Timeline for Wasserstein distance between rotated conditional distributions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2019 at 9:26 | comment | added | Steve | Tim Carson's counterexample seems to work well. $\rho(x | x \in R_\theta E)$ is basically the point mass at 0, so the rotation $R_\theta$ is not a valid transport. Note that $\rho(x | x \in R_\theta E)$ is very different from the pushforward $\rho(x | x \in E) \circ R_{\theta}^{-1}$. | |
| Aug 8, 2019 at 12:45 | comment | added | Terzo | Thank you, good point! So, since the mass is concentrating in the middle, you may have to pay $\operatorname{diam}(\operatorname{supp}(\rho))$ to spread it back along $E$. If this gives a lower bound on $W_1$, then your counterexample works. Still, there seems to be another natural action to transport $ \rho(x \mid x \in R_\theta E) $ onto $ \rho(x \mid x \in E) $, which is the rotation $R_\theta$ itself... I am still confused... :) | |
| Aug 8, 2019 at 12:17 | history | edited | Terzo | CC BY-SA 4.0 |
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| Aug 7, 2019 at 23:32 | comment | added | Tim Carson | What if we take the same $E$, but we take $\rho$ to be $(1-\delta)$ times the uniform distribution on $E$, plus $\delta$ times the uniform distribution on the circle of radius 2? Now for fixed $\epsilon$ and $\theta$ if you send $\delta$ to zero the mass of the rotated-conditional-distribution should concentrate in the intersection of $E$ and $R_\theta E$. So you may be able to get back to my previous comment. | |
| Aug 7, 2019 at 21:13 | comment | added | Tim Carson | Ah, I missed that assumption :) You're right. | |
| Aug 7, 2019 at 20:45 | comment | added | Terzo | Thanks for your comment. I am not sure to follow, though. In your example, $\operatorname{supp}(\rho)=E$, right? If so, $R_\theta E$ is not contained in $\operatorname{supp}(\rho)$. | |
| Aug 7, 2019 at 20:17 | comment | added | Tim Carson | "good factor of $\epsilon$" should read "good factor of $\theta$". | |
| Aug 7, 2019 at 17:45 | comment | added | Tim Carson | I don't think you can expect the good factor of $\epsilon$ in general. Consider $E = [-1, 1] \times [-\epsilon, \epsilon]$ and $\rho$ the uniform distribution on $E$. Note the diameter is independent of $\epsilon$. For any $\theta$, we can send $\epsilon$ to zero and the distribution of the rotated measure gets all in a little region around the origin (think of the intersection of a line and a rotated line). This should show we can make the LHS at least [something] independent of $\theta$. | |
| Aug 7, 2019 at 8:40 | review | First posts | |||
| Aug 7, 2019 at 12:05 | |||||
| Aug 7, 2019 at 8:36 | history | asked | Terzo | CC BY-SA 4.0 |