For a standard Brownian motion $\{W_t, t\geq 0\}$, find $\mathbb{P}(\max_{ t \in [0,1]}|W_t| <x)$.

Page 79-80 of Billingsley, P., Convergence of probability measures, New York-London-Sydney-Toronto: John Wiley and Sons, Inc. XII, 253 p. (1968). ZBL0172.21201. says:

$\mathbb{P}(\max_{ t \in [0,1]}|W_t| <x)=1-\frac{4}{\pi}\sum \frac{(-1)^{k}}{2k+1} \exp\left(-\frac{\pi^2 (2k+1)^2}{8 x^2}\right)$

I think it is not correct. I plotted the series $k=100$ and $x\in[0,10]$. It is really weird. The probability is always larger than $1$ and it goes to $1.2$ !!! Can you help me find the problem? enter image description here

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    $\begingroup$ Interesting. This formula does not appear in the second edition. $\endgroup$ – d.k.o. Feb 25 at 22:57
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    $\begingroup$ It is possible to compute the joint density of $M_t := \sup_{s \leq t} W_s$ and $m_t := \inf_{s \leq t} W_s$ explicitly, and this allows to compute the probability you are looking for; see e.g. Section 6.5 by the book by Schilling & Partzsch for the result about the density. $\endgroup$ – saz Feb 26 at 6:47
  • $\begingroup$ @saz thanks the name of the chapter is levy's triple law? I need |Ws| not Ws. $\endgroup$ – Susan_Math123 Feb 26 at 16:58
  • $\begingroup$ @d.k.o. Yes, it does not appear in the latest edition. It seems that they knew there was an error but they did not know the correct result at the end. The answer below gives the correct result without a good reference. $\endgroup$ – Susan_Math123 Feb 26 at 19:31

The formula should be $$P(\max_{t\in [0,1]} |w_t| < x)=\frac{4}{π} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \exp\left\{-\frac{π^2(2n+1)^2}{8x^2}\right\}$$

Here is a plot of the resulting function: enter image description here


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