Let $B$ a Brownian I want to compute $$\mathbb P(\sup_{0\leq s\leq t} |B_s|\leq \sqrt t).$$
I tried as follow : $$\mathbb P(\sup_{0\leq s\leq t} |B_s|\leq t)=\mathbb P(\sup_{0\leq s\leq t}B_s\leq \sqrt t, \inf_{0\leq s\leq t}B_s>-\sqrt t )\leq \mathbb P(\sup_{0\leq s\leq t}B_s\leq \sqrt t)=\mathbb P(|B_t|\leq \sqrt t),$$ and this can be easily computed. But the bounded is a bit to coarse. Is there a better way to have a tighter upper-bound ?