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LAN, Contiguity and Efficiency

1 · Prerequisites

2 · Summary

This page defines contiguity and pointwise LAN using mean-one likelihood proxies, then proves the matching-mean criterion for contiguity and Le Cam's third lemma. For dominated iid models, QMD yields LAN after the local alternative is restricted to the base-law support and normalized; the proof separately controls singular and zero-overlap mass, so it never asserts a full Radon–Nikodym derivative when one may not exist. The page also defines regularity under each fixed local alternative and proves the convolution bound under the stated LAN and regularity hypotheses.

The local asymptotic minimax theorem uses only pointwise LAN. Its proof passes through finite priors, approximates diffuse Gaussian Bayes values by finite-support priors, and approximates a bounded Borel bowl-shaped loss from below by bounded Lipschitz minorants. A separate posterior approximation lemma states an explicit compact-uniform likelihood-field condition; that stronger condition is not used by the minimax proof. Brunn–Minkowski and Gaussian centering are proved locally as supporting results.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Contiguous sequences of probability measures

Definition

Let $(X, A)$ be one measurable space and let $P_{n}, Q_{n}$ be probability measures on it (Probability measures and probability spaces). The sequence $(Q_{n})$ is contiguous with respect to $(P_{n})$ , written $Q_{n} ◃ P_{n}$ , if for every sequence $A_{n} \in A$ , $P_{n} (A_{n}) ⟶ 0 ⟹ Q_{n} (A_{n}) ⟶ 0.$ The sequences are mutually contiguous, written $Q_{n} ◃ ▹ P_{n}$ , if both $Q_{n} ◃ P_{n}$ and $P_{n} ◃ Q_{n}$ .

Contiguity is a relation between sequences of laws, not absolute continuity at each finite $n$ . It says that events negligible under one sequence remain negligible under the other in the indicated direction.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The boundary of a convex set is Lebesgue null

Statement

Assume countable choice. For every convex set $C \subseteq R^{d}$ , its topological boundary $\partial C = \overline{C} ∖ C^{\circ}$ has $d$ -dimensional Lebesgue measure zero. If $C$ has empty interior, then $C$ itself is contained in a proper affine subspace and is Lebesgue null.

Facts & Assumptions

Given: Countable choice, a finite integer $d \geq 1$ , and a convex set $C \subseteq R^{d}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable choice permits the countable approximations used by the Lebesgue measure suppliers.

[F1]

Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null: Every proper linear subspace of $R^{d}$ is Lebesgue null.

[F2]

The geometric simplex spanned by affinely independent vertices, Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ , and The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : Affine independence of $c_{0}, x_{1}, \dots, x_{d}$ means the differences $x_{i} - c_{0}$ are linearly independent. Their closed simplex contains the image of ${t_{i} > 0 : \sum_{i = 1}^{d} t_{i} < 1}$ under $t \mapsto c_{0} + \sum_{i} t_{i} (x_{i} - c_{0})$ . The linear part and its inverse are continuous, so this image is a nonempty open set; balls use the Euclidean norm.

[F3]

[F4]

Measure of a set difference when the smaller set has finite measure and Measures are monotone: For nested measurable sets of finite measure, the measure of the difference is the difference of their measures, and measure is monotone.

[F5]

Finite and countable subadditivity of measures: A countable union of null measurable sets is null.

[F6]

A finite-dimensional normed subspace is closed and Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: Finite-dimensional linear subspaces are closed, and translation preserves Lebesgue measurability and measure.

[F7]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line and Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure: A closed bounded subset of $R^{d}$ is compact and has finite Lebesgue measure.

Proof

technique · affine-hull reduction and homothetic contraction

1.1F1F2F6givenchoosealgebra

If $C = \emptyset$ , the claim is immediate. Otherwise choose $c_{0} \in C$ and let $W$ be the linear span of ${x - c_{0} : x \in C}$ . If $W = R^{d}$ , finite-dimensional linear algebra selects $d$ linearly independent differences $x_{i} - c_{0}$ from this spanning family. By definition these $d + 1$ points are affinely independent; their simplex lies in $C$ by convexity and contains a nonempty open set by [F2], contradicting empty interior. Thus $W$ is proper and $C \subseteq c_{0} + W$ . By [F1], $W$ is null; [F6] makes $c_{0} + W$ closed and null. Since $\overline{C} \subseteq c_{0} + W$ , both $C$ and its boundary are null. This proves the empty-interior case.

1.2A1givenalgebra

Suppose $C$ has nonempty interior. Choose $x_{0} \in C^{\circ}$ and $ρ > 0$ with $B (x_{0}, ρ) \subseteq C$ . If $y \in \overline{C}$ and $0 < t < 1$ , [A1] selects $y_{j} \in C$ with $y_{j} \to y$ . Convexity gives $B (x_{0} + t (y_{j} - x_{0}), (1 - t) ρ) \subseteq C .$ The centers converge to $x_{0} + t (y - x_{0})$ , so that limit point has a ball of radius $(1 - t) ρ / 2$ contained in $C$ . Therefore $x_{0} + t (\overline{C} - x_{0}) \subseteq C^{\circ}$ .

2.1F3F4F6F7step 1.2algebra

For each integer $r$ large enough that $K_{r} = \overline{C} \cap \overline{B} (0, r)$ contains $x_{0}$ , [F7] makes $K_{r}$ compact with finite measure. The contraction $x_{0} + t (K_{r} - x_{0})$ lies in $C^{\circ}$ by step 1.2 and in $K_{r}$ by convexity. Write this contraction as $t K_{r} + (1 - t) x_{0}$ . Applying [F3] to $t I_{d}$ and then translation invariance [F6] gives its measure as $t^{d} λ_{d} (K_{r})$ . Hence $\partial C \cap K_{r} \subseteq K_{r} ∖ (x_{0} + t (K_{r} - x_{0})),$ and [F4] gives $λ_{d} (\partial C \cap K_{r}) \leq (1 - t^{d}) λ_{d} (K_{r}) .$ Let $t ↑ 1$ to obtain $λ_{d} (\partial C \cap K_{r}) = 0$ .

3.1A1F5step 2.1∎

The bounded sets $K_{r}$ cover $\overline{C}$ as $r$ ranges over sufficiently large integers. Therefore $\partial C$ is a countable union of the null sets in step 2.1 and is null by [F5].

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Local asymptotic normality for local experiments

Definition

Let $P_{n, h}$ be probability laws on a common measurable space for each integer $n \geq 1$ and local parameter $h \in R^{d}$ , where $d < \infty$ and $P_{n, 0}$ is the base law. Suppose that for every fixed $h$ we are supplied a nonnegative measurable function $L_{n, h}$ with $\int L_{n, h} d P_{n, 0} = 1.$ It defines an absolutely continuous proxy ${\overset{ˉ}{P}}_{n, h} (A) : = \int_{A} L_{n, h} d P_{n, 0} .$ The actual local law is allowed to have a singular part; the approximation condition below is what connects it to this proxy.

The experiments are locally asymptotically normal (LAN) at $h = 0$ with information matrix $K$ and central sequence $Δ_{n}$ if $K$ is a symmetric positive semidefinite $d \times d$ matrix, $Δ_{n}$ is an $R^{d}$ -valued statistic, $Δ_{n} \Rightarrow N_{d} (0, K) under P_{n, 0},$ and for every fixed $h \in R^{d}$ , $∥ P_{n, h} - {\overset{ˉ}{P}}_{n, h} ∥_{T V} \to 0, \log L_{n, h} = h^{T} Δ_{n} - \frac{1}{2} h^{T} K h + o_{P_{n, 0}} (1),$ where $∥ \cdot ∥_{T V}$ is the event-supremum total variation distance (Total variation distance for probability laws) and $o_{P_{n, 0}} (1)$ means convergence to zero in $P_{n, 0}$ -probability. We set $\log 0 = - \infty$ ; the finite-valued expansion entails $P_{n, 0} (L_{n, h} = 0) \to 0$ . Convergence in distribution is as in Convergence in distribution of random elements, and $N_{d} (0, K)$ allows singular $K$ (Multivariate normal law, including singular covariance).

If $P_{n, h} = {\overset{ˉ}{P}}_{n, h}$ , this is the usual dominated LAN comparison. The proxy formulation also covers QMD paths whose local alternatives are not absolutely continuous with respect to the base law, provided the singular product mass tends to zero in total variation. This definition is pointwise in each fixed $h$ ; it does not assert uniformity over bounded sets of $h$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Brunn–Minkowski inequality for compact convex sets

Statement

Assume countable choice. Let $d \geq 1$ and let $A, B \subseteq R^{d}$ be nonempty compact convex sets. With $A + B = {a + b : a \in A, b \in B}$ and $λ_{d}$ denoting Lebesgue measure, $λ_{d} (A + B)^{1 / d} \geq λ_{d} (A)^{1 / d} + λ_{d} (B)^{1 / d} .$ Zero-volume and lower-dimensional sets are included.

Facts & Assumptions

Given: Countable choice, an integer $d \geq 1$ , and nonempty compact convex sets $A, B \subseteq R^{d}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable choice is sufficient for the Lebesgue-measure change-of-variables and Tonelli suppliers used here; full AC is not assumed.

[F1]

Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ : Lebesgue volume of a measurable set is its measure $λ_{d}$ .

[F2]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: For a nonnegative measurable function, integration over $R^{d}$ equals iterated integration over its one-dimensional slices.

[F3]

The weighted arithmetic-geometric mean inequality for real weights: For positive $x, y$ and $0 \leq λ \leq 1$ , $(1 - λ) + λ x \geq x^{λ}$ .

[F4]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: For a $C^{1}$ diffeomorphism of open sets, the nonnegative Lebesgue integral transforms with the absolute Jacobian.

[F5]

[F6]

Measures are monotone and Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: Lebesgue measure is monotone and translation invariant.

[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 and Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ : On each compact interval inside a projection interval, the fundamental theorem gives the derivative of the cumulative integral as its continuous integrand. The inverse theorem then gives $(F_{B}^{- 1})^{'} (u) = 1 / g (F_{B}^{- 1} (u))$ ; this derivative is continuous because $g$ is continuous and positive and $F_{B}^{- 1}$ is continuous.

[F8]

Monotone convergence for the integral: Increasing nonnegative integrals converge to the integral of their pointwise limit.

[F9]

Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure: Every bounded measurable subset of $R^{d}$ has finite Lebesgue measure.

[F10]

The boundary of a convex set is Lebesgue null: A convex set with empty interior has zero $d$ -dimensional Lebesgue measure.

[F11]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: Compact subsets of finite-dimensional Euclidean space are closed and bounded, and closed bounded subsets are compact. In particular, $A \times B$ is closed and bounded in $R^{2 d}$ , hence compact; continuous images of compact sets are compact, so $A + B$ , positive dilations of $A$ and $B$ , and every weighted convex sum used below are compact and convex (an open cover of the image pulls back to one of the compact domain, which has a finite subcover). When $d \geq 2$ , for a compact $C \subseteq R \times R^{d - 1}$ , each coordinate projection is compact (and is an interval when $C$ is convex) and each slice $C_{z} = {y : (z, y) \in C}$ is closed and bounded, hence compact. Thus these bodies and slices are Borel and have well-defined finite Lebesgue volumes; applying [F2] to $1_{C}$ makes $z \mapsto λ_{d - 1} (C_{z})$ measurable and gives its integral as $λ_{d} (C)$ .

Proof

Proof technique: induction on dimension using convex slices and a one-dimensional monotone transport.

1.1F1F6F9F11givenalgebra

If one of $A, B$ has zero volume, say $λ_{d} (A) = 0$ , choose $a \in A$ . By [F11], $A + B$ is compact and measurable. Then $a + B \subseteq A + B$ , so [F6] gives $λ_{d} (A + B) \geq λ_{d} (B)$ ; the stated inequality follows after taking $d$ th roots. If both volumes vanish the right side is zero. Thus assume both volumes are positive.

2.1A1F1F5F9F11step 1.1givenalgebra

Put $a = λ_{d} (A)^{1 / d}$ , $b = λ_{d} (B)^{1 / d}$ , $A^{'} = a^{- 1} A$ , $B^{'} = b^{- 1} B$ , and $λ = b / (a + b)$ . By [F5], $λ_{d} (A^{'}) = λ_{d} (B^{'}) = 1$ ; [F11] makes these dilates and their convex combinations compact. It suffices to show that every convex combination $(1 - λ) A^{'} + λ B^{'}$ has volume at least one, because $A + B = (a + b) ((1 - λ) A^{'} + λ B^{'}) .$ The cases $λ = 0, 1$ are immediate, so take $0 < λ < 1$ and rename $A^{'}, B^{'}$ as $A, B$ .

