A sufficient covariance condition for the Gaussian correlation inequality

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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}\geq \sigma^X_{i,j}\geq 0

issufficientfortheinequalityis sufficient for the inequality

\operatorname{Cov}(\log S_N({\mathbf G}),p_i({\mathbf G})p_j({\mathbf G}))\leq 0.

Thisconjectureproposesasufficientcovariancecomparisonextendingtheprovedindependentstandard−GaussiancasetoabroaderclassunderAssumptionThis 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.

References

Primary source

Chien-Hao Huang, “Nonsymmetric examples for Gaussian correlation inequalities”, arXiv:2110.11641 (2023).

Progress summary

Refreshed
Claimed solved

An unverified calculation claims the conjecture is false in three or more dimensions, while the published work had established only special cases.

Huang’s 2021 paper formulates the entrywise covariance comparison as Conjecture 1 and notes that it would imply a corresponding variance comparison. A reader-written calculation now claims a complete counterexample, but that claim has not been independently verified.

Known results

  • Huang (2021): for the standard Gaussian vector, the covariance inequality holds for N≥3N\geq 3 and i≠ji\ne j.
  • Huang (2021): the inequality holds for every bivariate Gaussian distribution.
  • Huang (2021): if XX is standard Gaussian and YY has only one nonzero off-diagonal covariance, then the associated variance comparison holds.

Posted attempt

An unverified calculation claims that, for N≥3N\geq 3, taking ΣX=IN\Sigma_X=I_N and ΣY=IN+ρ(E12+E21)\Sigma_Y=I_N+\rho(E_{12}+E_{21}) with 0<ρ<10<\rho<1 gives Cov⁡(log⁡SN(G),p1(G)p2(G))>0\operatorname{Cov}(\log S_N(G),p_1(G)p_2(G))>0 for sufficiently small positive β\beta. It therefore claims a complete disproof of the covariance conjecture; no independent verification is supplied.

Current status (as of August 2026): The published special cases are settled, but the full conjecture is unresolved because the proposed counterexample remains unverified.

Sources

Solutions 1

CounterexampleThis solution needs a summarySee full solutionHide full solution

The proposed covariance condition is not sufficient for the correlated Gaussian inequality. There are counterexamples in every dimension N≥3N\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≥3N\ge3 and 0<ρ<10<\rho<1, take

ΣX=IN,ΣY=IN+ρ(E12+E21),G∼N(0,ΣY).\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

σijY≥σijX≥0(i≠j).\sigma^Y_{ij}\ge\sigma^X_{ij}\ge0 \qquad(i\ne j).

Set

Lβ=log⁡∑ℓ=1NeβGℓ,Pβ=p1(G)p2(G),G‾=1N∑ℓ=1NGℓ.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 β=0\beta=0,

L0′=G‾,P0′=G1+G2−2G‾N2,L'_0=\overline G,\qquad P'_0=\frac{G_1+G_2-2\overline G}{N^2},

while L0L_0 and P0P_0 are deterministic. Therefore

Cov⁡(Lβ,Pβ)=β2N2Cov⁡(G‾,G1+G2−2G‾)+O(β3).\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

Cov⁡(G‾,G1+G2)=2(1+ρ)N,Var⁡(G‾)=N+2ρN2,\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

Cov⁡(Lβ,Pβ)=2ρ(N−2)N4β2+O(β3)>0\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 β>0\beta>0. The same calculation applies at every positive interpolation parameter by replacing ρ\rho with θρ\theta\rho.

A completely explicit choice is

N=3,ρ=12,β=120000,ΣY=(11/201/210001).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∣Gi∣R=\max_i|G_i|. Direct differentiation gives

∣Lβ−L0−βL0′∣≤β2R22,|L_\beta-L_0-\beta L'_0| \le\frac{\beta^2R^2}{2},

and

∣Pβ−P0−βP0′∣≤9β2R2,∣P0′∣≤4R9.|P_\beta-P_0-\beta P'_0| \le9\beta^2R^2,\qquad |P'_0|\le\frac{4R}{9}.

Indeed Lβ′′L_\beta'' is a variance of values in [−R,R][-R,R], hence at most R2R^2, while

Pβ′′=Pβ[(G1+G2−2μβ)2−2Lβ′′],μβ=∑ipi(G)Gi,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 ∣Pβ′′∣≤18R2|P_\beta''|\le18R^2.

The exact Gaussian moment bounds are

ER4≤(tr⁡ΣY)2+2tr⁡(ΣY2)=16,ER3≤8.\mathbb E R^4 \le (\operatorname{tr}\Sigma_Y)^2+ 2\operatorname{tr}(\Sigma_Y^2) =16, \qquad \mathbb E R^3\le8.

Using

∣Cov⁡(F,H)∣≤2cd ERa+bif∣F∣≤cRa, ∣H∣≤dRb|\operatorname{Cov}(F,H)| \le2cd\,\mathbb E R^{a+b} \quad\text{if}\quad |F|\le cR^a,\ |H|\le dR^b

yields

∣Cov⁡(Lβ,Pβ)−β281∣≤13289β3+144β4.\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 β\beta, it follows that

Cov⁡(Lβ,Pβ)≥10059271810000000000000000>0.\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.