arXiv:2511.17822cs.LGcs.DS2025-11

提出高精度列表解码均值估计方法,突破误差随噪声比例衰减的瓶颈。

High-Accuracy List-Decodable Mean Estimation

  • 设计新识别性证明与算法,不依赖平方和层次结构。
  • 输出大小为 exp(O(log²(1/α)/ε²)) 的候选列表,其一元素误差≤ε。
  • 适用于高精度要求的鲁棒统计学习,尤其适合小污染率场景。

在列表解码学习中,给定的数据点中仅 α(α≪1)来自良好分布 D,目标是输出一个短候选列表,其中至少一个解能恢复关于 D 的非平凡信息。现有算法虽可实现最优列表大小,但误差随 1/α 衰减,常表现不佳。本文研究能否通过扩大列表大小换取更高精度:给定 ε>0,是否可输出略大列表,使其中至少一个元素与真实均值的 ℓ₂ 距离不超过 ε?我们证明在身份协方差高斯分布的均值估计问题中,存在大小至多为 L = exp(O(log²(1/α)/ε²)) 的候选列表,其中至少一项与真均值的 ℓ₂ 距离 ≤ ε。同时设计出运行时间与样本复杂度为 n = d^{O(log L)} + exp(exp(~O(log L))) 的算法。核心贡献在于全新的可识别性证明及无需使用平方和层次结构的新算法框架,具有独立技术价值。

原文摘要 · Abstract (English)

In list-decodable learning, we are given a set of data points such that an $α$-fraction of these points come from a nice distribution $D$, for some small $α\ll 1$, and the goal is to output a short list of candidate solutions, such that at least one element of this list recovers some non-trivial information about $D$. By now, there is a large body of work on this topic; however, while many algorithms can achieve optimal list size in terms of $α$, all known algorithms must incur error which decays, in some cases quite poorly, with $1 / α$. In this paper, we ask if this is inherent: is it possible to trade off list size with accuracy in list-decodable learning? More formally, given $ε> 0$, can we can output a slightly larger list in terms of $α$ and $ε$, but so that one element of this list has error at most $ε$ with the ground truth? We call this problem high-accuracy list-decodable learning. Our main result is that non-trivial high-accuracy guarantees, both information-theoretically and algorithmically, are possible for the canonical setting of list-decodable mean estimation of identity-covariance Gaussians. Specifically, we demonstrate that there exists a list of candidate means of size at most $L = \exp \left( O\left( \tfrac{\log^2 1 / α}{ε^2} \right)\right)$ so that one of the elements of this list has $\ell_2$ distance at most $ε$ to the true mean. We also design an algorithm that outputs such a list with runtime and sample complexity $n = d^{O(\log L)} + \exp \exp (\widetilde{O}(\log L))$. We do so by demonstrating a completely novel proof of identifiability, as well as a new algorithmic way of leveraging this proof without the sum-of-squares hierarchy, which may be of independent technical interest.

鲁棒估计列表解码高精度统计学习

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