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A one-sided DQM path yields an iid LAN proxy

Statement

Assume AC. Let $t \mapsto P_{t}$ be a one-sided path ( $0 \leq t < ϵ$ ) of probability laws on $(X, A)$ dominated by one supplied sigma-finite measure $μ$ , with supplied densities $p_{t}$ and $P_{0} = P$ . Suppose the path is differentiable in quadratic mean at $0$ with score $s \in L^{2} (P)$ , in the sense of A dominated differentiable-in-quadratic-mean submodel. Put $I = P [s^{2}]$ . For a fixed $h \geq 0$ , let $t_{n} = h / \sqrt{n}$ and let $X_{1}, \dots, X_{n}$ be iid $P$ . On $S = {p_{0} > 0}$ set $α_{t} = P_{t} (S^{c})$ and, for sufficiently small $t$ , define

$q_{t} (x) = \frac{1_{S} (x) p_{t} (x)}{(1 - α_{t}) p_{0} (x)} on S, q_{t} (x) = 0 on S^{c} .$

Then $q_{t}$ is a probability density relative to $P$ . With $L_{n} = \prod_{i = 1}^{n} q_{t_{n}} (X_{i})$ and $Δ_{n} = n^{- 1 / 2} \sum_{i = 1}^{n} s (X_{i})$ ,

${∥ P_{t_{n}}^{\otimes n} - L_{n} P^{\otimes n} ∥}_{T V} ⟶ 0, \log L_{n} = h Δ_{n} - \frac{1}{2} h^{2} I + o_{P} (1), Δ_{n} \Rightarrow N (0, I) .$

For $h = 0$ take $L_{n} = 1$ and use the convention $\log 0 = - \infty$ . The proxy $L_{n} P^{\otimes n}$ is the normalized absolutely continuous component; the conclusion does not assert a full likelihood ratio for $P_{t_{n}}^{\otimes n}$ .

Facts & Assumptions

Given: AC; the common-dominator one-sided DQM path and supplied density family; iid $X_{i} \sim P$ ; and fixed $h \geq 0$ .

[F1]

A dominated differentiable-in-quadratic-mean submodel: $∥ \sqrt{p_{t}} - \sqrt{p_{0}} - \frac{t}{2} s \sqrt{p_{0}} ∥_{L^{2} (μ)} = o (t)$ as $t ↓ 0$ ; the path may be one-sided.

[F2]

A DQM score has mean zero and is unique: $P s = 0$ and $P s^{2} < \infty$ for this dominated one-sided DQM path.

[F3]

Cauchy-Schwarz inequality for $L^{2}$ : the $L^{2}$ pairing is bounded by the product of the $L^{2}$ norms.

[F4]

Khinchin weak law for integrable IID variables: averages of iid integrable real variables converge in probability to their expectation.

[F5]

Multivariate iid central limit theorem: under AC, iid square-integrable vectors have the centered Gaussian limit with their covariance, including the zero-variance case.

[F6]

Identical distribution and IID families: the sample coordinates are iid under $P^{\otimes n}$ .

[F7]

Total variation distance for probability laws: $∥ Q - R ∥_{T V} = \sup_{A \in A} ∣ Q (A) - R (A) ∣$ .

[F8]

Markov's inequality for random variables: for nonnegative integrable $Y$ and $c > 0$ , $P (Y \geq c) \leq E Y / c$ .

[F9]

Finite and countable subadditivity of measures: the measure of a finite union is at most the sum of its members' measures.

[F10]

The Lebesgue integral is linear on $L^{1} (μ)$ : the integral is linear on integrable functions, so the normalization identity for $q_{t} = (1 + Y_{t} / 2)^{2}$ expands into its three terms.

[F11]

A uniform derivative bound gives a uniform Taylor remainder bound: if the third derivative is bounded on an interval, the second-order Taylor polynomial has a remainder bounded by a constant times the cube of the displacement.

[F12]

$L^{1}$ convergence implies uniform integrability: on a probability space, convergence in $L^{1}$ to an integrable limit makes the convergent sequence uniformly integrable.

[F13]

The measure with density $f$ relative to $μ$ : for a nonnegative measurable function $f$ , $A \mapsto \int_{A} f d ν$ is a measure.

[F14]

The product sigma-algebra and its finite iterates: the finite product sigma-algebra is generated by measurable coordinate rectangles. Coordinate projections are measurable because the inverse image of a coordinate event is such a rectangle; finite products of nonnegative measurable coordinate functions are therefore product-measurable.

[F15]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: products of sigma-finite measures exist and have the rectangle formula; iterating this applies to finite products of probabilities.

[F16]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: nonnegative product-measurable functions have their product integral equal to the iterated integrals.

[F17]

Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system: measures agreeing on a generating pi-system agree everywhere if an increasing finite-measure exhaustion belongs to that pi-system.

[A1]

The Axiom of Choice: AC is the density-version convention in [F1] and is assumed by [F5].

Proof

Given: The path, densities, score, iid sample, and fixed $h$ in the statement.

