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.

The binomial sample proportion

Example

Let $X_{1}, \dots, X_{n}$ be independent Bernoulli $(p)$ observations, where $n \geq 1$ and $p \in [0, 1]$ . For $\overline{X} = n^{- 1} \sum_{i = 1}^{n} X_{i}$ , $E_{p} [\overline{X}] = p, {Var}_{p} (\overline{X}) = {MSE}_{p} (\overline{X}) = \frac{p (1 - p)}{n} .$ The equalities include the degenerate endpoints $p = 0$ and $p = 1$ .

Facts & Assumptions

Given: The finite product Bernoulli experiment, $n \geq 1$ , and $p \in [0, 1]$ .

[F1]

An estimator is unbiased when its expectation equals the estimand at every parameter; MSE is the squared-error expectation. (Unbiased estimators and UMVU estimators, Bias and mean squared error)

[F2]

Variance of a finite sum is the sum of individual variances and twice the sum of pairwise covariances. (Variance of a finite sum as the sum of all variances and covariances)

[F3]

Independent square-integrable variables have zero covariance. (Independence forces covariance to vanish)

[F4]

On a finite probability space, variance is the expectation of the centered square. (Variance, standard deviation, and covariance on a finite probability space)

[F5]

Expectation is linear on finite linear combinations of integrable random variables. (The Lebesgue integral is linear on $L^{1} (μ)$ )

[F6]

For an unbiased estimator, MSE equals variance. (Bias–variance decomposition)

[F7]

A measurable real-valued data map to the named target is an estimator. (Statistics and estimators)

[F8]

A Bernoulli $(p)$ variable takes values $1$ and $0$ with probabilities $p$ and $1 - p$ , and mutual independence of finite-valued variables means their joint cell probabilities factor. (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, Pairwise and mutual independence of finite-valued random variables)

Verification

technique · direct

1.1givenF1F4F5F7F8

Each $X_{i}$ and $\overline{X}$ is measurable on the finite product space, hence an estimator by [F7]. The Bernoulli law [F8] gives $P_{p} (X_{i} = 1) = p$ and $P_{p} (X_{i} = 0) = 1 - p$ ; summing these two values gives $E_{p} X_{i} = p$ and ${Var}_{p} (X_{i}) = (1 - p) p^{2} + p (1 - p)^{2} = p (1 - p)$ , including $p = 0, 1$ . Thus [F5] gives $E_{p} \overline{X} = p$ , so $\overline{X}$ is unbiased by [F1].

2.1givenF2F3F6step 1.1∎

The independent $X_{i}$ are square-integrable, so [F3] makes every off-diagonal covariance zero. Applying [F2] gives ${Var}_{p} (\sum_{i} X_{i}) = n p (1 - p)$ and hence ${Var}_{p} (\overline{X}) = p (1 - p) / n$ . Since the bias is zero, [F6] gives the same value for MSE.

Depends on

Used by

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Dependency tree · two levels

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Sources

George Deligiannidis, SB2.1 Foundations of Statistical Inference, Chapter 5, PDF pp. 29–31 (standard reference, not scraped)
Karl B. Gregory, STAT 713 Lecture 6: Best Unbiased Estimators, Rao-Blackwell Theorem, PDF pp. 2–8 (standard reference, not scraped)