How to prove $$\limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n) = +\infty$$ using Blumenthal zero-one law, where $(B(t))_{t \geq 0}$ is a Brownian motion?
2026-04-13 16:35:14.1776098114
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Blumenthal zero-one law
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Here are some hints:
1) Use the fact that $aB_t$ has the same distribution as $B_{a^2t}$. This means that at any given time $n$, your quantity inside the supremum has distribution $B(1)$.
2) For any $a>0$,
$$\mathbb{P}(B(n)>a\sqrt{n} \mbox{ occurs infinitely often})\geq \limsup_{n\rightarrow\infty}\mathbb{P}(B(n)>a\sqrt{n})=P(B(1)>a)>0$$
3) You now want to show that $B(n)>a\sqrt{n}$ occuring infinitely often is a 0-1 event. Do you see how to invoke the Blumenthal 0-1 law to show this? As a hint, let $T$ be the hitting time of $a\sqrt{n}$ and notice that $T$ satisfies the strong Markov property...
The idea of the proof is rather similar to the hints provided by AlexR., but using the time inversion it is easier to see how to apply Blumenthal's 0-1 law.
Hints: