Info
The Gaussian Correlation Inequality was proved in 2014 (arXiv: 1408.1028), see(Royen, 2014). The interesting story can be found in Quanta Magazine.
🏷️ Introduction
The inequality is to show the Gaussian measure on centrally symmetric convex sets and satisfies
That is to say, if a dart hits the wall with standard Gaussian distribution, suppose two targets are centrally symmetric convex sets , , then hitting both targets with one dart is easier than hitting with the first dart and hitting with the second, vice versa.
The proof of inequality is simple and elegant. I think there are a few keys in the proof which are insightful (that is why this note exists 😁). The first observation is the following.
Observation
A centrally symmetric convex closed set can be formed by the intersection of countable symmetric strips. (Schechtman, Schlumprecht & Zinn, 1998)
This observation of “strips” is natural, since a convex symmetric body can be approximated by a sequence of convex, symmetric polytopes. Moreover, convex, symmetric polytopes are just slices of the unit cube in a higher dimension satisfying the constraints for . Therefore, the problem can be reduced to proving
Gaussian Correlation Theorem
where , and .
The special case was proved by the following theorem (Khatri, 1967).
Theorem (Khatri)
Let be a jointly Gaussian random variables with mean zero. Then
🌊Multivariate Gamma-type distribution
The set is better described by chi-squared distribution or Gamma distribution. Surprisingly, the multivariate Gamma distributions on have several (non-equivalent) definitions🤣.
The distribution is defined as follows.
Definition
If the random vector satisfies the Laplace transform
then it obeys the distribution.
Example
If and or , then the covariance matrix and
Note, not all values of suffice to produce an admissible distribution. Some possible values of are given in (Krishnamoorthy & Parthasarathy, 1951).
Example
Suppose , then the Wishart matrix . The diagonal part of is by Hadamard product. Then, the Laplace transform (or equivalently moment generating function) is
Therefore, all are admissible values.
🌵Variational Technique
-
In order to distinguish the dependence and independence, it is very common to introduce the correlation matrix for the dimensional vector , that is, in the spirit of variational method, then the left-hand and right-hand sides of the desired inequality are referring the case and . It equivalently means the function
is non-decreasing in , where and . Let be the joint distribution’s density function of , then that is to show the derivative is non-negative.
-
The following claim is from Lebesgue’s dominated convergence theorem. The differentiation can be swapped with the Laplace transform.
which equals to
-
The rest is a linear algebra problem only. Here is not important anymore, we drop it as identity.
which should be decreasing in .
In the original proof by Thomas Royen, the inequality is extended to the distributions such that the Laplace transform is infinitely divisible.
💬 Further discussions
- If the convex sets are not quite symmetric (say up to some local perturbations), does the inequality still hold for Gaussian measure? Such question is raised naturally, some existing works (Cordero-Erausquin, 2002) (Lim & Luo, 2012) might be a starting point.