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Expected squared operator error of thresholded sparse covariance

Statement

Fix τ>0, 0≤q<1 and β>1. Let n≥1, p≥2, p≥nβ and a=log⁡(p)/n≤1. For iid observations from Pq(τ,c) of Weak-lq covariance class with directional sub-Gaussian tails, use Entrywise-thresholded sample covariance with λ=8192τa. There is C=C(q,τ)<∞ such that sup⁡P∈Pq(τ,c)EP⊗n∥Σ^−Σ(P)∥op2≤C{c2a1−q+a}. The estimator uses sample centering and divisor n, including unknown means; no independence between coordinates is assumed. This is the unmodified threshold estimator, not a clipped version.

Facts & Assumptions

Given: The sample, class, dimension and threshold hypotheses in the statement.

[F1]

The directional tail is P(∣vT(X−EX)∣>t)≤e−t2/(2τ). Weak-lq covariance class with directional sub-Gaussian tails.

[F2]

Under the n-fold sample law P⊗n, for every u>0 the centered sample covariance entrywise error is at most 192τ(u/n+u/n) except with probability 2pe−u+2p2e−2u. Entrywise deviation for sample-centered covariance under directional Gaussian tails.

[F3]

On entrywise error at most λ/2, thresholding has operator error at most Cq(cλ1−q+λ). Weak-lq thresholding error from an entrywise deviation event.

[F4]

Nonnegative moments can be integrated from their tails. Layer-cake formulas for random variables.

[F5]

The algorithm uses the sample-centered covariance with divisor n, keeps the diagonal and never increases the absolute value of an entry. Sample covariance estimators and squared operator loss Entrywise-thresholded sample covariance.

[F6]

Cauchy–Schwarz for square-integrable random variables gives ∣EUV∣≤(EU2)1/2(EV2)1/2, including covariance entries and an indicator of a bad event. Cauchy-Schwarz for random variables.

[F7]

Finite-sum Cauchy–Schwarz gives ∣∑jajbj∣2≤(∑jaj2)(∑jbj2). The Cauchy-Schwarz inequality for finite sums.

[F8]

Finite-sum Hölder with conjugate exponents 4 and 4/3 gives (n−1∑iyi)4≤n−1∑iyi4 for nonnegative yi. Hölder's inequality for finite sums (rational exponents).

Proof

1.1givenF2F3algebra

Put u=log⁡(8p10)≤13log⁡p. Since a≤1, 192τ(u/n+u/n)≤192τ(13+13)a<4096τa=λ/2. The failure bound in [F2] is 1/(4p9)+1/(32p18)≤p−8. Thus an event G has P⊗n-probability at least 1−p−8, and on it [F3] bounds squared loss by C(q,τ)(c2a1−q+a).

1.2givenF1F4F6F7F8algebra

Write Uij=Xij−EXij. By [F1] and [F4], EP⊗n∣Uij∣8≤(2τ)4 4!=384τ4. Also, for unit v, vTΣv≤∫0∞e−t/(2τ)dt=2τ; applying [F6] to centered coordinates gives ∣σjk∣≤2τ. The sample-centered diagonal obeys 0≤σ^jj∗≤n−1∑iUij2. Applying [F7] to the two centered coordinate sample vectors gives ∣σ^jk∗∣2≤σ^jj∗σ^kk∗. Therefore EP⊗n∣σ^jk∗∣4≤12EP⊗n[(σ^jj∗)4+(σ^kk∗)4]≤384τ4, where the final inequality uses [F8] with yi=Uij2 (and likewise for k). These bounds are independent of the unknown means.

2.1givenF5F6F7step 1.1step 1.2algebra∎

By [F5], EP⊗n∣σ^jk−σjk∣4≤8EP⊗n(∣σ^jk∗∣4+∣σjk∣4)≤Cτ4. For any matrix A, [F7] gives ∥A∥op2≤∑jk∣Ajk∣2 by applying finite Cauchy–Schwarz to each row against a unit vector; [F7] also gives (∑jk∣Ajk∣2)2≤p2∑jk∣Ajk∣4. Thus EP⊗n∥Σ^−Σ∥op4≤Cτ4p4. Applying [F6] to ∥Σ^−Σ∥op2 and 1Gc and using step 1.1 gives EP⊗n[∥Σ^−Σ∥op2;Gc]≤Cτ2p2p−4=Cτ2p−2. Since p≥nβ implies n≤p1/β≤p and 1/p≤log⁡p, we have p−2≤log⁡p/n=a. Adding the good-event bound proves the claim uniformly over the class.

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