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.

A confounder and a conditioned collider

Example

Consider first a confounding SCM with independent $U \sim B e r n o u l l i (1 / 2)$ and $E \sim B e r n o u l l i (1 / 4)$ , observed $A = U x o r E$ , and observed $Y = U$ . Its DAG is $U \to A$ and $U \to Y$ . The only path between $A$ and $Y$ is the backdoor path $A \leftarrow U \to Y$ .

For collider conditioning, let independent $B, C \sim B e r n o u l l i (1 / 2)$ , and set $D = B \lor C$ . Its DAG is $B \to D \leftarrow C$ .

Facts & Assumptions

Given: The finite-valued recursive models and mutually independent Bernoulli exogenous variables specified above.

[F1]

Mutual independence gives the product joint law for each finite exogenous family. Independent random elements, Independent random elements have product joint law

[F2]

A noncollider conditioned on blocks a path, while conditioning on a collider activates it. Paths, active paths, and d-separation in a finite DAG

[F3]

The causal response is obtained by replacing only the treatment equation. Recursive structural causal models on finite DAGs

Verification

technique · calculate the two finite joint laws and compare their conditional probabilities

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

1.1F1F2F3algebra

In the confounding model, $P (A = 1) = P (U = 0, E = 1) + P (U = 1, E = 0) = \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4} = \frac{1}{2}$ . Also $P (U = 1, A = 1) = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$ and $P (U = 1, A = 0) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$ . Thus $P (Y = 1 ∣ A = 1) = 3 / 4$ and $P (Y = 1 ∣ A = 0) = 1 / 4$ , although under $d o (A = a)$ the variable $Y = U$ remains Bernoulli $(1 / 2)$ . The difference between the observational conditionals and the intervention law is caused by the open backdoor path through $U$ ; conditioning on $U$ blocks that path and the adjusted mean is $\sum_{u \in {0, 1}} P (Y = 1 ∣ A = a, U = u) P (U = u) = 1 / 2$ .

2.1F1F2algebra∎

For the collider model, $P (B = 1, C = 1) = 1 / 4 = P (B = 1) P (C = 1)$ by [F1], so the variables are independent before conditioning. But $D = 1$ on exactly the three equiprobable pairs $(1, 1), (1, 0), (0, 1)$ . Hence $P (B = 1 ∣ D = 1) = P (C = 1 ∣ D = 1) = 2 / 3$ while $P (B = 1, C = 1 ∣ D = 1) = 1 / 3 \neq 4 / 9$ . Conditioning on the collider $D$ therefore creates dependence, exactly as the active-path rule [F2] predicts.

Depends on

Used by

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Dependency tree · two levels

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Sources

Judea Pearl, Causal Diagrams for Empirical Research, §2-1 and §3-1, pp. 671–675 (standard reference, not scraped)
Miguel A. Hernán and James M. Robins, Causal Inference: What If, Fine Point 6.1 and §7.1, pp. 78, 85–86 (standard reference, not scraped)