Timeline for The infinity Wasserstein distance $W_\infty$ and the weak topology
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2021 at 11:27 | vote | accept | Baruch Spinoza | ||
| Sep 24, 2021 at 9:24 | answer | added | Steve | timeline score: 3 | |
| Sep 23, 2021 at 20:16 | comment | added | Ronnie Pavlov | Of course you're right. I figured I was missing something easy. Interesting question! | |
| Sep 23, 2021 at 14:12 | comment | added | Baruch Spinoza | Also, I guess what I really should have said instead of "without isolated points" is "connected". | |
| Sep 23, 2021 at 14:00 | comment | added | Baruch Spinoza | But doesn't adding the Lebesgue measure in fact give $W_\infty$ convergence? My intuition for what goes wrong in the example I gave is that any small neighbourhood $U$ of $1$ must be extended all the way to $0$ to get $\delta_0(U_r)>\mu_n(U)$. But with Lebesgue available, you can get away with extending by a small amount. Thanks for your response, and sorry if I am way off the mark. | |
| Sep 23, 2021 at 13:24 | comment | added | Ronnie Pavlov | And it shouldn't hold even if you assume your measures are monatomic; you could just replace your delta-measures by normalized Lebesgue measure over a pair of disjoint closed intervals, and it should have weak but not $W_\infty$ convergence for the same reason as yours. It seems that $W_\infty$ is just a drastically stronger notion of convergence. | |
| Sep 23, 2021 at 13:20 | comment | added | Ronnie Pavlov | Couldn't you just add Lebesgue measure to both $\mu_n$ and $\delta_0$ (and normalize if you want), and then you have two fully supported measures with weak but not $W_\infty$ convergence for the same reason? | |
| S Sep 23, 2021 at 12:20 | review | First questions | |||
| Sep 23, 2021 at 12:51 | |||||
| S Sep 23, 2021 at 12:20 | history | asked | Baruch Spinoza | CC BY-SA 4.0 |