TheoremStatement: AI-adaptedProof: 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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Within-batch covariance and replicate-mean variance

Statement

Let $I$ and each $J_{i}$ be nonempty finite sets, and let $m_{i j} \geq 1$ be an integer for each $i \in I$ and $j \in J_{i}$ . On one probability space, suppose the real random variables $B_{i j}$ and $ϵ_{i j k}$ , for $1 \leq k \leq m_{i j}$ , are jointly independent and square-integrable, with $E [B_{i j}] = 0, Var (B_{i j}) = σ_{B}^{2} < \infty, E [ϵ_{i j k}] = 0, Var (ϵ_{i j k}) = σ_{ϵ}^{2} < \infty,$ where these two variances are common over their indices. For fixed real constants $μ$ and $τ_{i}$ , set $Y_{i j k} = μ + τ_{i} + B_{i j} + ϵ_{i j k}, {\overset{ˉ}{Y}}_{i j \cdot} = \frac{1}{m_{i j}} \sum_{k = 1}^{m_{i j}} Y_{i j k} .$ Then for every $i, j$ and distinct $k, ℓ \in {1, \dots, m_{i j}}$ , $Cov (Y_{i j k}, Y_{i j ℓ}) = σ_{B}^{2},$ and for every $i, j$ , $Var ({\overset{ˉ}{Y}}_{i j \cdot}) = σ_{B}^{2} + \frac{σ_{ϵ}^{2}}{m_{i j}} .$ No Gaussian assumption is needed.

Facts & Assumptions

Given: The finite nested model and all moment and independence hypotheses in the Statement.

[F1]

Batch and measurement-process variation: all $B_{i j}$ and $ϵ_{i j k}$ are square-integrable and jointly independent, have mean zero and common finite variances $σ_{B}^{2}$ and $σ_{ϵ}^{2}$ , and $Y_{i j k} = μ + τ_{i} + B_{i j} + ϵ_{i j k}$ .

[F2]

Moments, variance, and covariance on a probability space: for square-integrable real variables, variance is the expectation of the centered square and covariance is the expectation of the centered product.

[F3]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear on finite linear combinations of integrable real variables.

[F4]

Expectations factor over finite products of independent random variables: for a finite independent family and integrable coordinate functions, the product is integrable and its expectation is the product of the expectations.

[F5]

Expectation of a nonnegative or integrable random variable: expectation of an integrable real random variable is its Lebesgue integral.

[F6]

Probability measures and probability spaces: the probability measure of the whole sample space is one, so the expectation of a constant $c$ is $c$ .

Proof

1.1givenF1F2F3F5F6algebra

Each variable in [F1] is integrable because it is square-integrable on a probability space and $2 ∣ X ∣ \leq 1 + X^{2}$ ; finite sums are square-integrable by $(x_{1} + \dots + x_{n})^{2} \leq n \sum_{r = 1}^{n} x_{r}^{2}$ . Every product used below is integrable by $2 ∣ X Z ∣ \leq X^{2} + Z^{2}$ . Thus all expectations are finite and the moment definitions and expectation linearity apply.

2.1F1F2F3F4F5F6step 1.1algebra

Fix distinct $k, ℓ$ in one batch. By [F1], [F3], [F5], and [F6], $E [Y_{i j k}] = E [Y_{i j ℓ}] = μ + τ_{i}$ . Thus [F2] gives $Cov (Y_{i j k}, Y_{i j ℓ}) = E [(B_{i j} + ϵ_{i j k}) (B_{i j} + ϵ_{i j ℓ})]$ . Expand the product and apply [F3] to get $E [B_{i j}^{2}] + E [B_{i j} ϵ_{i j ℓ}] + E [ϵ_{i j k} B_{i j}] + E [ϵ_{i j k} ϵ_{i j ℓ}]$ . By [F1] and [F4] each cross-product expectation factors into the product of two zero means; the remaining term is $E [B_{i j}^{2}] = Var (B_{i j}) = σ_{B}^{2}$ by [F1] and [F2].

3.1F1F2F3F4F5F6step 1.1algebra∎

For fixed $i, j$ , [F1], [F3], [F5], and [F6] give $E [{\overset{ˉ}{Y}}_{i j \cdot}] = μ + τ_{i}$ , so its centered form is $B_{i j} + m_{i j}^{- 1} \sum_{k} ϵ_{i j k}$ . By [F2], its variance is the expectation of the square of this centered form; expanding that square and repeatedly applying [F3] gives $E [B_{i j}^{2}] + (2 / m_{i j}) \sum_{k} E [B_{i j} ϵ_{i j k}] + (1 / m_{i j}^{2}) \sum_{k, ℓ} E [ϵ_{i j k} ϵ_{i j ℓ}]$ . By [F1] and [F4], all cross-product expectations vanish and each diagonal term equals $E [ϵ_{i j k}^{2}] = σ_{ϵ}^{2}$ . Also $E [B_{i j}^{2}] = σ_{B}^{2}$ . Therefore the result is $σ_{B}^{2} + σ_{ϵ}^{2} / m_{i j}$ , including $m_{i j} = 1$ and zero variance components.

Remarks

The result is a model identity, not an estimate or an empirical claim about a particular instrument. Its proof uses finite sums and no choice axiom.

Depends on

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Sources

NIST/SEMATECH Engineering Statistics Handbook, Chapter 2: Measurement Process Characterization (standard reference, not scraped)
Fan Li, STA610 Lecture 5 Notes: Estimation of Group Effects (standard reference, not scraped)
Fan Li, STA610 Lecture 6 Notes: Linear Mixed-Effects Models (standard reference, not scraped)