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Let $(B_t)_{t \geq 0}$ be a Brownian motion on an arbitrary probability space.

Then Levy's modulus of continuity says that

$\limsup_{h \to 0} \sup_{0 \leq t-s \leq h} \frac{B_{t} - B_s}{\sqrt{2h \log(h^{-1})}} = 1$,

as a result for small increments for the Brownian motion for moving starting points (see here).

I'm wondering whether we can find such results for large increments, say something like

$\limsup_{n \to \infty} \sup_{0 \leq t-s \leq n} \frac{B_{t} - B_s}{\sqrt{2n \log(n)}} = 1$.

Can we obtain such a result from time inversion, i.e. the fact the $t B_{1/t}$ is also a Brownian motion? Or is it even easier to obtain such a result? Or is it not possible at all?

Student1369321
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1 Answers1

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About the large increments $B(t+h)-B(t)$ of BM(Brownian motion), please refer to the book: M. Csörgo and P.Révész, Strong Approximations in Probability and Statistics, Academic Press, 1981. $\S1.2$, p.29--.

The Corollary 1.2.2(ibid) give the following result,
\begin{equation*} \varlimsup_{n\to\infty}\sup_{\substack{0\le s,t\le n\\ 0\le t-s\le n}} \frac{B_t-B_s}{\sqrt{2n\log\!\log n}}=1. \end{equation*}

JGWang
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