TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Laplace asymptotic under a unique nondegenerate interior maximum

Statement

Let $d \geq 1$ and let $K \subseteq R^{d}$ be compact, Lebesgue measurable, and of positive Lebesgue measure. Let $a : K \to [0, + \infty]$ be measurable with $\int_{K} a d λ_{d} < \infty$ . Let $h : K \to R$ be continuous and have a unique maximizer $\hat{θ}$ in the interior of $K$ . Assume $a$ is continuous at $\hat{θ}$ with $0 < a (\hat{θ}) < \infty$ . Suppose $h$ is $C^{3}$ on an open neighborhood $U \subseteq K$ of $\hat{θ}$ , and put $H : = - D^{2} h (\hat{θ})$ using the Hessian convention of The Hessian matrix and critical points of a scalar field. Assume $H$ is symmetric positive definite in the sense of Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal. For $x \in R^{d}$ and $r > 0$ , write $B (x, r) : = {θ \in R^{d} : ∥ θ - x ∥_{2} < r}$ and $\overline{B} (x, r) : = {θ \in R^{d} : ∥ θ - x ∥_{2} \leq r}$ . Assume there are $r_{0}, c, r_{1}, η > 0$ such that $\overline{B} (\hat{θ}, r_{0}) \subseteq U \cap K$ , $h (θ) \leq h (\hat{θ}) - c ∥ θ - \hat{θ} ∥_{2}^{2} (θ \in \overline{B} (\hat{θ}, r_{0})),$ and $h (θ) \leq h (\hat{θ}) - η (θ \in K ∖ B (\hat{θ}, r_{1})) .$ All integrals are the nonnegative Lebesgue integral of The nonnegative Lebesgue integral. For $I_{n} : = \int_{K} a (θ) e^{n h (θ)} d λ_{d} (θ)$ , as the positive integer $n \to \infty$ , $I_{n} \sim a (\hat{θ}) e^{n h (\hat{θ})} {(\frac{2 π}{n})}^{d / 2} \frac{1}{\sqrt{\det H}},$ where $u_{n} \sim v_{n}$ means $u_{n} / v_{n} \to 1$ . No relative $O (n^{- 1})$ remainder is asserted.

Facts & Assumptions

Given: Countable Choice; the compact measurable set $K$ , amplitude $a$ , continuous phase $h$ , unique interior maximizer $\hat{θ}$ , open $C^{3}$ neighborhood $U$ , positive-definite Hessian $H$ , and the two displayed local and global gap bounds.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function.

[F1]

$C^{k}$ maps and multi-index derivative notation in Euclidean space: $C^{3}$ means the partial derivatives through order three exist and are continuous on the open neighborhood; The Hessian matrix and critical points of a scalar field defines $D^{2} h$ as the matrix of second partial derivatives and the gradient as the vector of first partial derivatives.

[F2]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ v ∥_{2}^{2} = v^{T} v$ on $R^{d}$ ; Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal defines real positive definiteness by symmetry and $v^{T} H v > 0$ for every nonzero $v$ .

[F3]

Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ : for a $C^{2}$ scalar field, $h (\hat{θ} + v) = h (\hat{θ}) + \nabla h (\hat{θ}) \cdot v + \frac{1}{2} v^{T} D^{2} h (\hat{θ}) v + o (∥ v ∥_{2}^{2})$ .

[F4]

Multivariable Taylor formula with a Lagrange remainder along a line segment and A convex subset of $R^{m}$ contains every line segment between two of its points: on an open convex neighborhood, a $C^{k + 1}$ scalar field equals its Taylor polynomial of degree $k$ plus the stated Lagrange remainder along the segment.

[F5]

A closed subset of a compact metric space is compact and A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a closed subset of compact metric $K$ is compact, and a continuous real-valued function on a nonempty compact metric space attains its maximum.

[F6]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: $∥ \cdot ∥_{2}$ is a norm and satisfies the triangle inequality, so $x \mapsto ∥ x - \hat{θ} ∥_{2}$ is continuous.

[F7]

Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume: under Countable Choice, $λ_{d}$ is a complete measure on the Lebesgue measurable sets.

[F8]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: under Countable Choice, an affine $C^{1}$ diffeomorphism changes a nonnegative Lebesgue integral by its absolute Jacobian determinant.

