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LAN, Contiguity and Efficiency — Examples

1 · Prerequisites

2 · Summary

The known-variance normal location model has an exact Gaussian local likelihood ratio, an exact normal central sequence, and a regular sample mean. A two-point sequence with a shrinking base probability and fixed alternative probability shows that a likelihood-ratio limit with the wrong mean can fail contiguity despite every finite likelihood ratio having mean one.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Gaussian shift LAN and the regular sample mean

Statement

Let $X_{1}, \dots, X_{n}$ be IID $N (θ, 1)$ and take $θ_{0} = 0$ . This one-dimensional model is QMD at zero with score $s (x) = x$ and information $I = 1$ . For every fixed $h$ , the local laws $P_{n, h} = N (h / \sqrt{n}, 1)^{\otimes n}$ are exactly dominated by $P_{n, 0}$ and $\log \frac{d P_{n, h}}{d P_{n, 0}} (X_{1}, \dots, X_{n}) = h Δ_{n} - \frac{1}{2} h^{2}, Δ_{n} = \frac{1}{\sqrt{n}} \sum_{i = 1}^{n} X_{i} .$ The sample mean estimator is regular, with $\sqrt{n} ({\overset{ˉ}{X}}_{n} - \frac{h}{\sqrt{n}}) \Rightarrow N (0, 1)$ under $P_{n, h}$ for each fixed $h$ .

Facts & Assumptions

Given: AC; the standard normal density $ϕ (x) = (2 π)^{- 1 / 2} e^{- x^{2} / 2}$ ; independent observations; and the definitions of QMD, LAN, and regularity.

[A1]

The Axiom of Choice: AC is assumed by the normal-law and characteristic-function uniqueness suppliers.

[F1]

Standard normal and normal laws and The standard normal density has total mass one: The density $ϕ$ defines the probability law $N (0, 1)$ .

[F2]

Characteristic function of a normal law: If $X \sim N (m, σ^{2})$ , then $E X = m$ , $Var (X) = σ^{2}$ , and its characteristic function is $\exp (i m t - σ^{2} t^{2} / 2)$ .

[F3]

Characteristic functions under affine maps and independent sums: The characteristic function of an independent sum is the product of the component characteristic functions, with the corresponding affine scaling.

[F4]

Uniqueness of a law from its characteristic function: Probability laws with the same characteristic function are equal.

[F5]

Dominated convergence: A pointwise limit dominated in absolute value by one integrable function passes through the Lebesgue integral.

[F6]

The exponential function is smooth and $(\exp)^{'} = \exp$ and The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : The real exponential differentiates by itself, and derivatives of compositions obey the chain rule.

[F7]

The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive: Integrating a continuous derivative along an interval recovers the endpoint difference.

[F9]

Quadratic-mean differentiability at a parameter, Local asymptotic normality for local experiments, and Regular estimator under local alternatives: QMD is the square-root density $L^{2}$ expansion, LAN is the local likelihood expansion with Gaussian central sequence, and regularity requires one common centered limit law for each fixed local $h$ .

[F10]

Quadratic-mean differentiability implies iid LAN: A QMD IID model has the LAN expansion with its score and information, using mean-one proxies when needed.

[F11]

Likelihood densities: The Gaussian densities used here are measurable densities with respect to Lebesgue measure and give the stated probability laws by integration.

Proof

Proof technique: verify square-root differentiability and compute the local likelihood ratio exactly.

1.1F1F2F5F6F7F11givenalgebra

Let $g_{θ} (x) = \sqrt{ϕ (x - θ)} = (2 π)^{- 1 / 4} e^{- (x - θ)^{2} / 4}$ . By [F6] its derivative is $\partial_{θ} g_{θ} (x) = \frac{1}{2} (x - θ) g_{θ} (x)$ , so $\partial_{θ} g_{0} (x) = \frac{1}{2} x g_{0} (x)$ . For $∣ θ ∣ \leq 1$ , $∣ \partial_{θ} g_{θ} (x) ∣ \leq C (1 + ∣ x ∣) e^{- ((∣ x ∣ - 1)_{+})^{2} / 4} = : G (x),$ whose square is integrable: it is bounded on $[- 2, 2]$ , and for $∣ x ∣ \geq 2$ it is bounded by $C^{'} (1 + x^{2}) e^{- x^{2} / 8}$ . Substituting $x = 2 y$ reduces the tail integral to a constant times $\int (1 + 4 y^{2}) e^{- y^{2} / 2} d y$ , finite by the second moment statement in [F2]. By [F7], $(g_{t} - g_{0}) / t = \int_{0}^{1} \partial_{θ} g_{r t} d r$ ; this converges pointwise to $\frac{1}{2} x g_{0}$ . Its difference is bounded by $2 G$ , so [F5] applied to the squared difference gives convergence in $L^{2} (d x)$ . Thus the QMD residual has norm $o (∣ t ∣)$ , which is exactly the scalar QMD condition.

