Timeline for Does log-concave approximable distribution satisfy transportation-cost inequality?
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| Sep 20, 2018 at 18:30 | comment | added | dohmatob | OK, ya makes sense, and it indeed is a counterexample. Upvoted. There was a strong inductive bias in my previous comment, as I thought your comment was just a recycling of previous comments based on the misunderstanding (language). My bad. | |
| Sep 18, 2018 at 16:03 | comment | added | Gabe K | It's a counter-example because while it is not log-strongly concave, it can be approximated by log-strongly concave functions (that's what Lemma 2.1 states). Therefore, it is an example of a distribution that can be approximated by a log strongly-concave function that does not satisfy a TCI. | |
| Sep 18, 2018 at 10:17 | history | edited | Mark Meckes | CC BY-SA 4.0 |
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| Sep 18, 2018 at 6:33 | comment | added | dohmatob | Thanks for the effort. I think we simply don't understand each other and it won't get any better. No offence taken. The exponential distribution $d\mu \propto e^{-\lambda x}1_{x \ge 0}dx$ is not a counterexample. It's not log strongly-concave. | |
| Sep 18, 2018 at 0:32 | comment | added | Mark Meckes | Okay, so I've removed the use of terminology which is inconsistent with yours from my answer, and the answer to your question is unambiguously no, with the exponential distribution as a counterexample. | |
| Sep 18, 2018 at 0:31 | history | edited | Mark Meckes | CC BY-SA 4.0 |
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| Sep 17, 2018 at 20:12 | comment | added | dohmatob | the last edit I made to the question got screwed and the definition was inserted at the end / middle instead of the beginning. Fixed. So the question is whether a distribution $\nu$ can approximated by a log-strongly concave distribution $\mu$, does $\nu$ itself satisfy TCI ? Not that $\nu$ satisfies TCI (consequence of results by Bobkov et Goetze 1999; Otto et Villani 2000). The question is whether this transfers to $\nu$ (of course with perhaps a degraded TCI constant). | |
| Sep 17, 2018 at 16:37 | comment | added | Mark Meckes | On the other hand, if you mean to assume that $\mu$ is itself log strongly-concave, then yes, it does satisfy a TCI. This was first proved by Otto and Villani, as you stated yourself in your other question linked here. | |
| Sep 17, 2018 at 16:36 | comment | added | Mark Meckes | @dohmatob: Please clarify the question. As stated, you did not assume $\mu$ to be log-concave in any sense (although you cite a result that it can be approximated in some sense by a log-concave measure), and then asked if it must satisfy a TCI. The exponential distribution is an example of a measure $\mu$ which satisfies the assumptions you stated and does not satisfy a TCI. | |
| Sep 17, 2018 at 14:51 | comment | added | dohmatob | Well, I don't see the point repeating the same inapplicable example over and over. As I said, by log-concave, I actually mean log strongly-concave. In particular the exponential distribution is not log-strongly concave. Please see previous comments. I'll include this precision explicitly in the question in hope that it maybe clearer. | |
| Sep 17, 2018 at 13:27 | comment | added | Mark Meckes | @dohmatob: Yes: an exponential distribution has finite moments and density not supported on an affine subspace, but does not satisfy a TCI. | |
| Sep 17, 2018 at 3:31 | comment | added | dohmatob | Well, this presumption is just a restatement of the question, in negative form. Any counterexample ? | |
| Sep 16, 2018 at 21:48 | comment | added | Mark Meckes | But this is beside the point for the question you asked: a distribution $\mu$ on $\mathbb{R}^d$ which has finite moment and density not supported on an affine subspace typically will not satisfy a TCI. | |
| Sep 16, 2018 at 21:47 | comment | added | Mark Meckes | That is not the standard terminology in any source I've seen, including the paper you linked in the question, which defines log-concave to mean of the form $e^{-V(x)} dx$ for $V$ convex (not necessarily strictly convex). | |
| Sep 14, 2018 at 19:24 | comment | added | dohmatob | I think there might be some misunderstanding here. By log-concave, what is actually meant is log strongly-convave. This is usual language. That is, this are measures with density of the form $e^{-V(x)}dx$ for some $\mathcal C^2$ potential $V:\mathbb R^d \rightarrow \mathbb R$ with Hessian $\operatorname{Hess}V(x) \succeq \rho I_d$ for all $x \in \mathbb R^d$. See Bobkov et Goetze 1999. No doubt, the exponential disitrbution on $\mathbb R^p$ doesn't have the TCI, which is fine since it's not log-concave in the sense abovE. | |
| Sep 14, 2018 at 17:01 | history | answered | Mark Meckes | CC BY-SA 4.0 |