[F9]

Dominated convergence: almost-everywhere convergence under one integrable majorant permits passage of the limit through the Lebesgue integral.

[F10]

Gaussian integral for a positive-definite quadratic form: under Countable Choice, $\int_{R^{d}} e^{- u^{T} H u / 2} d λ_{d} = (2 π)^{d / 2} / \sqrt{\det H}$ for every real symmetric positive-definite $H$ .

[F11]

The exponential function is strictly increasing: the real exponential is continuous and strictly increasing; for fixed $n$ , $t \mapsto n t$ is continuous, so $t \mapsto e^{n t}$ is continuous.

[F12]

A compact subset of a metric space is closed and bounded, A continuous map has Borel preimages of Borel sets, The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra, and Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable: $K$ is closed, its relative Borel sets are traces of ambient Borel sets, and continuous $h$ is Borel on $K$ ; under Countable Choice these sets are Lebesgue measurable.

[F13]

A measurable function between measurable spaces, A C^1 diffeomorphism maps Lebesgue measurable sets to Lebesgue measurable sets, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs, Arithmetic and lattice operations preserve measurability whenever they are defined, and Closure properties of measurable functions used by the integral: measurability is tested by measurable preimages; under Countable Choice, a $C^{1}$ diffeomorphism and its inverse preserve Lebesgue measurability, so the affine pullback $g_{n}$ in step 3.3 is measurable; $e^{n h}$ and its product with $a$ are measurable, and extension by zero off a measurable set preserves measurability.

[F14]

The nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, and Additivity of the nonnegative Lebesgue integral: the nonnegative integral is the supremum of the simple integrals of its nonnegative simple minorants, is monotone, homogeneous for nonnegative finite scalars, and additive on nonnegative measurable functions.

Proof

1.1A1givenF11F12F13F14

Put $h_{0} : = h (\hat{θ})$ . By [F12], $h$ is Lebesgue measurable on $K$ ; by [F11] and [F13], $a e^{n h}$ is nonnegative and measurable for each $n$ . Since $h \leq h_{0}$ and the exponential is increasing, $0 \leq a e^{n h} \leq e^{n h_{0}} a$ . Thus [F14] and $\int_{K} a < \infty$ show that $I_{n}$ is finite.

2.1givenstep 1.1choose

Choose $0 < δ < \min {r_{0}, r_{1}}$ small enough that $a (θ) \leq a (\hat{θ}) + 1$ whenever $∥ θ - \hat{θ} ∥_{2} < δ$ ; this is possible by continuity of $a$ at $\hat{θ}$ . Then $\overline{B} (\hat{θ}, δ) \subseteq U \cap K$ , and the local quadratic bound holds there.

3.1givenstep 2.1F5F6

The set $A : = K \cap {θ : δ \leq ∥ θ - \hat{θ} ∥_{2} \leq r_{1}}$ is closed in $K$ by [F6], hence compact by [F5]. If $A$ is nonempty, [F5] gives a maximizer of $h$ on $A$ ; it is not $\hat{θ}$ , so uniqueness of the maximizer on $K$ gives $\sup_{A} h < h_{0}$ . Combining this gap with the assumed gap on $K ∖ B (\hat{θ}, r_{1})$ yields some $η^{'} > 0$ such that $h \leq h_{0} - η^{'}$ throughout $K ∖ B (\hat{θ}, δ)$ ; if $A$ is empty, take $η^{'} = η$ .

3.2givenF1step 1.1step 2.1

For each coordinate vector $e_{j}$ , the one-variable function $t \mapsto h (\hat{θ} + t e_{j})$ has a local maximum at $t = 0$ and is differentiable there. Its right difference quotients are nonpositive and its left difference quotients are nonnegative, so the derivative is zero. Thus every component of $\nabla h (\hat{θ})$ vanishes.