3.1A1F1F2F9F10F11step 2.1givenbaseihinduction

We prove the normalized claim by induction on $d$ . In dimension one, $A$ and $B$ are intervals of length one, so their convex combination is an interval of length $(1 - λ) + λ = 1$ . Now let $d \geq 2$ and assume the normalized claim in dimension $d - 1$ . The normalization calculation in step 2.1 is dimension-independent, so this induction hypothesis also gives the full weighted inequality in dimension $d - 1$ for positive-volume compact convex sets. By [F10], positive volume implies each body has nonempty interior and contains an interior ball. By [F11], their coordinate projections are compact intervals and their slices are compact and measurable; [F2] makes the slice-volume functions measurable. Project each convex body onto the first coordinate; its projection is a nondegenerate compact interval. For an interior coordinate $x$ , let $A_{x} \subseteq R^{d - 1}$ be its slice and $f (x) = λ_{d - 1} (A_{x})$ . Define $B_{y}$ and $g (y)$ likewise. These slice volumes are positive and finite in the open projection intervals. Write the projections as $[a_{-}, a_{+}]$ and $[b_{-}, b_{+}]$ . An interior ball in $A$ has center with first coordinate $x_{0}$ . If $a_{-} < x < x_{0}$ , choose $p \in A$ with first coordinate $p_{1} < x$ and combine $p$ with the ball centered at $x_{0}$ ; the convex combination with weight $λ = (x - p_{1}) / (x_{0} - p_{1}) \in (0, 1)$ on the ball lies in the slice at $x$ and contains a $(d - 1)$ -ball of radius $λ ρ > 0$ . If $x_{0} < x < a_{+}$ , choose $p_{1} > x$ and use weight $(p_{1} - x) / (p_{1} - x_{0})$ on the interior ball. At $x = x_{0}$ , the original ball supplies a positive-radius slice. The same argument applies to $B$ .

4.1F2F11step 3.1step 2.1inductionalgebra

Convexity gives $A_{(1 - t) x_{0} + t x_{1}} \supseteq (1 - t) A_{x_{0}} + t A_{x_{1}}$ for interior $x_{0}, x_{1}$ . The compact slices in [F11] let the induction hypothesis apply to these sums. It therefore makes $f^{1 / (d - 1)}$ concave on the open projection interval; likewise $g^{1 / (d - 1)}$ . To see continuity of $ψ = f^{1 / (d - 1)}$ , fix an interior point $x_{0}$ and $a < x_{0} < b$ in the interval. For $x ↓ x_{0}$ , concavity gives the chord lower bound $ψ (x) \geq \frac{b - x}{b - x_{0}} ψ (x_{0}) + \frac{x - x_{0}}{b - x_{0}} ψ (b) .$ Writing $x_{0} = η a + (1 - η) x$ with $η = (x - x_{0}) / (x - a)$ gives the upper bound $ψ (x) \leq \frac{ψ (x_{0}) - η ψ (a)}{1 - η} .$ Both bounds tend to $ψ (x_{0})$ ; exchanging left and right proves continuity from the other side. Thus $f, g$ are continuous and strictly positive on their open intervals. By [F2], $\int f = \int g = 1$ .

5.1F6F7F9step 3.1step 2.1step 4.1algebra

Define the cumulative functions $F_{A} (x) = \int_{a_{-}}^{x} f (t) d t$ and $F_{B} (y) = \int_{b_{-}}^{y} g (t) d t$ on their projection intervals. Since $A, B$ lie in bounded boxes, all their slices lie in one fixed bounded box in $R^{d - 1}$ ; [F6, F9] therefore bound $f, g$ uniformly. Their cumulative integrals consequently extend continuously to the projection endpoints with values zero and one, using $\int f = \int g = 1$ from step 4.1. Positivity of $f, g$ on the open intervals makes $F_{A}, F_{B}$ strictly increasing there, so they map those intervals onto $(0, 1)$ . On every compact interior subinterval [F7] gives derivatives $F_{A}^{'} = f$ and $F_{B}^{'} = g$ . The inverse theorem then gives that $φ = F_{B}^{- 1} \circ F_{A}$ is $C^{1}$ on the open interval and $φ^{'} (x) = f (x) / g (φ (x)) .$ Let $z (x) = (1 - λ) x + λ φ (x)$ ; then $z^{'} (x) = (1 - λ) + λ f (x) / g (φ (x)) > 0$ .

6.1F2F3F11step 5.1step 4.1step 2.1step 3.1inductionalgebra

Let $C = (1 - λ) A + λ B$ , which is compact by [F11], and let $c_{z}$ be the $(d - 1)$ -dimensional volume of its compact slice at coordinate $z$ . The slice-volume function is measurable by [F2] and [F11]. The slice of $C$ at $z (x)$ contains $(1 - λ) A_{x} + λ B_{φ (x)}$ . The induction hypothesis yields $c_{z (x)} \geq ((1 - λ) f (x)^{1 / (d - 1)} + λ g (φ (x))^{1 / (d - 1)})^{d - 1} .$ Set $r = g (φ (x)) / f (x) > 0$ . Multiplying this bound by $z^{'} (x)$ and using [F3] twice gives $c_{z (x)} z^{'} (x) \geq f (x) ((1 - λ) + λ r^{1 / (d - 1)})^{d - 1} ((1 - λ) + λ / r) \geq f (x) .$ The last inequality follows by applying [F3] to both positive brackets.

7.1F1F2F4F7F8F11step 6.1step 5.1step 4.1step 2.1step 3.1givenalgebra

Restrict $x$ to an open interval $J$ strictly inside $(a_{-}, a_{+})$ . The map $z$ is $C^{1}$ with positive derivative by step 5.1, and [F7] makes it a $C^{1}$ diffeomorphism onto its open image $z (J)$ . Thus [F4] and step 6.1 give $λ_{d} (C) = \int c_{z} d z \geq \int_{z (J)} c_{z} d z = \int_{J} c_{z (x)} z^{'} (x) d x \geq \int_{J} f (x) d x .$ Here $c_{z}$ is measurable and the equality is the one-dimensional change of variables. For all sufficiently large integers $m$ , take $J_{m} = (a_{-} + 1 / m, a_{+} - 1 / m)$ ; these intervals increase to $(a_{-}, a_{+})$ . By [F8] and $\int f = 1$ , $λ_{d} (C) \geq 1$ . This proves the normalized claim in dimension $d$ .

8.1F1step 1.1step 2.1step 3.1step 7.1algebradischarge-induction∎

The induction proves the normalized claim in every finite dimension. Undoing the normalization in step 2.1 gives $λ_{d} (A + B)^{1 / d} \geq a + b = λ_{d} (A)^{1 / d} + λ_{d} (B)^{1 / d},$ including the zero-volume cases from step 1.1.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A mean-one likelihood-ratio limit implies contiguity

Statement

For each $n$ , let $P_{n}, Q_{n}$ be probability measures on one measurable space, and let $L_{n} \geq 0$ satisfy $Q_{n} (A) = \int_{A} L_{n} d P_{n}$ for every measurable $A$ . Suppose $E_{P_{n}} L_{n} = 1$ , $L_{n} \Rightarrow L$ under $P_{n}$ , $L \geq 0$ , and $E L = 1$ . Then $(Q_{n})$ is contiguous with respect to $(P_{n})$ .

Facts & Assumptions

Given: Probability measures $P_{n}, Q_{n}$ , mean-one densities $L_{n}$ , and the stated nonnegative distributional limit $L$ with matching mean.

[F1]

Contiguous sequences of probability measures: $Q_{n} ◃ P_{n}$ means $P_{n} (A_{n}) \to 0$ implies $Q_{n} (A_{n}) \to 0$ for every event sequence $A_{n}$ .

[F2]

Convergence in distribution of random elements: $L_{n} \Rightarrow L$ means expectations converge for every bounded continuous function.

Proof

technique · direct uniform-integrability argument

1.1F2givenalgebra

For fixed $M > 0$ , $x \mapsto x \land M$ is bounded and continuous on $[0, \infty)$ , so [F2] and the mean assumptions give $E (L_{n} - M)_{+} = E L_{n} - E (L_{n} \land M) ⟶ E L - E (L \land M) = E (L - M)_{+} .$

2.1step 1.1givenalgebra

If $x > 2 M$ , then $x \leq 2 (x - M)$ ; hence ${lim sup}_{n} E [L_{n} 1_{{L_{n} > 2 M}}] \leq 2 E (L - M)_{+} .$ The right side tends to zero as $M \to \infty$ because $L$ is integrable. Each of the finitely many initial integrable $L_{n}$ also has vanishing tail integral, so $(L_{n})$ is uniformly integrable.

3.1F1F2step 2.1givenalgebra∎

For any event sequence $A_{n}$ and any $M > 0$ , $Q_{n} (A_{n}) = E [L_{n} 1_{A_{n}}] \leq M P_{n} (A_{n}) + E [L_{n} 1_{{L_{n} > M}}] .$ If $P_{n} (A_{n}) \to 0$ , the first term tends to zero for fixed $M$ , and uniform integrability makes the limsup of the second term arbitrarily small as $M \to \infty$ . Thus $Q_{n} (A_{n}) \to 0$ , which is $Q_{n} ◃ P_{n}$ by [F1].

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Regular estimator under local alternatives

Definition

In a local experiment ${P_{n, h} : h \in R^{d}}$ at $h = 0$ (Local asymptotic normality for local experiments), let $T_{n}$ be an $R^{d}$ -valued statistic. It is regular at zero with limit law $W$ if, for every fixed $h \in R^{d}$ , $T_{n} - h \Rightarrow W under P_{n, h},$ where the same probability law $W$ works for all $h$ . In a parametric model with true parameter $θ_{0}$ , this is the definition for $T_{n} = \sqrt{n} ({\hat{θ}}_{n} - θ_{0})$ and local laws at $θ_{0} + h / \sqrt{n}$ ; equivalently the centered error $\sqrt{n} ({\hat{θ}}_{n} - θ_{0} - h / \sqrt{n})$ has a limit independent of the fixed local displacement $h$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Compact-uniform LAN approximates compact-prior posteriors

Statement

Assume AC.

Let $H \subset R^{d}$ be compact, let $π$ be a probability measure on $H$ , and let $P_{n, 0}$ be base laws. Suppose $L_{n} (h, x)$ is jointly measurable in $(h, x)$ , nonnegative, and satisfies $\int L_{n} (h, x) P_{n, 0} (d x) = 1$ for every $h \in H$ . Let $Δ_{n} \Rightarrow Z \sim N_{d} (0, K)$ under $P_{n, 0}$ , where $K$ is positive semidefinite, and put $g_{h} (z) = \exp (h^{T} z - \frac{1}{2} h^{T} K h) .$ Assume the random variable $a_{n} (x) : = \sup_{h \in H} ∣ L_{n} (h, x) - g_{h} (Δ_{n} (x)) ∣$ is measurable and $E_{P_{n, 0}} a_{n} \to 0$ . Let the prior-predictive law be $M_{n} (d x) = m_{n} (x) P_{n, 0} (d x)$ with $m_{n} (x) = \int_{H} L_{n} (h, x) π (d h)$ , and define posteriors $Π_{n} (d h ∣ x) = \frac{L_{n} (h, x) π (d h)}{m_{n} (x)}, Π^{G} (d h ∣ z) = \frac{g_{h} (z) π (d h)}{m (z)}, m (z) = \int_{H} g_{h} (z) π (d h),$ where an arbitrary probability measure is assigned if a denominator is zero. Then $E_{M_{n}} \int_{H} ∣ \frac{L_{n} (h, X)}{m_{n} (X)} - \frac{g_{h} (Δ_{n} (X))}{m (Δ_{n} (X))} ∣ π (d h) ⟶ 0.$ In particular the posterior laws converge in expected total variation, because their event-probability differences are bounded by this $L^{1} (π)$ distance. The convention for total variation is Total variation distance for probability laws.

This lemma assumes an $L^{1}$ uniform likelihood-field approximation. Pointwise LAN alone does not supply that hypothesis.

Facts & Assumptions

Given: AC; the compact set and prior, the mean-one likelihood field, the central sequence and its Gaussian limit, and the measurable $L^{1}$ uniform approximation above.

[A1]

The Axiom of Choice: AC is assumed for the multivariate normal law.

[F1]

Convergence in distribution of random elements: Weak convergence of $Δ_{n}$ gives convergence of expectations for bounded continuous functions.

[F2]

Dominated convergence: For a nonnegative integrable $U$ , $E (U - Q)_{+} \to 0$ as $Q \to \infty$ , since $(U - Q)_{+} \leq U$ .

[F3]

Multivariate normal law, including singular covariance and Standard normal and normal laws: For $Z \sim N_{d} (0, K)$ , $h^{T} Z \sim N (0, h^{T} K h)$ ; completing the square in the standard normal density gives $E \exp (h^{T} Z) = \exp (h^{T} K h / 2)$ , so $E g_{h} (Z) = 1$ .

[F4]

Measurability of integration against a kernel: Integrating a jointly measurable nonnegative function against the prior gives measurable predictive densities and posterior integrals.

