# $\mathbb P(\sup_{t\in[0,1]}|W_t|\le1)$ for Brownian motion

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.

• This is equivalent to asking what is $\Bbb{P}(\tau \ge 1)$, where $\tau$ is the first time that $W_{t}$ hits $\pm 1$. See math.stackexchange.com/questions/907126/… for some information on this. May 20, 2020 at 19:57

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 $$\P(E_u)=\P(E_d)=\P(1 Unfortunately, the answer is not as simple as $$\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 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 $$\boxed{\P\big(\sup_{t\in [0,1]} |W_t|\le 1\big) =\P(|W_1|\le 1)+2\sum_{k\ge 1}(-1)^k\;\P(2k-1 Put another way, let $$f(w)$$ be a the unqiue function on the reals satisfying $$f(w)=1$$ for $$-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 $$W_1$$.
• 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 $W_t$? Jul 22, 2022 at 13:39
• 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? Jul 26, 2022 at 15:17