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I am doing a homework question. But I get confused.

$\{B_t: t \geqslant 0\}$ is a standard Brownian motion. Show that there exists $t_{1}<t_{2}<\cdots$ with $t_{n} \rightarrow \infty$ such that with probability one, $$ \limsup _{n \rightarrow \infty} \frac{B_{t_{n}}}{\sqrt{t_{n} \log \log t_{n}}}=0 $$ But there is a theorem:

(Law of the Iterated Logarithm for Brownian motion) Suppose $\{B_t: t \geqslant 0\}$ is a standard Brownian motion. Then, almost surely, $$ \limsup _{t \rightarrow \infty} \frac{B(t)}{\sqrt{2 t \log \log (t)}}=1 $$ is it a contradiction? Actually I tried $t_n=\exp(\exp(n))$ and apply the borel cantelli lemma, it seems to have: for any $\epsilon>0$

$$ \limsup _{n \rightarrow \infty} \frac{B_{t_{n}}}{\sqrt{t_{n} \log \log t_{n}}}< \epsilon $$

But $t_n$ always go to infinite, so the theorem should give us $\sqrt{2}$, really confused...

Ian
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PaulWH
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  • Are $t_n$ non-random? If so then this doesn't seem right to me. – Ian Oct 15 '20 at 23:06
  • @Ian I don't see the problem. $\mathbb{P} (B(t) > \log \log (t)^{-1/3} \sqrt{t \log \log (t)})$ converges to $0$, so if $(t_n)$ grows fast enough, you can conclude by the Borel-Cantelli lemma. That's more or less PaulWH's argument, and it is perfectly fine. – D. Thomine Oct 15 '20 at 23:12
  • @D.Thomine Yeah I guess I see what you mean. That's awfully counterintuitive, though, since when the excursions up to $\pm \sqrt{2t \log(\log(t))}$ occur is random. – Ian Oct 15 '20 at 23:23
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    $$ \lim \inf_{t \rightarrow \infty} \frac{B(t)}{\sqrt{2 t \log \log (t)}}=-1 $$ and $$ \limsup _{t \rightarrow \infty} \frac{B(t)}{\sqrt{2 t \log \log (t)}}=1 $$. So there is no contradiction. – Kavi Rama Murthy Oct 15 '20 at 23:29
  • @Ian Right, I meant to put an absolute value for $B(t)$. – Kavi Rama Murthy Oct 15 '20 at 23:35

2 Answers2

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There is no contradiction. First if the $t_n$ could depend on the BM, the we can find $t_n \to \infty$ where $B(t_n)=0$, so let's assume the question requires the $t_n$ to be deterministic. The LIL ensures that almost surely there is a random sequence $\tau_n \to \infty$ along which the ratio $$\frac{B_{\tau_{n}}}{\sqrt{\tau_{n} \log \log \tau_{n}}}$$ tends to $\sqrt{2}$, but this sequence is quite sparse and it will intersect only finitely often a rapidly growing deterministic sequence such as $t_n=\exp(e^n)$. Your Borel-Cantelli calculation is right, indeed, it can give you a law of triple-iterated logarithm along this sequence: Almost surely, $$\limsup_n \frac{B_{t_{n}}}{\sqrt{t_{n} \log \log \log t_{n}}}=\sqrt{2} \,.$$

Yuval Peres
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I hope this help you, is related to you question: I was trying to plot an envelope function to paths of standard Brownian motion as a classic Wiener process, and I found that instead of using $$ \lim \frac{B(t)}{\sqrt{2t\log\log t}}$$ it works much better using $$ \lim \frac{B(t)}{|\text{Modulus of Continuity}|}$$ as is defined in https://en.wikipedia.org/wiki/Wiener_process#Modulus_of_continuity.

I have left the explanation with an example in Plotting tight bounds for simple Wiener Brownian motion - problems with classic definitions (Law of iterated logarithm).

There is said that experimentally (not mathematically proved), it works better as an envelope: $$ \lim \frac{B(t)}{\sqrt{2t\sqrt{\pi^2+(\log(\log(t+1)+1))^2}}}$$ Specially at the beginning, since at $t \rightarrow \infty$ is equivalent to the Law of the iterated Logarithm. Hope this will help.

Added later

After some trials I think the found bound $f_{\pm}(t)$ could be improved just adjusting by just one displacement the mentioned Modulus of continuity of the Wiener process as: $$h(t)_\pm=\left|\sqrt{2t\log\left(\log\left(\frac{1}{t\color{red}{+1}}\right)\right)}\right|=\pm\sqrt{2t\sqrt{\pi^2+{\log(\log(1+t))}^2}}$$

Joako
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