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Lasso prediction and coefficient bounds under incoherence

Statement

Let n,d,s be integers with n≥2, d≥1, 1≤s≤d, and let σ>0, δ∈(0,1). In the model Y=Xθ∗+ε of Fixed-design regression with sub-Gaussian noise, suppose ∣{j:θj∗≠0}∣≤s and the deterministic design satisfies the incoherence condition of Fixed-design incoherence condition. Define τ=4σ(log⁡(2d)n+log⁡(1/δ)n). For every response realization the minimizer set in Lasso estimator for fixed design is nonempty by The Lasso objective has a nonempty compact convex minimizer set. The measurable score event has probability at least 1−δ, and on that event every minimizer θ^ in the set satisfies ∥Xθ^−Xθ∗∥22n≤576 sσ2log⁡(2d/δ)n,∥θ^−θ∗∥1≤962 sσlog⁡(2d/δ)n. In particular, the prediction and coefficient rates are respectively O(sσ2log⁡(2d/δ)/n) and O(sσlog⁡(2d/δ)/n). Incoherence is sufficient for these bounds, but not necessary. Without any design restriction, the coefficient bound can fail.

Facts & Assumptions

Given: The hypotheses in the Statement, the fixed design, response and noise, and an arbitrary minimizer θ^ for each response realization.

[F1]

For every response the Lasso objective has a nonempty minimizer set (Lasso estimator for fixed design, The Lasso objective has a nonempty compact convex minimizer set).

[F2]

On the event ∥XTε/n∥∞≤τ/2, the basic inequality gives a2+τ∥ΔSc∥1≤3τ∥ΔS∥1, where Δ=θ^−θ∗ and a=∥XΔ∥2/n (Lasso basic inequality and cone constraint).

[F3]

If ∥ΔSc∥1≤3∥ΔS∥1, the exact incoherence bound implies a2≥12∥ΔS∥22 for the support S with ∣S∣≤s (The stated incoherence bound implies the Lasso restricted eigenvalue condition).

[F4]

With the displayed τ, the score event in [F2] has probability at least 1−δ (Sub-Gaussian score bound under fixed-design incoherence).

[F5]

For ∣S∣≤s, ∥ΔS∥1≤s∥ΔS∥2 (The Cauchy-Schwarz inequality for finite sums).

Proof

1.1givenF1F2F3F4F5cases

Work on the event in [F4], whose probability is at least 1−δ, and choose any minimizer, which exists by [F1]. Let S={j:θj∗≠0} and Δ=θ^−θ∗. Since a2≥0 and τ>0, [F2] gives ∥ΔSc∥1≤3∥ΔS∥1, so [F3] applies. If a=0, [F3] gives ΔS=0; then [F2] gives ΔSc=0, and both conclusions hold. If a>0, [F3] and [F5] imply ∥ΔS∥1≤2s a. From [F2], a2≤3τ∥ΔS∥1≤32sτa, hence a2≤18sτ2.

2.1F2F3F5step 1.1algebra

The cone inequality from [F2] gives ∥Δ∥1≤4∥ΔS∥1. Using [F3], [F5] and a≤32sτ from step 1.1 yields ∥Δ∥1≤42sa≤24sτ.

3.1step 1.1step 2.1algebra

Put L=log⁡(2d/δ)=log⁡(2d)+log⁡(1/δ). Since (u+v)2≤2(u2+v2), τ2≤32σ2L/n and τ≤4σ2L/n. Substitution into steps 1.1 and 2.1 gives the two displayed bounds; the event probability is at least 1−δ.

4.1givenF1F2step 3.1constructalgebra∎

To check the design caveats, take n=2, d=s=1 and ε=0, which satisfies the noise condition for every σ>0 and makes the score event certain. Choose X with XTX/n=4/5. Its diagonal deviation 1/5>1/14 violates incoherence, yet [F2] gives (4/5)∣Δ∣2≤3τ∣Δ∣, hence ∣Δ∣≤15τ/4 and ∥XΔ∥22/n≤45τ2/4 for every minimizer and every θ∗. Step 3.1 then gives both displayed bounds with smaller constants. Conversely, for X=0 and ε=0, the objective is 2τ∥θ∥1, whose unique minimizer is zero. Taking θ∗=M∈R with M>962σL/n violates the coefficient bound.

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