Timeline for Question about Wasserstein metric
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2017 at 8:56 | comment | added | user111097 | I've found the increasing rearrangement in Villani's book. Thank you very much! | |
| Jun 15, 2017 at 12:36 | comment | added | Benoît Kloeckner | @MB2009: there are many possible $f(x,\cdot)$, just don't try to determine it by the constraints you are given. In diemension $1$, increasing rearrangement works (i.e. you add the requirement that $f(x,\cdot)$ be increasing, which then makes it unique). | |
| Jun 15, 2017 at 12:08 | comment | added | user111097 | Could you please give more details on this construction of $f_x$? | |
| Jun 15, 2017 at 9:36 | comment | added | user111097 | Could you please specify a bit more for the case of dimension $1$? Actually, let $\rho$ be a standard Gaussian with normal distribution $F$, so we are looking for $f_x(\cdot)$ s.t. $\mathbb P[f_x(G)\le y]=\lambda_x((-\infty,y])$ for all $y\in\mathbb R$. Notice that $\lambda_x((-\infty,y])=\mathbb P[f_x(G)\le y]=\mathbb P[G\le f_x^{-1}(y)]=F\circ f_x^{-1}(y)$, so basically we should take $f_x^{-1}(y)=F^{-1}(\lambda_x((-\infty,y]))$. But how to define properly $f_x$? Thank you very much! | |
| Jun 14, 2017 at 14:25 | comment | added | user111097 | Merci pour la réponse et je vais verifier moi-meme. Je vais peut-être retourner vers vous (car je suis ne suis pas très familier avec la théorie de mesure). | |
| Jun 14, 2017 at 14:09 | history | undeleted | Benoît Kloeckner | ||
| Jun 14, 2017 at 14:07 | history | deleted | Benoît Kloeckner | via Vote | |
| Jun 14, 2017 at 14:05 | history | answered | Benoît Kloeckner | CC BY-SA 3.0 |