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.

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Weighted means from two independent subject groups

Example

Assume AC. Let $X_{1 i}$ be IID Bernoulli $(p_{1})$ and $X_{2 i}$ IID Bernoulli $(p_{2})$ , with the two subject groups independent. Let deterministic sample sizes satisfy $n_{1} + n_{2} = n$ , $n_{1} / n \to ρ \in (0, 1)$ , and $n_{2} / n \to 1 - ρ$ . For $g \in {1, 2}$ define ${\overset{ˉ}{X}}_{g} = n_{g}^{- 1} \sum_{i = 1}^{n_{g}} X_{g i}$ . For the fixed target $θ = ρ p_{1} + (1 - ρ) p_{2}$ , use the fixed-allocation-weight estimator ${\hat{θ}}_{n} = ρ {\overset{ˉ}{X}}_{1} + (1 - ρ) {\overset{ˉ}{X}}_{2}$ .

Facts & Assumptions

Given: The two independent Bernoulli groups and estimator in the Example. Find its exact group influence expansion and limit.

[A1]

In group $g$ , $P (X_{g i} = 1) = p_{g}$ and $P (X_{g i} = 0) = 1 - p_{g}$ ; the observations are IID within groups and the two groups are independent.

[A2]

Assume AC. The Axiom of Choice

[F1]

Identical distribution and IID families: IID means a common law and mutual independence of the whole family.

[F2]

Independent families of event classes: For every finite selection of distinct independent indices and one event from each selected class, the probability of their intersection is the product of their probabilities.

[F3]

Moments, variance, and covariance on a probability space: For a square-integrable real random variable $X$ , $Var (X) = E [(X - E X)^{2}]$ .

[F4]

Independent-group terminal influence limit: For a fixed finite family of independent IID groups, the covariance of the terminal normal limit is $\sum_{g} ρ_{g}^{- 1} Σ_{g}$ , provided the stated group expansion and hypotheses hold.

Verification

1.1A1F1F2F3F4algebra

Put $ϕ_{1} (x) = ρ (x - p_{1})$ and $ϕ_{2} (x) = (1 - ρ) (x - p_{2})$ . The measurable maps preserve within-group identical laws and the finite event-product identities for independence across groups, so their sampled values meet [F1] and [F2], as required by [F4]. By [A1] and [F3], they are centered, square-integrable group contributions with variances $Σ_{1} = ρ^{2} p_{1} (1 - p_{1})$ and $Σ_{2} = (1 - ρ)^{2} p_{2} (1 - p_{2})$ . The identity $\sqrt{n} ({\hat{θ}}_{n} - θ) = \sqrt{n / n_{1}} n_{1}^{- 1 / 2} \sum_{i = 1}^{n_{1}} ϕ_{1} (X_{1 i}) + \sqrt{n / n_{2}} n_{2}^{- 1 / 2} \sum_{i = 1}^{n_{2}} ϕ_{2} (X_{2 i})$ is exact, so its remainder is zero.

2.1A2F4step 1.1algebra∎

The covariance in [F4] is $ρ^{- 1} Σ_{1} + (1 - ρ)^{- 1} Σ_{2} = ρ p_{1} (1 - p_{1}) + (1 - ρ) p_{2} (1 - p_{2})$ . Thus $\sqrt{n} ({\hat{θ}}_{n} - θ)$ converges to the centered normal with this variance. The formula includes $p_{g} \in {0, 1}$ , where the corresponding group variance is zero.

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