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First-order equivalence of Wald, score, and likelihood-ratio statistics

Statement

Assume AC and the hypotheses of Wald, score, and likelihood-ratio asymptotic chi-square calibration. Under the null, the three statistics satisfy $W_{n} - S_{n} = o_{P} (1)$ , $S_{n} - L_{n} = o_{P} (1)$ , and $L_{n} - W_{n} = o_{P} (1)$ . Hence they have the same first-order null calibration, namely $χ_{q}^{2}$ .

Facts & Assumptions

Given: AC and the hypotheses of Wald, score, and likelihood-ratio asymptotic chi-square calibration.

[F1]

Wald, score, and likelihood-ratio asymptotic chi-square calibration: all three statistics equal one common quadratic score form up to $o_{P} (1)$ , and that form converges to $χ_{q}^{2}$ .

[A1]

The Axiom of Choice: AC is inherited from [F1], whose iid score CLT uses the multivariate iid CLT.

Proof

1.1A1F1algebra

By [F1], there is a common random quadratic form $T_{n}$ with $W_{n} = T_{n} + o_{P} (1)$ , $S_{n} = T_{n} + o_{P} (1)$ , and $L_{n} = T_{n} + o_{P} (1)$ . Subtracting the first two equalities and applying the triangle inequality gives $W_{n} - S_{n} = o_{P} (1)$ ; subtracting the second and third gives $S_{n} - L_{n} = o_{P} (1)$ ; subtracting the third and first gives $L_{n} - W_{n} = o_{P} (1)$ .

2.1F1step 1.1∎

Theorem [F1] states $T_{n} \Rightarrow χ_{q}^{2}$ and each statistic has that same limit under the theorem's hypotheses. The differences proved in step 1.1 therefore express first-order equivalence, not exact finite-sample equality.

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Sources

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 26, PDF pp. 11–17 (standard reference, not scraped)
Yen-Chi Chen, STAT 581 Chapter 4: Hypothesis tests in likelihood models, lines 73–76 (standard reference, not scraped)