Supremum of absolute value of Brownian Motion

Question

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 $P\left[\sup_{0 < s < t} |B_s|> a \right]$?

Answer 1

Let $\Phi_t(x)$ be the cdf for $B_t$. Then $$ \mathbb P\left(\sup_{0< s< t} |B_s|\le a\right)=\sum_{k=-\infty}^\infty(-1)^k\Big(\Phi_t\big(a(2k+1)\big)-\Phi_t\big(a(2k-1)\big)\Big). $$