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.

Pairwise uncorrelated data can be jointly dependent

Example

Let $Ω = {(- 1, - 1, 1), (- 1, 1, - 1), (1, - 1, - 1), (1, 1, 1)}$ with each outcome assigned probability $1 / 4$ . Let $X_{1}, X_{2}, X_{3}$ be the three coordinate random variables. This finite data vector is pairwise uncorrelated, but it is not mutually independent.

Facts & Assumptions

Given: The four-outcome uniform probability space $Ω$ above and its three coordinate random variables.

[F1]

A finite probability space assigns nonnegative outcome weights summing to one; events are subsets of outcomes with probability the sum of their weights (Finite probability spaces, outcome weights, events, and event probabilities).

[F2]

A finite-valued real random variable is a function on the finite outcome space (Real random variables on finite probability spaces and their finite distributions).

[F3]

Expectation on a finite probability space is the weighted sum (Expectation of a real random variable on a finite probability space).

[F4]

Covariance is the expectation of the centered product (Variance, standard deviation, and covariance on a finite probability space).

[F5]

Mutual independence of finite-valued random variables requires factorization of every joint point probability (Pairwise and mutual independence of finite-valued random variables).

Verification

technique · enumerate the four outcomes

1.1F1F2F3givenalgebra

Each coordinate takes the value $1$ twice and $- 1$ twice. By [F1]–[F3], $E [X_{i}] = 0$ and $E [X_{i}^{2}] = 1$ for each $i \in {1, 2, 3}$ .

2.1F3F4step 1.1algebra

For each of the three pairs, the four products $X_{i} X_{j}$ contain two $1$ values and two $- 1$ values. Thus $E [X_{i} X_{j}] = 0$ ; using the means from step 1.1 in [F4] gives $Cov (X_{i}, X_{j}) = 0$ whenever $i \neq j$ .

3.1F1F5step 1.1algebra∎

The event ${X_{1} = 1, X_{2} = 1, X_{3} = 1}$ contains only the outcome $(1, 1, 1)$ , so its probability is $1 / 4$ . Each event ${X_{i} = 1}$ has probability $1 / 2$ by step 1.1. Mutual independence would give joint probability $(1 / 2)^{3} = 1 / 8$ by [F5], contradicting $1 / 4$ . Therefore the variables are not mutually independent.

Depends on

Used by

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Dependency tree · two levels

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