Let $W_t$ be a standard Brownian motion. Show that almost surely we have:

$$ \limsup_{t \to \infty} \frac{W_t}{\sqrt{t \log t}}<\infty $$

The approach suggested by the writer of the problem is to consider sets: $$ V_{n,r} = \left \{\max_{0 \leq t \leq 2^n} W_t \geq r\cdot 2^{n/2} \cdot n^{1/2} \right\} $$ And showing that for $r$ sufficiently large but finite, we have that: $$ \sum_{n = 1}^\infty \mathbb{P}(V_n) < \infty $$ Applying the Borel-Cantelli lemma to the events $V_n$ finishes the proof.

This is all well and good, and I understand how the hint proves the result. However, does anyone have any idea how I may get any kinds of bounds on $\mathbb{P}(V_{n,r})$?


1 Answer 1


Hint: By the reflection principle, $\sup_{0 \leq t \leq T} W_t$ equals in distribution $|W_T|$ for each $T>0$. Apply Markov's inequality to get suitable bounds for $\mathbb{P}(V_{n,r})$.

Remark: In case that you are not familiar with the reflection principle, you can combine instead Markov's inequality with Doob's maximal inequality for martingales; it allows you to get rid of the supremum: $$\mathbb{E} \left( \sup_{t \leq T} |W_t|^p \right) \leq c_p \mathbb{E}(|W_T|^p), \qquad p>1.$$


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