[F5]

Markov's inequality for random variables: For nonnegative integrable $U$ and $b > 0$ , $P (U > b) \leq E U / b$ .

[F6]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: A nonnegative jointly measurable function may be integrated in either order.

[F7]

Continuous mapping theorem: Applying a continuous map to weakly convergent random elements preserves convergence in distribution.

[F8]

Dominated convergence: Pointwise convergence under a common integrable majorant passes through the integral.

[F9]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ z ∥_{2}$ is the Euclidean norm.

[F10]

Weakly convergent sequences are tight: A weakly convergent sequence of probability laws on $R^{d}$ is tight.

Proof

Proof technique: normalize the two posterior densities and control their denominators on tight central-sequence sets.

1.1A1F1F2F3F4F6F7F8givenalgebra

By [F4], integration against the prior makes $m_{n}$ measurable; [F6] gives $E_{P_{n, 0}} m_{n} = 1$ . Also $∣ m_{n} (x) - m (Δ_{n} (x)) ∣ \leq a_{n} (x),$ so $E_{P_{n, 0}} m (Δ_{n}) \to 1$ . The function $m$ is continuous: if $z_{j} \to z$ , then $g_{h} (z_{j}) \to g_{h} (z)$ for each $h$ , and compactness of $H$ gives a common finite bound on $g_{h} (z_{j})$ ; [F8] passes the limit through the prior integral. Thus [F1] and [F7] give $m (Δ_{n}) \Rightarrow m (Z)$ . By [F3], $E m (Z) = 1$ . Write $U_{n} = m (Δ_{n})$ and $U = m (Z)$ , and discard finitely many indices so $E U_{n} < \infty$ . For $Q > 0$ , weak convergence of the bounded continuous cutoff and convergence of the means give $E (U_{n} - Q)_{+} = E U_{n} - E (U_{n} \land Q) \to E U - E (U \land Q) = E (U - Q)_{+}$ . Since $u 1_{{u > 2 Q}} \leq 2 (u - Q)_{+}$ , the limsup of the tail expectations is at most $2 E (U - Q)_{+}$ , which tends to zero by [F2]. The finitely many initial integrable terms also have vanishing tails; thus $(U_{n})$ is uniformly integrable on this eventual index range.

1.2F9givenalgebra

Fix $R > 0$ . On the compact set ${z : ∥ z ∥_{2} \leq R}$ , the continuous function $m (z)$ has a positive minimum $c_{R}$ , since each $g_{h} (z) > 0$ and $π$ is a probability. On the event $∥ Δ_{n} ∥_{2} \leq R$ and $a_{n} \leq c_{R} / 2$ , one has $m_{n} \geq c_{R} / 2$ . For the two posterior densities with respect to $π$ , $\int_{H} ∣ \frac{L_{n} (h, x)}{m_{n} (x)} - \frac{g_{h} (Δ_{n} (x))}{m (Δ_{n} (x))} ∣ π (d h) \leq \frac{2 a_{n} (x)}{m_{n} (x)} \leq \frac{4 a_{n} (x)}{c_{R}} .$ Thus the posterior total variation distance is at most $4 a_{n} / c_{R}$ on this event.

2.1F1F2F10step 1.1givenalgebra

The sequence $Δ_{n}$ is tight by [F10] because it converges in distribution. For any $Q > 0$ , $E [m (Δ_{n}) 1_{{∥ Δ_{n} ∥_{2} > R}}] \leq E [m (Δ_{n}) 1_{{m (Δ_{n}) > Q}}] + Q P (∥ Δ_{n} ∥_{2} > R) .$ First choose $Q$ large using uniform integrability from step 1.1, then $R$ large using tightness. Since $m_{n} \leq m (Δ_{n}) + a_{n}$ and $E a_{n} \to 0$ , the predictive probability of $∥ Δ_{n} ∥_{2} > R$ is correspondingly small.

3.1F4F5step 1.2step 2.1givenalgebra

On ${∥ Δ_{n} ∥_{2} \leq R}$ , $m (Δ_{n}) \leq M_{R}$ for some finite $M_{R}$ . By [F5], $M_{n} (∥ Δ_{n} ∥_{2} \leq R, a_{n} > c_{R} / 2) \leq M_{R} \frac{2 E a_{n}}{c_{R}} + E a_{n} \to 0,$ using $m_{n} \leq m + a_{n}$ . On the complement of the exceptional events established in the preceding two steps, step 1.2 bounds the posterior $L^{1} (π)$ distance by $4 a_{n} / c_{R}$ , and $m_{n} \leq M_{R} + c_{R} / 2$ . Its predictive expectation is therefore at most $4 (M_{R} + c_{R} / 2) E a_{n} / c_{R} \to 0$ . The distance is at most two on the exceptional events, whose predictive probabilities vanish as first $n \to \infty$ and then $R \to \infty$ . This proves the first conclusion.

4.1F1step 1.1F3givenalgebra∎

For bounded continuous $f$ , the predictive law satisfies $E_{M_{n}} f (Δ_{n}) = E_{P_{n, 0}} [m_{n} f (Δ_{n})] = E_{P_{n, 0}} [m (Δ_{n}) f (Δ_{n})] + o (1) .$ For a cutoff $Q$ , $(m \land Q) f$ is bounded and continuous, so its expectation converges by [F1]. The tail error is at most $∥ f ∥_{\infty} E [m (Δ_{n}) 1_{{m (Δ_{n}) > Q}}]$ , which tends uniformly to zero as $Q \to \infty$ by step 1.1. Thus the limit is $E [m (Z) f (Z)]$ . This is the Gaussian-shift mixture with base law $N_{d} (0, K)$ and density $m (z)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A centered Gaussian minimizes bounded symmetric quasi-convex loss

Statement

Assume AC. Let $X \sim N_{d} (m, Σ)$ with positive-definite covariance $Σ$ . Let $L : R^{d} \to [0, M]$ be Borel measurable and symmetric, $L (- x) = L (x)$ , and quasi-convex, meaning each sublevel set ${x : L (x) \leq t}$ is convex. Then for every $a \in R^{d}$ , $E L (X - a) \geq E L (X - m) .$ Thus the Gaussian center $m$ minimizes expected loss. The result includes quasi-convex losses whose sublevel sets are not closed.

Facts & Assumptions

Given: AC; a finite-dimensional nondegenerate Gaussian vector; and a bounded Borel measurable symmetric loss with convex sublevel sets.

[A1]

The Axiom of Choice: AC is assumed for the finite-dimensional multivariate normal construction and characteristic laws used here.

[F1]

Brunn–Minkowski inequality for compact convex sets: For nonempty compact convex $U, V \subset R^{d}$ , $λ_{d} (U + V)^{1 / d} \geq λ_{d} (U)^{1 / d} + λ_{d} (V)^{1 / d} .$

[F2]

The boundary of a convex set is Lebesgue null: A convex set and its closure differ only by a Lebesgue-null boundary; lower-dimensional convex sets are null.

[F3]

Multivariate normal law, including singular covariance, Standard normal and normal laws, and The standard normal density has total mass one: A nondegenerate Gaussian is an invertible affine image of a standard Gaussian vector with density $(2 π)^{- d / 2} e^{- ∥ z ∥_{2}^{2} / 2}$ . The Euclidean norm is the one in The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ .

[F4]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: Nonnegative measurable integrands may be integrated in either order.

[F5]

A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: An invertible linear map scales Lebesgue measure by the absolute determinant. In particular, reflection $x \mapsto - x$ preserves volume and scaling by $1 / 2$ multiplies it by $2^{- d}$ . Thus the unweighted inequality in [F1], applied to $U, V$ , gives $λ_{d} ((U + V) / 2)^{1 / d} \geq \frac{λ_{d} (U)^{1 / d} + λ_{d} (V)^{1 / d}}{2} .$

[F6]

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: Translation preserves Lebesgue measure; accordingly, translating the variable in a Lebesgue integral does not change the integral after the integrand is translated.

[F7]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: Closed bounded subsets of $R^{d}$ are compact. Thus intersections of the closed set $C$ with the closed balls used below are compact and measurable.

Proof

Proof technique: prove the Gaussian shift inequality for convex level sets by Brunn–Minkowski, then integrate their indicators.

1.1F5F7givenalgebra

First let $C \subseteq R^{d}$ be closed, symmetric, and convex. For each radius $r > 0$ and shift $v$ , put $U = C \cap (v + \overline{B} (0, r)), V = C \cap (- v + \overline{B} (0, r)) .$ They are compact convex sets by [F7], and $V = - U$ , so they have equal volume by [F5]. For $u \in U$ and $w \in V$ , symmetry and convexity put $(u + w) / 2$ in $C$ , while $u - v$ and $w + v$ lie in the radius- $r$ ball, so $∥ (u + w) / 2 ∥_{2} = ∥ (u - v + w + v) / 2 ∥_{2} \leq r$ . Thus $(U + V) / 2 \subseteq C \cap \overline{B} (0, r)$ . This sets up the Brunn–Minkowski comparison.

2.1F1F5step 1.1algebra

If $U$ has positive volume, [F1] and its derived weighted form [F5] give $λ_{d} (C \cap \overline{B} (0, r)) \geq λ_{d} ((U + V) / 2) \geq λ_{d} (U),$ since $λ_{d} (V) = λ_{d} (U)$ . If $U$ has zero volume the same inequality follows from nonnegativity. Therefore $λ_{d} (C \cap (v + \overline{B} (0, r))) \leq λ_{d} (C \cap \overline{B} (0, r)) .$ This is the required shifted-volume inequality.

3.1F3F4F5F6step 2.1algebra

Let $Z \sim N_{d} (0, I_{d})$ . Its density is a positive constant times $e^{- ∥ z ∥_{2}^{2} / 2}$ . For any shift $v$ , translation invariance [F6] and symmetry of $C$ (using the reflection in [F5]) give $P (Z \in C + v) = c_{d} \int_{C} e^{- ∥ z + v ∥_{2}^{2} / 2} d z = c_{d} \int_{C} e^{- ∥ z - v ∥_{2}^{2} / 2} d z$ , where $c_{d} > 0$ is the normalizing constant. The identity $e^{- ∥ z - v ∥_{2}^{2} / 2} = \int_{0}^{\infty} e^{- t} 1_{{∥ z - v ∥_{2}^{2} / 2 \leq t}} d t$ and [F4] express the latter integral as $c_{d} \int_{0}^{\infty} e^{- t} λ_{d} (C \cap (v + \overline{B} (0, \sqrt{2 t}))) d t .$ By step 2.1, this is at most the same expression at shift zero. Hence $P (Z \in C + v) \leq P (Z \in C)$ . This proves the Gaussian shift inequality.

4.1F2F6step 3.1given

If $C$ is convex, symmetric, but not closed, replace it by $\overline{C}$ . The difference is contained in $\partial C$ , which is null by [F2]; its translate is also null by [F6]. Thus the probability inequality in step 3.1 holds for every convex symmetric Borel set. If $C$ has empty interior, [F2] makes both probabilities zero, which also covers this boundary case.

5.1A1F2F3step 4.1algebra

Write $X = m + A Z$ with $A = Σ^{1 / 2}$ invertible. For any symmetric convex Borel set $C$ , the event $X - a \in C$ is $Z \in A^{- 1} C + A^{- 1} (a - m)$ . The set $A^{- 1} C$ is again symmetric convex, so step 4.1 gives $P (X - a \in C) \leq P (X - m \in C) .$ Apply this to every sublevel set $C_{t} = {x : L (x) \leq t}$ . This gives the comparison for each loss sublevel.

6.1F4step 5.1givenalgebra∎

For any random variable $Y$ with $0 \leq Y \leq M$ , $E Y = \int_{0}^{M} P (Y > t) d t$ , by [F4]. Step 5.1 says $P (L (X - a) \leq t) \leq P (L (X - m) \leq t)$ for each $t$ , hence the corresponding tail probabilities are ordered in the reverse direction. Integrating them gives $E L (X - a) \geq E L (X - m)$ . This proves the statement without requiring closed loss sublevels.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Le Cam's third lemma for a Gaussian likelihood-ratio limit

Statement

Assume AC.

Let $P_{n}$ be probability measures and let $Q_{n} = L_{n} P_{n}$ , where $L_{n} \geq 0$ and $E_{P_{n}} L_{n} = 1$ . Let $T_{n}$ take values in $R^{k}$ . Suppose under $P_{n}$ that $(T_{n}, L_{n}) \Rightarrow (T, e^{Z})$ , where $Z$ is normal with mean $- σ^{2} / 2$ and variance $σ^{2}$ ; the joint law of $(T, Z)$ need not be Gaussian. Then under $Q_{n}$ , $T_{n}$ converges in distribution to the probability law determined by $E [f (T^{(Q)})] = E [e^{Z} f (T)]$ for every bounded continuous $f$ . In particular, $Q_{n}$ is contiguous with respect to $P_{n}$ . If $(T, Z)$ is jointly Gaussian with $T \sim N_{k} (μ, Σ)$ and $Cov (T, Z) = τ$ , this tilted law is $N_{k} (μ + τ, Σ)$ , including singular $Σ$ .

