Timeline for expectation of supremum
Current License: CC BY-SA 3.0
9 events
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Sep 28, 2011 at 16:06 | history | edited | user16215 | CC BY-SA 3.0 |
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Sep 28, 2011 at 15:18 | answer | added | Jeff Schenker | timeline score: 2 | |
Sep 28, 2011 at 14:07 | comment | added | kaleidoscop | I don't know if it can helps, but I imagine your assumption with $c(h)$ implies that the family $\{X_n\}$ is tight for the uniform convergence topology, meaning the law of $X_n$ converges weakly for this topology to the law of some random continuous process $X$. | |
Sep 28, 2011 at 13:34 | history | edited | user16215 | CC BY-SA 3.0 |
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Sep 28, 2011 at 13:01 | history | edited | user16215 | CC BY-SA 3.0 |
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Sep 28, 2011 at 12:59 | comment | added | user16215 | @Anthony Quas : Thank you for this nice counter-example. Do you think the implication is still false if one assumes that $X_n(t)$ is continuous on $[0,T]$, with a modulus of continuity independent of $n$? | |
Sep 28, 2011 at 12:05 | comment | added | The Bridge | Hi I wonder if for some (good) local martingales the implication can be proved using BDG inequlities. Regards | |
Sep 28, 2011 at 11:57 | comment | added | Anthony Quas | The implication is false: Let $X_n(t)$ be a stochastic process on $[0,1]$ with $X_n(t)=1$ for an interval $[(i-1)/2^n,i/2^n]$ where $i$ is chosen uniformly at random from the set $\{1,\ldots,2^n\}$. Then for any fixed $t$, $X_n(t)$ is 1 with probability $2^{-n}$ so that the left side is $2^{-n}$. On the other hand, the right side is 1. | |
Sep 28, 2011 at 9:57 | history | asked | user16215 | CC BY-SA 3.0 |