Two-sided hitting time of Brownian motion
Solution 1:
I will give this a try.
For simplicity let $T_a=\inf \{t: |W_t| = a \}$
\begin{align} Pr(|W(t)|>a)&=P(|W(t)|>a|T_a<t)Pr(T_a<t)+P(|W(t)|>a|T_a>t)Pr(T_a>t)\\ \end{align}
$P(|W(t)|>a|T_a>t)=0$ since the time that $|W(t)|$ hits $a$ for the first time has not arrived, hence $|W(t)|$ can not be bigger than $a$.
Also note that $P(|W(t)|>a|T_a<t)=\frac12+Pr(W(t)<-2a)$ since we know that $|W(t)|$ has hit $a$ before $t$ (we have $T_a<t$). Therefore the event $\{|W(t)|>a|T_a<t\}$ is equivalent to $\{|a+W(t)|>a\}$.
Thus $$Pr(T_a<t)=\frac{2P(W(t)>a)}{\frac12+P(W(t)<-2a)}.$$