Timeline for "Cut norm" of conditional expectation has supremum on products of sets in sub-$\sigma$-algebra, or not?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2021 at 3:46 | comment | added | MikeG | @NarutakaOZAWA I see! This makes a lot of sense. Thanks! | |
| Sep 10, 2021 at 0:11 | comment | added | Narutaka OZAWA | When $\mathcal{F}_n$ is nested and $\vee \mathcal{F}_n=\mathcal{F}$, then the Martingale convergence theorem says the corresponding conditional expectations $E_n$ converge to the identity in a suitable sense depending on the function space in consideration. | |
| Sep 9, 2021 at 14:48 | comment | added | MikeG | @NarutakaOZAWA I get your second thought! But by approximating a sigma algebra by a finite one, the approximation is in what sense? | |
| Sep 9, 2021 at 7:28 | comment | added | Narutaka OZAWA | You can always approximate a $\sigma$-subalgebra by a finite one. Alternatively, since the indicator functions $\{ 1_S : S\subset X\}$ are the extreme points of the compact convex set $K$ of the measurable functions from $X$ into $[0,1]$, one has $\| W \|_\Box=\sup_{f,g\in K}|\langle W, f\otimes g\rangle|$. | |
| Sep 9, 2021 at 3:34 | history | asked | MikeG | CC BY-SA 4.0 |