TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Expected mean squares in the balanced one-way normal random-effects model

Statement

Let $k, r \geq 2$ , let $μ \in R$ , $τ^{2} \geq 0$ , and $σ^{2} > 0$ . For $i = 1, \dots, k$ , let the group blocks $U_{i} = (a_{i}, ϵ_{i})$ be independent random elements on one probability space, where $ϵ_{i} = (ϵ_{i 1}, \dots, ϵ_{i r})$ . Within each block, $a_{i}$ and $ϵ_{i}$ are independent and the coordinates of $ϵ_{i}$ are mutually independent. Assume $a_{i} \sim N (0, τ^{2})$ and $ϵ_{i j} \sim N (0, σ^{2})$ . Put $Y_{i j} = μ + a_{i} + ϵ_{i j}$ , ${\overset{ˉ}{Y}}_{i \cdot} = r^{- 1} \sum_{j = 1}^{r} Y_{i j}$ , and ${\overset{ˉ}{Y}}_{\cdot \cdot} = k^{- 1} \sum_{i = 1}^{k} {\overset{ˉ}{Y}}_{i \cdot}$ . Define $M S B = \frac{r}{k - 1} \sum_{i = 1}^{k} ({\overset{ˉ}{Y}}_{i \cdot} - {\overset{ˉ}{Y}}_{\cdot \cdot})^{2}, M S W = \frac{1}{k (r - 1)} \sum_{i = 1}^{k} \sum_{j = 1}^{r} (Y_{i j} - {\overset{ˉ}{Y}}_{i \cdot})^{2} .$ Then $E [M S B] = σ^{2} + r τ^{2}, E [M S W] = σ^{2} .$ Equating these two population moments to the observed mean squares gives the raw method-of-moments estimates ${\hat{σ^{2}}}_{M O M} = M S W, {\hat{τ^{2}}}_{M O M} = \frac{M S B - M S W}{r} .$ The second estimate is not constrained to be nonnegative.

Facts & Assumptions

Given: A balanced one-way model with $k, r \geq 2$ , a fixed mean $μ$ , independent group blocks, independent random effects and errors within each block, and the stated normal laws and variances.

[A1]

AC is assumed because the library's standard normal law and the cited normal-moment result require it. The Axiom of Choice

[F1]

For $X \sim N (m, v)$ with $v \geq 0$ , $E [X] = m$ and $Var (X) = v$ . Location-scale normal laws have their stated mean and variance

[F2]

For a finite pairwise-independent family of square-integrable random variables, the variance of its sum is the sum of its variances. Variance adds for every finite pairwise-independent family

[F3]

Applying measurable functions separately to independent random elements preserves their independence. Measurable coordinatewise functions preserve independence

[F4]

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

[F5]

Expectation is integration on the underlying probability space. Expectation of a nonnegative or integrable random variable

[F6]

Expectation is linear for finite sums of integrable random variables. Linearity, monotonicity, and the modulus bound for expectation

[F7]

Random elements are independent when their generated sigma-algebras are independent. Independent random elements

Proof

technique · compute the centered group-mean and within-group sums of squares from their first two moments

Given: The hypotheses in the Statement.

1.1A1F1F2F3F4F5F6F7given

Put ${\overset{ˉ}{ϵ}}_{i} = r^{- 1} \sum_{j} ϵ_{i j}$ . By [F1], every error has mean zero and variance $σ^{2}$ ; [F6] gives $E [{\overset{ˉ}{ϵ}}_{i}] = 0$ . By [F2], the variance of $\sum_{j} ϵ_{i j}$ is $r σ^{2}$ , and the variance definition [F4] gives $Var ({\overset{ˉ}{ϵ}}_{i}) = r^{- 2} r σ^{2} = σ^{2} / r$ . Since $a_{i}$ is independent of the error vector, [F3] makes $a_{i}$ independent of ${\overset{ˉ}{ϵ}}_{i}$ . Thus [F2] gives $Var ({\overset{ˉ}{Y}}_{i \cdot}) = τ^{2} + σ^{2} / r$ , and [F6] gives $E [{\overset{ˉ}{Y}}_{i \cdot}] = μ$ . Applying [F3] to the independent group blocks shows that the group means are independent by [F7].

2.1F2F4F5F6step 1.1algebra

For any numbers $x_{1}, \dots, x_{k}$ with mean $\overset{ˉ}{x}$ , expansion gives $\sum_{i} (x_{i} - \overset{ˉ}{x})^{2} = \sum_{i} (x_{i} - μ)^{2} - k (\overset{ˉ}{x} - μ)^{2}$ . Apply this with $x_{i} = {\overset{ˉ}{Y}}_{i \cdot}$ . By [F2], $Var ({\overset{ˉ}{Y}}_{\cdot \cdot}) = (τ^{2} + σ^{2} / r) / k$ ; [F4], [F5], and [F6] therefore give $E \sum_{i} ({\overset{ˉ}{Y}}_{i \cdot} - {\overset{ˉ}{Y}}_{\cdot \cdot})^{2} = (k - 1) (τ^{2} + σ^{2} / r)$ . Multiplying by $r / (k - 1)$ yields $E [M S B] = r τ^{2} + σ^{2}$ .

2.2F1F4F5step 1.1algebra

Since $Y_{i j} - {\overset{ˉ}{Y}}_{i \cdot} = ϵ_{i j} - {\overset{ˉ}{ϵ}}_{i}$ , expansion gives $\sum_{j} (ϵ_{i j} - {\overset{ˉ}{ϵ}}_{i})^{2} = \sum_{j} ϵ_{i j}^{2} - r {\overset{ˉ}{ϵ}}_{i}^{2}$ . By [F1], [F4], and [F5], $E [\sum_{j} ϵ_{i j}^{2}] = r σ^{2}$ and $E [{\overset{ˉ}{ϵ}}_{i}^{2}] = Var ({\overset{ˉ}{ϵ}}_{i}) = σ^{2} / r$ ; hence each group contributes expected within sum of squares $(r - 1) σ^{2}$ . Summing over $k$ groups and dividing by $k (r - 1)$ proves $E [M S W] = σ^{2}$ .

3.1step 2.1step 2.2algebra∎

Solving $E [M S W] = σ^{2}$ and $E [M S B] = σ^{2} + r τ^{2}$ for the two variance components gives ${\hat{σ^{2}}}_{M O M} = M S W$ and ${\hat{τ^{2}}}_{M O M} = (M S B - M S W) / r$ . The algebra imposes no nonnegativity constraint on the second expression.

Depends on

Used by

A negative balanced-ANOVA moment estimate Example

Dependency tree · two levels

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Sources

Andrew Gelman et al., Bayesian Data Analysis, 3rd ed., §5.4 (standard reference, not scraped)
NIST/SEMATECH e-Handbook of Statistical Methods, §7.4.4, What are variance components? (standard reference, not scraped)