Diagonal dominance conjecture for squared components of a truncated normal vector

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.