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I know that by the reflection principle, $$ P\left[\sup_{0 < s < t} B_s > a \right] = 2P[B_t> a] $$ where $B_t$ is a Brownian Motion. But what is $P\left[\sup_{0 < s < t} |B_s|> a \right]$?

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    $\begingroup$ Hint: $B_t$ has the same distribution as $-B_t$ and $P(\sup_{0 < s < t} |B_s| > a) = P(\sup_{0 < s < t}B_t > a$ or $\inf_{0 < s < t} B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly. $\endgroup$ – Chris Janjigian Jul 30 '18 at 16:52
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Let $\Phi_t(x)$ be the cdf for $B_t$. Then $$ \mathbb P\left(\sup_{0< s< t} |B_s|\le a\right)=\sum_{k=-\infty}^\infty(-1)^k\Big(\Phi_t\big(a(2k+1)\big)-\Phi_t\big(a(2k-1)\big)\Big). $$

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