Brownian motion: law of iterated logarithm 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...
 A: 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,  $$\limsup_n \frac{B_{t_{n}}}{\sqrt{t_{n} \log \log \log t_{n}}}=\sqrt{2} \,.$$
A: 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.
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.
