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QMD controls singular and zero-overlap masses

Statement

Assume AC and the hypotheses of Quadratic-mean differentiability at a parameter. Write $p_{0} = p_{θ_{0}}$ , $p_{u} = p_{θ_{0} + u}$ , and $P_{0} = P_{θ_{0}}$ . Let $α (u) : = P_{θ_{0} + u} (p_{0} = 0), β (u) : = P_{0} (p_{u} = 0) .$ Then $α (u) = o (∥ u ∥_{2}^{2})$ and $β (u) = o (∥ u ∥_{2}^{2})$ as $u \to 0$ .

Facts & Assumptions

Given: AC; QMD at $θ_{0}$ with score $s$ square-integrable under $P_{0}$ , and density versions $p_{0}, p_{u}$ relative to the common measure $μ$ .

[A1]

The Axiom of Choice: AC supplies the density representatives in Likelihood densities.

[L1]

Likelihood densities: Each selected $p_{θ}$ is nonnegative and measurable with $P_{θ} (A) = \int_{A} p_{θ} d μ$ .

[F1]

Quadratic-mean differentiability at a parameter: The residual $r_{u} = \sqrt{p_{u}} - \sqrt{p_{0}} - \frac{1}{2} (u^{T} s) \sqrt{p_{0}}$ satisfies $∥ r_{u} ∥_{L^{2} (μ)} = o (∥ u ∥_{2})$ .

[F2]

A QMD score is centered and has finite information: $E_{0} ∥ s ∥_{2}^{2} < \infty$ .

[F3]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: $∣ u^{T} s ∣ \leq ∥ u ∥_{2} ∥ s ∥_{2}$ .

[F4]

Holder's inequality for integrals, including the endpoint cases and Minkowski's inequality for integrals, including $p = \infty$ : Cauchy–Schwarz bounds $L^{2}$ inner products and the $L^{2}$ norm satisfies the triangle inequality.

[F5]

Absolute continuity of the integral: If $g \in L^{1} (P_{0})$ and $P_{0} (B_{u}) \to 0$ , then $E_{0} [∣ g ∣ 1_{B_{u}}] \to 0$ .

Proof

Proof technique: restrict the QMD residual to the two support-complement sets.

1.1A1F1L1givenalgebra

Set $S = {p_{0} > 0}$ . On $S^{c}$ , both $\sqrt{p_{0}}$ and $(u^{T} s) \sqrt{p_{0}}$ vanish, so $r_{u} = \sqrt{p_{u}}$ there. Therefore $α (u) = \int_{S^{c}} p_{u} d μ = \int_{S^{c}} r_{u}^{2} d μ \leq ∥ r_{u} ∥_{2}^{2} = o (∥ u ∥_{2}^{2}) .$ This is the claimed singular-mass bound.

1.2F1F2F3F4L1givenalgebra

Let $B_{u} = S \cap {p_{u} = 0}$ . On $B_{u}$ , $\sqrt{p_{u}} - \sqrt{p_{0}} = - \sqrt{p_{0}}$ , so $P_{0} (B_{u}) \leq ∥ \sqrt{p_{u}} - \sqrt{p_{0}} ∥_{2}^{2} = O (∥ u ∥_{2}^{2}) \to 0,$ where the last bound follows from [F1], [F2], [F3], and [F4]. Thus the sets $B_{u}$ shrink to zero in $P_{0}$ measure.

2.1F1F2F3F4L1step 1.2given

On $B_{u}$ , the residual identity in [F1] gives $\sqrt{p_{0}} 1_{B_{u}} = - \frac{1}{2} (u^{T} s) \sqrt{p_{0}} 1_{B_{u}} - r_{u} 1_{B_{u}} .$ Taking $L^{2} (μ)$ norms and using [F3] yields $\sqrt{β (u)} \leq \frac{1}{2} ∥ u ∥_{2} (E_{0} [∥ s ∥_{2}^{2} 1_{B_{u}}])^{1 / 2} + o (∥ u ∥_{2}) .$ This is the required norm bound.

3.1F2F5step 1.2step 2.1algebra∎

Since $∥ s ∥_{2}^{2}$ is integrable and $P_{0} (B_{u}) \to 0$ , [F5] gives $E_{0} [∥ s ∥_{2}^{2} 1_{B_{u}}] \to 0$ . Step 2.1 therefore gives $\sqrt{β (u)} = o (∥ u ∥_{2})$ and $β (u) = o (∥ u ∥_{2}^{2})$ . Because $P_{0} (p_{u} = 0) = P_{0} (B_{u})$ , this is the claimed zero-overlap bound.

Depends on

Used by

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Sources

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 22, PDF pp. 8–10 (standard reference, not scraped)
John Duchi, A few notes on contiguity, asymptotics, and local asymptotic normality, §1.5 (standard reference, not scraped)