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Huber covariance modulus lower bound for unrestricted matrix-valued actions

Statement

Assume AC. Fix p,n≥1, M>0, 0≤ϵ≤1, and let Θ be either Fp(M) from Bounded-spectrum covariance class or Fp,s(M) from Single-block sparse covariance class and restricted directions with 1≤s≤p/2, equipped with its induced Borel sigma-algebra. Let En be the set of Borel maps (Rp)n→Rp×p. The clean family is PΣ=Np(0,Σ). For Σ∈Θ and matrix action A, put ℓ(Σ,A)=∥Σ−A∥op2. Its restriction to Θ×Θ defines the contamination modulus ω(ϵ,Θ) through Huber contamination modulus. Suppose ω(ϵ,Θ)<∞, 0≤M0<∞, 0<c0≤1, and inf⁡Σ^∈Ensup⁡Σ∈ΘPΣ⊗n{ℓ(Σ,Σ^)≥M0}≥c0. Then inf⁡Σ^∈Ensup⁡Σ∈Θsup⁡QPϵ,Σ,Q⊗n{ℓ(Σ,Σ^)≥116(M0∨ω(ϵ,Θ))}≥min⁡(c0,1/2).

Facts & Assumptions

Given: AC, the bounded covariance class, the Gaussian Huber model, its contamination modulus, and the clean tail-risk premise in the Statement.

[F1]

Gaussian Huber contamination model for covariance estimation: one observation has law (1−ϵ)Np(0,Σ)+ϵQ and the sample consists of iid observations from this law.

[F2]

Huber contamination modulus: for ϵ<1, pairs in the modulus satisfy TV(PΣ0,PΣ1)≤ϵ/(1−ϵ); at ϵ=1 all pairs are allowed.

[F3]

TV-close laws have identical Huber mixtures at the contamination threshold: under AC, every pair satisfying [F2] admits contaminants making the two one-observation Huber laws equal.

[F4]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: finite product laws are determined by their factor laws and rectangle values.

[F5]
[F6]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| and The norm ∥v∥=⟨v,v⟩ induced by a real or complex inner product define the matrix and vector norms. The inner-product norm is definite, homogeneous, and satisfies the triangle inequality gives the vector-norm triangle inequality: for every unit v, ∥(B+C)v∥2≤∥Bv∥2+∥Cv∥2≤∥B∥op+∥C∥op. Taking the supremum gives ∥B+C∥op≤∥B∥op+∥C∥op, without symmetry assumptions on B,C.

Proof

1.1F1givenalgebra

Fix any Σ^∈En. For each Σ∈Θ, choosing Q=PΣ makes the one-observation Huber law equal to the clean law by [F1]. If ω≤M0, then 116(M0∨ω)≤M0, so the robust supremum probability at the smaller threshold is at least the clean supremum probability at M0. The clean premise therefore gives the claimed bound c0 for this estimator. Taking the infimum preserves it.

1.2F1F2F3F4choose

Suppose ω>M0. By the definition of the finite supremum in [F2], there are Σ0,Σ1∈Θ with ℓ(Σ0,Σ1)>ω/2 and satisfying the total-variation condition there; when ϵ=1 every pair is admissible. By [F3], choose contaminants Q0,Q1 whose one-observation Huber laws coincide at a law R1. Their iid sample laws both equal R1⊗n by [F1] and [F4].

2.1F5F6step 1.2algebra

For every matrix action A, the triangle inequality gives max⁡i=0,1ℓ(Σi,A)≥14ℓ(Σ0,Σ1)>18ω(ϵ,Θ)>116(M0∨ω(ϵ,Θ)). Indeed, one of the two distances from A is at least half the distance between Σ0 and Σ1. The two measurable loss events at threshold 14ℓ(Σ0,Σ1) therefore cover the sample space under the common law; their probabilities sum to at least one. One has probability at least 1/2, and its event is contained in the event at the smaller displayed threshold. The robust supremum probability for this estimator is at least 1/2.

3.1step 1.1step 1.2step 2.1algebra∎

Step 1.1 applies when ω≤M0 and steps 1.2–2.1 apply when ω>M0. The bound holds for every Borel matrix-valued estimator, so taking their infimum proves the statement.

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