(u)=P(|W(s)|\leq u,0\leq s\leq t)$.

This arises from the problem to find the first time a random walk has travelled a given distance, or the first time a $M/M/1$ queue grows up to a given size.

Maybe this problem is closely related to the Rice's formula?

asked Apr 06 '24 at 11:43

陈进泽

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Possibly helpful: https://math.stackexchange.com/questions/2867162/supremum-of-absolute-value-of-brownian-motion – Jose Avilez Apr 06 '24 at 12:21
@JoseAvilez So the previous post seems to be correct. – 陈进泽 Apr 06 '24 at 12:29
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I haven't checked their work. Indeed, it was correct to propose looking at the process $M_t = \max_{0\leq s \leq t} |B_s|$, the distribution of which is known. This has been asked on the site several times, so I encourage you to look at the answers here – Jose Avilez Apr 06 '24 at 12:30
There is a semi-analytical solution that can be found here. – NN2 Apr 06 '24 at 12:36
Many thanks. Need some time to understand. – 陈进泽 Apr 06 '24 at 12:39

Let $W_t$ be the standard Wiener process, find probability $P_t(u)=P(|W(s)|\leq u,0\leq s\leq t)$.

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