T. Royen proved the Gaussian correlation inequality in the context of Gamma distributions back in 2014, which was since popularized by Latala and Matlak. The properties of Gaussian integration seem heavily exploited in the original proof: even the reduction from convex sets to cylindrical ones rely on the fact that projections of Gaussian vectors are again Gaussian. Has anyone attempted to prove the result for more general distributions beyond multivariate Gamma, or are there obvious counterexamples?
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$\begingroup$ is there any reason to hope/believe that the word "Gaussian" is superfluous? $\endgroup$– Carlo BeenakkerCommented Apr 10, 2017 at 15:07
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1$\begingroup$ @CarloBeenakker: The original paper of Royen addresses multivariate Gamma. The definition of the latter is in terms of its Laplace transform. The referenced papers from the 50s are pretty terse to understand. $\endgroup$– John JiangCommented Apr 11, 2017 at 4:00
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2 Answers
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I'm not that familiar with this area but there do exist some results for more general probability distributions but more restricted sets. For example, as explained by Li and Shao, the Gaussian correlation inequality may be reformulated as saying that if $(X_1, \ldots, X_n)$ is a centered, Gaussian random vector, then $$\mathbb{P}\left(\max_{1\le i\le n} |X_i| \le 1\right) \ge \mathbb{P}\left(\max_{1\le i\le k} |X_i| \le 1\right) \mathbb{P}\left(\max_{k+1\le i\le n} |X_i| \le 1\right)$$ for all $1\le k < n$. The case $k=1$ was proved by Khatri and Šidák (independently), and for this special case, Das Gupta et al. have proved a generalization to elliptically contoured distributions. Das Gupta et al. also give some counterexamples to overly strong generalizations of the conjecture.
answered Apr 11, 2017 at 16:12
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$\begingroup$ GCI is for general convex sets, this is more like a result on bounds of extreme values. But did learn sth, thanks! $\endgroup$– Henry.LCommented Apr 11, 2017 at 16:51
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2$\begingroup$ @Henry.L : Well, we certainly cannot expect any results about general convex sets for more general distributions because for general convex sets, even the Gaussian case was intractable until recently! The question is whether we intuitively expect the inequality to generalize to other distributions. In other words, is Royen's use of special properties of Gaussians an artifact of his method of proof? If we think of GCI as a general property of convex sets rather than as an amazing property of a very special distribution then we might expect it to generalize to other distributions. $\endgroup$ Commented Apr 11, 2017 at 18:32
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1$\begingroup$ The fact that more general distributions satisfy a related but weaker condition is, to me, evidence that something more general is going on, and that GCI is not just a very unusual property that is satisfied only by an extremely special class of probability distributions. $\endgroup$ Commented Apr 11, 2017 at 18:34
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$\begingroup$ Without any disrespect, I do not see there is anything more general going on since Royden's proof uses Gaussian-ity heavily; nor did I think it is a special property for Gaussian dist only, since we can use central limit theorem to extend the result to a wider range of applications I guess. $\endgroup$– Henry.LCommented Apr 11, 2017 at 18:56
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2$\begingroup$ @Henry.L : Regarding whether something more general is going on, I do not think that looking at the proof is a reliable indicator, because there are too many cases in mathematics where a result generalizes but the proof does not. Also, people suspected that GCI was true long before the proof existed, so whatever intuitions drove them to conjecture GCI could not have been based on the proof. $\endgroup$ Commented Apr 11, 2017 at 20:09
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Unlike Chow's answer, I do not think the results for elliptically contoured distributions is in the same spirit as GCI because they are controlling the bound of extreme values, which is more like results from U-statistics instead of the generality of GCI.
I think Royden's thinking is basically following Renyi's theorem [5](Or Cramer-Wold if you like) and consequential work in this direction is ongoing using Renyi's divergence applied on convex bodies[4].
...even the reduction from convex sets to cylindrical ones rely on the fact that projections of Gaussian vectors are again Gaussian.
According to the technique that Royden used, it relies heavily on the fact that the Gamma family is reproducing[1] (OR projection closed, which does not generalize to many other families). The key arguments in his proof, as pointed out by Latala and Matlak[2], is the repeatative use of rectangular sets and the projected images onto these sets.
So I am doubtful that the GCI can be generalize further to other families beyond Gamma. At least I do not believe that these set of techniques can be generalized directly for otherwise Latala and Matlak must have already done.:) There is also another discussion about the application of GCI [3].
Reference
[1]Teicher, Henry. "On the convolution of distributions." The Annals of Mathematical Statistics (1954): 775-778. https://projecteuclid.org/euclid.aoms/1177728664
[2]Latała, Rafał, and Dariusz Matlak. "Royen's proof of the Gaussian correlation inequality." arXiv preprint arXiv:1512.08776 (2015). https://arxiv.org/abs/1512.08776
[4]Kumar, M. A., & Sason, I. (2016). Projection Theorems for the Rényi Divergence on $\ alpha $-Convex Sets. IEEE Transactions on Information Theory, 62(9), 4924-4935.
[5]Renyi, Alfréd. "On projections of probability distributions." Acta Mathematica Hungarica 3.3 (1952): 131-142.
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$\begingroup$ Historically, didn't GCI (as a conjecture) arise out of considerations of things like U-statistics? $\endgroup$ Commented Apr 11, 2017 at 18:37
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$\begingroup$ @TimothyChow I think I did not express what I commented in your answer well. I do not expect the product inequality extend to any family without reproducing property, that is because the conjecture is first proposed to address the issue occur in multiple comparison as some comments in [3] pointed out. $\endgroup$– Henry.LCommented Apr 11, 2017 at 18:41
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$\begingroup$ What I want to express is that the reproducing property is the central thing that works in Royden's proof, and hence I tend to believe that Renyi's projection scheme seems like a more appropriate direction rather than U-statistics. $\endgroup$– Henry.LCommented Apr 11, 2017 at 18:42
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$\begingroup$ For the history part, it actually arise from multiple comparison. See pp.1-2 here: arxiv.org/pdf/1012.0676.pdf @TimothyChow $\endgroup$– Henry.LCommented Apr 11, 2017 at 18:51
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$\begingroup$ Consider $\mu=\frac{1}{4}( \delta_{(1,2)} + \delta_{(-1,-2)} + \delta_{(-2,0)} + \delta_{(2,0)})$, the measure that assigns probability .25 at each of four points. $\mu$ is symmetric. Trivially, $\mu(|X|<\frac{1}{2} \text{ and } |Y|<1) = 0 < \mu( |X|<\frac{1}{2} )* \mu(|Y|<1) = \frac{1}{2}*\frac{1}{2}$. So the correlation inequality does not hold for $\mu$. It seems that for the inequality to hold, most mass should be placed around $(0,0)$. $\endgroup$ Commented Apr 11, 2017 at 19:50