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 binary SCM where front-door adjustment works

Example

Let $U \sim B e r n o u l l i (1 / 2)$ , $E_{A} \sim B e r n o u l l i (1 / 4)$ , and $E_{M} \sim B e r n o u l l i (1 / 5)$ be mutually independent. Set $A = U x o r E_{A}, M = A x o r E_{M}, Y = M \lor U .$ The full graph is $U \to A$ , $U \to Y$ , $A \to M$ , $M \to Y$ . The only directed $A$ -to- $Y$ path passes through $M$ . The path $A \leftarrow U \to Y \leftarrow M$ is blocked by its collider $Y$ , so $A$ and $M$ are d-separated by the empty set in $G_{\underline{A}}$ . The path $M \leftarrow A \leftarrow U \to Y$ is blocked by $A$ when checking the other front-door condition. Both mediator values have positive probability conditional on either treatment value, so the nested conditional laws have the required support.

Facts & Assumptions

Given: The finite SCM and independent Bernoulli laws in the Example.

[F1]

The exogenous family has its finite product law. Independent random elements, Independent random elements have product joint law

[F2]

The front-door theorem states the observed functional and its graph and support hypotheses. Front-door adjustment identifies an interventional mean

[F3]

Interventions replace only the named structural equation. Recursive structural causal models on finite DAGs

Verification

technique · evaluate the observed front-door functional and the interventional model directly

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

1.1F1F2algebra

Since $P (A = 1) = 1 / 2$ , Bayes' rule gives $P (U = 1 ∣ A = 1) = 3 / 4$ and $P (U = 1 ∣ A = 0) = 1 / 4$ . Since $E_{M}$ is independent of $(U, A)$ , $P (M = 1 ∣ A = 1) = 4 / 5$ and $P (M = 1 ∣ A = 0) = 1 / 5$ . Thus for each $a \in {0, 1}$ both mediator outcomes have positive conditional probability; the nested-law support condition in [F2] holds.

2.1step 1.1F1F2algebra

If $M = 1$ , then $Y = 1$ . If $M = 0$ , then $Y = U$ , so $E (Y ∣ M = 0, A = 1) = 3 / 4$ and $E (Y ∣ M = 0, A = 0) = 1 / 4$ . Averaging over the marginal law of $A$ gives $\int E (Y ∣ M = 0, A = a^{'}) P_{A} (d a^{'}) = 1 / 2$ , while the same inner average for $M = 1$ is $1$ . The front-door formula [F2] therefore gives $\frac{4}{5} \cdot 1 + \frac{1}{5} \cdot \frac{1}{2} = \frac{9}{10} .$

3.1F1F3step 1.1step 2.1algebra∎

Under $d o (A = 1)$ , $M = 1$ with probability $4 / 5$ and $M = 0$ with probability $1 / 5$ . The exogenous $U$ remains Bernoulli $(1 / 2)$ and independent of $E_{M}$ , so $E (Y ∣ d o (A = 1), M = 1) = 1$ and $E (Y ∣ d o (A = 1), M = 0) = 1 / 2$ . Hence $E [Y^{1}] = \frac{4}{5} \cdot 1 + \frac{1}{5} \cdot \frac{1}{2} = 9 / 10$ , agreeing with the observed functional.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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

Judea Pearl, Causal Diagrams for Empirical Research, §3-2, pp. 675–676 (standard reference, not scraped)
Miguel A. Hernán and James M. Robins, Causal Inference: What If, Technical Point 7.4, p. 101 (standard reference, not scraped)