Diagonal dominance conjecture for squared components of a truncated normal vector

About 14 years old · traced to 2012

Let $X\sim\mathcal{N}_{v}(0,\Lambda)$ , where $\Lambda=\operatorname{diag}(\lambda)$ , and let

\mathcal{B}_{v}(\rho)=\{x\in\mathbb{R}^{v}:\|x\|\leq\rho\}

be the Euclidean ball of radius $\rho$ . Consider the covariance matrix of the vector $(X_1^2,\ldots,X_v^2)$ conditional on $X\in\mathcal{B}_{v}(\rho)$ . Diagonal dominance conjecture. This covariance matrix is diagonally dominant: for every $n$ ,

\operatorname{var}(X_n^2\mid X\in\mathcal{B}_{v}(\rho))\geq\sum_{m\ne n}\left|\operatorname{cov}(X_n^2,X_m^2\mid X\in\mathcal{B}_{v}(\rho))\right|,

for all $\rho\in\mathbb{R}_{+}$ . The claim is presented as an unproved exercise concerning the relative amplitudes of variances and covariances under weak truncation; no resolution is supplied in the source.

References

Primary source

Filippo Palombi and Simona Toti, “A note on the variance of the square components of a normal multivariate within a Euclidean ball”, arXiv:1211.1614 (2013).

covariance-matrices multivariate-normal-distributions probability

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A posted argument claims a complete proof of the conjecture, but no independent source has verified it, so the problem remains mathematically unsettled.

Palombi and Toti posed the diagonal-dominance conjecture in 2012 for squared coordinates of a centered Gaussian vector restricted to a Euclidean ball. Their source records no proof or counterexample for the row-sum inequality.

Known results

Palombi and Toti (2012) establish the separate bound $\operatorname{var}(X_n^2\mid X\in\mathcal{B}_v(\rho))\le 2\lambda_n\mathbb{E}[X_n^2\mid X\in\mathcal{B}_v(\rho)]$ in strong truncation and order by order for sufficiently large $\rho$ ; the intermediate regime remains open.

Posted attempt

A reader-posted argument claims a complete, strict diagonal-dominance proof using nonpositive off-diagonal covariances, log-concavity of weighted chi-square distribution functions, and a conditional covariance row-sum estimate. The argument has not been independently verified.

Current status (as of August 2026): a complete proof has been claimed in discussion but remains unverified, so diagonal dominance is not settled and no counterexample is recorded.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Proof (with a dimension-uniform strict improvement). Let $X_1,\ldots,X_d$ be independent centered Gaussian variables with variances $\lambda_i>0$ , write $U_i=X_i^2$ , $S=\sum_{i=1}^d U_i$ , and condition throughout on $E=\{S<R\}$ , where $R>0$ is the squared truncation radius. Set $C_{ij}=\operatorname{Cov}(U_i,U_j\mid E)$ . In fact, for every $i$ ,

\boxed{\displaystyle \sum_{j\ne i}|C_{ij}| \le \bigl(1-e^{-(d-1)/2}\bigr)C_{ii} <C_{ii}. }

Thus the conjectured weak diagonal dominance holds strictly, with a quantitative gap independent of $R$ and all the variances.

Known ingredients and attribution. Mukerjee and Ong, J. Multivariate Anal. 139 (2015), 1–6, DOI $10.1016/j.jmva.2015.02.010$ , proved the distinct pairwise inequalities $C_{ij}\le0$ for $i\ne j$ , and proved that the distribution function of any positive weighted sum of independent chi-square variables is log-concave. Their paper does not state or prove the diagonal-dominance conjecture; the following argument supplies the missing row-sum inequality.

Fix $i$ , assume $d\ge2$ , and put

V=\sum_{j\ne i}U_j,\qquad F(t)=\Pr(V\le t),\qquad r(t)=\frac{F'(t)}{F(t)},\qquad \mu(t)=\mathbb E[V\mid V\le t].

For $t>0$ , integration by parts gives

t-\mu(t)=\int_0^t\frac{F(s)}{F(t)}\,ds, \qquad \mu'(t)=r(t)\bigl(t-\mu(t)\bigr).

Log-concavity of $F$ implies

\frac{F(s)}{F(t)} \le \exp\bigl(-r(t)(t-s)\bigr) \quad(0<s<t),

and therefore

0<\mu'(t)\le 1-e^{-t r(t)}<1.

Independence now yields

g(u):=\mathbb E[S\mid U_i=u,E] =u+\mu(R-u),

so $g'(u)\ge e^{-(R-u)r(R-u)}>0$ . Consequently,

\sum_{j=1}^d C_{ij} =\operatorname{Cov}(U_i,S\mid E) =\operatorname{Cov}(U_i,g(U_i)\mid E)>0.

Together with the established $C_{ij}\le0$ for $j\ne i$ , this already proves strict diagonal dominance.

For the uniform quantitative improvement, let $m=d-1$ . Rescaling the complementary Gaussian integral gives

F(t) =K t^{m/2} \int_{\|y\|\le1} \exp\!\left( -t\sum_{j\ne i}\frac{y_j^2}{2\lambda_j} \right)dy,

where $K>0$ . The integral is decreasing in $t$ , and hence

0<t\,r(t)\le \frac m2.

It follows that

g'(u)\ge e^{-m/2}.

For independent copies $U,U'$ from the conditional law of $U_i$ ,

\begin{aligned} \operatorname{Cov}(U,g(U)) &=\frac12\mathbb E[(U-U')(g(U)-g(U'))]\\ &\ge e^{-m/2}\operatorname{Var}(U). \end{aligned}

Accordingly,

C_{ii}-\sum_{j\ne i}|C_{ij}| =\sum_{j=1}^d C_{ij} \ge e^{-(d-1)/2}C_{ii}>0,

proving the displayed bound. For $d=1$ , the off-diagonal sum is empty and $C_{11}>0$ .

Original conjecture: Palombi and Toti, J. Multivariate Anal. 122 (2013), Conjecture 4.1, https://arxiv.org/abs/1211.1614. Previously proved pairwise/log-concavity ingredients: Mukerjee and Ong, https://arxiv.org/abs/1311.6018.