Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Empirical mean and scale distance control Gaussian-regression likelihoods

Statement

Let 0<a<b<∞, let σ0,σ∈[a,b], and fix design points x1,…,xn. Under the true independent laws Yi∼N(w0(xi),σ02), the model has independent laws Yi∼N(w(xi),σ2). Write δi=w(xi)−w0(xi) and ∥δ∥n2=n−1∑iδi2. There is a finite constant Ca,b, depending only on a,b, such that whenever max⁡i∣δi∣≤1 and ∣σ−σ0∣≤1, 1n∑i=1nKL(Pi0∥Pi)≤Ca,b(∥δ∥n2+∣σ−σ0∣2), 1n∑i=1nEi0 ⁣[(log⁡pi(Yi)pi0(Yi))2]≤Ca,b(∥δ∥n2+∣σ−σ0∣2).

Facts & Assumptions

Given: The independent fixed-design normal model with true and candidate scales in [a,b], a>0, and the displayed local bounds on mean and scale differences.

[A1]

Assume AC for Kullback–Leibler divergence and the imported normal laws.

[F1]

Standard normal and normal laws: a N(m,σ2) variable with σ>0 has the translated and scaled standard-normal law.

[F2]

Gaussian even moments for Brownian increments: a centered N(0,σ02) variable has second and fourth moments σ02 and 3σ04; symmetry makes its third moment zero.

[F3]

A uniform derivative bound gives a uniform Taylor remainder bound: a uniform bound on the second derivative bounds the first-order Taylor remainder by a constant times the squared increment.

Proof

1.1F1F2givenalgebra

Put ei=Yi−w0(xi) and δi=w(xi)−w0(xi). Direct substitution of the normal densities from [F1] gives log⁡pi(Yi)pi0(Yi)=log⁡σ0σ−(ei−δi)22σ2+ei22σ02. Taking its negative expectation and using [F2] yields Ki:=KL(Pi0∥Pi)=log⁡σσ0+σ02+δi22σ2−12.

2.1A1F3step 1.1givenalgebra

Let s=σ−σ0 and g(σ)=log⁡(σ/σ0)+σ02/(2σ2)−1/2. Then g(σ0)=g′(σ0)=0. Its second derivative is bounded uniformly when σ,σ0∈[a,b], so [F3] gives ∣g(σ)∣≤Ca,bs2. Explicitly, g′(σ)=1/σ−σ02/σ3 vanishes at σ0, and ∣g′′(σ)∣=∣−1/σ2+3σ02/σ4∣≤a−2+3b2a−4. Also δi2/(2σ2)≤δi2/(2a2). The formula in step 1.1 therefore gives ∣Ki∣≤Ca,b(δi2+s2) and, in particular, the first claimed upper bound after averaging.

3.1F2step 1.1step 2.1givenalgebra∎

The log ratio in step 1.1 equals mi+Biei+A(ei2−σ02), where A=1/(2σ02)−1/(2σ2), Bi=δi/σ2, and mi=−Ki. On [a,b], ∣A∣≤Ca,b∣s∣ and ∣Bi∣≤a−2∣δi∣. By [F2], Eei2=σ02, E(ei2−σ02)2=2σ04, and E[ei(ei2−σ02)]=0. Therefore the second moment is Ki2+Bi2σ02+2A2σ04. Step 2.1 gives Ki2≤Ca,b(δi2+s2)2, which is at most a constant times δi2+s2 under the local bounds. Averaging proves the log-square inequality.

Source notes

The 2008 source states the fixed-covariate model but leaves its contraction proof as a sketch. The complete local KL and log-square calculations above make the scale, direction, empirical norm and moment bounds explicit; the independent-observation source is read as separate context, not substituted for the proof.

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