ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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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 $P_{0} = Bernoulli (1 / 4)$ and $P_{1} = Bernoulli (3 / 4)$ , with prior mass $1 / 2$ at each parameter. This finite dominated model is identifiable, has KL support at its true law $P_{0}$ , and admits exponentially consistent tests. Hence the posterior mass at parameter $1$ tends to zero almost surely when the data are IID from $P_{0}$ .

Facts & Assumptions

Given: The finite model, proper prior, IID observations from $P_{0}$ , 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 $0 \leq Y \leq 1$ , then $E \exp {λ (Y - E Y)} \leq \exp (λ^{2} / 8)$ 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 $a > 0$ , $P (Z \geq a) \leq E Z / a$ . Markov's inequality for random variables

[F8]

The series $\sum_{n \geq 1} r^{n}$ converges for $0 < r < 1$ . For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges

[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 $p_{0} (0) = 3 / 4$ , $p_{0} (1) = 1 / 4$ , $p_{1} (0) = 1 / 4$ , and $p_{1} (1) = 3 / 4$ ; 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 $P_{0} ({1}) = 1 / 4 \neq 3 / 4 = P_{1} ({1})$ , so [F1] makes the model identifiable. For every $ϵ > 0$ , the KL neighborhood contains $θ = 0$ because $d P_{0} / d P_{0} = 1$ and [F2] gives $D (P_{0} ∥ P_{0}) = \int \log 1 d P_{0} = 0$ ; its prior mass is therefore at least $1 / 2$ .

1.2F3F4F5F7F9givenalgebra

Let $d$ be library total variation and put $ϕ_{n} = 1 {n^{- 1} \sum_{i = 1}^{n} X_{i} > 1 / 2}$ . Since $d (P_{0}, P_{1}) = 1 / 2$ by [F3], the parameter complement is $V_{1 / 4} = {1}$ , whose law is $P_{1}$ . Under either law, [F9] rewrites the exponential of the centered sum as a product, and independence [F4] factors its MGF into at most $e^{n λ^{2} / 8}$ by [F5]. Under $P_{0}$ , rejection implies $\sum_{i} (X_{i} - 1 / 4) > n / 4$ ; Markov [F7] applied to the exponential with $λ = 1$ gives $P_{0}^{n} ϕ_{n} \leq e^{- n / 8}$ . Under $P_{1}$ , nonrejection implies $\sum_{i} (X_{i} - 3 / 4) \leq - n / 4$ ; Markov [F7] with $λ = - 1$ gives $P_{1}^{n} (1 - ϕ_{n}) \leq e^{- n / 8}$ .

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 $V_{η} = {1}$ (whose law is $P_{1}$ ); for $η \geq 1 / 2$ it is empty, so the same tests verify every required nonempty complement. The theorem [F6] therefore gives $Π_{n} ({1} ∣ X_{1 : n}) = Π_{n} (V_{1 / 4} ∣ X_{1 : n}) \to 0$ almost surely for any IID sequence with common law $P_{0}$ .

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