Facts & Assumptions

Given: AC; the sequences of laws, mean-one densities, and joint convergence above.

[A1]

The Axiom of Choice: AC supplies the standard-normal coordinate representation and the finite-dimensional Gaussian law facts used in the jointly Gaussian tilt calculation, and implies the Countable Choice premise of the $C^{1}$ change-of-variables supplier used below.

[F1]

A mean-one likelihood-ratio limit implies contiguity: If $L_{n} \geq 0$ , $E_{P_{n}} L_{n} = 1$ , $L_{n} \Rightarrow L \geq 0$ , and $E L = 1$ , then $L_{n} P_{n}$ is contiguous with respect to $P_{n}$ .

[F2]

Multivariate normal law, including singular covariance, Characteristic function of a multivariate normal law, Standard normal and normal laws, and The standard normal density has total mass one: A jointly Gaussian vector is an affine image of independent standard normal coordinates, and its characteristic function is determined by its mean and covariance. The standard normal density is $ϕ (x) = (2 π)^{- 1 / 2} e^{- x^{2} / 2}$ .

[F3]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: Translation by $b$ on $R^{m}$ is a $C^{1}$ diffeomorphism with Jacobian one, so the change-of-variables formula applies to the positive and negative parts of the bounded integrand in the Gaussian tilt calculation.

[F4]

Contiguous sequences of probability measures: $Q_{n} ◃ P_{n}$ means $P_{n} (A_{n}) \to 0$ implies $Q_{n} (A_{n}) \to 0$ for every measurable event sequence.

[F5]

Convergence in distribution of random elements and Dominated convergence: Bounded continuous expectations converge under weak convergence, and integrable majorants permit passage to limits.

[F6]

Independent random elements have product joint law, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, and On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}: Independent coordinates have product joint law; repeated scalar integration and the Borel product-Lebesgue identity identify its product density. AC supplies the countable choice required by the last supplier.

Proof

Proof technique: uniform integrability transfers expectations, then a Gaussian exponential tilt computes the shifted limit.

1.1A1F2givenalgebra

If $σ > 0$ , write $Z = - σ^{2} / 2 + σ G$ with $G \sim N (0, 1)$ . Completing the square gives $E e^{σ G} = \int e^{σ x} ϕ (x) d x = e^{σ^{2} / 2} \int ϕ (x - σ) d x = e^{σ^{2} / 2},$ so $E e^{Z} = 1$ . If $σ = 0$ , then $Z = 0$ and the same equality holds. The hypothesis gives the joint likelihood-ratio convergence.

2.1F5step 1.1givenalgebra

For $M > 0$ , [F5] and step 1.1 give $E (L_{n} - M)_{+} = 1 - E (L_{n} \land M) \to 1 - E (e^{Z} \land M) = E (e^{Z} - M)_{+}$ . Since $x 1_{{x > 2 M}} \leq 2 (x - M)_{+}$ , the limsup of the tail expectations is at most $2 E (e^{Z} - M)_{+} \to 0$ by dominated convergence [F5]. Each finite initial set of integrable $L_{n}$ has vanishing tails as well, so $(L_{n})$ is uniformly integrable. For bounded continuous $f$ and $M > 0$ , the function $(t, ℓ) \mapsto f (t) (ℓ \land M)$ is bounded and continuous. Its expectations converge by [F5] and the given joint convergence; uniform integrability lets $M \to \infty$ , giving $E_{Q_{n}} f (T_{n}) = E_{P_{n}} [L_{n} f (T_{n})] ⟶ E [e^{Z} f (T)] .$ The equality on the left follows from $Q_{n} = L_{n} P_{n}$ , first for indicators and then for bounded measurable functions by simple approximation. Thus the right side is the distributional limit under $Q_{n}$ .

3.1A1F2F3F6step 2.1algebra

Suppose $(T, Z)$ is jointly Gaussian. By [F2], represent its law as $T = μ + B G$ and $Z = - σ^{2} / 2 + b^{T} G$ , where $G$ has $m$ independent standard-normal coordinates, $B B^{T} = Σ$ , $B b = τ$ , and $b^{T} b = σ^{2}$ . By [F6] and the scalar density in [F2], $G$ has Borel density $ϕ_{m} (g) = \prod_{j = 1}^{m} (2 π)^{- 1 / 2} e^{- g_{j}^{2} / 2}$ with respect to $m$ -dimensional Lebesgue measure (the one-coordinate case uses the scalar law directly). Completing the square coordinatewise gives $e^{b^{T} g - ∥ b ∥^{2} / 2} ϕ_{m} (g) = ϕ_{m} (g - b)$ . Thus for every bounded continuous $f$ , the translation change of variables justified by [F3] gives $E [e^{Z} f (T)] = \int_{R^{m}} f (μ + B g) ϕ_{m} (g - b) d g = \int_{R^{m}} f (μ + B (g + b)) ϕ_{m} (g) d g .$ The integrals are absolutely convergent because $f$ is bounded and the shifted Gaussian density integrates to one. The final integral is $E [f (μ + B b + B G)]$ , so the tilted law from step 2.1 is the affine Gaussian law $N_{k} (μ + τ, Σ)$ by [F2]. No inverse of $Σ$ is used, and the argument includes singular $Σ$ and $b = 0$ .

4.1F1F4step 1.1∎

Contiguity is the first conclusion of [F1] under the present hypotheses; [F4] states its event-sequence meaning.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Bounded bowl-shaped losses have Gaussian-centered Lipschitz minorants

Statement

Assume AC. Let $L : R^{d} \to [0, M]$ be Borel measurable, symmetric and quasi-convex, with $M > 0$ . For integers $N \geq 2$ and $m \geq 1$ , put $δ = M / N$ , $C_{k} = \overline{{x : L (x) \leq k δ}}$ and $g_{N, m} (x) = δ \sum_{k = 1}^{N - 1} \min (1, m d (x, C_{k})) .$ For $C_{k} = \emptyset$ the corresponding summand is defined to be one, and for $C_{k} = R^{d}$ it is zero. Then $0 \leq g_{N, m} \leq L$ , and $g_{N, m}$ is bounded and Lipschitz. For every nondegenerate Gaussian $H \sim N_{d} (b, S)$ , $\inf_{a} E g_{N, m} (a - H) = E g_{N, m} (H - b), \lim_{N \to \infty} \lim_{m \to \infty} E g_{N, m} (H - b) = E L (H - b) .$ Each summand is symmetric and quasi-convex; no quasi-convexity of their sum is asserted.

Facts & Assumptions

Given: AC, the bounded bowl-shaped loss, integers and Gaussian law.

[F1]

$∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz: distance to a nonempty set is one-Lipschitz.

[F2]

The boundary of a convex set is Lebesgue null: the boundary of a convex set is Lebesgue-null, including lower-dimensional convex sets.

[F3]

A centered Gaussian minimizes bounded symmetric quasi-convex loss: a bounded Borel symmetric quasi-convex loss has least Gaussian shift risk at the Gaussian center.

[F4]

Multivariate normal law, including singular covariance: A nondegenerate Gaussian can be realized as $b + A G$ , where $A = S^{1 / 2}$ is invertible and $G$ has independent standard-normal coordinates.

[F5]

Dominated convergence passes limits of functions bounded by $M$ through the Gaussian expectation.

[F6]

Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Greatest lower bound (infimum), and Every nonempty set bounded below has an infimum: $d (x, C) = \inf_{c \in C} ∥ x - c ∥_{2}$ , and for every $ϵ > 0$ a point of nonempty $C$ can be chosen within $ϵ$ of this infimum.

[F7]

The inner-product norm is definite, homogeneous, and satisfies the triangle inequality: The Euclidean norm satisfies $∥ u + v ∥_{2} \leq ∥ u ∥_{2} + ∥ v ∥_{2}$ .

[F8]

The Lebesgue integral is linear on $L^{1} (μ)$ : The expectation of a finite sum of bounded measurable functions is the sum of their expectations.

[F9]

Borel measurable and Lebesgue measurable functions on $R^{n}$ : Borel functions on $R^{d}$ are measurable for the Gaussian and Lebesgue integrals.

[F10]

A convex set and its closure have the same interior and boundary: Under AC and countable choice, a nonempty convex subset and its closure have the same interior and boundary, and the closure is convex. AC supplies countable choice here.

[F11]

A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed: A point in the closure of a subset of Euclidean space is the limit of a sequence from that subset.

[A1]

The Axiom of Choice is retained for the Gaussian and centering suppliers.

[F12]

Standard normal and normal laws, Independent random elements have product joint law, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, and On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}: The scalar standard normal has density $ϕ (t) = (2 π)^{- 1 / 2} e^{- t^{2} / 2}$ ; independent coordinates have product law, repeated integration is valid, and product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets.

[F13]

A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: Under countable choice, an invertible linear map sends Lebesgue-null sets to null sets. AC supplies countable choice in [F12] and [F13].

Proof

1.1A1F1F6F7F9F10F11givenchooseconstructalgebra

Each sublevel $A_{k} = {L \leq k δ}$ is convex and symmetric by hypothesis. Its closure $C_{k}$ is closed; it is convex by [F10] when $A_{k}$ is nonempty, and is convex by vacuity when $A_{k} = \emptyset$ . If $x \in C_{k}$ , [F11] supplies $x_{j} \in A_{k}$ with $x_{j} \to x$ ; then $- x_{j} \in A_{k}$ and $- x \in C_{k}$ , so $C_{k}$ is symmetric (the empty case is symmetric as well). For a nonempty closed convex $C$ , choose $c_{x}, c_{y} \in C$ within $ϵ$ of $d (x, C), d (y, C)$ using [F6]. Their convex combination lies in $C$ , and [F7] gives $d ((1 - t) x + t y, C) \leq (1 - t) d (x, C) + t d (y, C) + ϵ$ . Let $ϵ ↓ 0$ to prove convexity of the distance function. When $C = - C$ , replacing each $c$ by $- c$ in the infimum shows $d (- x, C) = d (x, C)$ . Thus for a nonempty proper $C_{k}$ , $q (x) = \min (1, m d (x, C_{k}))$ is symmetric; its sublevel at $0 \leq t < 1$ is ${d (x, C_{k}) \leq t / m}$ , which is convex, and its sublevel for $t \geq 1$ is the whole space. It is Borel and $m$ -Lipschitz by [F1] and [F9]. For empty $C_{k}$ or $C_{k} = R^{d}$ , the stated constant conventions have the same properties.

2.1F3F8A1step 1.1algebra

If $L (x) \leq k δ$ , then $x \in C_{k}$ and the $k$ th term is zero. Every remaining term is at most one, so $0 \leq g_{N, m} (x) \leq δ # {1 \leq k < N : k δ < L (x)} \leq L (x)$ . The finite sum is bounded by $M$ and has Lipschitz constant at most $M m$ . Apply F3 separately to each summand. Every summand's expected loss is minimized at $a = b$ (use its symmetry to replace $b - H$ by $H - b$ ), so the same action minimizes their sum by [F8]. This proves the first displayed identity without assuming the sum quasi-convex.

3.1A1F2F4F5F9F10F12F13step 2.1algebra∎

Since $C_{k}$ is closed, its $k$ th term increases to $1_{C_{k}^{c}}$ as $m \to \infty$ . If $C_{k} = R^{d}$ , then [F10] gives $int {L \leq k δ} = R^{d}$ , so the level set is all of $R^{d}$ and both indicators vanish. If $C_{k} = \emptyset$ , the level set is empty and both indicators equal one. Otherwise the difference between ${L > k δ}$ and $C_{k}^{c}$ is contained in the boundary of the convex set ${L \leq k δ}$ , which is null by [F2]. To justify the Gaussian nullity, [F4] realizes $H - b = A G$ . By [F12], repeated integration of the scalar densities gives $P (G \in E) = \int_{E} \prod_{j = 1}^{d} ϕ (z_{j}) d λ_{d} (z)$ for every Borel $E$ (for $d = 1$ this is the scalar definition). The closed boundary $D$ is Borel and null, and $A^{- 1} D$ is null by [F13], so $P (H - b \in D) = P (G \in A^{- 1} D) = 0$ . Therefore F5 gives $\lim_{m} E g_{N, m} (H - b) = E [δ \sum_{k = 1}^{N - 1} 1_{{L (H - b) > k δ}}]$ . The expression inside the expectation differs from $L (H - b)$ by at most $δ$ : count the grid thresholds strictly below each value in $[0, M]$ . Let $N \to \infty$ to obtain the second identity.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A QMD score is centered and has finite information

Statement

Assume AC and a parametric model QMD at $θ_{0}$ with score $s_{θ_{0}}$ . Then $E_{θ_{0}} s_{θ_{0}} = 0,$ and the information matrix $I_{θ_{0}} = E_{θ_{0}} [s_{θ_{0}} s_{θ_{0}}^{T}]$ has finite entries and is positive semidefinite. The QMD convention and square-integrability of the score are those of Quadratic-mean differentiability at a parameter.

