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Exact confidence and prediction intervals in Gaussian regression

Statement

Use the fixed full-column-rank Gaussian regression model and notation of Exact t and partial F inference in full-rank Gaussian regression, with ν=n−p. Let 0<α<1, let t∗=tν,1−α/2 be the unique (1−α/2) quantile of tν, and put vc=cT(XTX)−1c for c≠0. Then Cc(Y)=[cTβ^−t∗svc, cTβ^+t∗svc] has exact coverage 1−α for cTβ.

For any x0∈Rp, put h0=x0T(XTX)−1x0. The mean-response interval Cmean(Y;x0)=[x0Tβ^−t∗sh0, x0Tβ^+t∗sh0] has exact coverage 1−α for x0Tβ when x0≠0. If x0=0, it is the singleton {0} and has coverage 1.

If Y0∗=x0Tβ+ϵ0∗ is one future response with ϵ0∗∼N(0,σ2) independent of the training data, then Cpred(Y;x0)=[x0Tβ^−t∗s1+h0, x0Tβ^+t∗s1+h0] satisfies Pβ,σ2{Y0∗∈Cpred(Y;x0)}=1−α. The added 1 is the variance contribution of the future response's own error.

Facts & Assumptions

Given: The Axiom of Choice, the Gaussian regression experiment in Exact t and partial F inference in full-rank Gaussian regression, 0<α<1, a contrast c≠0, a predictor vector x0∈Rp, and, for prediction, an independent future Gaussian error ϵ0∗∼N(0,σ2).

[A1]

The Axiom of Choice is inherited from the Gaussian model and its normal-law suppliers. The Axiom of Choice

[F1]

In this model, β^∼Np(β,σ2(XTX)−1), β^ is independent of RSSF, and the stated nonzero-contrast pivot has law tν. Exact t and partial F inference in full-rank Gaussian regression

[F2]

A pivot is a parameter-indexed family of measurable statistics with one common sampling law. Pivotal quantities

[F3]

A confidence-set procedure is measurable when each parameter-membership section is measurable, and coverage is the probability that the true parameter belongs to the set. Confidence sets and coverage

[F4]

Inverting a pivot acceptance set of probability 1−α gives a confidence set with exact coverage 1−α. Coverage from pivot inversion and exact calibration

[F5]

The tν CDF is symmetric, continuous, and strictly increasing, so [−tν,1−α/2,tν,1−α/2] has probability 1−α. Student t and Fisher F CDFs are continuous and strictly increasing on their supports

[F6]

For a multivariate normal vector, every linear projection is normal with variance given by its covariance quadratic form; independent normal coordinates give the corresponding product normal law. Multivariate normal law, including singular covariance

[F7]

Independent random elements have product joint law, and measurable functions of independent random elements remain independent. Independent random elements have product joint law Measurable coordinatewise functions preserve independence

[F8]

The future error has the centered normal law with variance σ2, independently of the training data. Standard normal and normal laws

[F9]

For ν≥1, tν is the law of an independent standard normal divided by χν2/ν. Student t and Fisher F laws Chi-square laws, including zero degrees

[F10]

A positive-degree chi-square denominator is positive almost surely. Positive-degree chi-square laws have a positive density and regular CDF

Proof

Given: The hypotheses in the Statement.

1.1A1F1F2F3F4F5algebra

Fix c≠0. Then vc>0 because XTX is positive definite. For each full parameter θ=(β,σ2), the statistic qθ(Y)=(cTβ^−cTβ)/(svc), assigned value 0 when s=0, is a pivot by [F1] and [F2], with common law tν. By [F5], B=[−t∗,t∗] has probability 1−α. Define Dc(Y)={(β′,σ′2)∈Θ:cTβ′∈Cc(Y)}. For each fixed (β′,σ′2)∈Θ, its membership event is ∣cTβ^−cTβ′∣≤t∗svc, a closed, hence Borel, set because β^ and s are continuous functions of Y. Thus [F3] applies to Dc. Pivot inversion [F4] gives exact coverage for Dc′(Y)={θ′∈Θ:∣qθ′(Y)∣≤t∗}. On s>0, Dc(Y)=Dc′(Y); the exceptional event s=0 has probability zero for every parameter by [F1]. Therefore Dc has exact coverage 1−α, which is precisely Pβ,σ2{cTβ∈Cc(Y)}=1−α.

2.1F1step 1.1algebra

If x0≠0, the same argument with c=x0 gives the displayed mean-response interval because h0=vx0. If x0=0, then x0Tβ=x0Tβ^=h0=0 for every parameter and data vector, so the interval is {0} and contains the true mean with probability 1.

3.1A1F1F5F6F7F8F9F10step 2.1algebra∎

Write Z=(β^−β)/σ and G∗=ϵ0∗/σ. By [F1] and [F8], Z and G∗ are independent centered normal vectors/scalars with covariance (XTX)−1 and variance 1. Their joint law is the product law by [F7], so [F6] gives Y0∗−x0Tβ^σ1+h0=G∗−x0TZ1+h0∼N(0,1), since the numerator has variance 1+x0T(XTX)−1x0=1+h0. Because G∗ is independent of the training data, [F7] gives L(Z,RSSF,G∗)=L(Z,RSSF)⊗L(G∗). Since Z is a measurable affine function of β^, [F1] and [F7] imply that Z is independent of RSSF; the joint law therefore factors as L(Z)⊗L(RSSF)⊗L(G∗), so (Z,G∗) is independent of RSSF. The displayed standard normal is a measurable function of (Z,G∗), so it is independent of RSSF/σ2∼χν2 by [F7]. By [F9] and [F10], the ratio (Y0∗−x0Tβ^)/(s1+h0) has law tν with positive denominator almost surely. Its central acceptance interval has probability 1−α by [F5], which is exactly the event Y0∗∈Cpred(Y;x0).

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