Is Hitting time a finite stopping time?

Question

Let $ B=(B_t)_{t\geq 0}$ be a Brownian Motion and $a \in \mathbb R$. $$\tau_a:=\inf \{ t\geq 0 :B_t =a \} $$ How can i prove that $\tau_a$ is a finite stopping time?

Thanks in advance for any help!

Answer 1

By the Law of the Iterated Logarithm, $\sup_t B_t=\infty$ and $\inf_t B_t=-\infty$ almost surely. By continuity, if there exists $T_+,T_-$ such that $B_{T_+}>|a|$ and $B_{T_-}<-|a|,$ then $\tau_a$ is finite.

Answer 2

More directly: If $a>0$ then $\Bbb P[\tau_a \le t]\ge\Bbb P[B_t>a]=1-\Phi(a/\sqrt{t})$. Here $\Phi$ is the standard normal distibution function. This lower bound tends to $1/2$ as $t\to\infty$, so $\Bbb P[\tau_a<\infty]\ge 1/2$. Crucially, this lower bound doesn't depend on $a$. Temporarily fix a positive integer $n$. For the Brownian path to hit $na>0$, it must first hit $a$, then $2a$, ... , then $(n-1)a$, and finally $a$. From before, the probability of hitting $a$, starting at $0$, (let's call it $\theta$) is at least $1/2$. Likewise the probability of hitting $2a$, starting at $a$ is also $\theta$, and so on. It follows that $$ 1/2\le\Bbb P[\tau_{na}<\infty]=\theta^n, $$ so $\theta\ge(1/2)^{1/n}$. But the positive integer $n$ was arbitrary, and this last lower bound tends to $1$ as $n\to\infty$. It follows that $\theta=\Bbb P[\tau_a<\infty]=1$. The argument for $a<0$ is similar.