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Exact Gaussian regression inference and prediction — Examples

1 · Prerequisites

2 · Summary

Objective

Work out a full-rank Gaussian regression interval calculation and show by a finite counterexample why the exact Gaussian $t$ pivot does not follow from zero means and equal variances alone.

The numerical example uses the same center for the mean-response and prediction intervals; the prediction interval is wider because it includes the future response's independent noise. The Rademacher example keeps the design full rank and errors independent with the same first two moments, then computes all possible outcomes directly.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Independent Rademacher errors need not give the Gaussian regression t pivot

Statement refuted

Full column rank, $n > p$ , independent errors with mean zero and common positive variance suffice for the exact Gaussian-regression Student $t_{n - p}$ contrast pivot.

Facts & Assumptions

Given: The Rademacher-error model described below, and the Student $t_{1}$ law.

[A1]

The Axiom of Choice is assumed for the normal and Student $t$ law definitions; the finite enumeration below uses no choice. The Axiom of Choice

[F1]

The $t_{1}$ CDF is continuous and strictly increasing on $R$ , so $t_{1}$ is not a point mass at zero. Student t and Fisher F CDFs are continuous and strictly increasing on their supports

Counterexample

Take $n = 2$ , $p = 1$ , $X = (1, 1)^{T}$ , $β = 0$ , and independent errors $ϵ_{1}, ϵ_{2}$ with $P (ϵ_{i} = 1) = P (ϵ_{i} = - 1) = 1 / 2$ . They have mean zero and variance $σ^{2} = 1$ , while $X$ has rank one and $n > p$ . Let $\hat{β} = (X^{T} X)^{- 1} X^{T} Y$ , $R S S = ∥ (I - P_{X}) Y ∥^{2}$ , $s^{2} = R S S / (n - p)$ , and use the nonzero contrast $c = 1$ with pivot set to zero when $s = 0$ .

Given: All four equally likely pairs $(ϵ_{1}, ϵ_{2}) \in {(- 1, - 1), (- 1, 1), (1, - 1), (1, 1)}$ .

1.1givenconstructalgebra

Since $X^{T} X = 2$ , $\hat{β} = (ϵ_{1} + ϵ_{2}) / 2$ and $R S S = (ϵ_{1} - ϵ_{2})^{2} / 2$ . If the errors agree, then $R S S = 0$ and the pivot is zero by its definition. If they disagree, then $\hat{β} = 0$ , $R S S = 2$ , and the pivot denominator is $s \sqrt{(X^{T} X)^{- 1}} = \sqrt{2} \sqrt{1 / 2} = 1$ , so the pivot is again zero. These cases exhaust the four outcomes.

2.1A1F1step 1.1algebra∎

Thus the pivot equals zero almost surely. By [F1], a $t_{1}$ random variable has a continuous law and hence assigns probability zero to the singleton ${0}$ , whereas this pivot assigns it probability one. Its law is not $t_{1}$ , refuting the statement.

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

Mean-response and prediction intervals from one Gaussian regression fit

Example

Take $n = 4$ , $p = 2$ and $X = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{array}), Y = (1, 2, 2, 4)^{T} .$ Under the exact Gaussian regression model, compare the 95% confidence interval for the slope $β_{1}$ with the 95% mean-response and future-response prediction intervals at $x_{0} = (1, 2)^{T}$ . Also distinguish the Gaussian maximum-likelihood estimate of $σ^{2}$ from the residual mean square.

Facts & Assumptions

[F1]

The exact coefficient, mean-response, and future-response intervals use the formulas in Exact confidence and prediction intervals in Gaussian regression.

Verification

Given: The displayed fixed design and response, confidence level $0.95$ , and the interval theorem [F1].

1.1givenalgebra

Direct multiplication gives $X^{T} X = (\begin{array}{cc} 4 & 6 \\ 6 & 14 \end{array})$ , $(X^{T} X)^{- 1} = \frac{1}{20} (\begin{array}{cc} 14 & - 6 \\ - 6 & 4 \end{array})$ , and $X^{T} Y = (9, 18)^{T}$ . Hence $\hat{β} = (0.9, 0.9)^{T}$ , the fitted values are $(0.9, 1.8, 2.7, 3.6)^{T}$ , and $R S S_{F} = {0.1}^{2} + {0.2}^{2} + (- 0.7)^{2} + {0.4}^{2} = 0.70$ . Thus $ν = 2$ , $s^{2} = 0.35$ , while ${\hat{σ}}_{M L}^{2} = R S S_{F} / n = 0.175$ .

2.1F1step 1.1algebra∎

For $x_{0} = (1, 2)^{T}$ , multiplication by $(X^{T} X)^{- 1}$ gives $h_{0} = x_{0}^{T} (X^{T} X)^{- 1} x_{0} = 0.30$ and the fitted mean $x_{0}^{T} \hat{β} = 2.7$ . With $q = t_{2, 0.975}$ , [F1] gives the slope interval $0.9 \pm q \sqrt{0.35 \cdot 0.20}$ , the mean-response interval $2.7 \pm q \sqrt{0.35 \cdot 0.30}$ , and the prediction interval $2.7 \pm q \sqrt{0.35 \cdot 1.30}$ . The prediction half-width is $\sqrt{13 / 3}$ times the mean-response half-width, so it is wider because it includes the future response error.

Exact Gaussian regression inference and prediction — Examples

1 · Prerequisites

2 · Summary

Objective

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Independent Rademacher errors need not give the Gaussian regression t pivot

Statement refuted

Facts & Assumptions

Counterexample

Mean-response and prediction intervals from one Gaussian regression fit

Example

Facts & Assumptions

Verification

Sources