Skip to main content

Timeline for Blackbox Theorems

Current License: CC BY-SA 3.0

7 events
when toggle format what by license comment
Jun 17, 2012 at 3:32 comment added Nate Eldredge Every book I've ever seen that proves the existence of BM goes on to prove it is nowhere differentiable.
Jun 16, 2012 at 21:44 comment added Alexander Shamov The proof of continuity usually follows from the "Kolmogorov Criterion" ... which itself is a little technical, but nevertheless fairly intuitive chaining argument.
Jun 15, 2012 at 23:54 comment added Felipe Olmos The proof of continuity usually follows from the "Kolmogorov Criterion": If there exists strictly positive constants $\varepsilon$, $p$ and $C$ such that $$\mathbb{E}|X_t - X_s|^p \leq C|t-s|^{1+\varepsilon}$$ then almost surely $X$ has a modification which has $\alpha$-Hölder continuous paths for any $\alpha \in (0,\frac{\varepsilon}{p})$
Jun 15, 2012 at 16:12 comment added Zsbán Ambrus Okay, you're right. I accept that you have to prove existence and continuity together.
Jun 14, 2012 at 22:03 comment added George Lowther @Zsbán: Continuity of BM is part of the standard definition, so proving that is the same as proving that it exists. However, the proof that BM is almost-surely nowhere differentiable is probably less well known.
Jun 14, 2012 at 9:49 comment added Zsbán Ambrus I think it's not so much the existence that's hard to prove but that it being continuous everywhere has probability 1.
Jun 14, 2012 at 1:12 history answered weakstar CC BY-SA 3.0