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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Integrated score error can miss every Euler evaluation

Example

In the finite-dimensional Gaussian credible-ball question reproduced below, choose d=n=1, R=1, σ=1, T=1, β(t)=1, and γ=1/4, with arbitrary positive p,s,a. Take the numerical operator equal to the true operator, so δA=0. For every integer K≥1 there is a jointly Borel finite score field with integrated score error exactly zero, whose Euler terminal law is N(1,K−1) for every observation. With M=K independent terminal draws, the resulting empirical ball has frequentist coverage tending to zero at truth θ=0, although the stated uniform boundary-margin condition holds with exponent κ=1. The exact credible-ball baseline satisfies δ1,1/40<3/4. Consequently no bound in that question can have Φ→0 uniformly over its original score class as (δA,εs,K−1,M−1/2)→(0,0,0,0). The original formulation is retained verbatim under Original question.

Original question

Fix integers d,n≥1, parameters p,s,a,R,σ>0, a credibility error γ∈(0,1), and a continuous schedule β:[0,T]→(0,∞) with T>0. Let Θs(R)={θ∈Rd:∑j=1dj2sθj2≤R2},Ap=diag⁡(1−p,2−p,…,d−p). Observe Yi=Apθ+εi, 1≤i≤n, with independent εi∼N(0,σ2Id). Use the proper Gaussian prior Π0=N(0,diag⁡(j−1−2a)j=1d), and let Πn(⋅∣Y) be its exact posterior under the stated forward operator and noise law. Let mn(Y) be the posterior mean and let rn,γ(Y) be the unique radius for which the Euclidean ball Cn,γ(Y)={u∈Rd:∥u−mn(Y)∥2≤rn,γ(Y)} has posterior probability 1−γ. Define the exact frequentist coverage cn,γ(θ)=Pn,θ{θ∈Cn,γ(Y)} and its baseline error δn,γ0=sup⁡θ∈Θs(R)∣cn,γ(θ)−(1−γ)∣.

For a deterministic numerical matrix A^n∈Rd×d with δA≥0 and ∥A^n−Ap∥op≤δA, let Π^n(⋅∣Y) be the Gaussian posterior formed with A^n and the same prior and noise variance. Write p^n,tY for the density obtained by applying the forward variance-preserving diffusion dXt=−12β(t)Xt dt+β(t) dWt to X0∼Π^n(⋅∣Y), and set s^n,tY=∇log⁡p^n,tY. A jointly Borel learned score field sn,tY is required to satisfy sup⁡θ∈Θs(R)EY∼Pn,θ[1T∫0T∫Rd∥sn,tY(x)−s^n,tY(x)∥22p^n,tY(x) dx dt]≤εs2. For h=T/K, define the reverse-time Euler chain by Z0∼p^n,TY and, for k=0,…,K−1, Zk+1=Zk+h[12β(T−kh)Zk+β(T−kh)sn,T−khY(Zk)]+hβ(T−kh) ζk, where the ζk are independent N(0,Id) vectors independent of Z0 and the observations. Draw M independent copies of this chain conditionally on Y and denote their terminal values by X(1),…,X(M). Let Xˉ=M−1∑iX(i), let r^ be the empirical 1−γ quantile of {∥X(i)−Xˉ∥2}i=1M (the smallest order statistic whose empirical cumulative mass is at least 1−γ), and put C^n,K,M(Y)={u:∥u−Xˉ∥2≤r^}. Suppose the exact credible-ball boundary satisfies the uniform margin condition sup⁡θ∈Θs(R)Pn,θ(∣∥θ−mn(Y)∥2−rn,γ(Y)∣≤u)≤Cacuκ(0<u≤u0) for fixed Cac,u0>0 and κ>0. Determine whether a minimax-sharp nonasymptotic function Φ exists uniformly over all such score fields and, if it does, find it as a function of d,n,p,s,a,R,σ,T,γ,Cac,u0,κ,δA,εs,K−1, and M−1/2 such that sup⁡θ∈Θs(R)∣Pn,θ,sampler{θ∈C^n,K,M(Y)}−(1−γ)∣≤δn,γ0+Φ, with matching lower bounds that identify when Φ→0. The probability includes observation noise and sampler randomness. The target concerns the additional coverage error introduced by the approximate forward operator, learned score, finite reverse discretization, and finite posterior sample; it does not assume that the exact Bayesian credible ball is already calibrated on every truth class.

