Assume I've two hitting times for a Brownian Motion $W_t$ as the following:
$\tau_a=inf\{t>0:W_t=a\}$ and $\tau_{-a}=inf\{t>0:W_t=-a\}$ for $a>0$
Now I want to find $P(\tau_a\le T,\tau_{-a}\le T)$ for some $T>0$.
I was thinking writing above as
$P(\tau_a\le T,\tau_{-a}\le T)=P(\tau_a\le\tau_{-a}\le T)+P(\tau_{-a}\le\tau_a\le T)$.
Then I am not sure how to proceed. I know if we write $\tau=min(\tau_a,\tau_{-a})$ then $\tau$ is the exit time of brownian motion. There is a famous result about $P(W_\tau=a)=P(\tau_a<\tau_{-a})=\frac{a}{a+a}=1/2$, and also $E(\tau^2)=a^2$. However, I am not sure if this helps, because somehow to solve this problem we need a specific expression of distribution of the exit time.
Another possible way out is to apply reflection principle to $P(\tau_a\le\tau_{-a}\le T)$, and in this note:http://www.math.nyu.edu/faculty/avellane/band.ps it says that we may get an iterative relation for this by using reflection principle but I don't see how.
Here is one way. We have $\mathbb{P}(\{\tau_a, \tau_{-a} \leq T\}^C) = \mathbb{P}(\tau_a > T) + \mathbb{P}(\tau_{-a} > T) - \mathbb{P}(\tau_a, \tau_{-a} > T)$. Each of the probabilities on the right hand side can be computed (see two-sided hitting times for the third summand).