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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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Ideal-marginal score error allows unbounded finite-Euler Wasserstein error

Example

For the separated-mixture question reproduced below, there are admissible fixed one-dimensional parameters with L=2, fixed positive score-error allowance εs, and every fixed integer J≥2, for which the supremum of the displayed expected fidelity loss over the original finite-valued Borel score class is +∞, for every finite M≥1. Each individual field in the construction has a finite-first-moment Euler chain and obeys the required score bound at every time. Thus the original ideal-marginal L2 hypothesis supplies no finite uniform bound for the requested loss, however large the fixed budget JM is. The original question is retained verbatim under Original question.

Original question

Fix dimension d≥1, component count L≥2, noise scale σ>0, diffusion horizon T>0, and a continuous schedule β:[0,T]→(0,∞). Assume the parameter choices below admit at least one component configuration and mixture weight vector. Consider a posterior density on Rd of the form π(x)=∑ℓ=1Lwℓ N(x;μℓ,Σℓ),∑ℓ=1Lwℓ=1,wℓ≥wmin⁡>0, Here N(⋅;μℓ,Σℓ) denotes the normal density with mean μℓ and covariance matrix Σℓ. The component means satisfy ∥μℓ∥2≤M0 and ∥μℓ−μr∥2≥Δ>0 for ℓ≠r, and every eigenvalue of each positive-definite Σℓ lies in [vmin⁡,vmax⁡] for fixed 0<vmin⁡≤vmax⁡<∞. Assume 0<wmin⁡≤1/L and Δ≥8dvmax⁡. For t∈[0,T], let pt be the density of Xt under dXt=−12β(t)Xt dt+β(t) dWt,X0∼π, where Wt is a standard d-dimensional Brownian motion, and let st=∇log⁡pt be its exact score. Let s^t be a supplied jointly Borel, finite-valued score field satisfying the integrated error bound sup⁡0≤t≤T∫Rd∥s^t(x)−st(x)∥22pt(x) dx≤εs2. Consider fields whose Euler chain has finite first moment. For h=T/J, use the reverse-time Euler convention Z0∼pT and Zk+1=Zk+h[12β(T−kh)Zk+β(T−kh)s^T−kh(Zk)]+hβ(T−kh) ζk,k=0,…,J−1, with ζk∼N(0,Id) independent of Z0 and of one another. Generate M≥1 independent terminal values X(1),…,X(M) and let π^J,M=M−1∑i=1MδX(i); one score evaluation per reverse step gives online budget B=JM. Define disjoint mode regions Rℓ={x:∥x−μℓ∥2<Δ/4}, target masses qℓ=π(Rℓ), and empirical masses q^ℓ=π^J,M(Rℓ). For probability measures on Rd, let W1 be the infimum of E∥X−Y∥2 over all couplings with the specified marginals. Determine whether the integrated score-error assumption alone yields finite minimax-sharp nonasymptotic upper and lower bounds, uniformly over the stated score-field class, for E W1(π^J,M,π)+Δ∑ℓ=1LE∣q^ℓ−qℓ∣+Δ P{min⁡1≤ℓ≤Lq^ℓ=0} as a function of d,L,wmin⁡,Δ,vmin⁡,vmax⁡,M0,T,β,εs,J,M. In particular, characterize the compute regimes B=JM that guarantee simultaneous control of distributional error and missed-mode probability uniformly over this family. This asks about score-based reverse diffusion with finite computation; it does not assume that the posterior is unimodal or that a good posterior mean certifies its mode weights.

Facts & Assumptions

Given: Choose d=1, π=(N(−a,v)+N(a,v))/2 with a,v>0 and 2a≥8v. Choose L=2, wmin⁡=1/2, M0=a, Δ=2a, vmin⁡=vmax⁡=v, T>0, β=1, εs>0, J≥2 and h=T/J. The extraneous positive parameter called σ in the original formulation may be arbitrary; it is not used by its displayed sampler.

[A1]

The integrating factor et/2 in the scalar forward equation gives Xt=e−t/2X0+e−t/2∫0teu/2 dWu. The stochastic integral is the mean-square limit of deterministic dyadic sums of independent Gaussian increments; these sums are centered Gaussian, with variances tending to ∫0teu du=et−1, and are independent of X0. The limit is therefore Gaussian with that variance and independent of X0, as follows by taking limits of its Gaussian characteristic functions. At time t, the noised target is the symmetric Gaussian mixture with means ±e−t/2a and common variance vt=e−tv+1−e−t>0. Its exact score is a posterior-weighted average of −(x−e−t/2a)/vt and −(x+e−t/2a)/vt, so ∣st(x)∣≤(∣x∣+a)/min⁡{v,1} and is continuous. These formulas follow directly by differentiating the sum of the two Gaussian densities.

