Let $B$ stand for a Brownian motion on a finite interval $[0,1]$. If I am not wrong, I think that there exists a positive constant $c$, such that almost surely, for $h$ small enough , for all $0< t < 1- h$

\begin{align} |B(t+h)-B(t)| < c\sqrt{h\log(1/h)} \end{align} or something like this. As a result

\begin{align} \bigg|\frac{B(t+h)-B(t)}{h}\bigg| < K(h) \end{align}

Am I correct ?

  • $\begingroup$ The inequality cannot hold with probability 1 since the random variable B(T')-B(t) is unbounded. Please clarify. $\endgroup$ – Did Sep 13 '14 at 10:33
  • $\begingroup$ Oops, yes of course, what a nonsense I wrote! $\endgroup$ – herrsimon Sep 13 '14 at 20:48

Yes, the result is known as Lévy's modulus of continuity:

Let $(B_t)_{t \geq 0}$ a (one-dimensional) Brownian motion. Then $$\mathbb{P} \left( \limsup_{h \to 0} \frac{\sup_{0 \leq t \leq 1-h} |B(t+h)-B(t)|}{\sqrt{2h |\log h|}} = 1 \right)=1.$$

See e.g. René Schilling/Lothar Partzsch: Brownian motion - An introduction to stochastic processes for a proof (Section 10.3).


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