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.

IV observed contrasts as potential-outcome contrasts

Statement

In an IV setup satisfying consistency, exclusion, and joint instrument independence, let τ=Y(1)−Y(0). Then E[A∣Z=1]−E[A∣Z=0]=E[A(1)−A(0)] and E[Y∣Z=1]−E[Y∣Z=0]=E[(A(1)−A(0))τ]. No monotonicity or relevance condition is needed for these equalities; positivity of both instrument atoms and integrability are part of the setup.

Facts & Assumptions

Given: The IV setup of Binary instrumental-variable potential-outcome setup, consistency A=A(Z) and Y=Y(A,Z), exclusion Y(a,z)=Y(a), and joint independence Z⊥(A(0),A(1),Y(0),Y(1)). Show the two displayed contrasts.

[A1]

The setup supplies binary Z,A(z) and integrable Y(a,z), with positive P(Z=z) for z=0,1. (Binary instrumental-variable potential-outcome setup)

[A2]

The setup assumes consistency, exclusion and joint instrument independence. (Binary instrumental-variable potential-outcome setup)

[F1]

Measurable functions of independent random elements remain independent. (Measurable coordinatewise functions preserve independence)

[F2]

For independent integrable real random variables, the expectation of their product factors as the product of their expectations. (Expectations factor over finite products of independent random variables)

[F3]

Finite sums, scalar multiples and products of measurable real functions are measurable whenever pointwise defined. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[F4]

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

Proof

1.1givenA1A2F1F3

Put W=(A(0),A(1),Y(0),Y(1)) and Iz=1{Z=z}. The functions A(z) and Y(0)+A(z)(Y(1)−Y(0)) are measurable functions of W by [F3], and are integrable because A(z) is binary and Y(0),Y(1) are integrable. Independence in [A2] and [F1] make each such function independent of Z.

2.1A1F2step 1.1

By [F2], for each of these integrable functions H(W) we have E[IzH(W)]=E[Iz]E[H(W)]=P(Z=z)E[H(W)]. Dividing by the positive P(Z=z) and using the event-conditioned mean in [A1] gives E[H(W)∣Z=z]=E[H(W)].

3.1A1A2step 2.1

On {Z=z}, consistency and exclusion in [A2] give A=A(z) and Y=Y(A(z))=Y(0)+A(z)(Y(1)−Y(0)). Thus step 2.1 yields E[A∣Z=z]=E[A(z)] and E[Y∣Z=z]=E[Y(0)+A(z)τ].

4.1F4step 3.1algebra∎

Subtracting the z=0 identities from the z=1 identities in step 3.1, and using [F4], gives E[A(1)−A(0)] and E[(A(1)−A(0))τ], respectively.

Depends on

Used by

Dependency tree · two levels

31 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