Two-sided hitting time of Brownian motion I am trying to compute the hitting time of a linear Brownian motion on a two-sided boundary. More specifically, let $W_t$ be a (one-dimensional) Wiener process. Let $T = \inf \{t: |W_t| = a \}$ for some $ a > 0$. I want to find $\mathbb{P}\{ T > t\}$. 
I know that probability distribution hitting time of a positive level, $\inf \, \{t: W_t = b\,, \  b > 0 \}$ can be computed quite easily, but I am not sure how to deal with it when dealing with the two-sided hitting time, i.e. with the absolute value. I am thinking of the minimum of hitting times of level $a$ and $-a$, but I can't get a promising conclusion.
 A: 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 $$Pr(T_a<t)=\frac{2P(W(t)>a)}{\frac12+P(W(t)<-2a)}.$$
A: The distribution of hitting times for a Brownian Motion, $S$ starting at $0$ with barriers at $c$ and $-c$ and step size $l=1$ is given by
$$
P(T>x)=\left(1-\sum_{i=-\infty}^\infty(-1)^{i+1}\left[\Phi\left(\frac{(2i+1)c}{\sqrt{x}}\right)-\Phi\left(\frac{(2i-1)c}{\sqrt{x}}\right)\right]\right),
$$
where $\Phi()$ is the CDF of the standard normal distribution.
I am looking for a tidier representation myself in this question.
The proof comes from the reflection principle. The probability of the $S$ hitting the lower bound $-c$ ($c$) and returning back to the region $[-c,c]$ before time $x$ is equal to $S$ being in $[-3c,-c]$ ($[c,3c]$) region at time step $x$. For one sided boundary, this would be it. However, it is also possible for $S$ to hit $-c$, return back to $[-c,c]$ and then hit $c$. $S$ being in $[-3c,-c]\cup [c,3c]$ counts $S$ hitting both barriers twice. So, we need to take $S$ hitting both barriers separately.
The probability of $S$ hitting first $-c$ and hitting $c$ and then returning back to $[-c,c]$ is equal to $S$ being in $[-5c,-3c]$.
You can extend this approach to infinity to reach the correct distribution.
