CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A collinear design defeats a sparsity-only coefficient guarantee

Statement refuted

Sparsity alone guarantees a positive restricted eigenvalue constant for every fixed design.

Facts & Assumptions

Given: $n = 1$ , $d = 2$ , $X = (1 1)$ , $s = 1$ , and $S = {1}$ .

[F1]

The restricted-eigenvalue cone condition is $∥ v_{S^{c}} ∥_{1} \leq 3 ∥ v_{S} ∥_{1}$ and requires a positive lower bound for $∥ X v ∥_{2}^{2} / (n ∥ v_{S} ∥_{2}^{2})$ (Restricted eigenvalue condition on the Lasso cone).

[F2]

Incoherence at $s = 1$ requires $\max_{j, k} ∣ (X^{T} X / n - I)_{j k} ∣ \leq 1 / 14$ (Fixed-design incoherence condition).

Proof

1.1F2algebra

The Gram matrix is $X^{T} X / n = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ , so its maximum entrywise deviation from $I$ is $1 > 1 / 14$ and the incoherence condition fails.

2.1F1algebra∎

Set $v = (1, - 1)$ . Then $∥ v_{S^{c}} ∥_{1} = 1 \leq 3 ∥ v_{S} ∥_{1} = 3$ , but $X v = 0$ while $∥ v_{S} ∥_{2}^{2} = 1$ . The ratio required to be bounded below by a positive restricted-eigenvalue constant is therefore zero. The two distinct one-sparse parameters $(1, 0)$ and $(0, 1)$ also have the same fitted mean $X θ = 1$ . Thus sparsity by itself gives neither the RE condition nor coefficient identifiability.

Depends on

Used by

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Dependency tree · two levels

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Sources

Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, Chapter 2, Lemma 2.17 and Theorem 2.18 (standard reference, not scraped)
Peter J. Bickel, Ya'acov Ritov and Alexandre B. Tsybakov, Simultaneous Analysis of Lasso and Dantzig Selector, Section 3 (standard reference, not scraped)