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This question already has an answer here:

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 ?

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marked as duplicate by Did probability Sep 14 '14 at 21:38

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  • $\begingroup$ Are you user @peter5? $\endgroup$ – Did Sep 14 '14 at 21:42
  • $\begingroup$ yes, i wanted a cool nickname :P $\endgroup$ – peter5 Sep 14 '14 at 21:46
  • $\begingroup$ "yes, i wanted a cool nickname :P" Are you aware that the site forbids to use several user accounts? Let me suggest to close all your accounts except one and to avoid reposting duplicates of your own questions. $\endgroup$ – Did Sep 14 '14 at 21:48
  • $\begingroup$ in fact i didn't $\endgroup$ – peter5 Sep 14 '14 at 21:49