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 holdout estimate targets a smaller training sample

Example

Take $n = 4$ IID observations $Z_{i} \sim Bernoulli (1 / 2)$ , squared loss $L (a, z) = (a - z)^{2}$ , and the learning rule $A_{k} (z_{1}, \dots, z_{k}) = k^{- 1} \sum_{q = 1}^{k} z_{q}$ . Hold out $Z_{4}$ and train on $Z_{1}, Z_{2}, Z_{3}$ . The expected holdout score estimates $R_{3}$ , while training on all four observations has risk $R_{4}$ .

Facts & Assumptions

Given: The four IID Bernoulli observations, the sample-mean rule, squared loss, and holdout split above. Compute the training-size risks and their difference.

[F1]

Cross-validation estimates risk at its training sizes: the expected holdout score with $k$ training observations is $R_{k}$ .

[F2]

Cross-validation splits and training-size risk targets: $R_{k} = E [L (A_{k} (Z_{1}, \dots, Z_{k}), Z_{k + 1})]$ for an independent test observation.

[F3]

Identical distribution and IID families: the observations are mutually independent and have one common law.

[F4]

Moments, variance, and covariance on a probability space: variance is $E [(X - E X)^{2}]$ for square-integrable $X$ .

[F5]

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

[F6]

Holdout, leave-out, V-fold, and repeated cross-validation: a holdout score averages losses on the validation indices after training on their complement.

[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

1.1givenF3F4F5F7algebra

For $Z \sim Bernoulli (p)$ , [F7] gives $E Z = p$ and the two-point expectation gives $Var (Z) = p (1 - p)$ . Expanding the variance of ${\overset{ˉ}{Z}}_{k} = k^{- 1} \sum_{q = 1}^{k} Z_{q}$ , the diagonal terms sum to $k p (1 - p)$ and each cross covariance vanishes by [F3] and [F5]; hence $Var ({\overset{ˉ}{Z}}_{k}) = p (1 - p) / k$ .

2.1F2F3F4F5step 1.1algebra

The independent test variable $Z_{k + 1}$ has mean $p$ and variance $p (1 - p)$ . Expanding $E [({\overset{ˉ}{Z}}_{k} - Z_{k + 1})^{2}]$ around the common mean $p$ , the centered cross term is zero by independence, so [F2] gives $R_{k} = Var ({\overset{ˉ}{Z}}_{k}) + Var (Z_{k + 1}) = p (1 - p) (1 + 1 / k)$ .

3.1F1F2F6step 2.1algebra∎

At $p = 1 / 2$ , $R_{3} = \frac{1}{4} (1 + \frac{1}{3}) = \frac{1}{3}$ and $R_{4} = \frac{1}{4} (1 + \frac{1}{4}) = \frac{5}{16}$ . The split in [F6] holds out one observation, so [F1] gives expected holdout score $R_{3}$ ; its bias for full-sample risk is $R_{3} - R_{4} = \frac{1}{48} > 0$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Sylvain Arlot and Alain Celisse, A survey of cross-validation procedures for model selection, §5.1.1 (standard reference, not scraped)
Gareth James, Daniela Witten, Trevor Hastie, and Robert Tibshirani, An Introduction to Statistical Learning, 7th printing, Chapter 5 §5.1.1 (standard reference, not scraped)