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Subspace Uncertainty and Sharp Sampling Thresholds on the Boolean Cube

Thomas Weinberger

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Source: arXiv

Published: Oct 8, 2026

arXiv: 2610.12358

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Source 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 q0<1/2q_0<1/2, 1≤k≤q0d1\le k\le q_0d, and sufficiently large fixed AA, the worst-subspace sample threshold for minimax error Aσ2(m+t)/nAσ^2(m+t)/n with confidence 1−e−t1-e^{-t}, t≥log⁡4t\ge\log4, is N=(m+t)exp⁡{Ed,k+O(k1/3)},Ed,k=dΨ(k/d), 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⌊k1/3⌋)m\le\binom d{\lfloor k^{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−ρ1-ρ of a nonzero degree-at-most-kk polynomial's energy has probability exp⁡{−Ed,k+Oρ,q0(k1/3)}\exp\{-E_{d,k}+O_{ρ,q_0}(k^{1/3})\}. An Airy-kernel construction proves that the remainder cannot be o(k1/3)o(k^{1/3}) in general. Second, we construct a subspace of dimension (d⌊k1/3⌋)\binom d{\lfloor k^{1/3}\rfloor} such that every function in the subspace has at least a fraction 1−ρ1-ρ of its energy on the same set, whose probability is at most exp⁡{−Ed,k+Cρ,q0k1/3}\exp\{-E_{d,k}+C_{ρ,q_0}k^{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)4kexp⁡{−O(k1/3)}(m+t)4^k\exp\{-O(k^{1/3})\} samples, whereas O((m+t)2k)O((m+t)2^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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