Timeline for wasserstein distance between distributions with bounded ratio
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2018 at 15:28 | comment | added | dohmatob | Yeah the "both" was a thinko, what I really meant was "or". :) | |
| Aug 30, 2018 at 14:20 | comment | added | Gabe K | For sure. In fact, you only need to assume that either $p$ or $q$ (not both) satisfies a log-Sobolev inequality to get this. This estimate depend on the $\rho$ in the log-Sobolev inequality, so this introduces another ingredient while eliminating the diameter. | |
| Aug 30, 2018 at 4:20 | comment | added | dohmatob | Actually, in case of infinite diameter assuming that $p$ and $q$ both satisfy the Talagrand transportation-cost inequality, i.e $W_{d,2}(p,r)\le \sqrt{(2/\rho)\operatorname{kl}(r\|p)}$ for any distribution $r$ on $X$ relatively continuous w.r.t $p$, with a similar story for $q$ (e.g for standard Gaussian, this holds with $\rho=1$), allows us to still get an upper bound, namely $w_{d,2}(p,q) \le \sqrt{2\min (\beta\log(\beta),(1/\alpha)\log(1/\alpha))/\rho}$. | |
| Aug 29, 2018 at 17:19 | comment | added | dohmatob | Also this document provides numerous useful inequalities between metrics on probability measures math.hmc.edu/~su/papers.dir/metrics.pdf. In general to upper bound $W$ metric with another metric, you need an upper bound on the diameter of the metric space. | |
| Aug 29, 2018 at 17:16 | comment | added | dohmatob | OK, i can workout $TV(p,q) \le \max(1-\alpha,\beta-1)$ and so $W_d(p,q) \le \max(1-\alpha,\beta-1)\operatorname{diam}(X)$ | |
| Aug 29, 2018 at 15:03 | comment | added | dohmatob | sure. Thanks for the response (upvoted). Sure, I suspected it involve the diameter of $X$ w.r.t to $p$. See my comment section above. More precisely, it would involve something like $\sup_{x_0} \mathbb E_{x \in p} d(x,x_0)$ (which is $\le D$). | |
| Aug 29, 2018 at 14:48 | history | answered | Gabe K | CC BY-SA 4.0 |