Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A two-member Gaussian ensemble has a fluctuating covariance

Example

Let X1,X2 be independent N(0,1) forecast members. The divisor-N sample covariance used by Perturbed-observation ensemble Kalman filter at N=2 is random and differs from the true forecast variance with positive probability.

Facts & Assumptions

Given: Independent X1,X2∼N(0,1) and the divisor-N covariance convention of Perturbed-observation ensemble Kalman filter.

[A1]

AC is assumed in the standard-normal construction. The Axiom of Choice

[F1]

The standard normal has density (2π)−1/2e−x2/2, which is strictly positive on every real interval. Standard normal and normal laws

[F2]

N(0,1) has mean zero and variance one under the specified multivariate-normal convention. Multivariate normal law, including singular covariance

[F3]

The EnKF covariance is N−1∑i(Xi−Xˉ)(Xi−Xˉ)T. Perturbed-observation ensemble Kalman filter

Verification

Given: The hypotheses in the Example.

1.1F3givenalgebra

For N=2, Xˉ=(X1+X2)/2, so X1−Xˉ=(X1−X2)/2 and X2−Xˉ=−(X1−X2)/2. By [F3], C2=12((X1−Xˉ)2+(X2−Xˉ)2)=(X1−X2)24.

2.1A1F1F2givenstep 1.1algebra∎

The event A={−0.1<X1<0.1, 3<X2<3.1} has positive probability: each interval has positive probability by [F1], and the variables are independent. On A, X2−X1>2.9, so step 1.1 gives C2>2.92/4=2.1025>1. The true variance of N(0,1) is 1 by [F2]; hence the finite-ensemble covariance differs from it with positive probability.

Depends on

Used by

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Dependency tree · two levels

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Sources