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Gaussian Orthogonal Decomposition and Quadratic Forms — Examples

1 · Prerequisites

2 · Summary

These examples calculate an exact one-sample t interval and a fixed-design regression contrast and nested-model F statistic. A Rademacher counterexample shows that an orthogonal change of coordinates does not preserve independence for arbitrary product distributions and does not give exact normal-theory t/F laws.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

An exact unknown-variance t interval for a four-point sample

Example

Assume the Axiom of Choice and an iid sample of size $4$ from $N (μ, σ^{2})$ with $σ > 0$ . For the observed values $(1, 2, 4, 5)$ , the sample mean is $\overset{ˉ}{x} = 3$ and the unbiased sample variance is $s^{2} = 10 / 3$ . The exact $95 %$ t interval produced by the sample is

[3 - t_{3, 39 / 40} \frac{\sqrt{10 / 3}}{2}, 3 + t_{3, 39 / 40} \frac{\sqrt{10 / 3}}{2}] .

The repeated-sampling coverage of this interval procedure is exactly $0.95$ under the stated normal model; this is a coverage statement about the procedure, not a probability assigned to the fixed parameter after observing these data.

Facts & Assumptions

Given: The stated normal sample model and observed values $(1, 2, 4, 5)$ .

[F1]

For an iid normal sample with unknown variance and sample size $n \geq 2$ , the t pivot has law $t_{n - 1}$ and its central quantile interval has exact coverage $1 - α$ . Normal sample mean and residual are independent, giving the exact t pivot

[F2]

The $u$ -quantile $t_{k, u}$ is the unique point where the continuous strictly increasing $t_{k}$ CDF equals $u$ . Student t and Fisher F CDFs are continuous and strictly increasing on their supports

Verification

technique · direct

1.1givenalgebra

The arithmetic mean is $(1 + 2 + 4 + 5) / 4 = 3$ . The deviations from $3$ are $- 2, - 1, 1, 2$ , whose squared sum is $4 + 1 + 1 + 4 = 10$ .

2.1F1F2step 1.1algebra∎

Dividing this sum by $n - 1 = 3$ gives $s^{2} = 10 / 3$ , hence $s = \sqrt{10 / 3}$ . Substituting $n = 4$ , $α = 5 / 100$ , and the observed sample into [F1] yields the displayed interval with the unique quantile $t_{3, 39 / 40}$ from [F2]; its repeated-sampling coverage is $1 - α = 95 / 100$ .

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A fixed-design slope contrast and nested-model F statistic

Example

Assume the Axiom of Choice. Use the fixed design with rows $(1, - 1), (1, - 1), (1, 1), (1, 1)$ and observations $y = (1, 2, 4, 5)^{T}$ . The full model has intercept and slope; the reduced model has only an intercept, so the rank difference is $q = 1$ . The least-squares estimates are $({\hat{β}}_{0}, {\hat{β}}_{1}) = (3, 3 / 2)$ , the full residual sum is $R S S_{U} = 1$ , and the reduced residual sum is $R S S_{R} = 10$ . Thus the observed slope t statistic under $H_{0} : β_{1} = 0$ is $3 \sqrt{2}$ , and the nested-model F statistic is

F = \frac{(10 - 1) / 1}{1 / (4 - 2)} = 18.

Under fixed-design normal errors with the null true, the theorem gives exact laws $t_{2}$ for the contrast and $F_{1, 2}$ for the nested-model statistic.

Facts & Assumptions

Given: The displayed deterministic design, response vector, and the fixed-design normal error model under the null.

[F1]

For a fixed full-rank Gaussian design, every nonzero coefficient contrast has an exact t pivot and a nested reduced/full model has the stated exact F pivot. QR decomposition gives exact t and F pivots in the fixed Gaussian linear model

Verification

technique · direct

1.1givenalgebra

The two design columns are orthogonal: their inner product is $1 (- 1) + 1 (- 1) + 1 (1) + 1 (1) = 0$ , and each has squared norm $4$ . Hence $X^{T} X = 4 I_{2}$ . The displayed least-squares coefficients are ${\hat{β}}_{0} = (1 + 2 + 4 + 5) / 4 = 3$ and ${\hat{β}}_{1} = (- 1 - 2 + 4 + 5) / 4 = 3 / 2$ . The full fitted values are $(3 / 2, 3 / 2, 9 / 2, 9 / 2)$ , so the residuals are $(- 1 / 2, 1 / 2, - 1 / 2, 1 / 2)$ and $R S S_{U} = 1$ .