Facts & Assumptions

Given: AC; an open parameter set in $R^{d}$ , a common sigma-finite dominator with density versions $p_{θ}$ , an interior point $θ_{0}$ , and the QMD score $s = s_{θ_{0}}$ with $E_{θ_{0}} ∥ s ∥_{2}^{2} < \infty$ .

[A1]

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

[F1]

Likelihood densities: Each $p_{θ}$ is a measurable density version satisfying $P_{θ} (A) = \int_{A} p_{θ} d μ$ .

[F2]

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

[F3]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $u^{T} s = \sum_{j = 1}^{d} u_{j} s_{j}$ and $∥ u ∥_{2}^{2} = u^{T} u$ are the Euclidean inner product and norm.

[F4]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: For Euclidean vectors, $∣ u^{T} v ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ .

[F5]

Holder's inequality for integrals, including the endpoint cases and Minkowski's inequality for integrals, including $p = \infty$ : On a probability space, $∣ E U V ∣ \leq (E U^{2})^{1 / 2} (E V^{2})^{1 / 2}$ and the $L^{2}$ norm satisfies the triangle inequality.

Proof

technique · differentiate the squared $L^2(\mu)$ norm along a line

1.1A1F1F2givenalgebra

Fix $v \in R^{d}$ and put $f_{t} = \sqrt{p_{θ_{0} + t v}}$ , $f_{0} = \sqrt{p_{θ_{0}}}$ . By [F1]–[F2], $f_{t} - f_{0} = \frac{t}{2} (v^{T} s) f_{0} + r_{t v}, ∥ r_{t v} ∥_{2} = o (∣ t ∣) .$ Since each density integrates to one, $∥ f_{t} ∥_{2} = ∥ f_{0} ∥_{2} = 1$ , and $0 = 2 ⟨ f_{0}, f_{t} - f_{0} ⟩ + ∥ f_{t} - f_{0} ∥_{2}^{2} .$ This identity uses the unit norms of the square-root densities.

2.1F2F5step 1.1givenalgebra

The increment has norm $O (∣ t ∣)$ , so its squared norm divided by $t$ tends to zero. By Cauchy–Schwarz, the residual inner product $⟨ f_{0}, r_{t v} ⟩ / t$ also tends to zero. Divide the identity in step 1.1 by $t$ and let $t \to 0$ ; the remaining term is $E_{θ_{0}} (v^{T} s) = 0$ . Since this holds for every $v$ , each coordinate of $E_{θ_{0}} s$ is zero.

3.1F2F3F4F5givenalgebra∎

Each $s_{j}$ is square-integrable under a probability law, so $E ∣ s_{j} ∣ < \infty$ by Cauchy–Schwarz. Also $E ∣ s_{j} s_{k} ∣ \leq (E s_{j}^{2})^{1 / 2} (E s_{k}^{2})^{1 / 2} < \infty$ ; hence every entry of $I_{θ_{0}}$ is finite. For any $a \in R^{d}$ , $a^{T} I_{θ_{0}} a = E (a^{T} s)^{2} \geq 0$ , proving positive semidefiniteness, with no nonsingularity assertion.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Convolution bound for regular estimators

Statement

Assume AC. Let ${P_{n, h} : h \in R^{d}}$ be LAN at $h = 0$ with positive-definite information matrix $K$ and central sequence $Δ_{n}$ . Let $T_{n} = \sqrt{n} ({\hat{θ}}_{n} - θ_{0})$ be regular under every fixed local alternative, with common limit law $W$ as in Regular estimator under local alternatives. Then there is a probability law $ν$ on $R^{d}$ such that $L (W) = N_{d} (0, K^{- 1}) * ν .$ Here $*$ is convolution, the law of the sum of independent random vectors with the two indicated laws. In particular, the efficient Gaussian component with covariance $K^{- 1}$ is a convolution factor of every regular limit.

Facts & Assumptions

Given: AC; the LAN experiment, positive-definite $K$ , central sequence, and regular estimator with common limit law $W$ .

[A1]

The Axiom of Choice: AC is assumed by Prokhorov tightness, the Gaussian laws, and characteristic-function uniqueness.

[F1]

Local asymptotic normality for local experiments: For each fixed $h$ , an absolutely continuous proxy has log likelihood $h^{T} Δ_{n} - \frac{1}{2} h^{T} K h + o_{P_{n, 0}} (1)$ and is total-variation close to the true local law.

[F2]

Regular estimator under local alternatives: Under every fixed $P_{n, h}$ , $T_{n} - h \Rightarrow W$ with the same $W$ .

[F3]

Le Cam's third lemma for a Gaussian likelihood-ratio limit: A joint limit tilted by the mean-one Gaussian likelihood ratio describes limits under the proxy local alternative; total-variation closeness transfers that law to the true alternative.

[F4]

Weakly convergent sequences are tight and Prokhorov tightness theorem on polish spaces: Weakly convergent laws on Euclidean space are tight, and under AC every tight sequence of laws on a Polish space has a weakly convergent subsequence.

[F5]

Vector Slutsky theorem: If one component converges in distribution and another differs in probability by $o_{P} (1)$ from a continuous function of it, the joint vector converges to the corresponding joint limit.

[F6]

Multivariate normal law, including singular covariance and Characteristic function of a multivariate normal law: If $Δ \sim N_{d} (0, K)$ , then $E e^{h^{T} Δ} = e^{h^{T} K h / 2}$ and its characteristic function is $e^{- u^{T} K u / 2}$ .

[F7]

Identity theorem for holomorphic functions: Two holomorphic functions on a connected domain that agree on a set with an interior accumulation point agree throughout the domain.

[F8]

Uniqueness of a law from its characteristic function: Under AC, probability laws on $R$ with identical characteristic functions are equal.

[F9]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : Tightness and compact balls use the Euclidean norm on $R^{d}$ .

[F10]

Continuous mapping theorem: Continuous maps preserve convergence in distribution at points where they are continuous.

[F11]

Cramer wold device: Under AC, Borel probability laws on a finite-dimensional Euclidean space are determined by all their real linear projection laws.

[F12]

The product measure of two sigma-finite measure spaces, For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: The product of two Borel probability laws exists, and bounded product integrals factor into the marginal integrals, by applying Tonelli to the positive and negative parts of their real and imaginary parts.

Proof

Proof technique: extract a joint weak limit, use the third lemma for every fixed local shift, and analytically continue the Gaussian tilt identity.

1.1A1F1F2F4F9givenchoosealgebra

Under $P_{n, 0}$ , $T_{n} \Rightarrow W$ by regularity at $h = 0$ and $Δ_{n} \Rightarrow N_{d} (0, K)$ by LAN. By [F4], both marginal sequences are tight. Their joint laws are tight: for any $ϵ > 0$ , choose compact balls $C_{T}, C_{Δ}$ with marginal outside probabilities below $ϵ / 2$ ; the union bound puts the joint law on the compact product with probability at least $1 - ϵ$ .

2.1A1F4step 1.1

By [F4] and Prokhorov's theorem, select a subsequence along which $(T_{n}, Δ_{n}) \Rightarrow (T, Δ)$ . Its marginals are $T \sim W$ and $Δ \sim N_{d} (0, K)$ , since the full marginal sequences already converge. This supplies the joint-limit subsequence used below.

3.1F1F2F3F5F10step 2.1givenalgebra

Fix $h \in R^{d}$ . Along this subsequence [F1] and [F5] give $(T_{n}, Δ_{n}, \log L_{n, h}) \Rightarrow (T, Δ, h^{T} Δ - \frac{1}{2} h^{T} K h) .$ The final coordinate is normal with mean $- h^{T} K h / 2$ and variance $h^{T} K h$ . By [F1], $P_{n, 0} (L_{n, h} = 0) \to 0$ . Assign any finite value to the log on that vanishing event; [F10] then gives $(T_{n}, Δ_{n}, L_{n, h}) \Rightarrow (T, Δ, e^{h^{T} Δ - h^{T} K h / 2})$ . By [F3], under the true $P_{n, h}$ law, the first coordinate has limit $T^{(h)}$ given by tilting the joint law of $(T, Δ)$ by $e^{h^{T} Δ - h^{T} K h / 2}$ . Regularity [F2] also gives $T_{n} \Rightarrow W + h$ under $P_{n, h}$ . Therefore for every $t \in R^{d}$ and every real $h$ , $E [e^{h^{T} Δ - h^{T} K h / 2} e^{i t^{T} T}] = e^{i t^{T} h} φ_{W} (t),$ where $φ_{W}$ is the characteristic function of $W$ .

4.1F6F7step 3.1algebra

Rearranging step 3.1 gives, for real $h$ , $F_{t} (h) : = E [e^{i t^{T} T + h^{T} Δ}] = e^{h^{T} K h / 2 + i t^{T} h} φ_{W} (t) .$ For complex $z \in C^{d}$ , $F_{t} (z) = E [e^{i t^{T} T + z^{T} Δ}]$ is entire in each coordinate: on every compact set of $z$ , its exponential power series and coordinate derivatives are dominated by $e^{R ∥ Δ ∥_{1}}$ for some finite $R$ , whose expectation is finite because each Gaussian coordinate has finite exponential moments. The right-hand side is also entire. Apply [F7] one coordinate at a time, holding the other coordinates real and then successively allowing them to be complex. The equality therefore holds for every complex $z$ .

5.1F6step 4.1givenalgebra

Set $z = i u$ in step 4.1. Then $E [e^{i t^{T} T + i u^{T} Δ}] = φ_{W} (t) e^{- t^{T} u - u^{T} K u / 2} .$ Let $R = T - K^{- 1} Δ$ . Replacing $u$ by $u - K^{- 1} t$ gives $E [e^{i t^{T} R + i u^{T} Δ}] = (φ_{W} (t) e^{t^{T} K^{- 1} t / 2}) e^{- u^{T} K u / 2} .$ The first factor equals $E e^{i t^{T} R}$ by setting $u = 0$ ; the second is the characteristic function of $Δ$ . Thus the joint characteristic function of $(R, Δ)$ factors into its marginals.

6.1A1F6F8F11F12step 5.1algebra∎

By [F12], form the product law $ρ = L (R) \otimes L (Δ)$ on $R^{2 d}$ . Step 5.1 and factorization of the bounded product integrals show that $ρ$ and $L (R, Δ)$ have identical joint characteristic functions. For every $a, b \in R^{d}$ and real $v$ , evaluate these functions at $(v a, v b)$ ; this gives equality of the scalar characteristic functions of the projection $(r, z) \mapsto a^{T} r + b^{T} z$ under the two laws. Scalar uniqueness [F8] identifies each projection law, and Cramér–Wold [F11], under the existing AC hypothesis [A1], gives $L (R, Δ) = ρ$ . Thus $R$ and $Δ$ are independent. Since $K$ is invertible, $K^{- 1} Δ \sim N_{d} (0, K^{- 1})$ , and $T = K^{- 1} Δ + R$ . The law of $T$ , which is $W$ , is therefore $N_{d} (0, K^{- 1}) * L (R)$ , proving the claimed convolution with $ν = L (R)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

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.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite-prior Bayes envelopes converge under pointwise LAN

Statement

Assume AC. In a pointwise LAN experiment Local asymptotic normality for local experiments with $K ≻ 0$ , fix $h_{1}, \dots, h_{s} \in R^{d}$ and weights $w_{i} > 0$ summing to one. Let $g : R^{d} \to [0, M]$ be Lipschitz. The Bayes values for estimating $h$ with loss $g (a - h)$ and this finite prior, allowing all measurable randomized rules, converge to the corresponding Bayes value in the Gaussian shift $Y ∣ h \sim N_{d} (h, K^{- 1})$ . No likelihood-field uniformity or measurable-selection hypothesis is required.

Facts & Assumptions

Given: AC, the pointwise LAN experiment, finite prior and bounded Lipschitz loss.

[F1]

Local asymptotic normality for local experiments gives mean-one likelihood proxies $L_{n, h}$ , pointwise total-variation approximation, and $\log L_{n, h} = h^{T} Δ_{n} - h^{T} K h / 2 + o_{P} (1)$ with $Δ_{n} \Rightarrow N_{d} (0, K)$ . Its finite log expansion also gives $P_{n, 0} (L_{n, h} = 0) \to 0$ .

[F2]

Vector Slutsky theorem and Continuous mapping theorem give joint finite-dimensional convergence after adjoining finitely many remainders that tend to zero in probability.

[F3]

A mean-one likelihood-ratio limit implies contiguity proves uniform integrability for nonnegative mean-one variables converging in law to a mean-one limit.

