 Solution

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.