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Let $(B_t)_{t\geq 0}$ be a Standard Brownian motion. Show that

$$\mathbb{P}(\sup_{s\leq t} |B_s|\geq x)\leq 2\mathbb{P}(|B_t|\geq x).$$

My idea is: Let $S:=\sup_{s\leq t}|B_s|$. Then

\begin{align*} \mathbb{P}(S\geq x) &= \mathbb{P}(\sup_{s\leq t} B_s\geq x)+\mathbb{P}(\sup_{s\leq t}B_s\leq -x)\\ &= \mathbb{P}(\sup_{s\leq t} B_s\geq x)+\mathbb{P}(-\sup_{s\leq t}B_s\geq x)\\ &=\mathbb{P}(\sup_{s\leq t} B_s\geq x)+\mathbb{P}(\inf_{s\leq t}(-B_s)\geq x). \end{align*}

Is it true that $\inf(-B_s)\leq \sup(B_s)$? If yes, then I can say that

$\mathbb{P}(S\geq x)\leq 2\mathbb{P}(\sup_{s\leq t}B_s\geq x)=2\mathbb{P}(|B_t|\geq x).$

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Your idea is correct but your proof is not. Specifically:

  • Your first equality is incorrect - you will notice that the second probability is in fact equal to zero assuming $x>0$.
  • It is not true that $\inf(-B_s)\le\sup(B_s)$; indeed, $(-B_t)$ is also Brownian motion, so $\inf(-B_s)\,{\stackrel{d}{=}}\,\sup B_s$.

Instead, we have \begin{align} \mathbb P\left(\sup_{s\le t}|B_s|\ge x\right) &=\mathbb P\left(\left\{\sup_{s\le t}B_s\ge x\right\}\cup\left\{\inf_{s\le t}B_s\le-x\right\}\right)\\ &\le\mathbb P\left(\sup_{s\le t}B_s\ge x\right)+\mathbb P\left(\inf_{s\le t}B_s\le-x\right). \end{align} Now continue along the same lines as what you wrote, using the fact that $(-B_t)$ is also a Brownian motion, and then apply the reflection principle.

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