Given the most simple brownian motion:
$$ \dot x(t) = \sigma \eta(t)$$ where $\langle \eta(t)\eta(t')\rangle=\delta(t-t')$, I define as large event in a time-frame $\tau$ a portion of the trace $x(t)$ where
$$ x(t+\Delta t)-x(t)>\alpha, \ \mathrm{with}\; \Delta t < \tau. $$
Basically a large event happens when $x(t)$ increases by more than $\alpha$ in a time shorter than $\tau$.
QUESTION: I want to calculate the rate of these large events.
MY TENTATIVE ANSWER: Defining as $T_\alpha$ the time at which $x(t+T_\alpha)=x(t)+\alpha$, it is easy to calculate the probability that $T_\alpha<\tau$: $$ P(T_\alpha<\tau) = \frac{2}{\sqrt{2\pi\sigma^2 t}} \int_\alpha^\infty e^{-\frac{y^2}{2\sigma^2 t}} dy.$$
However, this gives me only the probability that $x(t)$ increases more than $\alpha$ in less than $\tau$. But I somehow want to convert this probability into the rate these large events happen. I first thought that the rate $r_{\alpha\tau}$ could be something like $$ r_{\alpha\tau}=\frac{P(T_\alpha<\tau)}{\tau},$$ but it is easy to show that this is a gross underestimation.
Anybody has a good idea?
Thank you in advance.