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.

Paired K-fold comparison retains correlated fold errors

Example

Let $n = 4$ , $K = 2$ , $I_{1} = {1, 2}$ , $I_{2} = {3, 4}$ , and $Z_{i} \overset{I I D}{\sim} Bernoulli (1 / 2)$ . Compare $A_{k} (z_{1}, \dots, z_{k}) = \prod_{q} z_{q}$ with the zero predictor $B_{k} = 0$ under loss $L (h, z) = h z$ , using the same folds for both procedures. Write $\hat{Δ}$ for the paired difference of their cross-validation scores.

Facts & Assumptions

Given: The two-fold partition, IID Bernoulli observations, product and zero rules, and loss above. Compute the covariance terms and the variance of the paired difference.

[F1]

Cross-validation splits and training-size risk targets: the equal-fold score is the average of the four held-out losses.

[F2]

Holdout, leave-out, V-fold, and repeated cross-validation: paired comparison applies the same split schedule and averages within-fold loss differences.

[F3]

Identical distribution and IID families: the four observations are mutually independent with a common Bernoulli law.

[F4]

Expectations factor over finite products of independent random variables: expectations of products of functions of distinct observations factor.

[F5]

Moments, variance, and covariance on a probability space: variance and covariance are the centered second moments.

[F6]

The covariance decomposition of a fixed K-fold score: for $K$ equal folds, the average-error variance is $σ^{2} / n + (m - 1) ω / n + (n - m) γ / n$ .

[F7]

Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: For $p \in [0, 1]$ , a Bernoulli $(p)$ random variable takes the value $1$ with probability $p$ and $0$ with probability $1 - p$ .

Verification

1.1givenF1F2

Since the zero rule has zero loss, the four paired errors are $e_{1} = Z_{1} Z_{3} Z_{4}$ , $e_{2} = Z_{2} Z_{3} Z_{4}$ , $e_{3} = Z_{3} Z_{1} Z_{2}$ , and $e_{4} = Z_{4} Z_{1} Z_{2}$ , and $\hat{Δ} = \frac{1}{4} \sum_{i = 1}^{4} e_{i}$ .

2.1F3F4F5F7step 1.1algebra

Each $e_{i}$ is a product of three independent Bernoulli $(1 / 2)$ variables, so [F3]–[F5] and [F7] give $E e_{i} = 1 / 8$ and $Var (e_{i}) = 1 / 8 - (1 / 8)^{2} = 7 / 64$ . For any distinct $i, j$ , the product $e_{i} e_{j}$ equals $Z_{1} Z_{2} Z_{3} Z_{4}$ because $Z_{q}^{2} = Z_{q}$ ; hence $E [e_{i} e_{j}] = 1 / 16$ and $Cov (e_{i}, e_{j}) = 1 / 16 - (1 / 8)^{2} = 3 / 64$ . Thus $ω = γ = 3 / 64$ .

3.1F5F6step 2.1algebra

Substituting $n = 4, m = 2, σ^{2} = 7 / 64, ω = γ = 3 / 64$ into [F6] gives $Var (\hat{Δ}) = \frac{7}{256} + \frac{3}{256} + \frac{6}{256} = \frac{1}{16}$ . Treating the four errors as independent would keep only $7 / 256$ , omitting the two positive covariance contributions.

4.1F3F4F5F7step 3.1algebra∎

Directly, if all four observations are one then $\hat{Δ} = 1$ ; if exactly three are one then $\hat{Δ} = 1 / 4$ ; with at most two ones it is zero. By [F7] these events have probabilities $1 / 16$ , $4 / 16$ , and $11 / 16$ , respectively, so $E \hat{Δ} = 1 / 8$ , $E {\hat{Δ}}^{2} = 5 / 64$ , and the direct variance is $5 / 64 - (1 / 8)^{2} = 1 / 16$ , confirming step 3.1.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Yoshua Bengio and Yves Grandvalet, No Unbiased Estimator of the Variance of K-Fold Cross-Validation, JMLR 5 (2004), §§2–3 (standard reference, not scraped)
Sylvain Arlot and Alain Celisse, A survey of cross-validation procedures for model selection, §5.2 (standard reference, not scraped)