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Hellinger geometry and posterior consistency — Examples

1 · Prerequisites

2 · Summary

Objective

The examples verify the consistency hypotheses in a finite identifiable Bernoulli model and display the exact factor between density L1 distance and event-supremum total variation. The main page's counterexample gives an identifiable dominated model with KL support but no uniformly consistent tests.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

A finite identifiable model with posterior consistency

Example

Under the library AC convention for KL divergence and posterior laws, take Θ={0,1}, data space {0,1}, and laws P0=Bernoulli⁡(1/4) and P1=Bernoulli⁡(3/4), with prior mass 1/2 at each parameter. This finite dominated model is identifiable, has KL support at its true law P0, and admits exponentially consistent tests. Hence the posterior mass at parameter 1 tends to zero almost surely when the data are IID from P0.

Facts & Assumptions

Given: The finite model, proper prior, IID observations from P0, and the test constructed in the Verification.

[A1]

AC is included through the imported library KL definition and the posterior-consistency theorem. The Axiom of Choice

[F1]

A parametric model is identifiable exactly when its parameter-to-law map is injective. Identifiability of an estimand

[F2]

KL divergence is directional and D(P∥P)=0 for a probability law with itself, by its density-ratio definition. Kullback–Leibler divergence

[F3]

The library total variation distance is sup⁡A∣P(A)−Q(A)∣. Total variation distance for probability laws

[F4]

A finite vector of independent random elements has the product of its marginal laws. Independent random elements have product joint law

[F5]

If 0≤Y≤1, then Eexp⁡{λ(Y−EY)}≤exp⁡(λ2/8) for every real λ. The Hoeffding MGF bound for a variable in a unit interval

[F6]

In a well-specified dominated IID model, prior KL support and summable type-I plus uniform type-II testing errors imply almost-sure posterior consistency on every measurable metric-separated complement. Posterior consistency from KL support and uniformly summable tests

[F7]

For nonnegative Z and a>0, P(Z≥a)≤EZ/a. Markov's inequality for random variables

[F9]

The exponential is positive and strictly increasing, satisfies exp⁡(0)=1 and exp⁡(−x)=1/exp⁡(x), and obeys exp⁡(x+y)=exp⁡(x)exp⁡(y). The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y), The exponential function is strictly increasing, The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x)

[F10]

A parameter-measurable model is a probability kernel, domination supplies a sigma-finite common measure, and a likelihood density is a supplied measurable representative. Parameter-measurable statistical models as kernels, Dominated statistical experiments, Likelihood densities

[F11]

A probability measure has total mass one. Probability measures and probability spaces

Verification

Given: The definitions and hypotheses in the Example.

1.1A1F1F2F10F11given

The parameter and observation spaces have their full finite sigma-algebras. Counting measure is finite and hence sigma-finite. Its densities are p0(0)=3/4, p0(1)=1/4, p1(0)=1/4, and p1(1)=3/4; they normalize, are jointly measurable, and define a parameter-measurable kernel. The two prior masses sum to one, so [F10,F11] verify the dominated kernel and proper prior hypotheses. The parameter-to-law map is injective because P0({1})=1/4≠3/4=P1({1}), so [F1] makes the model identifiable. For every ϵ>0, the KL neighborhood contains θ=0 because dP0/dP0=1 and [F2] gives D(P0∥P0)=∫log⁡1 dP0=0; its prior mass is therefore at least 1/2.

1.2F3F4F5F7F9givenalgebra

Let d be library total variation and put ϕn=1{n−1∑i=1nXi>1/2}. Since d(P0,P1)=1/2 by [F3], the parameter complement is V1/4={1}, whose law is P1. Under either law, [F9] rewrites the exponential of the centered sum as a product, and independence [F4] factors its MGF into at most enλ2/8 by [F5]. Under P0, rejection implies ∑i(Xi−1/4)>n/4; Markov [F7] applied to the exponential with λ=1 gives P0nϕn≤e−n/8. Under P1, nonrejection implies ∑i(Xi−3/4)≤−n/4; Markov [F7] with λ=−1 gives P1n(1−ϕn)≤e−n/8.

2.1A1F6F8F9F10F11step 1.1step 1.2algebra∎

By [F9], r=e−1/8 lies in (0,1), so [F8] makes both test-error series converge. For every 0<η<1/2, the parameter complement is Vη={1} (whose law is P1); for η≥1/2 it is empty, so the same tests verify every required nonempty complement. The theorem [F6] therefore gives Πn({1}∣X1:n)=Πn(V1/4∣X1:n)→0 almost surely for any IID sequence with common law P0.

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

The factor of two between density L1 and library total variation

Example

Under the library AC convention for density representatives, on {0,1} with counting measure let P=δ0 and Q=δ1. Their density L1 distance is 2, while their library event-supremum total variation is 1 and their root-L2 Hellinger distance is 2. Thus the factor in h2≤2 TV is sharp.

Facts & Assumptions

Given: The two point-mass laws on the two-point measurable space and counting measure.

[A1]

AC is inherited from the likelihood-density convention used by the Hellinger definition; the two densities here are explicitly supplied. The Axiom of Choice

[F1]

For densities p,q against a common measure, h(P,Q)2=∫(p−q)2 dμ. Root-L2 Hellinger distance for dominated laws

[F2]

Library total variation is sup⁡A∣P(A)−Q(A)∣. Total variation distance for probability laws

[F3]

For this convention, TV(P,Q)≤h(P,Q) and h(P,Q)2≤2 TV(P,Q). Hellinger invariance and comparison with library total variation

Verification

Given: The two probability laws and their counting-measure densities.

1.1F2givenalgebra

The densities are p(0)=1,p(1)=0 and q(0)=0,q(1)=1, so ∑x∈{0,1}∣p(x)−q(x)∣=2. The four measurable events have probability differences 0,1,−1,0, hence the supremum in [F2] is 1.

2.1A1F1F3step 1.1algebra∎

By [F1], h(P,Q)2=(1−0)2+(0−1)2=2, so h(P,Q)=2. Therefore h(P,Q)2=2 TV(P,Q), attaining equality in the upper comparison in [F3]; the lower comparison reads 1≤2.