Facts & Assumptions

Given: The scalar parameter choices in the Example and the original score-error, Euler and empirical-ball definitions reproduced below it.

[A1]

Write ϕ(x)=(2π)−1/2e−x2/2 and Φ(x)=∫−∞xϕ(z) dz. A standard Gaussian has density ϕ, mean zero, variance one and tail probability tending to zero at infinity.

[A2]

Let q=3/4 and let r>0 solve P{∣G∣≤2r}=q for a standard Gaussian G. Continuity and strict positivity of ϕ give existence and uniqueness of r.

Verification

Given: The scalar model and [A1]–[A2].

1.1A1A2algebra

Completing the square in e−u2/2e−(Y−u)2/2 gives posterior N(Y/2,1/2), so its exact credible interval is [Y/2−r,Y/2+r]. Under truth θ∈[−1,1], Y=θ+G and the exact coverage is c(θ)=P{∣θ/2−G/2∣≤r}. This is a continuous strictly positive function of θ, and is at most one. By compactness of [−1,1] it has a positive minimum c∗>0. Therefore δ1,1/40≤max⁡{q−c∗,1−q}<q.

2.1givenstep 1.1constructalgebra

Put h=1/K and let GK={1−kh:0≤k<K}. Away from this finite set of times, use the exact noised-posterior score. At each t∈GK set stY(x)=(1−x)/h−x/2. This finite-valued jointly Borel field has zero integrated score error: its discrepancy from the exact score is supported on the finite, Lebesgue-null set GK in time. The noised posterior has mean e−t/2Y/2 and variance 1−e−t/2: the integrating-factor solution of the linear forward equation is e−t/2X0+e−t/2∫0teu/2 dWu, whose independent Gaussian integral has variance 1−e−t (use dyadic Brownian-increment sums and pass their Gaussian characteristic functions to the mean-square limit). Its exact score is therefore affine in x, so the inner Gaussian error integral is finite at every time. The zero-error conclusion holds for every observation and hence uniformly over all truths.

2.2A1step 1.1algebra

The variable θ/2−G/2 has a density bounded by 2/2π uniformly in θ. The event ∣∣θ/2−G/2∣−r∣≤u is contained in two intervals of total length at most 4u. Thus its probability is at most 8u/2π. The required margin holds, for example with κ=1, u0=1 and Cac=8/2π.

3.1A1step 2.1constructalgebra

At every Euler step the deterministic update becomes x+h(x/2+stY(x))=1. Thus each new state is 1+h ζk, independently of its input and of Y, and the terminal law is N(1,h). For M=K, write the independent terminal draws as 1+K−1/2Gi. If VK=max⁡i≤K∣Gi∣, then the empirical center differs from one by at most K−1/2VK and every empirical distance to that center is at most 2K−1/2VK. Hence the empirical radius has that same upper bound.

4.1A1step 1.1step 2.2step 3.1algebra∎

The Gaussian density gives P{∣G∣>v}≤2e−v2/2 for v≥1: integrate xϕ(x)/v on [v,∞) and use 1/(v2π)≤1. The union bound then gives P{VK>K1/4}≤2Ke−K/2→0. On the complementary event the center is at least 1−K−1/4 and the radius at most 2K−1/4, so for all sufficiently large K the empirical interval excludes truth zero. Its coverage therefore tends to zero, and its deviation from q tends to q>δ1,1/40. Since both error norms are zero and K−1,M−1/2→0, a uniformly vanishing Φ would contradict this strict gap.

Scope of the resolution

This resolves the requested uniformly vanishing-error guarantee negatively under the original score hypothesis. Time-integrated error does not control pointwise values at a deterministic solver grid. A constant worst-case coverage bound remains possible; no impossibility of all numerical functions is asserted. No Bayesian calibration claim is imported from a conformal theorem.

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