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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Exact worst-case risk of the truncated estimator

Statement

Assume Countable Choice and let ϵ≥0. In the Gaussian sequence inverse model with positive singular numbers (sj)j∈N>0, let Q≥0 and let (aj)j∈N>0 be positive and nondecreasing. For each θ∈Θa(Q), define the squared sequence risk of the cutoff estimator at D∈N0 by Rθ(D):=Eθ ⁣[∥θ^D(Z)−θ∥ℓ22]. Then its exact worst-case squared sequence risk is sup⁡θ∈Θa(Q)Rθ(D)=Q2aD+12+ϵ2∑j=1Dsj−2, where the sum is 0 for D=0. The supremum is attained by the one-coordinate truth θ⋆=(Q/aD+1)eD+1, with θ⋆=0 when Q=0.

Facts & Assumptions

[F1]

Gaussian sequence inverse model: For ϵ>0, Zj=θj+ϵsj−1ξj with Eξj=0 and Eξj2=1; at ϵ=0, Zj=θj deterministically.

[F2]

Weighted ellipsoid class for sequence inversion: θ∈Θa(Q) means θ∈ℓ2 and ∑j≥1aj2θj2≤Q2, with 0<a1≤a2≤⋯.

[F3]

Singular-value cutoff estimator: θ^D,j(Z)=Zj for 1≤j≤D and 0 for j>D.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is assumed only for the upstream compact SVD coordinate identification; it is not used in the finite-coordinate risk calculation.

Given: Countable Choice, ϵ≥0, positive singular numbers (sj), a cutoff D∈N0, and the ellipsoid Θa(Q) with positive nondecreasing weights.

Proof

1.1givenA1F1F2F3algebra

Under [A1], the model coordinates in [F1] are fixed by the countable-choice SVD branch. If ϵ>0, [F1] and [F3] give θ^D,j(Z)−θj=ϵsj−1ξj for j≤D and −θj for j>D, hence ∥θ^D(Z)−θ∥ℓ22=ϵ2∑j=1Dsj−2ξj2+∑j>Dθj2. If ϵ=0, the retained error is zero and the omitted error is −θj, so the squared error is ∑j>Dθj2. In both cases the error is in ℓ2 because the estimator has finite support and θ∈ℓ2.

2.1step 1.1F1algebra

For ϵ>0, taking expectation in the finite random sum of [step 1.1] and using Eξj2=1 from [F1] gives Rθ(D)=ϵ2∑j=1Dsj−2+∑j>Dθj2. For ϵ=0, the deterministic error in [step 1.1] gives the same formula with the first term zero. The deterministic tail is finite because θ∈ℓ2.

3.1givenF2step 2.1algebra

For every θ∈Θa(Q), monotonicity gives aj≥aD+1 for j>D, so ∑j>Dθj2=∑j>Daj2θj2aj−2≤aD+1−2∑j>Daj2θj2≤Q2/aD+12. Thus [step 2.1] is bounded above by the asserted right-hand side.

4.1F2step 2.1step 3.1constructalgebra

The sequence with θD+1⋆=Q/aD+1 and all other coordinates zero belongs to ℓ2 and satisfies ∑j≥1aj2(θj⋆)2=Q2, including Q=0. Its tail squared norm is exactly Q2/aD+12, so it attains the upper bound from [step 3.1]. The finite-coordinate variance term in [step 2.1] is independent of θ, hence the same sequence attains the full worst-case risk.

5.1step 2.1step 3.1step 4.1algebra∎

Taking the supremum in [step 2.1], using the universal upper bound [step 3.1] and the attaining sequence [step 4.1], yields the exact equality and the stated zero-cutoff convention.

Remarks

This is expected squared loss in the Gaussian sequence experiment, over a weighted ellipsoid of coefficient sequences. It is not a deterministic worst-case error bound for arbitrary norm-bounded data perturbations.

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