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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The same observed law can have different intervention laws

Example

Let $U \sim B e r n o u l l i (1 / 2)$ . In SCM 1 set $A = U$ and $Y = A$ ; its graph is $U \to A \to Y$ . In SCM 2 set $A = U$ and $Y = U$ ; its graph is $U \to A$ and $U \to Y$ . Degenerate independent exogenous errors at $A$ and $Y$ complete both Markovian SCMs.

Facts & Assumptions

Given: The two binary SCMs in the Example.

[F1]

Observational equivalence means equality of the induced sampling laws. Parametric statistical models, Observational equivalence

[F2]

An estimand is identifiable only if it is constant over parameter values with the same observed law. Estimands on a statistical model, Identifiability of an estimand

[F3]

Under an intervention only the named structural equation is replaced. Recursive structural causal models on finite DAGs

Verification

technique · compare the finite observational laws and then compute both truncated structural systems

Given: [F1]–[F3] and the two SCMs in the Example.

1.1givenF1algebra

In either model, $A = Y = U$ , so the observed joint law assigns probability $1 / 2$ to $(A, Y) = (0, 0)$ and $1 / 2$ to $(A, Y) = (1, 1)$ , and zero to the other pairs. The parameter values representing the two SCMs are observationally equivalent by [F1].

2.1step 1.1F2F3algebra∎

Under $d o (A = a)$ in SCM 1, the equation $Y = A$ remains and gives $Y = a$ with probability one. Under $d o (A = a)$ in SCM 2, the equation $Y = U$ remains and $U$ is unchanged, so $Y$ is Bernoulli $(1 / 2)$ . Thus for $a = 0$ the interventional success probabilities are respectively $0$ and $1 / 2$ ; the intervention-law estimand is not constant on the observational-equivalence class in step 1.1. It is therefore not identifiable on the two-model statistical model by [F2].

Depends on

Used by

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Dependency tree · two levels

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Sources

Judea Pearl, Causal Diagrams for Empirical Research, §3-1, pp. 674–675 (standard reference, not scraped)
Miguel A. Hernán and James M. Robins, Causal Inference: What If, §6.1, pp. 71–74 (standard reference, not scraped)