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Subspace Uncertainty and Sharp Sampling Thresholds on the Boolean Cube

Published 8 Oct 2026 in math.PR, cs.IT, cs.LG, and math.CO | (2610.12358v1)

Abstract: We study Gaussian regression under squared population L2L_2 loss in a known mm-dimensional subspace of degree-at-most-kk functions on the dd-dimensional Boolean cube. Random inputs can undersample regions essential for prediction, delaying the parametric rate even when the model is known. For fixed $q_0&lt;1/2$, 1≤k≤q0d1\le k\le q_0d, and sufficiently large fixed AA, the worst-subspace sample threshold for minimax error Aσ<sup>2(m+t)/nAσ<sup>2(m+t)/n with confidence 1−e<sup>−t1-e<sup>{-t}, t≥log⁡4t\ge\log4, is [ N=(m+t)\exp{E_{d,k}+O(k{1/3})}, \quad E_{d,k}=dΨ(k/d), ] where Ψ(q)=log⁡2−H(12−q(1−q))Ψ(q)=\log2-\mathsf H(\tfrac12-\sqrt{q(1-q)}) and H\mathsf H is binary entropy with natural logarithms. The upper bound holds for every feasible mm; the matching lower bound holds when m≤(d⌊k<sup>1/3⌋)m\le\binom d{\lfloor k<sup>{1/3}\rfloor} or t≥mt\ge m. We sharpen the Polyanskiy--Samorodnitsky uncertainty principle in two respects. First, for fixed leakage ρ∈(0,1)ρ\in(0,1), the smallest set carrying a fraction $1-ρ$ of a nonzero degree-at-most-kk polynomial's energy has probability exp⁡−Ed,k+Oρ,q0(k<sup>1/3)\exp{-E_{d,k}+O_{ρ,q_0}(k<sup>{1/3})}. An Airy-kernel construction proves that the remainder cannot be o(k<sup>1/3)o(k<sup>{1/3}) in general. Second, we construct a subspace of dimension (d⌊k<sup>1/3⌋)\binom d{\lfloor k<sup>{1/3}\rfloor} such that every function in the subspace has at least a fraction $1-ρ$ of its energy on the same set, whose probability is at most exp⁡−Ed,k+Cρ,q0k<sup>1/3\exp{-E_{d,k}+C_{ρ,q_0}k<sup>{1/3}}. For sufficiently large kk, this set is a Hamming ball. A striking consequence is an exponential cost of noise: the parametric rate can require (m+t)4<sup>kexp⁡−O(k<sup>1/3)(m+t)4<sup>k\exp{-O(k<sup>{1/3})} samples, whereas O((m+t)2<sup>k)O((m+t)2<sup>k) suffice for noiseless identification. As k→∞k\to\infty with k/d→0k/d\to0, the noisy threshold is (m+t)exp⁡2k+o(k)(m+t)\exp{2k+o(k)}.

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