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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

saz
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Susan_Math123
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    Interesting. This formula does not appear in the second edition. –  Feb 25 '19 at 22:57
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    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. – saz Feb 26 '19 at 06:47
  • @saz thanks the name of the chapter is levy's triple law? I need |Ws| not Ws. – Susan_Math123 Feb 26 '19 at 16:58
  • @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. – Susan_Math123 Feb 26 '19 at 19:31

2 Answers2

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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

Riccardo Sven Risuleo
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  • Many thanks. Do you have a reference of support or proof? – Susan_Math123 Feb 25 '19 at 22:20
  • You should be able to find it in https://www.cambridge.org/core/journals/advances-in-applied-probability/article/calculation-of-noncrossing-probabilities-for-poisson-processes-and-its-corollaries/748E2D9EB82723E314ECB24EC3409C03 – Riccardo Sven Risuleo Feb 25 '19 at 22:29
  • I am sorry, can you be more specific? Where in the paper? I took a look at it twice and did not find it. – Susan_Math123 Feb 25 '19 at 22:33
  • I do not have access to the paper right now so I may be mistaken. You can check this website for a lot of references on the supremum of brownian motion http://homepages.ecs.vuw.ac.nz/~ray/Brownian/ I hope this helps! – Riccardo Sven Risuleo Feb 25 '19 at 22:43
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Feller obtains in his book An Introduction to Probability Theory. Vol II (p.342) the following result (... unfortunately he does not give a detailed proof):

Let $(B_t^x)_{t \geq 0}$ be a Brownian motion started at $x \in \mathbb{R}^d$ (i.e. $B_t^x = x+B_t$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion). For fixed $a>0$ define a stopping time $\tau^{x}_a$ by $$\tau^x_a := \inf\{t>0; B_t^x \notin (0,a)\}.$$ Then it holds for any $x \in [0,a]$ and $t \geq 0$ that $$\mathbb{P}(\tau^x_a > t) = \frac{4}{\pi} \sum_{n \geq 0} \frac{1}{2n+1} \exp \left( - \frac{(2n+1)^2 \pi^2}{2a^2} t \right) \sin \frac{(2n+1) \pi x}{a}. \tag{1}$$

Now let $(B_t)_{t \geq 0}$ be a standard Brownian motion and set $$M_t^* := \sup_{s \leq t} |B_s|.$$ Then $$\{M_t^* < r\} = \left\{ \sup_{s \leq t} |B_s| < r \right\} \stackrel{B_s^x=x+B_s}{=} \left\{ \sup_{s \leq t} |B_s^r| < 2r \right\} = \{\tau_{2r}^r >t\}$$ for any $r>0$, and so, by (1),

\begin{align*} \mathbb{P}(M_t^* < r) &= \frac{4}{\pi} \sum_{n \geq 0} \frac{1}{2n+1} \exp \left( - \frac{(2n+1)^2 \pi^2}{8r^2} t \right) \sin \frac{(2n+1) \pi}{2} \\ &= \frac{4}{\pi} \sum_{n \geq 0} (-1)^n \frac{1}{2n+1} \exp \left( - \frac{(2n+1)^2 \pi^2}{8r^2} t \right). \end{align*}

saz
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