Under Assumption , let sigma^Y_i,j and sigma^X_i,j denote the corresponding covariance entries of the Gaussian vectors Y and X. Covariance condition conjecture. The condition sigma^Y_i,j≥ sigma^X_i,j≥ 0 is sufficient for the inequality operatornameCov(log S_N(mathbf G),p_i(mathbf G)p_j(mathbf G))≤ 0. This conjecture proposes a sufficient covariance comparison extending the proved independent standard-Gaussian case to a broader class under Assumption ; its resolution is not indicated in the supplied text.

A sufficient covariance condition for the Gaussian correlation inequality

The conjecture asserts that a coordinatewise comparison of nonnegative off-diagonal Gaussian covariances is sufficient for the stated covariance inequality, and hence for a variance comparison. It is formulated as Conjecture 1 in the relevant paper, whose April 2023 version records it as unresolved.

Known results

Standard Gaussian case: the covariance inequality holds for $N\geq 3$ and $i\ne j$ (Theorem 1.3).
Bivariate case: it holds for every bivariate Gaussian distribution.
One-edge comparison: when $Y$ has only $\sigma^Y_{1,2}=\rho\in(0,1)$ as a nonzero off-diagonal covariance and $X$ is standard Gaussian, the associated variance comparison is proved (Theorem 1.4).

Current status (as of August 2026): The full covariance condition remains an open conjecture; the standard-Gaussian, bivariate, and one-edge cases are known, with no verified proof or counterexample recorded in the retrieved sources.

The proposed covariance condition is not sufficient for the correlated Gaussian inequality. There are counterexamples in every dimension $N\ge3$ , including an explicit strictly positive definite three-dimensional example.

There is a notation ambiguity in the source: equation (1.18) writes an independent standard Gaussian vector, for which the inequality is already proved. The substantive conjecture concerns the correlated Gaussian vectors occurring in the covariance interpolation. It is this proposed extension that fails.

For $N\ge3$ and $0<\rho<1$ , take

\Sigma_X=I_N,\qquad \Sigma_Y=I_N+\rho(E_{12}+E_{21}), \qquad G\sim\mathcal N(0,\Sigma_Y).

Both covariance matrices are strictly positive definite and satisfy every required comparison

\sigma^Y_{ij}\ge\sigma^X_{ij}\ge0 \qquad(i\ne j).

Set

L_\beta=\log\sum_{\ell=1}^Ne^{\beta G_\ell}, \qquad P_\beta=p_1(G)p_2(G), \qquad \overline G=\frac1N\sum_{\ell=1}^NG_\ell.

At $\beta=0$ ,

L'_0=\overline G,\qquad P'_0=\frac{G_1+G_2-2\overline G}{N^2},

while $L_0$ and $P_0$ are deterministic. Therefore

\operatorname{Cov}(L_\beta,P_\beta) = \frac{\beta^2}{N^2} \operatorname{Cov} (\overline G,G_1+G_2-2\overline G) +O(\beta^3).

Since

\operatorname{Cov}(\overline G,G_1+G_2) =\frac{2(1+\rho)}N, \qquad \operatorname{Var}(\overline G) =\frac{N+2\rho}{N^2},

we obtain

\boxed{\displaystyle \operatorname{Cov}(L_\beta,P_\beta) = \frac{2\rho(N-2)}{N^4}\beta^2 +O(\beta^3)>0}

for all sufficiently small $\beta>0$ . The same calculation applies at every positive interpolation parameter by replacing $\rho$ with $\theta\rho$ .

A completely explicit choice is

N=3,\qquad \rho=\frac12,\qquad \beta=\frac1{20000},\qquad \Sigma_Y= \begin{pmatrix} 1&1/2&0\\ 1/2&1&0\\ 0&0&1 \end{pmatrix}.

To certify the sign without numerical integration, put $R=\max_i|G_i|$ . Direct differentiation gives

|L_\beta-L_0-\beta L'_0| \le\frac{\beta^2R^2}{2},

and

|P_\beta-P_0-\beta P'_0| \le9\beta^2R^2,\qquad |P'_0|\le\frac{4R}{9}.

Indeed $L_\beta''$ is a variance of values in $[-R,R]$ , hence at most $R^2$ , while

P_\beta'' = P_\beta\left[ (G_1+G_2-2\mu_\beta)^2-2L_\beta'' \right], \quad \mu_\beta=\sum_i p_i(G)G_i,

so $|P_\beta''|\le18R^2$ .

The exact Gaussian moment bounds are

\mathbb E R^4 \le (\operatorname{tr}\Sigma_Y)^2+ 2\operatorname{tr}(\Sigma_Y^2) =16, \qquad \mathbb E R^3\le8.

Using

|\operatorname{Cov}(F,H)| \le2cd\,\mathbb E R^{a+b} \quad\text{if}\quad |F|\le cR^a,\ |H|\le dR^b

yields

\left| \operatorname{Cov}(L_\beta,P_\beta) -\frac{\beta^2}{81} \right| \le \frac{1328}{9}\beta^3+144\beta^4.

At the displayed rational value of $\beta$ , it follows that

\boxed{\displaystyle \operatorname{Cov}(L_\beta,P_\beta) \ge \frac{10059271}{810000000000000000}>0.}

This contradicts the conjectured nonpositive covariance while satisfying the source's full entrywise covariance assumptions. The counterexample even belongs to the positively correlated-pair family treated separately by the source's Theorem 1.4.

Source: C.-H. Huang, Nonsymmetric examples for Gaussian correlation inequalities, Statistics & Probability Letters 201 (2023), 109885, doi:10.1016/j.spl.2023.109885; arXiv:2110.11641, Conjecture 1.

A sufficient covariance condition for the Gaussian correlation inequality

Progress summary

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Solutions 1