Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

The monotone IV first stage is the complier proportion

Statement

Under consistency, joint instrument independence and monotonicity in the binary IV setup, E[A∣Z=1]−E[A∣Z=0]=P(C), where C={A(1)>A(0)} is the complier event. If the instrument is relevant, then P(C)>0.

Facts & Assumptions

Given: A binary IV setup with consistency, joint instrument independence, monotonicity and relevance. Prove the first-stage identity and positivity of the complier stratum.

[A1]

Relevance means the observed first-stage contrast is nonzero. (Binary instrumental-variable potential-outcome setup)

[F1]

Independent random elements have independent generated sigma-algebras; events from these sigma-algebras satisfy P(B∩D)=P(B)P(D). (Independent random elements, Independent sigma-algebras and independent events)

[F2]

Under monotonicity the defier event is null, and A(1)−A(0) is 1 on C and 0 on the other principal strata. (Binary monotonicity excludes defiers, Binary-instrument principal strata and LATE)

[F3]

The integral is linear on integrable functions. (The Lebesgue integral is linear on L1(μ))

Proof

1.1givenF2

On the complier event C, (A(0),A(1))=(0,1), so A(1)−A(0)=1; on the always-taker and never-taker events the difference is 0. The defier event is null by [F2]. Hence A(1)−A(0)=1C almost surely.

2.1givenF1F3step 1.1algebra

For z=0,1, the event {A(z)=1} belongs to the sigma-algebra of the potential-value tuple. Joint instrument independence and [F1] give P(A(z)=1,Z=z)=P(A(z)=1)P(Z=z). Since A=A(z) on {Z=z} by consistency and P(Z=z)>0, the event-conditioned mean in the setup gives E[A∣Z=z]=P(A(z)=1,Z=z)/P(Z=z)=P(A(z)=1)=E[A(z)]. Subtracting these identities and using [F3] and step 1.1 yields E[A∣Z=1]−E[A∣Z=0]=E[A(1)−A(0)]=P(C), without an exclusion assumption.

3.1A1step 2.1∎

Relevance [A1] says this first-stage contrast is nonzero. Since P(C) is nonnegative, step 2.1 implies P(C)>0.

Depends on

Used by

Dependency tree · two levels

15 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