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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.

Depends on

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Sources

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 21, PDF pp. 14–19 (standard reference, not scraped)
John Duchi, A few notes on contiguity, asymptotics, and local asymptotic normality, §§1.2–1.3 (standard reference, not scraped)