0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.