Let $B_t$ be Brownian Motion $T_a$ the time of first passge for point $a$ for $B_t$. From the reflection principle we know that that:
$P(T_a<t)=P(M_t>a)=2P(B_t>a)$
So, from what I can tell, for any $t>0$, $P(T_0 < t) = 2P(B_t>0)=1$ since $B_t$ is distributed $N(0, t)$. Is this right?
The reason I'm asking is that this seems to say that the set of all paths which area always negative has probability zero - which feel intuitively suspect to me.