[F4]

Bayes risk and minimax risk defines the Bayes risk of a rule $δ$ as $r_{π} (δ) = \int R (h, δ) π (d h)$ , while Measure kernel and probability kernel describes randomized rules. In this proof the Bayes value means the infimum of $r_{π} (δ)$ over all measurable randomized rules; the infimum is local notation and is not part of the supplier's definition.

[F5]

Total variation distance for probability laws and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product imply $∣ E_{P} f - E_{Q} f ∣ \leq M ∥ P - Q ∥_{T V}$ for $0 \leq f \leq M$ , by integrating the tail probabilities of $f$ .

[F6]

Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable gives a countable dense rational grid in finite-dimensional Euclidean action space by taking finite products.

[F7]

Multivariate normal law, including singular covariance, Standard normal and normal laws, The standard normal density has total mass one, and A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: For $Z \sim N_{d} (0, K)$ with $K ≻ 0$ , its density is proportional to $\exp (- z^{T} K^{- 1} z / 2)$ , and completing the square shows $E \exp (h^{T} Z - h^{T} K h / 2) = 1$ .

[F8]

The Lebesgue integral is linear on $L^{1} (μ)$ : Expectation commutes with the finite weighted sums defining $U_{n}$ and $U$ .

[A1]

The Axiom of Choice is retained for the LAN and Gaussian suppliers.

Proof

1.1F4F6givenconstructalgebra

For $l \in [0, \infty)^{s}$ , set $F (l) = \inf_{a} \sum_{i} w_{i} l_{i} g (a - h_{i})$ . For any $l, l^{'}$ , each action's objective changes by at most $M \sum_{i} w_{i} ∣ l_{i} - l_{i}^{'} ∣$ ; taking infima proves the same bound for $F$ . Also $0 \leq F (l) \leq M \sum_{i} w_{i} l_{i}$ . For fixed $l$ the objective is continuous in $a$ , so F6 makes its infimum equal to the infimum over a fixed enumeration $(q_{j})$ of the rational grid. For $ϵ > 0$ , choose the first $j$ whose objective is at most $F (l) + ϵ$ . This index is Borel: its defining set and the finitely many failures at smaller indices are Borel. Thus this is an explicit measurable $ϵ$ -optimal action function of $l$ .

1.2F1F2F3F7F8A1givenconstruct

Under the base law, F1 and F2 imply $(L_{n, h_{i}})_{i = 1}^{s} \Rightarrow (l_{i} (Z))_{i = 1}^{s}$ , where $Z \sim N_{d} (0, K)$ and $l_{i} (z) = \exp (h_{i}^{T} z - h_{i}^{T} K h_{i} / 2)$ . The log expansions imply that zero likelihoods have vanishing probability, so exponentiating the finite vector is legitimate. By [F7], completing the square in the Gaussian density gives $E l_{i} (Z) = 1$ . Hence $U_{n} = \sum_{i} w_{i} L_{n, h_{i}}$ converges in law to $U = \sum_{i} w_{i} l_{i} (Z)$ , and [F8] gives $E U_{n} = E U = 1$ . [F3] makes $(U_{n})$ uniformly integrable.

2.1F2F3step 1.1step 1.2algebra

For each cutoff $b$ , $F \land b$ is bounded continuous, so its expectations converge by step 1.2. The error from this cutoff is bounded by $M E [U_{n} 1_{{M U_{n} > b}}]$ , which vanishes uniformly as $b \to \infty$ ; the limit variable has the same integrable bound. Therefore $E F ((L_{n, h_{i}})_{i}) \to E F ((l_{i} (Z))_{i})$ .

3.1F1F4F5step 1.1step 2.1algebra∎

Under the proxy mixture, the predictive density is $U_{n} = \sum_{i} w_{i} L_{n, h_{i}}$ and, when $U_{n} > 0$ , the posterior weights are $w_{i} L_{n, h_{i}} / U_{n}$ . Conditional on the data, every randomized action rule has posterior risk at least $F ((L_{n, h_{i}})_{i}) / U_{n}$ , because $F$ is the infimum over deterministic actions and randomization averages those action risks. The measurable selector from step 1.1 has posterior risk at most this envelope plus $ϵ$ . Multiplying by the predictive density and integrating shows that the proxy Bayes value differs from $E_{P_{n, 0}} F ((L_{n, h_{i}})_{i})$ by at most $ϵ$ ; the set $U_{n} = 0$ has zero predictive probability. Letting $ϵ ↓ 0$ identifies the proxy Bayes value with that expectation. The same argument with $l_{i} (Z)$ identifies the limiting envelope expectation as the finite-prior Gaussian Bayes value: tilting the base Gaussian by $l_{i}$ gives $Z \sim N_{d} (K h_{i}, K)$ , and $Y = K^{- 1} Z$ has law $N_{d} (h_{i}, K^{- 1})$ . Finally [F5] bounds the actual/proxy risk difference uniformly over rules by $M \sum_{i} w_{i} ∥ P_{n, h_{i}} - L_{n, h_{i}} P_{n, 0} ∥_{T V} \to 0$ . Infima preserve this bound, proving the assertion.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite priors approximate Gaussian-shift Bayes values

Statement

Assume AC. Let $Q_{h} = N_{d} (h, S)$ with $S ≻ 0$ and let $g : R^{d} \to [0, M]$ have Lipschitz constant $B < \infty$ . For every Borel probability prior $π$ on $R^{d}$ and $ϵ > 0$ , there is a finite-support prior $π_{f}$ whose Bayes value for loss $g (a - h)$ differs from that of $π$ by less than $ϵ$ . The comparison is uniform over all measurable randomized rules. In particular this applies to every diffuse Gaussian prior; no posterior selector assumption is needed.

Facts & Assumptions

Given: AC, the Gaussian shift, bounded Lipschitz loss, prior and accuracy.

[F1]

Multivariate normal law, including singular covariance: For positive finite dimension, $N_{d} (h, S)$ is realized as $h + S^{1 / 2} G$ with independent standard-normal coordinates. For $S ≻ 0$ , the square root is invertible.

[F2]

Dominated convergence passes pointwise limits under an integrable envelope through an integral.

[F3]

Bayes risk and minimax risk defines the prior risk $r_{π} (δ)$ of a rule, and Measure kernel and probability kernel defines randomized rules. Here the Bayes value means $\inf_{δ} r_{π} (δ)$ over those rules; this infimum is the local value under comparison, not a separate definition supplied by the linked item.

[F4]

Total variation distance for probability laws and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product give the bound $∣ E_{P} f - E_{Q} f ∣ \leq M ∥ P - Q ∥_{T V}$ for $0 \leq f \leq M$ . The event-supremum distance satisfies the triangle inequality by the triangle inequality for each event difference.

[F5]

[F6]

Continuity from below for measures: If $B_{r} ↑ R^{d}$ , then $π (B_{r}) ↑ 1$ .

[F7]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ h - h^{'} ∥_{2}$ is the Euclidean distance used for the ball cover and Lipschitz bound.

[F8]

Measurability of integration against a kernel: Integrating a jointly measurable nonnegative function against a probability kernel gives a measurable function of the parameter.

[F9]

A definite quadratic form has a uniform signed bound on the Euclidean unit sphere: A positive definite quadratic form on $R^{d}$ has a positive lower bound on the Euclidean unit sphere.

[A1]

The Axiom of Choice is retained for the Gaussian construction.

[F10]

Standard normal and normal laws, Independent random elements have product joint law, and On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}: Standard-normal coordinates have scalar density $ϕ (t) = (2 π)^{- 1 / 2} e^{- t^{2} / 2}$ and product joint law; product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets. Tonelli [F4] therefore permits repeated integration of their product density.

[F11]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: Under countable choice, a $C^{1}$ diffeomorphism transforms a nonnegative Lebesgue integral with the absolute Jacobian determinant. AC supplies countable choice here and in [F10].

Proof

1.1givenconstructalgebra

If dimension zero is admitted with the singleton empty-tuple Gaussian convention, there is only one parameter, observation and action. Every prior is the point mass at that parameter and is already finitely supported; every rule has risk $g (0)$ . Taking $π_{f} = π$ gives zero discrepancy. No positive-dimension density, sphere or compactness theorem is needed. Henceforth let $d \geq 1$ .

2.1F1F2F4F5F7F9F10F11A1step 1.1givenalgebra

Write $A = S^{1 / 2}$ . By [F1] and [F10], $Q_{h}$ is the law of $h + A G$ , and repeated scalar integration gives $G$ density $(2 π)^{- d / 2} e^{- ∥ z ∥_{2}^{2} / 2}$ . Applying [F11] to $z \mapsto h + A z$ gives the density $q_{h} (y) = (2 π)^{- d / 2} ∣ \det A ∣^{- 1} \exp [- (y - h)^{T} S^{- 1} (y - h) / 2]$ . The map $h \mapsto Q_{h}$ is continuous in total variation: when $h_{j} \to h$ , these densities converge pointwise. The matrix $S^{- 1}$ is positive definite because $x^{T} S^{- 1} x = (S^{- 1} x)^{T} S (S^{- 1} x) > 0$ for $x \neq 0$ . By [F9], $y^{T} S^{- 1} y \geq c ∥ y ∥_{2}^{2}$ for some $c > 0$ . On a fixed bounded set of shifts, the density formula and $∥ y - h ∥_{2}^{2} \geq ∥ y ∥_{2}^{2} / 2 - ∥ h ∥_{2}^{2}$ bound all these densities by $C \exp (- c^{'} ∥ y ∥_{2}^{2})$ for constants $C, c^{'} > 0$ . This envelope is integrable by repeated scalar Gaussian integration [F4, F10, F11], so [F2] gives $L^{1}$ convergence; for every event, its probability difference is at most that $L^{1}$ distance. On a compact ball this continuity is uniform: otherwise there are $h_{j}, h_{j}^{'}$ in the ball with $∥ h_{j} - h_{j}^{'} ∥_{2} \to 0$ but TV distances bounded below. Compactness gives a subsequence with both converging to the same h; pointwise continuity and the event-supremum triangle inequality from [F4] contradict that lower bound.

3.1F1F3F4F8step 2.1algebra

For a randomized rule $δ_{y} (d a)$ , write $R_{δ} (h) = \int \int g (a - h) δ_{y} (d a) Q_{h} (d y)$ . Change the loss argument first and the sampling law second. [F4] yields $∣ R_{δ} (h) - R_{δ} (h^{'}) ∣ \leq B ∥ h - h^{'} ∥_{2} + M ∥ Q_{h} - Q_{h^{'}} ∥_{T V}$ , independently of the rule. Each risk is between zero and $M$ . The inner kernel integral is jointly measurable in $(h, y)$ by [F8]; multiplying by the jointly measurable Gaussian density and integrating in y makes $R_{δ} (h)$ measurable.

4.1F3F5F6step 2.1step 3.1constructalgebra∎

The balls $B_{r} = {h : ∥ h ∥_{2} \leq r}$ increase to $R^{d}$ , so [F6] lets us choose a closed ball with discarded prior mass less than $τ$ . Cover this compact ball by open balls of radius $η / 2$ ; [F5] gives a finite subcover. Intersect the first ball with the closed ball, then successively remove the previous covering sets from each later intersection. This makes a finite Borel partition whose cells have diameter at most $η$ . Assign one point to each nonempty cell and send points outside the closed ball to zero. The resulting finite-valued map q is Borel, and $π_{f} = π \circ q^{- 1}$ . On the ball, step 3.1 bounds $∣ R_{δ} (h) - R_{δ} (q (h)) ∣$ by $B η + M ω (η)$ , where the compact-ball TV modulus satisfies $ω (η) \to 0$ by step 2.1. Outside, use $M$ . Thus prior risks differ uniformly over rules by at most $B η + M ω (η) + M τ$ . Choose the ball first for small tail, then the cell diameter small on that fixed ball. This makes the bound less than $ϵ$ ; taking infima over the same rules preserves it.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Normalized product likelihood expansion under QMD

Statement

Assume AC and Quadratic-mean differentiability at a parameter at $θ_{0}$ with score $s$ and information $I$ . Let $h_{n} \to h$ in $R^{d}$ , put $u_{n} = h_{n} / \sqrt{n}$ , and write $P_{0} = P_{θ_{0}}$ and $P_{u_{n}} = P_{θ_{0} + u_{n}}$ . On $S = {p_{θ_{0}} > 0}$ define, for all sufficiently large $n$ , $q_{n} (x) = \frac{1_{S} (x) p_{θ_{0} + u_{n}} (x)}{(1 - α_{n}) p_{θ_{0}} (x)}, α_{n} = P_{u_{n}} (S^{c}) .$ Then $q_{n}$ is a probability density relative to $P_{0}$ . For IID $X_{1}, \dots, X_{n} \sim P_{0}$ , let $L_{n} = \prod_{i = 1}^{n} q_{n} (X_{i})$ and $Δ_{n} = n^{- 1 / 2} \sum_{i = 1}^{n} s (X_{i})$ . Under $P_{0}^{\otimes n}$ , $\log L_{n} = h_{n}^{T} Δ_{n} - \frac{1}{2} h^{T} I h + o_{P} (1), P_{0}^{\otimes n} (L_{n} = 0) \to 0.$ Moreover, if ${\overset{ˉ}{P}}_{n} = q_{n} P_{0}$ is the normalized absolutely continuous component of $P_{u_{n}}$ relative to $P_{0}$ , then $∥ P_{u_{n}}^{\otimes n} - {\overset{ˉ}{P}}_{n}^{\otimes n} ∥_{T V} \to 0.$ No full likelihood ratio for $P_{u_{n}}^{\otimes n}$ is asserted.

