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Isotropic ridgeless risk limit when the signal is zero

Statement

Assume the Axiom of Choice. Let (ξij)i,j≥1 be one infinite iid array of real random variables with mean zero, variance one, and finite fourth moment. Let pn≥1 with pn/n→γ∈(0,∞)∖{1}, and let Xn=(ξij)1≤i≤n, 1≤j≤pn be its upper-left block. Let εn have independent centered components of variance σ2, independent of Xn, and let z0,n be independent of (Xn,εn), centered, and have covariance Ipn. Set Sn=XnTXn/n, yn=εn, β^n=Xn+yn, and Rn:=Eεn,z0,n[(z0,nTβ^n)2∣Xn]. Then almost surely Rn⟶{σ2γ1−γ,0<γ<1,σ2γ−1,γ>1.

Facts & Assumptions

Given: The nested real iid design, positive aspect ratio γ≠1, independent centered noise with covariance σ2In, centered test feature with covariance Ipn, and the Axiom of Choice from the Statement.

[A1]

The Axiom of Choice permits a choice function for every family of nonempty sets (The Axiom of Choice).

[F1]

For the nested real iid finite-fourth-moment design and every deterministic signal sequence of fixed norm r>0, the design-conditional risk tends almost surely to σ2γ/(1−γ) for γ<1 and to r2(1−1/γ)+σ2/(γ−1) for γ>1 (Isotropic minimum-norm risk limits under a positive spectral gap).

[F2]

The conditional risk for a deterministic signal is its squared projection onto ker⁡Xn plus the signal-independent noise term (Exact bias and noise decomposition of isotropic prediction risk).

[F3]

Iid entries are independent and share one common distribution (Identical distribution and IID families).

Proof

1.1givenchoose

For each n, let e1,n be the first standard basis vector in Rpn. For a∈{1,2}, let Rn(a) denote the conditional risk with signal βn(a)=ae1,n, and put Bn=e1,nT(I−Xn+Xn)e1,n and Vn=(σ2/n)tr⁡(Sn+).

2.1A1F1F2F3step 1.1algebra

Applying [F2] to each auxiliary signal gives Rn(1)=Bn+Vn and Rn(2)=4Bn+Vn. Applying [F1] with norms 1 and 2 gives almost-sure limits for both risks; the intersection of these two probability-one events still has probability one. The iid law is common in both applications by [F3].

3.1F1step 2.1algebra

The exact identities in step 2.1 give Vn=(4Rn(1)−Rn(2))/3. If γ<1, both risk limits in [F1] equal σ2γ/(1−γ), so Vn tends to that quantity. If γ>1, their limits are respectively 1−1/γ+σ2/(γ−1) and 4(1−1/γ)+σ2/(γ−1); their indicated combination is σ2/(γ−1).

4.1givenF2step 3.1∎

For the zero signal in the Statement, [F2] identifies Rn with the same Vn. Step 3.1 therefore gives both asserted limits.

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