Definition: Recall that a distribution $\mu$ on $\mathbb R^d$ is said to be log-convave with constant $c > 0$, if density $d\nu \propto e^{-V}dvol$ satisfying the curvature condition $$ \operatorname{Hess}_x(V) \succeq cI_d,\text{ for all }x \in \mathbb R^d. $$

Now, It is known that a distribution $\mu$ on $\mathbb R^d$ which has finite moment and density not supported on an affine subspace can be approximated with a log-concave distribution $\nu$ (Lemma 2.1 of this paper).

Question

Would such a distribution $\mu$ then satisfy a transportation-cost inequality for the Wasserstein $2$-distance (see here for definitions, just in case) ?

  • Could someone kindly explain why this question was downvoted ? Thanks. – dohmatob 2 days ago

Let's try again.

The theorem of Otto and Villani implies that every distribution $\nu$ which is log-concave in the sense you define satisfies a transportation-cost inequality.

There are many distributions $\mu$ with finite moments and density supported everywhere which do not satisfy a transportation-cost inequality. Since a TCI in particular implies subgaussian concentration, any distribution $\mu$ which has larger than Gaussian tails will fail to satisfy a TCI. One such distribution $\mu$ is the exponential distribution. But by the theorem of Otto and Villani, there is no such distribution $\mu$ which is also log-concave in the sense you define.

Whichever question you mean to ask, the answer is in the above two paragraphs.

  • 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. – dohmatob Sep 14 at 19:24
  • 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). – Mark Meckes Sep 16 at 21:47
  • 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. – Mark Meckes Sep 16 at 21:48
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    @dohmatob: Yes: an exponential distribution has finite moments and density not supported on an affine subspace, but does not satisfy a TCI. – Mark Meckes Sep 17 at 13:27
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    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. – Gabe K Sep 18 at 16:03

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