Facts & Assumptions

Given: AC; a common sigma-finite dominator $μ$ , the QMD density versions at $θ_{0}$ and $θ_{0} + u$ , a square-integrable score $s$ , and $h_{n} \to h$ .

[A1]

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

[L1]

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

[F1]

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

[F2]

A QMD score is centered and has finite information: $E_{0} s = 0$ and $I = E_{0} [s s^{T}]$ is finite.

[F3]

QMD controls singular and zero-overlap masses: If $α (u) = P_{θ_{0} + u} (p_{0} = 0)$ and $β (u) = P_{0} (p_{θ_{0} + u} = 0)$ , then both are $o (∥ u ∥_{2}^{2})$ .

[F4]

Khinchin weak law for integrable IID variables: For IID integrable real variables $Z_{i}$ , $n^{- 1} \sum_{i = 1}^{n} Z_{i} \to E Z_{1}$ in probability.

[F5]

Identical distribution and IID families: Under $P_{0}^{\otimes n}$ , the coordinate variables $X_{i}$ are IID with common law $P_{0}$ .

[F6]

The product measure of two sigma-finite measure spaces, For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system, and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: Finite product laws are determined by measurable rectangles, and nonnegative product integrals can be iterated.

[F7]

Total variation distance for probability laws: Total variation is $\sup_{A} ∣ P (A) - Q (A) ∣$ over measurable events.

[F8]

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

[F9]

$L^{1}$ convergence implies uniform integrability: An $L^{1}$ -convergent sequence of nonnegative random variables is uniformly integrable.

[F10]

Markov's inequality for random variables: For a nonnegative integrable variable $W$ and $a > 0$ , $P (W > a) \leq E W / a$ .

[F11]

The Lagrange and Cauchy forms of Taylor's remainder: A $C^{3}$ function with bounded third derivative on an interval has a quadratic Taylor remainder bounded by a constant times the cube of the increment.

Proof

Proof technique: square-root expansion, a triangular-array logarithm calculation, and a product-mixture bound.

1.1A1F1F2F3F8L1givenalgebra

Write $p_{u} = p_{θ_{0} + u}$ , $S = {p_{0} > 0}$ , and $α (u) = \int_{S^{c}} p_{u} d μ$ . For small $u$ , [F3] gives $α (u) < 1 / 2$ . Define $q_{u} = 1_{S} p_{u} / ((1 - α (u)) p_{0})$ on $S$ and $q_{u} = 0$ on $S^{c}$ . It is nonnegative and measurable, and $\int q_{u} d P_{0} = (1 - α (u))^{- 1} \int_{S} p_{u} d μ = 1,$ so it is a probability density. At $u = 0$ , $α (0) = 0$ and $q_{0} = 1$ $P_{0}$ -almost surely, since $P_{0} (S^{c}) = \int_{S^{c}} p_{0} d μ = 0$ . On $S$ , $\sqrt{q_{u}} = (1 - α (u))^{- 1 / 2} (1 + \frac{1}{2} u^{T} s + \frac{r_{u}}{\sqrt{p_{0}}}) .$ The residual term has $L^{2} (P_{0})$ norm $∥ r_{u} 1_{S} ∥_{L^{2} (μ)} = o (∥ u ∥_{2})$ . Also $∣ (1 - α)^{- 1 / 2} - 1 ∣ = \frac{α}{\sqrt{1 - α} (1 + \sqrt{1 - α})} = O (α),$ and [F2], [F8] give $∥ u^{T} s ∥_{L^{2} (P_{0})} = O (∥ u ∥_{2})$ . Therefore $Y_{u} : = 2 (\sqrt{q_{u}} - 1) = u^{T} s + R_{u}, ∥ R_{u} ∥_{L^{2} (P_{0})} = o (∥ u ∥_{2}) .$ This is the required square-root expansion.

1.2F5F6givenalgebra

The product function $L_{n} = \prod_{i} q_{u_{n}} (X_{i})$ is a density of ${\overset{ˉ}{P}}_{n}^{\otimes n}$ relative to $P_{0}^{\otimes n}$ . For a measurable rectangle $A_{1} \times \dots \times A_{n}$ , Tonelli gives $\int_{A_{1} \times \dots \times A_{n}} \prod_{i} q_{u_{n}} (x_{i}) d P_{0}^{\otimes n} = \prod_{i} \int_{A_{i}} q_{u_{n}} d P_{0} = \prod_{i} {\overset{ˉ}{P}}_{n} (A_{i}) .$ By [F6], agreement on rectangles identifies the product measures. Taking the whole space gives $E_{P_{0}^{\otimes n}} L_{n} = 1$ .

2.1F2F8step 1.1givenalgebra

Since $E_{0} q_{u} = 1$ , $E_{0} Y_{u} = - \frac{1}{4} E_{0} Y_{u}^{2} .$ By step 1.1, [F2], and [F8], $E_{0} Y_{u}^{2} = u^{T} I u + o (∥ u ∥_{2}^{2}), E_{0} Y_{u} = - \frac{1}{4} u^{T} I u + o (∥ u ∥_{2}^{2}) .$ Indeed, the difference between $Y_{u}$ and $u^{T} s$ has $L^{2} (P_{0})$ norm $o (∥ u ∥_{2})$ , so Cauchy–Schwarz makes the difference of their squared norms $o (∥ u ∥_{2}^{2})$ .

2.2F2F4F8step 1.1givenalgebra

Apply [F4] to $Z_{i} = ∥ s (X_{i}) ∥_{2}^{2}$ , whose mean is finite by [F2]. Then $\frac{1}{n} \sum_{i = 1}^{n} (h_{n}^{T} s (X_{i}))^{2} \to h^{T} I h$ in probability: for fixed $h$ this is [F4] applied to $(h^{T} s)^{2}$ , and the difference for $h_{n}$ is bounded by $∥ h_{n} - h ∥_{2} ∥ h_{n} + h ∥_{2} \frac{1}{n} \sum_{i = 1}^{n} ∥ s (X_{i}) ∥_{2}^{2} \to 0.$ Step 1.1 and [F8] also give $\sum_{i = 1}^{n} Y_{u_{n}} (X_{i})^{2} - \frac{1}{n} \sum_{i = 1}^{n} (h_{n}^{T} s (X_{i}))^{2} \to 0$ in probability, because its expected absolute value is at most $n ∥ R_{u_{n}} ∥_{2} (∥ Y_{u_{n}} ∥_{2} + ∥ u_{n}^{T} s ∥_{2}) = o (n ∥ u_{n} ∥_{2}^{2}) = o (1)$ . Consequently $\sum_{i} Y_{u_{n}} (X_{i})^{2} \to h^{T} I h$ .

2.3F3F8F9F10step 1.1givenalgebra

Put $V_{n} = \sqrt{n} Y_{u_{n}} (X_{1})$ . Step 1.1 implies $V_{n} - h_{n}^{T} s (X_{1}) \to 0$ in $L^{2} (P_{0})$ , so $V_{n} \to h^{T} s (X_{1})$ in $L^{2}$ . By [F8], $V_{n}^{2} \to (h^{T} s (X_{1}))^{2}$ in $L^{1}$ ; [F9] makes $(V_{n}^{2})$ uniformly integrable. For every $ε > 0$ , [F10] and the union bound give $P_{0}^{\otimes n} ⁣ (\max_{i \leq n} ∣ Y_{u_{n}} (X_{i}) ∣ > ε) \leq ε^{- 2} E_{0} [V_{n}^{2} 1_{{∣ V_{n} ∣ > ε \sqrt{n}}}] \to 0.$ Also $P_{0}^{\otimes n} (L_{n} = 0) \leq n β (u_{n}) \to 0$ by [F3] and the union bound.

3.1F2F5F10step 2.1givenalgebra

Let $R_{u_{n}} = Y_{u_{n}} - u_{n}^{T} s$ . Its centered IID sum has variance at most $n E_{0} R_{u_{n}}^{2} = o (n ∥ u_{n} ∥_{2}^{2}) = o (1)$ . By step 2.1, $n E_{0} R_{u_{n}} = n E_{0} Y_{u_{n}} = - \frac{1}{4} h_{n}^{T} I h_{n} + o (1) \to - \frac{1}{4} h^{T} I h .$ Markov's inequality [F10], applied to the square of the centered sum, gives $\sum_{i} Y_{u_{n}} (X_{i}) = h_{n}^{T} Δ_{n} - \frac{1}{4} h^{T} I h + o_{P} (1) .$ This is the required expansion of the sum.

4.1F11step 2.2step 3.1step 2.3given

On the event that every $q_{u_{n}} (X_{i}) > 0$ and $\max_{i} ∣ Y_{u_{n}} (X_{i}) ∣ \leq 1$ , Taylor's formula gives $\log q_{u_{n}} (X_{i}) = Y_{u_{n}} (X_{i}) - \frac{1}{4} Y_{u_{n}} (X_{i})^{2} + O (∣ Y_{u_{n}} (X_{i}) ∣^{3}),$ with one constant valid for all $i, n$ by [F11], since the third derivative of $y \mapsto 2 \log (1 + y / 2)$ is bounded on $[- 1, 1]$ . The total cubic remainder is bounded by a constant times $\max_{i} ∣ Y_{u_{n}} (X_{i}) ∣ \sum_{i} Y_{u_{n}} (X_{i})^{2}$ , which tends to zero by the square-sum limit in step 2.2 and the maximum bound in step 2.3. Combining the sum expansion in step 3.1 with the square-sum limit gives $\log L_{n} = \sum_{i} \log q_{u_{n}} (X_{i}) = h_{n}^{T} Δ_{n} - \frac{1}{2} h^{T} I h + o_{P} (1) .$ This is the normalized log-likelihood expansion.

5.1F3F7givenalgebra∎

The full one-observation law decomposes as $P_{u_{n}} = (1 - α_{n}) {\overset{ˉ}{P}}_{n} + α_{n} S_{n}$ , where $S_{n}$ is the normalized restriction of $P_{u_{n}}$ to $S^{c}$ when $α_{n} > 0$ . In the product mixture, the probability that at least one coordinate is drawn from $S_{n}$ is at most $n α_{n}$ . Conditioning on the all-absolutely-continuous component shows, for every event, its probability under the full product differs from that under ${\overset{ˉ}{P}}_{n}^{\otimes n}$ by at most $n α_{n}$ . Since $n α_{n} = o (n ∥ u_{n} ∥_{2}^{2}) = o (1)$ , [F7] proves the total-variation claim. If $α_{n} = 0$ , the singular component is absent and the same bound is zero.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Local asymptotic minimax bound

Statement

Assume AC and let $d \geq 1$ be finite. Let ${P_{n, h} : h \in R^{d}}$ be a finite-dimensional experiment that is LAN at $h = 0$ with positive-definite information $K$ and central sequence $Δ_{n} \Rightarrow N_{d} (0, K)$ as in Local asymptotic normality for local experiments. Let $M \geq 0$ be finite and $L : R^{d} \to [0, M]$ be Borel measurable, symmetric ( $L (- x) = L (x)$ ), and quasi-convex (each sublevel set is convex). Then ${lim inf}_{c \to \infty} {lim inf}_{n \to \infty} \inf_{{\hat{h}}_{n}} \sup_{∥ h ∥_{2} \leq c} E_{P_{n, h}} L ({\hat{h}}_{n} - h) \geq E L (Z), Z \sim N_{d} (0, K^{- 1}) .$ The infimum ranges over measurable, possibly randomized estimators. The experiment is pointwise LAN in each fixed $h$ ; no uniform likelihood-field assumption is added.

Facts & Assumptions

Given: AC; the pointwise LAN experiment with $K ≻ 0$ ; and the stated bounded, Borel, symmetric quasi-convex loss.

[A1]

The Axiom of Choice: AC is assumed by the Gaussian and LAN prerequisites used here.

[F1]

Local asymptotic normality for local experiments: For each fixed $h$ , there are mean-one likelihood proxies with pointwise log-likelihood expansion and total-variation approximation to the actual local law.

[F2]

Finite-prior Bayes envelopes converge under pointwise LAN: For a finite prior and a bounded Lipschitz loss, pointwise LAN makes the actual finite-prior Bayes values converge to those in the Gaussian shift.

