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Identifiability is constancy on observational classes

Statement

Let P:Θ→P(X,A) be a parametric model, let ψ:Θ→T0 be an estimand, and set M:=P(Θ), the set of laws in the model. For θ∈Θ, write

[θ]∼:={η∈Θ:η∼θ}.

Then the following are equivalent: (i) ψ is identifiable; (ii) ψ is constant on every observational equivalence class [θ]∼; and (iii) there is a unique function ψˉ:M→T0 such that ψ=ψˉ∘P.

The factorization is a statement about sets and functions. It asserts no measurable structure or measurable factor map.

Facts & Assumptions

Given: A parametric model P:Θ→P(X,A) and an estimand ψ:Θ→T0.

[F1]

Parametric statistical models: the model is the parameter map P:Θ→P(X,A), and its set of laws is M=P(Θ).

[F2]

Observational equivalence: θ∼η exactly when Pθ=Pη as measures.

[F3]

Estimands on a statistical model: an estimand is a function ψ:Θ→T0.

[F4]

Identifiability of an estimand: ψ is identifiable exactly when θ∼η implies ψ(θ)=ψ(η).

Proof

technique · Direct equivalence and factorization through the image of the model map
1.1givenF2algebra

Equality of measures is reflexive, symmetric and transitive; by [F2], ∼ is therefore an equivalence relation, so each class [θ]∼ is defined.

1.2givenF1F2F4algebra

If ψ=ψˉ∘P, then Pθ=Pη gives ψˉ(Pθ)=ψˉ(Pη) and hence ψ(θ)=ψ(η). Thus (iii) implies identifiability.

1.3givenF1F2F3F4construct

Suppose ψ is identifiable. Form the relation G:={(Q,a):there exists θ∈Θ with Q=Pθ and a=ψ(θ)}. Every Q∈M has at least one such parameter by [F1]. If both θ and η represent the same Q, then [F2] gives θ∼η, so [F4] gives ψ(θ)=ψ(η). Thus G has exactly one value for each Q∈M, defines a function ψˉ:M→T0 by [F3], and satisfies ψ=ψˉ∘P. This construction defines the graph directly and uses no choice axiom.

2.1F2F4step 1.1algebra

By [F2], η∈[θ]∼ exactly when η∼θ. If a,b∈[θ]∼, then a∼θ and b∼θ, so symmetry and transitivity from step 1.1 give a∼b; conversely, θ∼η gives η∈[θ]∼. Hence [F4] is exactly constancy of ψ on every class.

3.1givenF1F3step 1.2step 1.3step 2.1algebra∎

If ψˉ1 and ψˉ2 both factor ψ, each Q∈M equals Pθ for some θ by [F1], and then ψˉ1(Q)=ψ(θ)=ψˉ2(Q) by [F3]. The factor is unique. The implication in step 1.2 and the construction in step 1.3 give (i) iff (iii), while step 2.1 gives (i) iff (ii).

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources