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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.

Depends on

QR decomposition gives exact t and F pivots in the fixed Gaussian linear model

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT 18.S096, Regression Analysis, Lecture 6, Normal Regression Models (standard reference, not scraped)
Yen-Chi Chen, UW STAT 512 Lecture 8: Linear Models and Multivariate Normal Distributions (standard reference, not scraped)