**Update:** This conjecture has now been solved! See A simple proof of the Gaussian correlation conjecture extended to multivariate gamma distributions by T. Royen, and Royen’s proof of the Gaussian correlation inequality by Rafał Latała, and Dariusz Matlak.

The Gaussian correlation conjecture is a simple looking inequality which, nevertheless, remains unproven. The standard n-dimensional Gaussian measure is

for measurable subsets of n-dimensional space . The conjecture states that

(1) |

for all convex and symmetric sets . This inequality can be shown to be equivalent to the following statement. If are jointly Gaussian random variables with mean zero, and , then

(2) |

A less general version of this inequality was initially stated in 1955 in a paper by Dunnet and Sobel, and the full version as stated above was conjectured in 1972 by Das Gupta et al.

Various special cases of the conjecture are known to be true. In 1967, Khatri and Šidák independently proved (2) for or, equivalently, (1) in the case where is a *symmetric slab* (the region between two parallel hyperplanes). The two dimensional case was proven in 1977 by Pitt, and the case where is an arbitrary centered ellipsoid was proven in 1999 by Hargé.

In this entry we shall discuss the following interesting partial results due to Schechtman, Schlumprecht and Zinn in 1998.

- There is a positive constant such that the conjecture is true whenever the two sets are contained in the Euclidean ball of radius .
- If, for every , the conjecture is true whenever the sets are contained in the ball of radius , then it is true in general.

At first sight it would seem that this comes close to proving the conjecture. Unfortunately, the constant in the first statement is , which is strictly less than one, so the second statement cannot be applied. Furthermore, it does not appear that the proof can be improved to increase to one. Alternatively, we could try improving the second statement to only require the sets to be contained in the ball of radius for some but, again, it does not seem that the proof can be extended in this way.

For full details, see the paper by Schechtman, Schlumprecht and Zinn, which is freely available online. As the methods used are interesting in themselves, we briefly run through them now, and look at alternative approaches in a later post. The proof of 1 relies on applying a rotation, or change of variables, in the calculation of the measure of , and a result on log-concave functions is used to turn it into an inequality. The key idea employed in the proof of 2 is a tensor power trick along the same lines as those described by Terence Tao in his blog. This will give the following result.

Lemma 1Let be positive constants with as goes to infinity. Suppose that, for each , the inequality

(3)

holds for all symmetric convex sets in the Euclidean ball of radius . Then, for all such .

To prove this, apply inequality (3) to the powers , which will be in the Euclidean ball of radius in .

Raising to the power of and taking the limit as proves the lemma.

Let us start with statement 2 above. The proof begins by showing that the correlation inequality (1) holds when is a Euclidean ball. In fact, by the invariance of the Gaussian measure under rotations, this can be reduced to the trivial case where both and are spherically symmetric, so that one is a subset of the other. Then, writing for the unit n-dimensional Euclidean ball,

The second of these inequalities uses the hypothesis of statement 2, and the third uses the Gaussian correlation inequality in the case where one of the two sets is a Euclidean ball. The result will follow from Lemma 1 as long as we can prove that tends to 1 as goes to infinity. In fact, it can be shown that converges to 1/2 (this is an easy application of the Central Limit Theorem), which completes the proof.

Moving on to statement 1, the trick is to consider rotating the product by 45 degrees, , . Then are in respectively if and only if is in and is in . As the measure is invariant under rotations, this gives

(4) |

In order to bound the integrand on the right hand side, we want to show that its maximum is attained when . For this, the following result of Prékopa (available here) and Leindler on log-concave functions is used.

Theorem 2If is log-concave then so is .

Applied to the log-concave function , with being the standard -dimensional Gaussian density and the indicator function, this shows that

is log-concave in . Together with symmetry in , this implies that its maximum occurs at . So, (4) gives the inequality

This almost completes the proof. It is not difficult to show that is bounded by and, if and are contained in a ball of radius , then is bounded by multiplied by the volume of a ball of radius ,

The formula for the volume of a ball of radius together with Stirling’s approximation can be used to show that as goes to infinity. So, using the value , the result follows from Lemma 1.

This completes the sketch of the proofs of 1 and 2. I don’t know if it is possible to go much further with the methods described here. However, in a later post, we shall look at an alternative method which is strong enough to prove the conjecture in the special cases where or where one of the sets is a centered ellipsoid.

[…] Gaussian Correlation Conjecture 2 We continue investigating the Gaussian correlation conjecture in this post. This states that if is the standard Gaussian measure on […]

Pingback by The Gaussian Correlation Conjecture 2 « Almost Sure — 7 October 09 @ 5:49 PM |

It’s not unproven anymore: http://arxiv.org/abs/1408.1028

Comment by Tani — 2 January 16 @ 7:42 PM |

Thanks for that link. Sorry, I have been away from this blog for a while. I still have to check some of the details, but the proof looks good, and surprisingly straightforward!

Comment by George Lowther — 6 June 16 @ 12:28 AM |

And, I see another submission which is the same as Royen’s proof which you linked to, but explains it in a bit more detail (with auxiliary lemmas to fill in the gaps) http://arxiv.org/abs/1512.08776.

Comment by George Lowther — 6 June 16 @ 12:31 AM |