AI 中文总结
该研究提出检测分布偏移的矩尺度定律与核校准规则，证明其优于拓扑方法，真实嵌入流实验显示其AUC≥0.95，可对抗针对均值等统计量优化的对手。
AI 中文摘要
检测高维嵌入流是否发生变化通常被表述为统计量的选择问题。我们提出一种尺度定律，它对任何基于矩的选择都施加约束，并将其与拓扑替代方案进行对比测试。该定律规定：要验证承载质量分数f的空间尺度ε的特征，所需的多项式测试次数N*≥log(1/f)/(2ε)，该结论通过切比雪夫极值问题证明；高斯正交构造给出N*≥4b-1，其中b为尺度拓扑，因此计算成本由特征的细微程度而非特征数量决定。该定律是单侧的：我们构造了一个圆环，其均值、协方差和所有四阶矩都与实心圆盘的对应矩相等，但H₁非零。其实际应用是一种校准规则：上界由高斯测试函数达到，即RBF核的RKHS见证，因此该定律预测MMD测试应使用的带宽为特征尺度。在真实嵌入流上，我们在3种设置和3种尺度下测量σ*/ε的中位数为1.12（四分位距1.01-1.52，样本量n=26），且数据驱动的带宽达到AUC≥0.95。针对针对防御者的统计量（均值、协方差、k-NN、峰度）优化的对手，只有带宽匹配的核测试仍能检测到偏移。对于持续同调，结果喜忧参半，取决于通常未明确说明的选择：摘要的重要性超过滤波，总持续度在FPR为1%时达到召回率0.75，而第一持续度景观仅达到0.00。存在的是成本差距而非能力差距：持续同调在有效时的成本是峰度的116倍，而峰度的效果至少相当。我们的结论不是拓扑摘要无用，而是在该任务上，带宽由该定律设定的核测试优于拓扑方法。
英文摘要
Detecting that a stream of high-dimensional embeddings has changed is usually framed as a choice of statistic. We give a scale law that constrains any moment-based choice and test it against topological alternatives. The law: certifying a feature of spatial scale eps carrying mass fraction f requires polynomial tests of degree N* >= log(1/f)/(2 eps), proved via the Chebyshev extremal problem; a Gauss-quadrature construction gives N* >= 4b-1 for a b-scale topology, so cost is set by feature fineness, not feature count. The law is one-sided: we exhibit an annulus whose mean, covariance and all fourth-order moments equal those of a filled disk, yet H_1 is nonzero. Its practical content is a calibration rule. The upper bound is attained by Gaussian test functions, the RKHS witness of an RBF kernel, so the law predicts which bandwidth an MMD test should use: the feature scale. On real embedding streams we measure sigma*/eps with median 1.12 (IQR 1.01-1.52, n=26) over three settings and three scales, and a data-driven bandwidth reaches AUC >= 0.95. Against an adversary optimised against the defender's statistics (mean, covariance, k-NN, kurtosis), only a bandwidth-matched kernel test still detects. For persistent homology the verdict is mixed and depends on choices usually left implicit. The summary matters more than the filtration: total persistence attains recall 0.75 at FPR 1% where the first persistence landscape attains 0.00. What survives is a cost gap, not a power gap: where persistence works it costs 116x kurtosis, which works at least as well. We conclude not that topological summaries are useless, but that on this task a kernel test whose bandwidth the law sets dominates them.
Comments18 pages, 5 figures. Code and scripts: doi:10.5281/zenodo.21649324