3.3A1F2F8F10F13givenstep 2.1algebra

Define $F_{n} (θ) : = a (θ) e^{n (h (θ) - h_{0})}$ on $K \cap B (\hat{θ}, δ)$ and $F_{n} (θ) : = 0$ elsewhere in $R^{d}$ , and put $g_{n} (u) : = F_{n} (\hat{θ} + u / \sqrt{n})$ . By [F13], $F_{n}$ and $g_{n}$ are nonnegative measurable. The affine change of variables $θ = \hat{θ} + u / \sqrt{n}$ is a $C^{1}$ diffeomorphism with Jacobian $n^{- d / 2}$ , so [F8] gives $n^{d / 2} e^{- n h_{0}} \int_{K \cap B (\hat{θ}, δ)} a (θ) e^{n h (θ)} d λ_{d} (θ) = \int_{R^{d}} g_{n} (u) d λ_{d} (u)$ . Equivalently, $g_{n} (u) = a (\hat{θ} + u / \sqrt{n}) e^{n [h (\hat{θ} + u / \sqrt{n}) - h_{0}]}$ when $∥ u ∥_{2} < δ \sqrt{n}$ and $g_{n} (u) = 0$ otherwise. For all $u, n$ , [given] and step 2.1 give $0 \leq g_{n} (u) \leq (a (\hat{θ}) + 1) e^{- c ∥ u ∥_{2}^{2}}$ . This majorant is integrable by [F10] with $H = 2 c I$ , which is positive definite by [F2] and $c > 0$ .

4.1givenF3F4F5step 2.1step 3.2

Fix $u \in R^{d}$ and put $v_{n} : = u / \sqrt{n}$ . Eventually $∥ v_{n} ∥_{2} < δ / 2$ . By [F3] and step 3.2, $n (h (\hat{θ} + v_{n}) - h_{0}) \to - \frac{1}{2} u^{T} H u$ , because $n o (∥ v_{n} ∥_{2}^{2}) = o (1)$ . For the stronger remainder bound, the finitely many third partial derivatives are bounded on the compact set $K \cap \overline{B} (\hat{θ}, δ / 2)$ by [F5]; apply [F4] on the open convex ball $B (\hat{θ}, δ / 2)$ with $k = 2$ to get a remainder $O (∥ v_{n} ∥_{2}^{3})$ , which also vanishes after multiplication by $n$ . Continuity of $a$ gives $a (\hat{θ} + v_{n}) \to a (\hat{θ})$ .

4.2givenstep 1.1F11F13F14step 3.1algebra

On $K ∖ B (\hat{θ}, δ)$ , step 3.1 gives $0 \leq n^{d / 2} e^{- n h_{0}} \int_{K ∖ B (\hat{θ}, δ)} a e^{n h} d λ_{d} \leq n^{d / 2} e^{- n η^{'}} \int_{K} a d λ_{d} ⟶ 0$ . Indeed, the ratio of consecutive terms $n^{d / 2} e^{- n η^{'}}$ tends to $e^{- η^{'}} < 1$ , so the sequence is eventually bounded by a geometric sequence tending to zero.

5.1A1step 4.1step 3.3F7F9F10

Steps 4.1 and 3.3 show pointwise convergence of $g_{n}$ to $a (\hat{θ}) e^{- u^{T} H u / 2}$ and provide an integrable majorant. Applying dominated convergence [F9] and then [F10] yields $n^{d / 2} e^{- n h_{0}} \int_{K \cap B (\hat{θ}, δ)} a e^{n h} d λ_{d} ⟶ a (\hat{θ}) (2 π)^{d / 2} / \sqrt{\det H}$ .

6.1A1givenF10F13F14step 5.1step 4.2algebra∎

By [F13] the two indicator-restricted integrands are measurable, and by [F14] the integral over $K$ is the sum of its integrals over $K \cap B (\hat{θ}, δ)$ and $K ∖ B (\hat{θ}, δ)$ . Adding the local limit in step 5.1 and the vanishing off-mode term in step 4.2 gives $n^{d / 2} e^{- n h_{0}} I_{n} ⟶ a (\hat{θ}) (2 π)^{d / 2} / \sqrt{\det H} > 0$ . Dividing by this positive limit proves the stated asymptotic equivalence. The argument establishes only the leading relative equivalent; it gives no $O (n^{- 1})$ remainder.

Remarks

The compactness and explicit gap assumptions localize the mass. For a noncompact domain, an additional integrable-tail argument is required before applying this proof.
The proof gives a leading equivalent and does not claim a relative error rate.

Depends on

Used by

A full-rank local Hessian does not capture both posterior modes Counterexample
Two equal modes contribute two Gaussian terms Counterexample

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Sources

Laplace method for integrals, PHYS 2400, University of Connecticut (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §2.3(iii), Laplace's Method (standard reference, not scraped)
Francis Bach, Approximating Integrals with Laplace's Method (standard reference, not scraped)