[F3]

Finite priors approximate Gaussian-shift Bayes values: In the Gaussian shift, the Bayes value under any prior can be approximated arbitrarily closely by a finite-support prior, uniformly over decision rules.

[F4]

Bounded bowl-shaped losses have Gaussian-centered Lipschitz minorants: For every $N \geq 2$ and $m \geq 1$ it constructs a bounded Lipschitz minorant $g_{N, m} \leq L$ and proves its Gaussian shift risk is minimized at the center; it also gives $\lim_{N \to \infty} \lim_{m \to \infty} E g_{N, m} (Z) = E L (Z)$ .

[F5]

Loss functions and risk and Bayes risk and minimax risk: A rule's risk is its expected loss and its prior risk is $r_{π} (δ)$ . In this proof write $B_{E} (π, g) : = \inf_{δ} r_{π} (δ)$ for the Bayes value in experiment $E$ , with the infimum over measurable randomized estimators; this is local notation, since the supplier defines $r_{π}$ and the minimax value but does not name a Bayes-value infimum. For any prior supported in a parameter set, every rule's worst-case risk on that set is at least its prior risk. Write $B_{n}$ and $B_{G}$ for this value in the $n$ th experiment and the Gaussian shift, respectively.

[F6]

Multivariate normal law, including singular covariance, Standard normal and normal laws, The standard normal density has total mass one, and A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: A nondegenerate Gaussian has its usual Lebesgue density and the stated affine transforms.

[F7]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: The finite Gaussian first moment is bounded by its finite second moment.

[F8]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ \cdot ∥_{2}$ is the Euclidean norm.

[F9]

Borel measurable and Lebesgue measurable functions on $R^{n}$ : Borel measurability on $R^{d}$ uses the sigma-algebra generated by its open sets.

Proof

Proof technique: compare finite-prior Bayes risks, compute a Gaussian prior Bayes value, then approximate the loss from below.

1.1F4F5F6F8givenconstructalgebra

If $M = 0$ , then $L$ is identically zero and the claimed bound is equality. Assume $M > 0$ for the rest of the proof. Fix arbitrary integers $N \geq 2$ and $m \geq 1$ , and set $g = g_{N, m}$ from [F4]; it is bounded Lipschitz and $g \leq L$ . For each $ρ > 0$ , set $Γ = ρ^{- 1} K^{- 1}$ , and let $π_{Γ} = N_{d} (0, Γ)$ . Define $V_{n, c} (g) = \inf_{{\hat{h}}_{n}} \sup_{∥ h ∥_{2} \leq c} E_{P_{n, h}} g ({\hat{h}}_{n} - h)$ and $V (g) = {lim inf}_{c \to \infty} {lim inf}_{n \to \infty} V_{n, c} (g)$ . If a finite prior $π_{f}$ is supported in the radius- $c$ ball, then every worst-case risk is at least its prior average by [F5]. Taking infima gives $V_{n, c} (g) \geq B_{n} (π_{f}, g)$ , the finite-prior Bayes value.

2.1A1F4F6step 1.1givenalgebra

In the Gaussian shift, $Y ∣ h \sim N_{d} (h, K^{- 1})$ . Multiplying its likelihood by the prior density $N_{d} (0, Γ)$ and completing the square gives $H ∣ Y \sim N_{d} (S K Y, S), S = (K + Γ^{- 1})^{- 1} .$ For the fixed constructed loss $g = g_{N, m}$ , [F4] gives that the posterior risk is minimized at $S K Y$ , with value $E g (N_{d} (0, S))$ independent of $Y$ . The action $Y \mapsto S K Y$ is measurable and attains this value, so $B_{G} (π_{Γ}, g) = E g (N_{d} (0, S))$ .

2.2F1F2step 1.1givenalgebra

Let $C_{f} = \max {∥ h ∥_{2} : h \in supp (π_{f})}$ . For every $c \geq C_{f}$ , [F2] gives $B_{n} (π_{f}, g) \to B_{G} (π_{f}, g)$ , where $B_{G}$ is the Bayes value in the Gaussian shift. Hence $V (g) \geq B_{G} (π_{f}, g)$ . Since this holds for every finite prior, the bound also holds for any value approximated by finite-prior Bayes values.

3.1F3step 1.1step 2.2givenalgebra

For every $η > 0$ , [F3] supplies a finite prior $π_{f}$ whose Gaussian-shift Bayes value differs from $B_{G} (π_{Γ}, g)$ by less than $η$ . Step 2.2 therefore gives $V (g) \geq B_{G} (π_{Γ}, g) - η$ . Let $η ↓ 0$ to obtain $V (g) \geq B_{G} (π_{Γ}, g)$ .

4.1F6F7step 1.1step 2.1step 3.1algebra

Here $S = (1 + ρ)^{- 1} K^{- 1}$ . Couple $N_{d} (0, S)$ as $(1 + ρ)^{- 1 / 2} Z$ for $Z \sim N_{d} (0, K^{- 1})$ . If $B_{g}$ is a Lipschitz constant for $g$ , then $∣ E g ((1 + ρ)^{- 1 / 2} Z) - E g (Z) ∣ \leq B_{g} (1 - (1 + ρ)^{- 1 / 2}) E ∥ Z ∥_{2} \to 0$ as $ρ ↓ 0$ by [F7]. By steps 3.1 and 2.1, for each $ρ > 0$ we have $V (g) \geq E g ((1 + ρ)^{- 1 / 2} Z)$ . Since the construction applies for every $ρ > 0$ , taking $ρ ↓ 0$ yields $V (g) \geq E g (Z)$ .

5.1F4F9step 1.1step 4.1givenalgebra∎

Since $g \leq L$ , every estimator's risk under $L$ dominates its risk under this fixed $g$ , so the local minimax value for $L$ is at least $V (g)$ . Step 4.1 gives the lower bound $E g_{N, m} (Z)$ . The choices $N, m$ in step 1.1 were arbitrary; first let $m \to \infty$ and then $N \to \infty$ . By [F4] these lower bounds converge to $E L (Z)$ . This is the stated inequality.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Quadratic-mean differentiability implies iid LAN

Statement

Assume the Axiom of Choice. Let ${P_{θ} : θ \in Θ \subseteq R^{d}}$ be dominated by a common sigma-finite measure, with $Θ$ open, and suppose the model is QMD at an interior $θ_{0}$ with score $s$ and information $I = E_{θ_{0}} [s s^{T}]$ . Let $X_{1}, \dots, X_{n}$ be IID $P_{θ_{0}}$ , and for each fixed local sequence $h_{n} \to h$ set $u_{n} = h_{n} / \sqrt{n}$ . Then the local product laws $P_{θ_{0} + u_{n}}^{\otimes n}$ admit mean-one likelihood-ratio proxies $L_{n}$ relative to $P_{θ_{0}}^{\otimes n}$ such that ${∥ P_{θ_{0} + u_{n}}^{\otimes n} - L_{n} P_{θ_{0}}^{\otimes n} ∥}_{T V} \to 0,$ and under $P_{θ_{0}}^{\otimes n}$ , $\log L_{n} = h_{n}^{T} Δ_{n} - \frac{1}{2} h^{T} I h + o_{P} (1), Δ_{n} = n^{- 1 / 2} \sum_{i = 1}^{n} s (X_{i}) \Rightarrow N_{d} (0, I) .$ Thus the IID QMD experiment is LAN in the proxy sense of Local asymptotic normality for local experiments. This assertion does not claim that the full local product law has a Radon–Nikodym derivative relative to the base law.

Facts & Assumptions

Given: AC; the QMD model, common dominator, score, and information above; and IID observations from $P_{θ_{0}}$ .

[A1]

The Axiom of Choice: AC is the stated set-theoretic axiom. It is assumed here because the multivariate IID CLT supplier explicitly assumes it.

[L1]

Likelihood densities: The chosen density versions satisfy $P_{θ} (A) = \int_{A} p_{θ} d μ$ for every measurable $A$ .

[F1]

Quadratic-mean differentiability at a parameter: At $θ_{0}$ , the square-root densities have an $L^{2} (μ)$ first-order expansion with score $s$ .

[F2]

Normalized product likelihood expansion under QMD: For every $h_{n} \to h$ , the normalized absolutely continuous product proxy has mean-one likelihood $L_{n}$ , differs from the full product law in total variation by $o (1)$ , has $P_{θ_{0}}^{\otimes n} (L_{n} = 0) \to 0$ , and its log likelihood has expansion $h_{n}^{T} Δ_{n} - \frac{1}{2} h^{T} I h + o_{P} (1)$ .

[F3]

Multivariate iid central limit theorem: Under AC and for a finite integer $d \geq 1$ , IID vectors with finite second moment and covariance $I$ satisfy $n^{- 1 / 2} \sum_{i} (X_{i} - E X_{i}) \Rightarrow N_{d} (0, I)$ ; singular $I$ is allowed.

[F4]

Local asymptotic normality for local experiments: LAN means a Gaussian central-sequence limit, the stated log-likelihood expansion for each fixed local direction, and a total-variation-close mean-one absolutely continuous proxy.

[F5]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : The local directions and score sums use the Euclidean inner product and norm on $R^{d}$ .

[F6]

A QMD score is centered and has finite information: A QMD score has mean zero and finite positive-semidefinite information matrix.

Proof

Proof technique: combine the normalized QMD likelihood expansion with the score central limit theorem.

1.1F4F5givenconstructalgebra

If $d = 0$ , the parameter space containing $θ_{0}$ is the singleton empty tuple by [F5]. Thus $h_{n} = h = 0$ , $u_{n} = 0$ , and every local law equals the base law. The score and $Δ_{n}$ are empty vectors and $I$ is the empty matrix. Choose $L_{n} = 1$ : its mean is one, the total variation error is zero, and the log expansion is the exact identity $0 = 0$ . The law of $Δ_{n}$ is the unique law on the singleton, denoted $N_{0} (0, 0)$ under the empty-tuple convention; its weak convergence is immediate. These facts give LAN by [F4] without a CLT. Henceforth suppose $d \geq 1$ .

2.1A1F1F2L1step 1.1given

By [A1, L1, F1] the path is QMD with the specified density versions. [F2] applies to the chosen sequence $u_{n} = h_{n} / \sqrt{n}$ and supplies the normalized proxy $L_{n}$ , the total variation approximation, and the log expansion. Its normalizer is $1 - α_{n} > 0$ for all sufficiently large $n$ , since $α_{n} = o (∥ u_{n} ∥_{2}^{2})$ and $u_{n} \to 0$ .

2.2A1F1F3F5F6step 1.1given

By [F6], the score has $E_{θ_{0}} s = 0$ and covariance $E_{θ_{0}} [s s^{T}] = I$ with finite entries. QMD also supplies $E_{θ_{0}} ∥ s ∥_{2}^{2} < \infty$ by [F1]. Thus [F3] applies to $s (X_{i})$ and yields $Δ_{n} \Rightarrow N_{d} (0, I)$ , including singular information.

3.1F2F4step 1.1step 2.1step 2.2∎

For each fixed $h$ , steps 2.1 and 2.2 give the proxy approximation, the log expansion and the Gaussian central-sequence limit required in [F4]. Hence the local experiment is LAN. The proof used the full likelihood only for the normalized absolutely continuous component; singular mass is handled by the total-variation bound in [F2]. This proves the stated proxy-LAN conclusion.

5 · Examples, counterexamples and false statements

None yet.

LAN, Contiguity and Efficiency

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Contiguous sequences of probability measures

Definition

The boundary of a convex set is Lebesgue null

Statement

Facts & Assumptions

Proof

Local asymptotic normality for local experiments

Definition

Brunn–Minkowski inequality for compact convex sets

Statement

Facts & Assumptions

Proof

A mean-one likelihood-ratio limit implies contiguity

Statement

Facts & Assumptions

Proof

Regular estimator under local alternatives

Definition

Compact-uniform LAN approximates compact-prior posteriors

Statement

Facts & Assumptions

Proof

A centered Gaussian minimizes bounded symmetric quasi-convex loss

Statement

Facts & Assumptions

Proof

Le Cam's third lemma for a Gaussian likelihood-ratio limit

Statement

Facts & Assumptions

Proof

Bounded bowl-shaped losses have Gaussian-centered Lipschitz minorants

Statement

Facts & Assumptions

Proof

A QMD score is centered and has finite information

Statement

Facts & Assumptions

Proof

Convolution bound for regular estimators

Statement

Facts & Assumptions

Proof

QMD controls singular and zero-overlap masses

Statement

Facts & Assumptions

Proof

Finite-prior Bayes envelopes converge under pointwise LAN

Statement

Facts & Assumptions

Proof

Finite priors approximate Gaussian-shift Bayes values

Statement

Facts & Assumptions

Proof

Normalized product likelihood expansion under QMD

Statement

Facts & Assumptions

Proof

Local asymptotic minimax bound

Statement

Facts & Assumptions

Proof

Quadratic-mean differentiability implies iid LAN

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources