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Identifiability is constancy on observational classes
Statement
Let be a parametric model, let be an estimand, and set , the set of laws in the model. For , write
Then the following are equivalent: (i) is identifiable; (ii) is constant on every observational equivalence class ; and (iii) there is a unique function such that .
The factorization is a statement about sets and functions. It asserts no measurable structure or measurable factor map.
Facts & Assumptions
Given: A parametric model and an estimand .
Parametric statistical models: the model is the parameter map , and its set of laws is .
Observational equivalence: exactly when as measures.
Estimands on a statistical model: an estimand is a function .
Identifiability of an estimand: is identifiable exactly when implies .
Proof
Equality of measures is reflexive, symmetric and transitive; by [F2], is therefore an equivalence relation, so each class is defined.
If , then gives and hence . Thus (iii) implies identifiability.
Suppose is identifiable. Form the relation . Every has at least one such parameter by [F1]. If both and represent the same , then [F2] gives , so [F4] gives . Thus has exactly one value for each , defines a function by [F3], and satisfies . This construction defines the graph directly and uses no choice axiom.
By [F2], exactly when . If , then and , so symmetry and transitivity from step 1.1 give ; conversely, gives . Hence [F4] is exactly constancy of on every class.
If and both factor , each equals for some by [F1], and then 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
- A redundant Bernoulli parameterization has a nonidentifiable coordinate Counterexample
- Binding function and indirect-inference target Definition
- A non-injective binding map cannot recover a sign Example
- Identifiable Bernoulli model and parameter target Example
- Backdoor adjustment identifies an interventional law Theorem
- Conditional-class standardization identifies a causal mean Theorem
- Front-door adjustment identifies an interventional mean Theorem
- Inverse probability weighting identifies a causal mean Theorem
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
- Ryan Martin, Lecture Notes on Advanced Statistical Theory, §§1.3.5–1.3.6, PDF pp. 19–21 (standard reference, not scraped)
- Charles J. Geyer, Stat 5421 Lecture Notes: Exponential Families, §6, PDF pp. 6–7 (standard reference, not scraped)