---
title: "Separation Is a Riesz Energy: How Many Environments Does ICA Need?"
source: "arXiv AI/CL/LG（论文）"
source_tier: "t2"
url: "https://arxiv.org/abs/2611.06228v1"
permalink: "https://feed.astralor.com/items/itm_761aa4237c749852a587a38e"
published: "2026-11-04T09:35:43.000Z"
score: 59
channel: "ai"
tags: ["Separation","Riesz Energy","How Many Environments","Does ICA Need","Can","Gaussian","Yes"]
---

# Separation Is a Riesz Energy: How Many Environments Does ICA Need?

## 速览

Can independent components be recovered when all of them are Gaussian? Yes, if their variances change across environments; but how many environments, and how many samples as the number of components $d$ grows? We answer…

## 关键点

- Can independent components be recovered when all of them are Gaussian?
- Yes, if their variances change across environments; but how many environments, and how many samples as the number of components $d$ grows?
- We answer both for the joint diagonalization of $K$ covariances.

## 全文

Can independent components be recovered when all of them are Gaussian? Yes, if their variances change across environments; but how many environments, and how many samples as the number of components $d$ grows? We answer both for the joint diagonalization of $K$ covariances. Each component has a signature, its variances across the environments. Separating component $j$ costs the Riesz 2-energy $E_j$ of this signature, a sum, over the other components, of inverse pairwise Fisher informations. For Gaussian sources no estimator in $O(d)$ beats $cE_j/(n+E_j)$ at any $n$ and $d$, and above an explicit sample size a spectral estimator attains $E_j/n$ up to a constant, for every mixing and without the signatures (given a lower bound on the pairwise information). Since signatures live in $K-1$ dimensions, the rate obeys a dimension law: $d^2/n$, $d\log d/n$ and $d/n$ for $K=2$, $3$ and $K\ge4$ well-spread environments. What counts is the dimension the signatures span, not their number, and $\log d$ well-designed environments make $d\log d$ samples suffice. Projecting the covariances onto one linear combination pays $d^2/n$ at first order instead. In simulations up to $d=1000$ the predictions hold and joint diagonalization with $K\ge4$ attains the rate once $n/d$ is in the hundreds; on task fMRI the energy says which components are hard to separate.

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