Relation between $\sup_{0\leq s\leq t}|W_s|$ and $\sup_{0\leq s\leq t}W_s$ for Brownian motion $W$

Question

Let $W$ be a Brownian motion. In a calculation, I have to compute $$\mathbb P \left(\sup_{0\leq s\leq t}|W_s|>a\right).$$ My idea would be to use the reflexion principle that says that $$\mathbb P\left(\sup_{0\leq s\leq t}W_s>a \right)=2\mathbb P(W_t>a).$$ $\sup_{0\leq s\leq t}|W_s|$ instead of $\sup_{0\leq s\leq t}W_s$, I'm not sure how to pursue. Can I do as follow :

$$\mathbb P\left(\sup_{0\leq s\leq t}|W_s|>a\right)=\mathbb P\left(\sup_{0\leq s\leq t}W_s>a\right)+\mathbb P\left(\sup_{0\leq s\leq t}W_s<-a\right)$$ $$=2\mathbb P(W_t>a)+1-\mathbb P(W_t>-a)\tag{*}$$ where I used $$\mathbb P\left(\sup_{0\leq s\leq t}W_s<x\right)=1-\mathbb P\left(\sup_{0\leq s\leq t}W_s\geq x \right)=1-2\mathbb P(W_t\geq x).\tag{**}$$

My first doubt is on the correctness of the first equality in $(*)$. Also, for $(**)$ I'm not sure because in the reflexion principle inequality are strict, whereas I used non strict inequality.

Answer 1

The first "=" in (*) fails to be true. Note that

$$\left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} \neq \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}. \tag{1}$$

Say, for instance, we have a sample path with $\sup_{s \leq t} W_s(\omega)<a$ but $\inf_{s \leq t} W_s(\omega)<-a$, then $\omega \in \{\sup_{s \leq t} |W_s|>a\}$ but $\omega \notin \{ \sup_{0 \leq s \leq t} W_s > a\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}$.

It does hold true that

\begin{align*} \left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} &= \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \color{red}{\inf_{0 \leq s \leq t}} W_s < -a \right\} \\ &= \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} (-W_s) > a \right\}. \tag{2} \end{align*}

However, the events $$A:=\left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \qquad B := \left\{ \sup_{0 \leq s \leq t} (-W_s) > a \right\}$$ are not disjoint, and therefore $$\mathbb{P}(A \cup B) \neq \mathbb{P}(A) + \mathbb{P}(B),$$ i.e. $$ \mathbb{P} \left( \sup_{0 \leq s \leq t} |W_s|> a \right) \neq \mathbb{P} \left( \sup_{0 \leq s \leq t} W_s > a \right) + \underbrace{\mathbb{P} \left( \sup_{0 \leq s \leq t} (-W_s) > a \right)}_{=\mathbb{P} \left( \sup_{0 \leq s \leq t} W_s > a \right)}.$$ Remark: The distribution of $\sup_{0 \leq s \leq t} |W_s|$ has been discussed e.g. here and here.