Distribution of $\max_{t \in [0,1]} |W_t|$ for Brownian motion

Question

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?

Answer 1

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:

Answer 2

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*}