ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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 finite randomized encouragement example

Example

Take four equally likely unit types: one always-taker with $(A (0), A (1)) = (1, 1)$ and $(Y (0), Y (1)) = (0, 4)$ ; one never-taker with $(0, 0)$ and $(0, 4)$ ; and two compliers with $(0, 1)$ and outcome pairs $(0, 1)$ and $(0, 3)$ . Independently assign $Z$ with probabilities $1 / 2, 1 / 2$ , impose consistency and exclusion, and let $A = A (Z)$ and $Y = Y (A)$ . The model has no defiers, but the effects vary by type.

Facts & Assumptions

Given: The finite law described in the Example. Verify its first stage, reduced form, LATE, ATE and Wald ratio.

[F1]

The binary IV setup defines consistency, exclusion, independence, relevance, the event-conditioned means, and the four principal strata. (Binary instrumental-variable potential-outcome setup, Binary-instrument principal strata and LATE)

[F2]

Under the stated relevance, exclusion, independence, consistency and monotonicity assumptions, the Wald ratio equals the complier mean effect. (The Wald ratio identifies the complier average treatment effect)

Verification

1.1givenF1constructalgebra

On the product of the four equally likely types and the fair assignment, $Z$ is independent of the potential values. The two compliers have effects $1$ and $3$ , while the always-taker and never-taker both have treatment effect $4$ ; thus $P (C) = 1 / 2$ , $LATE = (1 + 3) / 2 = 2$ , and $ATE = (4 + 4 + 1 + 3) / 4 = 3$ .

2.1F1F2step 1.1algebra∎

The treatment rate is $3 / 4$ under $Z = 1$ and $1 / 4$ under $Z = 0$ , so $Δ_{A} = 1 / 2$ . The outcome mean is $2$ under $Z = 1$ and $1$ under $Z = 0$ , so $Δ_{Y} = 1$ . Therefore $ψ_{W} = 1 / (1 / 2) = 2$ , agreeing with the LATE in [F2] and differing from the ATE.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Hernán and Robins, Causal Inference: What If, Chapter 16, §16.4 and Technical Point 16.7, printed pp. 217–225 (standard reference, not scraped)
Hyunseung Kang, Causal Inference: Identification Under Unmeasured Confounding (Instrumental Variables), randomized encouragement example and LATE proof (standard reference, not scraped)