1.2F1F2F3F4F9F11givenalgebra

The density ratio is $\prod_{i = 1}^{n} \frac{ϕ (X_{i} - h / \sqrt{n})}{ϕ (X_{i})} = \exp ⁣ (\frac{h}{\sqrt{n}} \sum_{i = 1}^{n} X_{i} - \frac{h^{2}}{2}) .$ Each density is positive, so this is a full mean-one likelihood ratio. By [F2]–[F4], the characteristic function of $Δ_{n}$ is $\prod_{i = 1}^{n} \exp (- \frac{t^{2}}{2 n}) = e^{- t^{2} / 2},$ so $Δ_{n} \sim N (0, 1)$ exactly. Comparing with [F9] proves exact LAN with $K = 1$ and no remainder.

2.1F2F10step 1.1algebra

The score is $s (x) = x$ . By [F2], $E_{0} s = 0$ and $E_{0} s^{2} = 1$ , so the information is $I = 1$ . By [F10], the general QMD-to-LAN result applies and predicts the same score and information; the direct computation below also shows the proxy is the exact alternative law.

3.1A1F2F3F4F9givenalgebra∎

Under $P_{n, h}$ , the variables $X_{i} - h / \sqrt{n}$ are IID $N (0, 1)$ . The same characteristic-function product using [F3]–[F4] gives $\sqrt{n} ({\overset{ˉ}{X}}_{n} - \frac{h}{\sqrt{n}}) = \frac{1}{\sqrt{n}} \sum_{i} (X_{i} - h / \sqrt{n}) \sim N (0, 1)$ exactly, which implies the displayed convergence. The limiting law does not depend on fixed $h$ , which is regularity by [F9]. The calculation includes $h = 0$ and every finite $h$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Binary laws with a rare event are not contiguous

Statement

For $n \geq 2$ , define probability laws on ${0, 1}$ by $P_{n} {1} = 1 / n, Q_{n} {1} = 1 / 2.$ Then $Q_{n}$ is not contiguous with respect to $P_{n}$ . The likelihood ratios $L_{n} = d Q_{n} / d P_{n}$ exist, but their $P_{n}$ -distributional limit has mean less than one; therefore the matching-mean hypothesis in A mean-one likelihood-ratio limit implies contiguity is essential.

Facts & Assumptions

Given: The two-point measurable space, the probability laws above, and the event-supremum definition of contiguity.

[F1]

Contiguous sequences of probability measures: $Q_{n} ◃ P_{n}$ would require $Q_{n} (A_{n}) \to 0$ whenever $P_{n} (A_{n}) \to 0$ .

[F2]

A mean-one likelihood-ratio limit implies contiguity: A nonnegative likelihood ratio with mean one and a nonnegative distributional limit of mean one gives contiguity.

Proof

technique · exhibit a rare event and compute the likelihood ratio

1.1F1givenalgebra

Let $A_{n} = {1}$ . Then $P_{n} (A_{n}) = 1 / n \to 0$ , while $Q_{n} (A_{n}) = 1 / 2$ for every $n$ . By [F1], this disproves $Q_{n} ◃ P_{n}$ .

2.1F2givenalgebra∎

On the two atoms, the density is $L_{n} (1) = \frac{Q_{n} {1}}{P_{n} {1}} = \frac{n}{2}, L_{n} (0) = \frac{Q_{n} {0}}{P_{n} {0}} = \frac{n}{2 (n - 1)} .$ Under $P_{n}$ , the value $n / 2$ occurs with probability $1 / n$ and the other value tends to $1 / 2$ , so $L_{n} \Rightarrow 1 / 2$ . Each $E_{P_{n}} L_{n} = 1$ , but $E (1 / 2) = 1 / 2 \neq 1$ ; hence the matching-mean condition of [F2] fails.

LAN, Contiguity and Efficiency — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Gaussian shift LAN and the regular sample mean

Statement

Facts & Assumptions

Proof

Binary laws with a rare event are not contiguous

Statement

Facts & Assumptions

Proof

Sources