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Normal sample mean and residual are independent, giving the exact t pivot

Statement

Assume the Axiom of Choice. Use the experiment with data space (Rn,B(Rn)), parameter set R×(0,∞), and sampling law Pμ,σ2=N(μ,σ2)⊗n. Let n≥2, σ>0, and let X1,…,Xn be the independent coordinates of the observation. Define Xˉ=1n∑i=1nXi,S2=1n−1∑i=1n(Xi−Xˉ)2,S=S2. The map x↦xˉ is a statistic. The sample mean Xˉ and residual sum S2 are independent, (n−1)S2σ2∼χn−12,S>0 almost surely,T~:={Xˉ−μS/n,S>0,0,S=0∼tn−1. For every 0<α<1, the interval [Xˉ−qn−1(1−α/2)Sn, Xˉ+qn−1(1−α/2)Sn] has exact coverage 1−α, where qn−1(u) is the unique u-quantile of tn−1 established in Student t and Fisher F CDFs are continuous and strictly increasing on their supports.

Facts & Assumptions

Given: The Axiom of Choice, n≥2, σ>0, and an independent sample Xi∼N(μ,σ2).

[A1]

The Axiom of Choice is assumed by the normal-law and orthogonal-invariance suppliers. The Axiom of Choice

[A2]

Under the Axiom of Choice, the sequence-selection conclusion of the countable-choice lemma gives Countable Choice, the hypothesis of the continuous-map measurability corollary. The Axiom of Countable Choice (ACω) The Axiom of Choice implies countable choice Continuous functions on Euclidean spaces are Borel measurable

[F1]

The law N(μ,σ2) is the pushforward of N(0,1) by the map z↦μ+σz; measurable functions of independent random elements remain independent. Standard normal and normal laws Measurable coordinatewise functions preserve independence

[F2]

A statistic is a measurable data map; under Countable Choice, continuous maps between Euclidean spaces are Borel measurable. Statistics and estimators Continuous functions on Euclidean spaces are Borel measurable

[F3]

For Z∼Nn(0,In) and a rank-r orthogonal projection P, PZ and (I−P)Z are independent and their squared norms have laws χr2 and χn−r2. Orthogonal projections split a standard Gaussian into independent chi-square sums

[F4]

The joint law of independent standard-normal coordinates is Nn(0,In), and each linear projection of that law is normal with the corresponding projected mean and variance. Independent random elements have product joint law Multivariate normal law, including singular covariance

[F5]

Positive-degree chi-square denominators are positive almost surely; tk has a continuous symmetric CDF with unique quantiles, and its law is the piecewise ratio G/Qk/k set to zero when Qk=0, for independent G∼N(0,1) and Qk∼χk2. Positive-degree chi-square laws have a positive density and regular CDF Student t and Fisher F laws Student t and Fisher F CDFs are continuous and strictly increasing on their supports

Proof

Given: The hypotheses in the Statement.

1.1A1A2F1F2F4

By [A2], Countable Choice holds, so [F2] applies to the linear continuous map x↦xˉ; hence it is a statistic on the stated Borel data space. Put Zi=(Xi−μ)/σ. The affine pushforward definition in [F1] gives each Zi∼N(0,1), and [F1] preserves their independence. Thus Z=(Z1,…,Zn)∼Nn(0,In) by [F4].

2.1step 1.1algebra

Define Zˉ=n−1∑i=1nZi, u=n−1/2(1,…,1)T, and P=uuT. Since uTu=1, P is a rank-one orthogonal projection. We have uTZ=n Zˉ and X−Xˉ(1,…,1)T=σ(In−P)Z. Therefore Xˉ=μ+σuTZ/n and S2=σ2∥(In−P)Z∥2/(n−1).

3.1A2F1F2F3F4algebrastep 2.1

By [F3], PZ and (In−P)Z are independent and Q:=∥(In−P)Z∥2∼χn−12. The continuous linear map z↦uTz is Borel by [A2] and [F2], and uTZ=uTPZ; the continuous squared-norm map on the complementary vector is likewise Borel. Thus G:=uTZ and Q are measurable functions of the independent vectors in [F3], so they are independent by [F1]. By [F4], G∼N(0,1).

4.1step 2.1step 3.1F1F5algebra

The formulas in step 2.1 express Xˉ as an affine function of G and S2 as σ2Q/(n−1). Hence Xˉ and S2 are independent, (n−1)S2/σ2=Q∼χn−12, and S>0 almost surely by [F5].

5.1step 2.1step 3.1step 4.1F5algebra

The set {S>0} is Borel and the displayed quotient is continuous there; assigning 0 on {S=0} therefore makes T~ measurable. Since S=0 exactly when Q=0, step 2.1 gives T~=G/Q/(n−1) on {Q>0} and both versions are 0 on {Q=0}. The variables G and Q are independent with laws N(0,1) and χn−12, so [F5] gives T~∼tn−1.

6.1F5step 4.1step 5.1algebra∎

Let q=qn−1(1−α/2). By [F5], the tn−1 CDF is continuous and symmetric, so Ftn−1(0)=1/2<1−α/2 and strict increase gives q>0; symmetry then gives Ftn−1(−q)=α/2. Step 5.1 therefore gives P(−q≤T~≤q)=1−α. On the full-probability event {S>0} from step 4.1, this event is exactly {μ lies in the displayed interval} by multiplying by S/n and rearranging. The events agree almost surely, so the interval has exact coverage 1−α.

Depends on

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