The stopping time when $|B_t|=1$, $B_t$ is one dimensional standard Brownian motion I meet a problem when I analyze the stopping time of Brownian motion.
Suppose $B_t$ is the one dimensional standard Brownian motion, define $\tau_1 = \min\{t\geq 0: |B_t|= 1\}$.
Claim 1: $P(\tau_1 < \infty) = 1$
Claim 2: Define $\tau_n = \min\{t> \tau_{n-1}: |B_t| = 1\}$ for $n\geq 2$, then $P(\tau_n < \infty) = 1$ for all $n$
I tried Borel-Cantelli lemma and it did not work since $\sum_{n=1}^\infty P(\tau_1 \geq n) = \infty$.
But I believe that at least Claim 1 is correct otherwise the Brownian motion would stay in the strip between [-1, 1] forever with some positive probability.
Any suggestions are welcomed!
 A: Claim 1 can be confirmed in a number of ways.


*

*Let me give an argument based on Durrett's textbook. From the reverse Fatou's lemma, for any $\lambda > 0$ we have
$$ \Bbb{P}(|B_n| > \lambda \text{ i.o.})
= \Bbb{E}\big[ \limsup_{n\to\infty} \mathbf{1}_{\{|B_n| \geq \lambda\}} \big]
\geq \limsup_{n\to\infty} \Bbb{E}\big[ \mathbf{1}_{\{|B_n| \geq \lambda\}} \big]
= 1, $$
where the last equality follows from $\Bbb{E}\big[ \mathbf{1}_{\{|B_n| \geq \lambda\}} \big] = \Bbb{P}(|B_n| \geq \lambda) = \Bbb{P}(|B_1| \geq \lambda/\sqrt{n})$, which converges to 1.

*If the optional stopping theorem is available, you can draw a stronger conclusion. Notice that $\tau_1 \wedge n$ is a bounded stopping time for each $n \in \Bbb{N}$. Applying the optional stopping theorem to the martingale $B_t^2 - t$, we have
$$\Bbb{E}[B_{\tau_1\wedge n}^2 - (\tau_1 \wedge n)] = 0. $$
Since $|B_t| \leq 1$ for all $t \leq \tau_1$, we have $|B_{\tau_1 \wedge n}| \leq 1$ and $\Bbb{E}[\tau_1 \wedge n] = \Bbb{E}[B_{\tau_1\wedge n}^2] \leq 1$. Taking limit as $n\to\infty$ to this bound, monotone convergence theorem yields $\Bbb{E}[\tau_1] \leq 1$. This is a much stronger statement that $\tau_1$ is finite $\Bbb{P}$-a.s.
For Claim 2, I am not sure if $\tau_n$ is what you really want to look at, since $\tau_1 = \tau_2 = \tau_3 = \cdots$ with the definition as stated.

EDIT. With the modified definition of $\tau_n$'s, we still have $\tau_1 = \tau_2 = \tau_3 = \cdots$ with probability one.
Before showing that, however, let me show how each $\tau_n$ is finite with probability one. In view of the intermediate value theorem, it suffices to show that
$$ \limsup_{t\to\infty} B_t = +\infty, \qquad \liminf_{t\to\infty} B_t = -\infty $$
with probability one. By symmetry, it is enough to prove only one of them. Write $Z_n = B_n - B_{n-1}$ and notice that $Z_1, Z_2, \cdots $ are i.i.d. standard normal variables and $B_n = Z_1 + \cdots + Z_n$. Then by the Kolmogorov 0-1 law, the tail event $\{ \limsup_{n\to\infty} B_n = +\infty \}$ is $\Bbb{P}$-trivial. On the other hand, adopting a similar trick as before,
$$ \Bbb{P}(B_{n} > n^{1/3} \text{ i.o.})
\geq \limsup_{n\to\infty} \Bbb{P}( B_{n} > n^{1/3} )
= \lim_{n\to\infty} \Bbb{P}( B_{1} > n^{-1/6} )
= \tfrac{1}{2}. $$
So it follows that $\Bbb{P}(\limsup_{n\to\infty} B_n = +\infty) = 1$ and the claim is trow.
Next we prove that $\tau_1 = \tau_2 = \tau_3 = \cdots$ with probability one. In order to see this, we write $W_t = B_{\tau_1 + t} - B_{\tau_1}$. Strong Markov property shows that $W_t$ is again a Brownian motion and
$$ \tau_2 = \tau_1 + \inf\{t > 0 : W_t = 0 \}. $$
Again by our good ol' trick,
$$ \Bbb{P}(W_{1/n} > 0 \text{ i.o.})
\geq \limsup_{n\to\infty} \Bbb{P}( W_{1/n} > 0 )
= \tfrac{1}{2}. $$
But since the event $\{W_{1/n} > 0 \text{ i.o.}\}$ lies in the $\sigma$-field $\mathcal{F}_0^+$ which is $\Bbb{P}$-trivial by the Blumenthal 0-1 law, we have $\Bbb{P}(E) = 1$. Similar consideration shows that $W_{1/n} < 0$ infinitely often $\Bbb{P}$-a.s. In view of the intermediate value theorem, $0$ is the accumulation point of the zero-set of the Brownian path $t\mapsto W_t(\omega)$. Therefore $\inf\{t > 0 : W_t = 0 \} = 0$ and $\tau_2 = \tau_1$.
This argument applies to all of $\tau_n$, proving the equality.
