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 two-stratum standardization and IPW calculation

Example

Let $L \in {0, 1}$ with equal probabilities, let $P (A = 1 ∣ L = 0) = 1 / 4$ and $P (A = 1 ∣ L = 1) = 3 / 4$ , and set $Y^{1} = 1 + 2 L$ and $Y^{0} = 0$ . Generate $A$ using the stated conditional law and define $Y = Y^{A}$ . Because both potential outcomes are functions of $L$ , conditional exchangeability holds; positivity holds in both strata.

Facts & Assumptions

Given: The finite probability law and potential outcomes in the Example, with consistency and AC for the conditional-expectation conventions.

[A1]

The example uses the point-treatment potential-outcome convention of Potential outcomes and consistency.

[F1]

Standardization averages the treatment-specific observed mean over the law of $L$ . (Standardization for a causal mean)

[F2]

IPW averages $I_{a} Y / e_{a} (L)$ under the observed population law. (Inverse probability weighting for a causal mean)

[F3]

Under the stated consistency, exchangeability, integrability and positivity assumptions, the standardization functional equals $E [Y^{a}]$ . (Conditional-class standardization identifies a causal mean)

[F4]

Under the same assumptions, the IPW functional equals $E [Y^{a}]$ . (Inverse probability weighting identifies a causal mean)

Verification

1.1givenA1F1algebra

The two propensity values are $e_{1} (0) = 1 / 4$ and $e_{1} (1) = 3 / 4$ . Since $Y^{1} = 1 + 2 L$ , the observed treated means are $m_{1} (0) = 1$ and $m_{1} (1) = 3$ . Thus [F1] gives $Ψ_{s t d} = \frac{1}{2} (1) + \frac{1}{2} (3) = 2$ .

1.2givenA1F2algebra

The treated joint masses are $P (L = 0, A = 1) = 1 / 8$ and $P (L = 1, A = 1) = 3 / 8$ . Therefore [F2] gives $Ψ_{I P W} = \frac{1}{8} \frac{1}{1 / 4} + \frac{3}{8} \frac{3}{3 / 4} = \frac{1}{2} + \frac{3}{2} = 2$ .

2.1givenA1F3F4step 1.1step 1.2∎

The target mean is $E [Y^{1}] = \frac{1}{2} (1) + \frac{1}{2} (3) = 2$ . Consequently the two displayed population functionals agree with the target, as asserted by [F3] and [F4].

Depends on

Used by

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