2.1givenstep 1.1algebra

The reduced fit is the sample mean $3$ , with residuals $(- 2, - 1, 1, 2)$ and $R S S_{R} = 10$ . For $c = (0, 1)^{T}$ , $c^{T} (X^{T} X)^{- 1} c = 1 / 4$ and $S^{2} = R S S_{U} / (4 - 2) = 1 / 2$ , so the observed null t statistic is $(3 / 2) / \sqrt{(1 / 2) (1 / 4)} = 3 \sqrt{2}$ .

3.1F1step 1.1step 2.1algebra∎

The rank difference is one and the denominator degrees of freedom are $4 - 2 = 2$ . Therefore $F = ((R S S_{R} - R S S_{U}) / 1) / (R S S_{U} / 2) = 18$ . Under the normal model and the null $β_{1} = 0$ , [F1] gives the exact $F_{1, 2}$ law; the t contrast has the exact $t_{2}$ law and its square agrees with $F$ here.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Rademacher errors show orthogonality alone gives neither independence nor exact t and F laws

Statement refuted

"An orthogonal transformation preserves independence for arbitrary independent coordinates, and orthogonality alone guarantees the exact normal-theory t and F laws."

Facts & Assumptions

Given: The Rademacher variables and the t and F laws from positive-degree normal and chi-square ratios.

[A1]

The Axiom of Choice is assumed by the imported t and F definitions and the positive-degree chi-square lemma used to establish positivity of their denominators; the finite Rademacher calculations themselves use no choice. The Axiom of Choice Student t and Fisher F laws Positive-degree chi-square laws have a positive density and regular CDF

[F1]

A $t_{k}$ law uses an independent standard normal divided by a positive-degree chi-square square root, and an $F_{r, s}$ law is a ratio of positive-degree chi-square variables; both definitions assign zero on denominator-zero sets, which have probability zero for these Gaussian laws. Student t and Fisher F laws Positive-degree chi-square laws have a positive density and regular CDF

Counterexample

technique · direct

Given: Assume the Axiom of Choice for the imported $t$ and $F$ laws. Take $Ω = {- 1, 1}^{3}$ with the uniform probability measure, assigning mass $1 / 8$ to each point, and define $R_{i} (ω) = ω_{i}$ for $i = 1, 2, 3$ . Each sign triple has probability $(1 / 2)^{3}$ , so the coordinate maps are independent Rademacher variables.

1.1givenalgebra

For two coordinates, apply $U = 2^{- 1 / 2} (\begin{array}{cc} 1 & 1 \\ 1 & - 1 \end{array})$ to $(R_{1}, R_{2})^{T}$ . The events $A = {∣ (U R)_{1} ∣ = \sqrt{2}}$ and $B = {∣ (U R)_{2} ∣ = \sqrt{2}}$ each have probability $1 / 2$ , while $A \cap B = \emptyset$ : $A$ occurs exactly when $R_{1} = R_{2}$ and $B$ exactly when $R_{1} = - R_{2}$ . If the transformed coordinates were independent, $P (A \cap B)$ would be $1 / 4$ , contradicting its actual value $0$ .

1.2givenA1F1algebra

For three independent Rademacher observations, the event $E = {R_{1} = R_{2} = R_{3}}$ has probability $2 / 8 = 1 / 4$ . On $E$ , $\overset{ˉ}{R} = \pm 1$ and $S^{2} = \frac{1}{2} \sum_{i = 1}^{3} (R_{i} - \overset{ˉ}{R})^{2} = 0$ , so the usual studentized mean $\overset{ˉ}{R} / (S / \sqrt{3})$ is not a real-valued statistic on an event of positive probability. It therefore cannot have the $t_{2}$ law, which is a probability law on $R$ with a positive chi-square denominator almost surely by [A1, F1].

2.1givenA1F1algebra∎

For $n = 2$ , take the full fixed design with one intercept column $(1, 1)^{T}$ and the reduced zero-dimensional design whose mean space is ${0}$ . Under the null intercept $0$ , the errors are $(R_{1}, R_{2})^{T}$ . On $E_{+} = {R_{1} = R_{2}}$ , which has probability $1 / 2$ , the full-model residual sum is $R S S_{U} = 0$ while the reduced residual sum is $R S S_{R} = 2$ . Thus the nested F ratio has positive numerator and zero denominator (or is $+ \infty$ under extended-value convention), so it is not an almost surely finite $F_{1, 1}$ variable.

Gaussian Orthogonal Decomposition and Quadratic Forms — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

An exact unknown-variance t interval for a four-point sample

Example

Facts & Assumptions

Verification

A fixed-design slope contrast and nested-model F statistic

Example

Facts & Assumptions

Verification

Rademacher errors show orthogonality alone gives neither independence nor exact t and F laws

Statement refuted

Facts & Assumptions

Counterexample

Sources