提出新型去噪方法，显著提升分布级去噪精度。
Distributional Shrinkage I: Universal Denoiser Beyond Tweedie's Formula
- 基于最优传输理论设计通用去噪器，不依赖信号与噪声分布
- 实现密度和广义矩匹配的 $O(σ^4)$ 与 $O(σ^6)$ 精度
- 适合关注分布恢复而非具体样本的场景，如生成建模
研究仅知噪声水平、不知噪声分布时的去噪问题。独立噪声 $Z$ 以 $Y = X + σZ$ 形式污染信号 $X$，其中 $σ∈(0,1)$ 已知。本文提出无需依赖信号与噪声分布的通用去噪器，从观测分布 $P_Y$ 恢复原始信号分布 $P_X$。当目标为分布级重建而非单个样本还原时，新方法相较基于 Tweedie 公式的贝叶斯最优去噪器（精度 $O(σ^2)$）实现数量级提升，在匹配广义矩和密度上分别达到 $O(σ^4)$ 与 $O(σ^6)$ 精度。方法基于最优传输理论，高阶逼近 Monge-Ampère 方程，并通过得分匹配高效实现。定义 $q$ 为 $P_Y$ 的密度，将传统贝叶斯去噪器 $\mathbf{T}^*(y) = y + σ^2 \nabla \log q(y)$ 替换为更温和的分布收缩形式：$\mathbf{T}_1(y) = y + \frac{σ^2}{2} \nabla \log q(y)$ 与 $\mathbf{T}_2(y) = y + \frac{σ^2}{2} \nabla \log q(y) - \frac{σ^4}{8} \nabla \left( \frac{1}{2} \| \nabla \log q(y) \|^2 + \nabla \cdot \nabla \log q(y) \right)$。
原文摘要 · Abstract (English)
We study the problem of denoising when only the noise level is known, not the noise distribution. Independent noise $Z$ corrupts a signal $X$, yielding the observation $Y = X + σZ$ with known $σ\in (0,1)$. We propose \emph{universal} denoisers, agnostic to both signal and noise distributions, that recover the signal distribution $P_X$ from $P_Y$. When the focus is on distributional recovery of $P_X$ rather than on individual realizations of $X$, our denoisers achieve order-of-magnitude improvements over the Bayes-optimal denoiser derived from Tweedie's formula, which achieves $O(σ^2)$ accuracy. They shrink $P_Y$ toward $P_X$ with $O(σ^4)$ and $O(σ^6)$ accuracy in matching generalized moments and densities. Drawing on optimal transport theory, our denoisers approximate the Monge--Ampère equation with higher-order accuracy and can be implemented efficiently via score matching. Let $q$ denote the density of $P_Y$. For distributional denoising, we propose replacing the Bayes-optimal denoiser, $$\mathbf{T}^*(y) = y + σ^2 \nabla \log q(y),$$ with denoisers exhibiting less-aggressive distributional shrinkage, $$\mathbf{T}_1(y) = y + \frac{σ^2}{2} \nabla \log q(y),$$ $$\mathbf{T}_2(y) = y + \frac{σ^2}{2} \nabla \log q(y) - \frac{σ^4}{8} \nabla \!\left( \frac{1}{2} \| \nabla \log q(y) \|^2 + \nabla \cdot \nabla \log q(y) \right)\!.$$
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