I was curious about the following question:
Let $\{B_t\}_{t\ge 0}$ be a standard Brownian motion and $\mu, a\in\Re$ be arbitrarily given real numbers. Let $L=\{\tau\in\Re_+\,|\,X_\tau=a\}$, where (assume that $\mu\ne 0$) $$X_t\equiv \mu t+B_t.$$ Then is it true that for every $T\in\Re_+$, the set $[0,T]\cap L$ is almost surely finite?
By intuition, I think the answer should be yes, but I cannot figure out a convincing logic. Would anyone give me some hint or reference? Thanks!
No. The set of zeros of Brownian motion is a a.s. a perfect set (closed and has no isolated points).
In particular, since $B_0=0$, for the choice $a=0$, each of the sets $[0,T]\cap L$ is a.s. uncountable.
Now for $a\ne 0$, on the event $\{\tau_a \le T\}$, where $\tau_a$ is hitting time of $a$, the same conclusion holds (note that the probability that $\tau_a = T$ is zero). For each $T$ this event has probability less than $1$, but increasing $1$ as $T\to\infty$ because of recurrence.