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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 tn−p contrast pivot.

Facts & Assumptions

Given: The Rademacher-error model described below, and the Student t1 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 t1 CDF is continuous and strictly increasing on R, so t1 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 ϵ1,ϵ2 with P(ϵi=1)=P(ϵi=−1)=1/2. They have mean zero and variance σ2=1, while X has rank one and n>p. Let β^=(XTX)−1XTY, RSS=∥(I−PX)Y∥2, s2=RSS/(n−p), and use the nonzero contrast c=1 with pivot set to zero when s=0.

Given: All four equally likely pairs (ϵ1,ϵ2)∈{(−1,−1),(−1,1),(1,−1),(1,1)}.

1.1givenconstructalgebra

Since XTX=2, β^=(ϵ1+ϵ2)/2 and RSS=(ϵ1−ϵ2)2/2. If the errors agree, then RSS=0 and the pivot is zero by its definition. If they disagree, then β^=0, RSS=2, and the pivot denominator is s(XTX)−1=21/2=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 t1 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 t1, refuting the statement.

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