AI & ML
Exact regions for the symmetrized Blest coefficient
arXiv stat.TH0 views
arXiv:2611.08026v1 Announce Type: new Abstract: Spearman's rho treats all ranks alike, yet in many comparisons of two rankings an agreement at the top matters more than one at the bottom. Blest's rank correlation builds this preference into Spearman's rho, and its symmetrized version $\eta$, introduced by Genest and Plante, does so without favoring either of the two rankings. How far can such a top-weighted coefficient stray from other rank correlations? We answer this question for Spearman's rho, Blomqvist's beta and Chatterjee's rank correlation $\xi$ by determining the exact set of values that $\eta$ and each of these coefficients can attain jointly. Given Spearman's rho, $\eta$ can differ from it by at most $27/128$, and the extreme values of $\eta$ are attained only when the two rankings agree perfectly on the top ranks and are completely reversed on the bottom ranks, or vice versa. Given Blomqvist's beta, $\eta$ obeys exactly the same sharp bounds as Spearman's rho. For Chatterjee's rank correlation, both boundary curves are explicit and are attained by copulas with densities. Here symmetrization has a surprising effect. Compared with Blest's original coefficient, $\eta$ has a narrower range at fixed Spearman's rho, but near independence a wider range at fixed Chatterjee's rank correlation, where $|\eta|$ can come close to $\sqrt{11\xi/5}$ while Blest's coefficient stays below $\sqrt{32\xi/15}$. The proofs combine explicit Kantorovich potentials from optimal transport with a convex relaxation that turns out to be exact.
Source:https://arxiv.org/abs/2611.08026PreviousA Note on the Cut-Locus-Corrected Fr\'echet Mean Central Limit TheoremNextSharp moderate deviations and uniform Berry--Esseen bounds for Hotelling's $T^2$-statistic via Stein's method
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