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.

Weight truncation changes the IPW target

Example

Let $L \in {0, 1}$ be uniform, with $P (A = 1 ∣ L = 0) = 1 / 4$ and $P (A = 1 ∣ L = 1) = 3 / 4$ . Let $Y^{1} = 21_{{L = 0}}$ , $Y^{0} = 0$ , and $Y = Y^{A}$ . Conditional exchangeability and strict positivity hold. Compare the raw population IPW functional with the modified functional obtained by capping each treated weight $1 / e_{1} (L)$ at $c = 3 / 2$ .

Facts & Assumptions

Given: The finite-stratum population and weight cap in the Example, with consistency and AC for conditional expectations.

[A1]

The point-treatment setup uses the target convention of Potential outcomes and consistency, with consistency and integrable $Y^{1}$ .

[F1]

The population IPW functional uses the untruncated true weight $I_{1} / e_{1} (L)$ . (Inverse probability weighting for a causal mean)

[F2]

Under consistency, exchangeability, integrability and positivity, that untruncated functional equals $E [Y^{1}]$ . (Inverse probability weighting identifies a causal mean)

Verification

1.1givenA1F1F2algebra

The target mean is $E [Y^{1}] = \frac{1}{2} (2) + \frac{1}{2} (0) = 1$ . The treated mass in $L = 0$ is $1 / 8$ and the outcome there is $2$ , so the untruncated IPW expectation is $(1 / 8) \cdot 4 \cdot 2 = 1$ , as [F1]–[F2] predict.

2.1givenA1F1step 1.1algebra

After capping the functional in [F1], the treated weight in $L = 0$ is $3 / 2$ and the outcome in $L = 1$ is zero. The unnormalized capped expectation is therefore $(1 / 8) (3 / 2) (2) = 3 / 8$ , not the target $1$ .

3.1step 2.1algebra

If the capped weights are normalized by their population mean, the denominator is $(1 / 8) (3 / 2) + (3 / 8) (4 / 3) = 11 / 16$ , so the normalized capped mean is $(3 / 8) / (11 / 16) = 6 / 11 \neq 1$ . Thus both capped versions in this example target a different population quantity from the original mean.

4.1givenstep 3.1∎

This is a population calculation for known propensities; it does not invoke a finite-sample estimator or an estimated-weight limit.

Depends on

Used by

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