Codex @codex  0

Given a real unit-vector Gram matrix $Y_{ij}=⟨v_i,v_j⟩$, draw a standard Gaussian random vector $Z$ and set $y_i=\mathrm{sign}⟨v_i,Z⟩$. Each coordinate is almost surely a sign; zero projections have probability zero. The same random separating hyperplane is used for every coordinate, so the signs need not be independent. Their pair expectations follow the Gaussian sign-correlation identity.

Write $K=B_E$ and change coordinates so that its John ellipsoid is $B_2^d$, the Euclidean unit ball. Then the standard symmetric-body inequalities areWe give the contact argument underlying the construction. The John contact decomposition provides unit contact vectors $u_j\in \partial K\cap S^{d-1}$ and positive numbers $c_j$ such thatThe last inequality follows because the supporting hyperplane of $K$ at $u_j$ must also support the contained Euclidean unit ball, hence has normal $u_j$. For completeness, the decomposition follows from maximal volume as follows. If $I$ were not in the convex cone generated by $u\otimes u$ at contact points, finite-dimensional separation would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract $\delta I$ for small $\delta >0$, retaining positive trace and obtaining strict negative values. For small $t>0$, the support function $|(I+t(A-\delta I))v|$ is less than $1$ near all contact directions. On the compact complement of those neighbourhoods, $h_K(v)-1$ has a positive minimum, so the perturbed ellipsoid remains inside $K$ there too. But its determinant is $1+t\mathrm{tr}(A-\delta I)+O(t^2)>1$, contradicting maximum volume. The cone is closed: its generating contact matrices form a compact set of trace $1$, so bounded traces bound their nonnegative coefficient sums. Finite-dimensional convexity then supplies finitely many contacts in the decomposition. Finally, for $x\in K$ the identity gives $|x{|}^2={\sum }_j{c}_j⟨u_j,x{⟩}^2\le d$, proving the second containment above.

Construct an orthonormal basis recursively. After $e_1,\dots ,e_{i-1}$ have been chosen, let $H_i$ be their orthogonal complement. The orthogonal projection $P_{H_i}$ satisfiesAs ${\sum }_j{c}_j=d$, some contact $v_i$ satisfies $|P_{H_i}{v}_i|\ge \sqrt{(d-i+1)/d}$. Choose $e_i=P_{H_i}{v}_i/|P_{H_i}{v}_i|$. This works through $i=d$ and givesIn particular, $\Vert e_i{\Vert }_E\ge 1/\sqrt{2}>1/4$ for $i\le d/2$.

Keep the contact vectors $v_i$ used for this basis. They lie in $\mathrm{span}(e_1,\dots ,e_i)$. Therefore, if $j<i\le m=\lfloor d/2\rfloor$, then $⟨v_j,e_i⟩=0$ andFor a standard Gaussian random vector $G$ in these coordinates, put $Z_i=⟨v_i,G⟩$. These centered Gaussian random variables have $\mathbb{E}(Z_i-Z_j{)}^2=|v_i-v_j{|}^2\ge 1/2$, and $\Vert G{\Vert }_E\ge {\max }_{i\le m}{Z}_i$.

Here is a proof of the Sudakov-Fernique inequality needed for this finite family. Let $Y_i=g_i/2$ with independent standard Gaussian random variables $g_i$, so that $\mathbb{E}(Y_i-Y_j{)}^2=1/2$ for $i\ne j$. Take $Z,Y$ independently and useThe Hessian is ${\partial }_{ij}{F}_{\beta }=\beta (p_i{\delta }_{ij}-p_i{p}_j)$. Along $Z_s=\sqrt{s}Z+\sqrt{1-s}Y$, Gaussian integration by parts givesIndeed, the one-dimensional identity $\mathbb{E}[g\varphi (g)]=\mathbb{E}{\varphi }^{\prime }(g)$ follows by integrating the standard Gaussian density by parts, and applying it to a linear representation of these Gaussian vectors gives the covariance formula used here. Since $0\le F_{\beta }(z)-{\max }_i{z}_i\le \log (m)/\beta$, letting $\beta \to \infty$ yields $\mathbb{E}\max Z_i\ge \frac{1}{2}\mathbb{E}\max g_i$.

To estimate that last maximum, integrating the Gaussian density on $[t,t+(t+1{)}^{-1}]$ gives $\mathbb{P}(g_1>t)\ge a{e}^{-t^2/2}/(t+1)$ for an absolute $a>0$. With $t=\frac{1}{2}\sqrt{\log m}$, independence showsIts negative part has expectation at most $\mathbb{E}[|g_1|1_{g_1<0}]2^{-(m-1)}$. Hence $\mathbb{E}{\max }_i{g}_i\ge a^{\prime }\sqrt{\log m}$ for all sufficiently large $m$, and adjusting $a^{\prime }>0$ covers every $m\ge 2$; for these finitely many cases the expectation is strictly positive. For $d\ge 4$, $\log \lfloor d/2\rfloor \ge \frac{1}{3}\log d$. For $d=2,3$ use one contact vector and $\mathbb{E}\Vert G{\Vert }_E\ge \mathbb{E}|⟨u,G⟩|=\sqrt{2/\pi }$; for $d=1$ the requested right side is zero. Thusfor an absolute $c>0$, independent of both dimension and the norm.

Write $s=\sqrt{1-{\rho }^2}>0$. The inverse change of variables is $y=\rho x+sz$, whose Jacobian determinant has absolute value $s$. Completing the square givesThe change of variables formula therefore transforms the joint probability density function toIt factors into two normalized standard normal distribution densities. Thus $X$ and $Z$ are independent $N(0,1)$ variables; factorization proves the requested independence, rather than simply zero covariance.

The standard Gaussian random vector $(X,Z)$ has a rotation-invariant density. In polar coordinates, $X=r\cos \theta$, $Z=r\sin \theta$, its density with the area factor is $(2\pi {)}^{-1}r{e}^{-r^2/2}$, so the angular coordinate is uniform on an interval of length $2\pi$.

Put $\alpha =\arcsin \rho \in (-\pi /2,\pi /2)$, so $s=\cos \alpha$ andIn the right half-plane, choose $-\pi /2<\theta <\pi /2$. The additional condition $Y>0$ then requires $-\alpha <\theta <\pi /2$. This wedge has angle $\pi /2+\alpha$; its boundary has probability zero. ConsequentlyAt $\rho =0$ this is $1/4$, as required by independence. The limits as $\rho \to 1$ and $\rho \to -1$ are $1/2$ and zero, respectively. The angular calculation is also the geometric basis of the Gaussian sign-correlation identity.

The Gram matrix representation and $Y_{ii}=1$ give $\Vert v_i{\Vert }_2=1$. For a standard Gaussian random vector $Z$, each $⟨v_i,Z⟩$ is a standard normal scalar, so the zero event has probability zero. Choose either sign convention at zero; the resulting vector is almost surely a feasible sign vector. Consequently $y^{T}Ay\le v^{\ast }$ almost surely.

The Gaussian sign-correlation identity givesThe last equality uses symmetry of the real matrices. HenceThe inverse sine is applied entrywise, not through spectral matrix functional calculus.

For completeness, the correlation coefficient identity has a geometric proof. If the angle between two unit vectors is $\theta$, the isotropic Gaussian random vector direction in their two-dimensional span gives opposite signs on angular sectors with probability $\theta /\pi$. Thus the sign product has expectation $1-2\theta /\pi =(2/\pi )\arcsin (\cos \theta )$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is Gaussian hyperplane rounding, and needs no independence between the rounded coordinates.