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Statistical Experiments, Models and Identifiability — Examples

1 · Prerequisites

2 · Summary

The Bernoulli family identifies its success probability because that probability is the mass assigned to the event ${1}$ . A redundant two-parameter Bernoulli model shows that a coordinate can fail to be identifiable when the sampling law depends only on a sum of the coordinates.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Identifiable Bernoulli model and parameter target

Example

Let $X = {0, 1}$ with $A = 2^{X}$ , let $Θ = [0, 1]$ , and, for $p \in Θ$ , define

P_{p} (A) : = p 1_{{1 \in A}} + (1 - p) 1_{{0 \in A}} (A \in A),

where each indicator is $1$ when its stated membership holds and $0$ otherwise. This is the Bernoulli model with success probability $p$ . Its parameter is identifiable, and the target $ψ (p) = p$ is identifiable.

Facts & Assumptions

Given: The parameter space is $[0, 1]$ and the sample space is ${0, 1}$ .

[F1]

Sigma-algebras: a sigma-algebra contains the whole space and is closed under complements and countable unions; the full power set satisfies these conditions.

[F2]

Measurable spaces and measurable sets: a measurable space is a set together with a sigma-algebra on it.

[F3]

Probability measures and probability spaces: a probability measure is a measure of total mass $1$ on a measurable space.

[F4]

Statistical experiments: an experiment assigns a probability measure $P_{p}$ on the sample measurable space to each parameter $p$ .

[F5]

Parametric statistical models: the model map sends $p$ to its sampling law $P_{p}$ .

[F6]

Estimands on a statistical model: $ψ (p) = p$ is a function from the parameter set to the target set $[0, 1]$ .

[F7]

Identifiability of an estimand: identifiability means equal sampling laws imply equal target values.

[F8]

Identifiability is constancy on observational classes: an identifiable target factors uniquely through the set of sampling laws.

Verification

Given: The finite Bernoulli family and target specified above.

1.1givenF1F2F3F4algebra

The full power set contains $X$ and is closed under finite unions and complements, so it is an algebra of subsets. Every countable union of its members is still a subset of $X$ , so it is a sigma-algebra by [F1] and gives a measurable space by [F2]. For any $p \in [0, 1]$ , both singleton masses $p$ and $1 - p$ are nonnegative, and $P_{p} (X) = p + (1 - p) = 1$ . If $(A_{n})$ is pairwise disjoint, each of $0$ and $1$ belongs to at most one $A_{n}$ , so $\sum_{n} P_{p} (A_{n}) = p 1_{{1 \in \cup_{n} A_{n}}} + (1 - p) 1_{{0 \in \cup_{n} A_{n}}} = P_{p} (\cup_{n} A_{n})$ . Hence every $P_{p}$ is a probability measure by [F3].

2.1F5F6F7F8algebra∎

If $P_{p} = P_{p^{'}}$ , evaluation on ${1}$ gives $p = P_{p} ({1}) = P_{p^{'}} ({1}) = p^{'}$ . Thus the model map is injective and equal laws force $ψ (p) = ψ (p^{'})$ ; the target is identifiable by [F7]. By [F8] it factors uniquely through the set of Bernoulli laws.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A redundant Bernoulli parameterization has a nonidentifiable coordinate

Statement refuted

For every parametric statistical model with parameter $(u, v) \in Θ \subseteq [0, 1]^{2}$ , the first-coordinate target $ψ (u, v) = u$ is identifiable.

Facts & Assumptions

Given: The parameter set $Θ$ , finite sample space, law family and target stated above.

[F1]

Sigma-algebras: a sigma-algebra contains the whole space and is closed under complements and countable unions; the full power set satisfies these conditions.

[F2]

Measurable spaces and measurable sets: a measurable space is a set together with a sigma-algebra on it.

[F3]

Probability measures and probability spaces: a probability measure has total mass $1$ and is countably additive on measurable sets.

[F4]

Statistical experiments: an experiment is a parameter-indexed family of probability measures on the sample measurable space.

[F5]

Parametric statistical models: each parameter is sent to its sampling law $P_{u, v}$ .

[F6]

Estimands on a statistical model: the first coordinate $ψ (u, v) = u$ is a function to $[0, 1]$ .

[F7]

Identifiability of an estimand: an estimand is identifiable only if equal sampling laws imply equal target values.

[F8]

Identifiability is constancy on observational classes: identifiability is equivalent to a unique factorization of the target through the set of model laws.

Counterexample

Take $X = {0, 1}$ with $A = 2^{X}$ and

Θ : = {(u, v) \in [0, 1]^{2} : u + v \leq 1} .

For each $(u, v) \in Θ$ , define

P_{u, v} (A) : = (u + v) 1_{{1 \in A}} + (1 - u - v) 1_{{0 \in A}} (A \in A),

and set $ψ (u, v) = u$ . The law depends only on the sum, while the target is the first coordinate.

Given: The redundant two-parameter Bernoulli family and first-coordinate target stated above.

1.1givenF1F2F3F4algebra

The full power set contains $X$ and is closed under finite unions and complements, so it is an algebra of subsets. Every countable union of its members is still a subset of $X$ , so it is a sigma-algebra by [F1] and gives a measurable space by [F2]. For every $(u, v) \in Θ$ , both masses $u + v$ and $1 - u - v$ are nonnegative and sum to $1$ . In a pairwise disjoint sequence of events, each singleton ${0}$ and ${1}$ appears at most once, so the displayed formula is countably additive. Thus $P_{u, v}$ is a probability measure by [F3], and the family is an experiment by [F4].

2.1givenF5F6F7F8algebra∎

The points $(u, v) = (1 / 5, 1 / 2)$ and $(u^{'}, v^{'}) = (3 / 10, 2 / 5)$ have coordinates in $[0, 1]$ , and their sums are $1 / 5 + 1 / 2 = 2 / 10 + 5 / 10 = 7 / 10$ and $3 / 10 + 2 / 5 = 3 / 10 + 4 / 10 = 7 / 10 \leq 1$ , so both belong to $Θ$ . Both measures assign $7 / 10$ to ${1}$ and $3 / 10$ to ${0}$ ; the four events $\emptyset, {0}, {1}, X$ consequently have probabilities $0, 3 / 10, 7 / 10, 1$ under both laws, so $P_{u, v} = P_{u^{'}, v^{'}}$ . But $ψ (u, v) = 1 / 5 \neq 3 / 10 = ψ (u^{'}, v^{'})$ . By [F7], $ψ$ is not identifiable; by [F8], it cannot factor through the model-law set.