1.1A1F1F2givenalgebra

For $t > 0$ let $r_{t} = \sqrt{p_{t}} - \sqrt{p_{0}} - \frac{t}{2} s \sqrt{p_{0}}$ . On $S^{c}$ , $r_{t} = \sqrt{p_{t}}$ , so $α_{t} = \int_{S^{c}} p_{t} d μ \leq ∥ r_{t} ∥_{2}^{2} = o (t^{2})$ . Let $B_{t} = S \cap {p_{t} = 0}$ . Then $P (B_{t}) \leq ∥ \sqrt{p_{t}} - \sqrt{p_{0}} ∥_{2}^{2} = O (t^{2})$ . On $B_{t}$ , the DQM identity gives $\sqrt{p_{0}} = - \frac{t}{2} s \sqrt{p_{0}} - r_{t}$ , hence $\sqrt{P (B_{t})} \leq \frac{t}{2} {P [s^{2} 1_{B_{t}}]}^{1 / 2} + o (t)$ . Since $P (B_{t}) \to 0$ and $s^{2}$ is integrable, truncating $s^{2}$ at a fixed bound and then increasing that bound shows $P [s^{2} 1_{B_{t}}] \to 0$ . Thus $β_{t} : = P (p_{t} = 0) = P (B_{t}) = o (t^{2})$ .

2.1F1step 1.1givenalgebra

For small $t$ , $1 - α_{t} > 0$ . The definition of $q_{t}$ gives $P q_{t} = (1 - α_{t})^{- 1} \int_{S} p_{t} d μ = 1$ . On $S$ , divide the DQM expansion by $\sqrt{p_{0}}$ and include the normalizer $(1 - α_{t})^{- 1 / 2} = 1 + o (t^{2})$ . With $Y_{t} = 2 (\sqrt{q_{t}} - 1)$ , this yields $Y_{t} = t s + R_{t}$ in $L^{2} (P)$ with $∥ R_{t} ∥_{2} = o (t)$ .

3.1F2F3F10step 2.1algebra

Since $P q_{t} = 1$ and $q_{t} = (1 + Y_{t} / 2)^{2}$ $P$ -almost surely, linearity [F10] expands this normalization identity to $P Y_{t} = - \frac{1}{4} P [Y_{t}^{2}]$ . By [F2], [F3], and step 2.1, $P [Y_{t}^{2}] = t^{2} I + o (t^{2})$ , so $P Y_{t} = - \frac{1}{4} t^{2} I + o (t^{2})$ .

3.2A1F2F3F4F5F6F8F10step 2.1given

Suppose $h > 0$ . By [F4] applied to $s^{2}$ and [F5] applied to $s (X_{i})$ , $n^{- 1} \sum_{i} s (X_{i})^{2} \to I$ in probability and $Δ_{n} \Rightarrow N (0, I)$ . Step 2.1 and [F3] give $E ∣ Y_{t_{n}}^{2} - t_{n}^{2} s^{2} ∣ \leq ∥ R_{t_{n}} ∥_{2} (∥ Y_{t_{n}} ∥_{2} + t_{n} ∥ s ∥_{2}) = o (t_{n}^{2})$ . Since $n t_{n}^{2} = h^{2}$ , linearity [F10] and identical distribution [F6] show that the expected sum of these absolute differences is $o (1)$ , so Markov's inequality [F8] makes that sum converge to zero in probability. Also $t_{n}^{2} \sum_{i} s (X_{i})^{2} = (h^{2} / n) \sum_{i} s (X_{i})^{2} \to h^{2} I$ in probability; therefore $\sum_{i} Y_{t_{n}} (X_{i})^{2} \to h^{2} I$ .

4.1F2F8step 3.1givenalgebra

The centered iid sum of $R_{t_{n}} (X_{i})$ has variance at most $n ∥ R_{t_{n}} ∥_{2}^{2} = o (n t_{n}^{2}) = o (1)$ , so [F8] (applied to its square) gives $\sum_{i} R_{t_{n}} (X_{i}) = n P R_{t_{n}} + o_{P} (1)$ . By [F2], $P R_{t_{n}} = P Y_{t_{n}}$ ; step 3.1 therefore gives $\sum_{i} Y_{t_{n}} (X_{i}) = h Δ_{n} - \frac{1}{4} h^{2} I + o_{P} (1)$ .

