# Supremum of absolute value of Brownian Motion

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

• 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. – Chris Janjigian Jul 30 '18 at 16:52

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