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What is $\mathbb P(\sup_{t\in[0,1]}|W_t|\le1)$ for $W_t$ a Brownian motion? Without the absolute value, we have $\mathbb P(\sup_{t\in[0,1]}W_t\le c)=1-\sqrt{2/\pi}\int_c^\infty e^{-x^2/2}dx$ for all $c\ge0.$ (The proof I know of uses the strong Markov property.) However, I have no idea how to proceed when we have the absolute value.

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There is a nice solution to this using the reflection principle. To find the probability that $\sup_{t\in [0,1]} |W_t|\le 1$, let \begin{align} E&=\{|W_1|< 1\} \\ E_u&=E\cap \{\sup_{t\in [0,1]} W_t>1\} \\ E_d&=E\cap \{\inf_{t\in [0,1]} W_t<-1\} \end{align} In words, $E$ is the easier to calculate event that $W_t$ ends between $-1$ and $1$, while $E_u$ (resp. $E_d$) is the bad event whose probability we must subtract where $W_t$ crosses the upper (resp. lower) boundary sometime before $t=1$.

Fortunately, $\def\P{\mathbb P}\P(E_u)=\P(E_d)$ can be computed easily with the reflection principle. If you take a path in $E_u$, and reflect the portion of the path after it first hits the horizontal line of heigh one across that line, the the result is an arbitrary Brownian motion path $\hat W_t$ for which $1<\hat W_1<3$. Since this this process is reversible and probability preserving, we have $$\P(E_u)=\P(E_d)=\P(1<W_1<3)$$ Unfortunately, the answer is not as simple as $$ \P(E)-\P(E_u)-\P(E_d) $$ because in subtracting out the bad events, the events where a path crosses both barriers has been doubly subtracted, so they must be added back in. These events come in two flavors: let $$ E_{ud}=E\cap \{W_t \text{ first hits $1$, then later hits $-1$}\}\\ E_{du}=E\cap \{W_t \text{ first hits $-1$, then later hits $1$}\} $$ Now, applying the reflection principle twice (see image for illustration of the $E_{ud}$ case), you can show $$ \P(E_{ud})=\P(E_{du})=\P(3<W_1< 5) $$ These two events must be added back in, so we currently are at $$ \P(E)-\P(E_u)-\P(E_d)+\P(E_{ud})+\P(E_{du}) $$ But it does not stop there: the triply bad events $E_{udu}$ and $E_{dud}$ now must be subtracted out, and then the qudaruply subtracted events must be subtracted out, and so on to infinty. This is a variation of the princple of inclusion exclusion.

In summary, we have $$ \boxed{\P\big(\sup_{t\in [0,1]} |W_t|\le 1\big) =\P(|W_1|\le 1)+2\sum_{k\ge 1}(-1)^k\;\P(2k-1<W_1<2k+1)\;\;} $$ Put another way, let $f(w)$ be a the unqiue function on the reals satisfying $f(w)=1$ for $-1<w\le 1$ and for all $w\in \mathbb R$, $f(w+2)=-f(w)$. Then $$ \P\big(\sup_{t\in [0,1]} |W_1|\le 1\big)=\mathbb E[f(W_1)]=\int_{-\infty}^\infty f(x)\phi(x)\,dx $$ where $\phi(x)$ is the pdf of $W_1$.

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  • $\begingroup$ Shouldn't it be $\mathbb{P}(3<W_1<5)$ instead of $W_t$? And similarly in the boxed final equation, $W_1$ everywhere on the RHS instead of $W_t$? $\endgroup$ Jul 22, 2022 at 13:39
  • $\begingroup$ Indeed it should. As a thank you, here's a cookie: 🍪 $\endgroup$ Jul 22, 2022 at 18:34
  • $\begingroup$ Thanks for the cookie! I am having difficulty in computing $\mathbb{P}(E_{ud})=\mathbb{P}(|W_1|\leq 1, \tau_1<\tau_{-1}\leq 1)$. But then I am not able to apply reflection principle twice to obtain the answer. Could you please help me on this? $\endgroup$ Jul 26, 2022 at 15:17
  • $\begingroup$ Does the picture help? $\endgroup$ Jul 27, 2022 at 16:20
  • $\begingroup$ Thank you very much! This was very helpful. I understand it now. $\endgroup$ Jul 28, 2022 at 12:24

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