[A2]

A convolution with a nondegenerate Gaussian has a strictly positive continuous density: continuity follows from the bounded Gaussian derivative, and positivity from its positive kernel. Gaussian densities have finite first moments and exponentially decaying tails.

Verification

Given: The fixed parameters and Gaussian identities in [A1]–[A2].

1.1A1A2construct

Use the exact score at all grid times before the last two updates. All those states have finite first moment because each exact-score drift has linear growth by [A1]. Let U be the state before the penultimate update, corresponding to ideal time 2h. Its law has a strictly positive continuous density ρ: for J=2 it is exactly pT, and otherwise the last preceding nondegenerate Euler noise gives [A2]. Since p2h is also positive and continuous, the density ratio ρ/p2h has positive minimum c and finite maximum C on [−1,1].

2.1A1step 1.1constructalgebra

For every sufficiently small q>0, choose the centered interval Aq⊆[−1,1] with ∫Aqp2h(x) dx=q. Such an interval exists by continuity and positivity of that density. Then cq≤P{U∈Aq}≤Cq. Put bq=εs/(2q) and add bq1Aq to the score at time 2h. Its squared ideal-marginal error is exactly εs2/4. Set f(x)=(1+h/2)x+hs2h(x) and H=sup⁡∣x∣≤1∣f(x)∣<∞. After this update the state is V=f(U)+hbq1Aq(U)+h G.

3.1A1A2step 2.1algebra

Let Iq=[hbq−H−h,hbq+H+h], and let rq=∫Iqph(x) dx>0. On {U∈Aq,∣G∣≤1} the state V lies in Iq, so P{V∈Iq}≥cqP{∣G∣≤1}. The ideal density ph is a fixed Gaussian mixture. Since the interval width is constant and its center hbq tends to infinity, its Gaussian density formula gives constants C1,c1>0, independent of small q, such that rq≤C1e−c1bq2. For example, once hbq≥2(H+h+a), every point of the interval is at distance at least hbq/2 from either component mean; use the largest fixed variance in its Gaussian exponent and the constant interval length.

4.1A1step 2.1step 3.1constructalgebra

At the final evaluation time h add dq1Iq to the exact score, where dq=εs/(2rq). This second squared ideal-marginal error is also εs2/4. At all other times use the exact score. These modifications at two distinct times define a jointly Borel finite-valued field with the original supremum-in-time score bound. Every individual field has finite first moments at all Euler steps: it is an affine-growth exact score plus two bounded, finite-amplitude additions.

5.1A1step 2.1step 3.1step 4.1algebra

The first moment of V is bounded uniformly as q↓0: the added term has expectation at most hbqCq=hCεsq/2, while E∣f(U)∣ and E∣hG∣ are fixed and finite. The terminal state is X=(1+h/2)V+hsh(V)+hdq1Iq(V)+hG′. By [A1] the absolute expectation of the first two terms is uniformly bounded; EG′=0. Therefore EX≥hεscqP{∣G∣≤1}/(2rq)−C2→+∞, because bq2=εs2/(4q) and the exponential bound in step 3.1 dominates every power of q.

6.1givenstep 4.1step 5.1algebra∎

Every coupling of two finite-first-moment probabilities satisfies ∣EX−EY∣≤E∣X−Y∣ by the integral triangle inequality. Thus W1(π^J,M,π)≥∣M−1∑iX(i)∣ because the target mean is zero. Taking expectations gives EW1≥∣EX∣, which tends to infinity by step 5.1 for every finite M. The other two terms in the requested fidelity loss are nonnegative. Its supremum is consequently infinite despite fixed positive score-error allowance and fixed admissible target parameters.

Scope of the resolution

The negative answer concerns every positive error allowance and every finite J>=2. It does not assert the same obstruction at J=1 or at exactly zero error. The mechanism is mismatch between ideal marginal mass and actual chain mass, not merely missing a mode by iid sampling. No numerical observation or source-specific failure experiment is used as the proof.

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