5.1F3F6F8F9F11F12step 1.1step 2.1step 3.2step 4.1algebra

The variables $V_{n} = \sqrt{n} Y_{t_{n}} (X_{1})$ converge in $L^{2} (P)$ to $h s (X_{1})$ by step 2.1. By [F3], $E ∣ V_{n}^{2} - h^{2} s (X_{1})^{2} ∣ \leq ∥ V_{n} - h s (X_{1}) ∥_{2} (∥ V_{n} ∥_{2} + h ∥ s ∥_{2}) \to 0$ , since the second factor is bounded. Thus [F12] makes $V_{n}^{2}$ uniformly integrable. For every $η > 0$ , [F8] and [F9] give $P^{\otimes n} (\max_{i \leq n} ∣ Y_{t_{n}} (X_{i}) ∣ > η) \leq η^{- 2} E [V_{n}^{2} 1_{{∣ V_{n} ∣ > η \sqrt{n}}}] \to 0$ , because the tail threshold for $V_{n}^{2}$ tends to infinity. Also $P^{\otimes n} (L_{n} = 0) \leq n β_{t_{n}} = o (1)$ by step 1.1. On the event all factors are positive and $\max_{i} ∣ Y_{t_{n}} (X_{i}) ∣ \leq 1$ , let $g (y) = 2 \log (1 + y / 2)$ ; its third derivative is $g^{'''} (y) = \frac{1}{2} (1 + y / 2)^{- 3}$ , so $∣ g^{'''} (y) ∣ \leq 4$ on $[- 1, 1]$ . [F11] gives $g (y) = y - \frac{1}{4} y^{2} + O (∣ y ∣^{3})$ there, with a uniform constant. Thus $\log q_{t_{n}} (X_{i}) = Y_{t_{n}} (X_{i}) - \frac{1}{4} Y_{t_{n}} (X_{i})^{2} + O (∣ Y_{t_{n}} (X_{i}) ∣^{3})$ . The sum of cubic terms is bounded by a constant times $\max_{i} ∣ Y_{t_{n}} (X_{i}) ∣ \sum_{i} Y_{t_{n}} (X_{i})^{2} = o_{P} (1)$ . Steps 3.2–4.1 now yield $\log L_{n} = h Δ_{n} - \frac{1}{2} h^{2} I + o_{P} (1)$ .

6.1F1F6F7F9F13F14F15F16F17step 1.1step 2.1step 5.1given∎

Decompose $P_{t_{n}} = (1 - α_{t_{n}}) {\overset{ˉ}{P}}_{t_{n}} + α_{t_{n}} S_{t_{n}}$ , where ${\overset{ˉ}{P}}_{t_{n}} = q_{t_{n}} P$ and, when $α_{t_{n}} > 0$ , $S_{t_{n}}$ is the normalized restriction to $S^{c}$ . Expanding the product mixture, the all- ${\overset{ˉ}{P}}_{t_{n}}$ term has weight $(1 - α_{t_{n}})^{n}$ and law ${\overset{ˉ}{P}}_{t_{n}}^{\otimes n}$ . If $α_{t_{n}} > 0$ , normalize the remaining mixture terms to a probability law $R_{n, t_{n}}$ ; then $P_{t_{n}}^{\otimes n} = (1 - α_{t_{n}})^{n} {\overset{ˉ}{P}}_{t_{n}}^{\otimes n} + (1 - (1 - α_{t_{n}})^{n}) R_{n, t_{n}}$ . When $α_{t_{n}} = 0$ , the two product laws are equal. Otherwise, this mixture identity and the event-supremum definition [F7] give $∥ P_{t_{n}}^{\otimes n} - {\overset{ˉ}{P}}_{t_{n}}^{\otimes n} ∥_{T V} \leq 1 - (1 - α_{t_{n}})^{n} \leq n α_{t_{n}}$ by finite subadditivity [F9]. Since $α_{t_{n}} = o (t_{n}^{2})$ and $n t_{n}^{2} = h^{2}$ for fixed $h > 0$ , this is $o (1)$ . For the asserted product-density identity, the coordinate factors $q_{t_{n}}$ are nonnegative measurable, so [F14] makes $L_{n} (x_{1}, \dots, x_{n}) = \prod_{i} q_{t_{n}} (x_{i})$ product-measurable. Define the measure $Q_{n} (A) = \int_{A} L_{n} d P^{\otimes n}$ by [F13]. Repeated Tonelli [F16] on a measurable rectangle $A_{1} \times \dots \times A_{n}$ gives $Q_{n} (A_{1} \times \dots \times A_{n}) = \prod_{i = 1}^{n} \int_{A_{i}} q_{t_{n}} d P = \prod_{i = 1}^{n} {\overset{ˉ}{P}}_{t_{n}} (A_{i})$ , where the one-coordinate identity is the definition ${\overset{ˉ}{P}}_{t_{n}} = q_{t_{n}} P$ . In particular, step 2.1 gives $Q_{n} (X^{n}) = (P q_{t_{n}})^{n} = 1$ . By [F15], ${\overset{ˉ}{P}}_{t_{n}}^{\otimes n}$ is also a probability and has the same rectangle values. The measurable rectangles form a pi-system (their intersections are coordinate intersections) generating the finite product sigma-algebra by [F14]; the constant exhaustion $X^{n}$ belongs to it and both measures have mass one there. Thus [F17] proves $Q_{n} = {\overset{ˉ}{P}}_{t_{n}}^{\otimes n}$ on every product-measurable event. Consequently $L_{n} P^{\otimes n} = {\overset{ˉ}{P}}_{t_{n}}^{\otimes n}$ , identifying exactly the proxy in the statement with the product law compared above. If $h = 0$ , both product laws equal $P^{\otimes n}$ and $L_{n} = 1$ . This proves every assertion.

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Sources

John Duchi, A few notes on contiguity, asymptotics, and local asymptotic normality, §§2 and 4.3.1 (2021) (standard reference, not scraped)
Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 22, PDF pp. 8–11 (2013) (standard reference, not scraped)