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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!

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  • $\begingroup$ Just for clarity, is your Brownian motion 1-dimensional? $\endgroup$ Commented Apr 29, 2017 at 2:19
  • $\begingroup$ @SangchulLee Yes, it's one dimensional standard BM. Thanks, I will make the problem more clear. $\endgroup$
    – Jason
    Commented Apr 29, 2017 at 2:20
  • $\begingroup$ No worries! I just wanted to know as the answer to the second claim depends on the dimension. $\endgroup$ Commented Apr 29, 2017 at 2:31

1 Answer 1

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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.

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  • $\begingroup$ Thank you so much for your solution. I will check them later. $\endgroup$
    – Jason
    Commented Apr 29, 2017 at 3:06
  • $\begingroup$ Yes, there are some typo in Claim 2. In the definition, it should be $t>\tau_{n-1}$ $\endgroup$
    – Jason
    Commented Apr 29, 2017 at 3:07
  • $\begingroup$ Just one question for the proof of claim 2, can we just apply the optional stopping theorem to $\tau_n$ for $n\geq 2$ ? $\endgroup$
    – Jason
    Commented Apr 29, 2017 at 19:55
  • $\begingroup$ Besides, the I don't understand why $\tau_2 = \tau_1 + \inf\{t>0: W_t = 0\}$. Could it be the case $B_{\tau_1} = 1$ and $B_{\tau_2} = -1$ ? $\endgroup$
    – Jason
    Commented Apr 29, 2017 at 20:28

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