E Evidence Press

Press release · 10 October 2026 · version 1.1.0-candidate

Every projection passes. The joint comparison fails.

A centred Gaussian and a three-component Gaussian mixture pass every one-dimensional convex-order test, yet an explicit joint convex function separates them.

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Summary

A comparison can hold from every viewing direction and still fail for a cost that depends on both coordinates. This candidate gives an exact example: one centred Gaussian distribution and a mixture of three centred Gaussian components.

Every one-dimensional projection is ordered in the required convex order. Nevertheless, a specified convex function has a larger expectation under the target than under the mixture. The proof also gives a positive definite version, so the failure survives beyond distributions supported on lines.

Convex order compares distributions by asking how they behave under every convex cost. Projecting onto a line simplifies the question, but removes information about how coordinates act together. Here all such projected comparisons pass—not merely a large collection of sampled directions. A joint cost detects what those comparisons miss.

The three-component mixture places probability on a horizontal line and the two diagonals. Its scales make the directional comparisons work. A small shift in a piecewise-linear cost exposes a different allocation of probability between regions of the plane. This is a mathematical limit on an inference, not an empirical claim about forecasting or financial performance.

Summary for specialists

Let $X$ have covariance $\operatorname{diag}(1/3,1/9)$ and mean zero. Let $Y$ be the centred Gaussian mixture with weights $(3/4,1/8,1/8)$ and covariance factors

$ v_0=(4/9,0)^\top,\qquad v_1=(4/3,4/3)^\top,\qquad v_2=(4/3,-4/3)^\top, $

so each component covariance is $v_i v_i^\top$. Then $u^\top X\leq_{\mathrm{cx}}u^\top Y$ for every real direction $u$, but $X\not\leq_{\mathrm{cx}}Y$. The witness

$ \phi(x,y)=\max\{|x|+|y|-1/24,\;2|y|\} $

satisfies

$ \mathbb E\phi(X)-\mathbb E\phi(Y)>\frac{37}{18432}>0. $

Adding $10^{-8}I$ to all four covariance matrices preserves every projected comparison and leaves a witness gap greater than $16213/11520000$. All resulting covariance matrices are positive definite.

Technical account

The scalar Gaussian-mixture criterion reduces every projected comparison to

$ \sqrt{3a^2+b^2}\leq |a|+\max\{|a|,|b|\}. $

The cases $|a|\geq|b|$ and $|b|\geq|a|$ prove this inequality exhaustively. Equality directions guide the construction.

Write $\phi_t(x,y)=\max\{|x|+|y|-t,2|y|\}$. At zero threshold, the expectations agree. Initially the target expectation decreases at rate $2/3$, while the mixture expectation decreases at rate $3/4$. A bounded half-normal density controls the second-order remainder. This yields the rational gap above without certified numerical integration.

Proposition 3 extends the mechanism to positive weights $p,q,r$ summing to one with $p>2/3$, using the specified reciprocal-weight component scales. It is a sufficient counterexample family, not a classification of all weights.

Evidence, assurance and limitations

The revised package includes the written all-directions argument, exact algebraic checks and two separately written numerical diagnostics. Normal and optimised Python runs pass; deliberate wrong-weight and exaggerated-gap inputs are rejected. Numerical quadrature reports a gap near 0.00333, but that number is not a certified error interval or a practical effect size.

The supplied model-assisted review has an itemised response. The manuscript distinguishes projected order, full centred order and the stronger sufficient Gaussian-block condition. The final journal body of Jourdain and Pagès was not inspected; the comparison is explicitly tied to its inspected arXiv version.

Relationship to earlier work

The distinction between projected and joint convex order is classical: see Koshevoy and Mosler, and Pinelis’s explicit finite-support example. The contribution here is the specified three-centred-Gaussian construction, not the discovery of that general distinction.

Guéant’s Theorem 4 establishes projection sufficiency for two centred Gaussian components, including singular covariances. The present example answers the larger-mixture question negatively at three components. Minimality is relative to that external theorem. The package does not classify all mixtures, claim failure for every three-component mixture, or establish absolute priority.

This is an unrefereed candidate. Internal replay, archive availability and numerical agreement are separate from external peer review, formal verification and independent reproduction; none of those latter assurances is claimed.

Who should care, and why

AudiencePotential useRequired caution
Probability and stochastic-order researchersAn exact boundary for Gaussian-mixture projection criteriaThe claim concerns this construction, not all three-component mixtures
Researchers using projected convex-order comparisonsA test case for an inference from scalar to joint costsNo empirical performance or practical effect size is established
Verification and formalisation researchersA short analytic argument with rational inputs and corruption controlsNumerical replay does not prove a continuum of directions

Why the problem matters

Two components and three components behave differently. The result identifies a concrete point where projection-based comparison stops being a sufficient guarantee for all joint convex costs. That structural boundary is the contribution; no measured downstream impact is claimed.

How to inspect or reproduce the recorded checks

Read Theorem 1, the common-noise corollary and Proposition 3. With the package's pinned NumPy and SciPy dependencies, run python code/run_checks.py and python -O code/run_checks.py. The entry point checks exact identities, runs both diagnostics and rejects two deliberately corrupted inputs. Inspect VERIFICATION.md for the precise finite scope and the distinction between numerical agreement and the written proof.

The most valuable next projects

An external audit of the scalar criterion and expectation-gap estimate would address assurance directly. Characterising which other component geometries make projections sufficient is a separate research problem, not a result already established here. A certified numerical integrator could bound diagnostic error, but is unnecessary for the analytic rational lower bound.

What is in the evidence package

Start with the machine-readable research index, then follow the claim map, manuscript and verification instructions. The package records exact inputs, limitations, source comparisons and the review response. Original prose and data are CC0; original code is MIT.

Media

The audio briefing is provided in the header above. Download the MP3 briefing · read the transcript.

Verification status

Unrefereed written-proof candidate with internal finite replay. No external peer review, unaffiliated reproduction, formal verification or absolute priority is established.

Cite

Anonymous (2026). Projection tests fail for three centred Gaussian components. Version 1.1.0-candidate. Evidence Press. 10.5281/zenodo.23284191.
BibTeX
@misc{threegaussianprojectioncounterexample2026,
  title        = {Every projection passes. The joint comparison fails.},
  author       = {Anonymous},
  year         = {2026},
  doi          = {10.5281/zenodo.23284191},
  url          = {https://doi.org/10.5281/zenodo.23284191},
  version      = {1.1.0-candidate},
  howpublished = {Zenodo},
  note         = {Unrefereed; internally replayed evidence package. Press page: https://evidencepress.org/releases/three-gaussian-projection-counterexample/}
}

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