I know how to prove this equation: $$ \forall a>0 \ \ P(\sup_{s \leq t}B(s) \geq a) \leq e^{-\frac{a^2}{2t}} , $$ where $B(t)$ is a Brownian Motion
Firstly, I have to prove that $ e^{c B(t)} $ is submartingale (c is constant), next I use Doob's Inequality and I have my thesis.
Now I have similar inequality like this: $$ \forall a>0 \ \ P(\sup_{s \leq t}|B(s)| \geq a) \leq 2 e^{-\frac{a^2}{2t}} $$
Can I do something like that:
$$ P(\sup_{s \leq t}|B(s)| \geq a) = P(\sup_{s \leq t}B(s) \geq a) + P(\inf_{s \leq t}B(s) \leq -a) = P(\sup_{s \leq t}B(s) \geq a) + P(-\inf_{s \leq t}B(s) \geq a) = P(\sup_{s \leq t}B(s) \geq a) + P(\sup_{s \leq t}B(s) \geq a) = 2 P(\sup_{s \leq t}B(s) \geq a) \leq 2 e^{-\frac{a^2}{2t}} $$