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Exact Recovery Thresholds for Weighted Data Selection in Vector-Valued Linear Regression

Aug 2026 · 1 citation · 15 references
Computer Science Mathematics

TL;DR

The threshold part of Question 4 of the COLT 2025 open problem "Data Selection for Regression Tasks" is resolved, and an erroneous claim circulating in a recent unrefereed preprint is correct.

Abstract

We resolve the threshold part of Question 4 of the COLT 2025 open problem"Data Selection for Regression Tasks"of Hanneke, Moran, Shlimovich and Yehudayoff. We study vector-valued linear regression with square loss $\ell_{(x,y)}(W)=\lVert Wx-y\rVert_2^2$, where $x\in\mathbb{R}^d$ and $y\in\mathbb{R}^m$. The learner returns the minimum-Frobenius-norm empirical risk minimizer. We prove that the minimal budget of weighted examples for recovering the full-data loss on every finite dataset is exactly $n^{\star}(d,m)=(m+1)d$. We determine the weighted selection profile $F_{\mathrm{weighted}}(d,m,n)$ at the near-threshold budget: $F_{\mathrm{weighted}}(d,m,(m+1)d-1)=1+1/(dm^2)$. We recover the known spanning-budget value $F_{\mathrm{weighted}}(d,m,d)=d+1$ for every $m$, and $F_{\mathrm{weighted}}(d,m,n)=\infty$ for $n<d$. For the smallest open intermediate cell $(d,m)=(2,2)$ we prove $F_{\mathrm{weighted}}(2,2,3)\in[13/8,15/8]$ and $F_{\mathrm{weighted}}(2,2,4)\in[5/4,3/2]$. We reduce the conjectured exact values $13/8$ and $5/4$ to a finite moment problem on the circle with at most seven atoms and assemble structural evidence for it. The upper bounds use a fixed-basis conic compression lemma, a determinant--facet rigidity theorem for maximal certificates, and sharp sparsification lemmas for zero-mean weighted point systems. These tools may be of independent interest. We also exhibit an explicit six-point integer dataset with $d=m=2$ on which no weighted selection of $2d$ points recovers the optimal loss. Thus the scalar sufficient budget $2d$ does not extend to vector-valued outputs. Our new regression-profile results for $m\ge2$ extend the scalar theory for $m=1$.

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