Pipeline-generated

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.

✓ 4 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Standardization and IPW identification — Examples

1 · Prerequisites

2 · Summary

Objective

These examples compute both population functionals in a finite-stratum model, show how a positive propensity near zero creates an extreme weight, and calculate how capping weights changes the population target. A final pair of compatible full-data laws shows why structural nonpositivity can prevent identification even when consistency and conditional exchangeability hold.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Positivity failure prevents standardization and IPW identification

Example

Let $L \in {0, 1}$ with $P (L = 0) = P (L = 1) = 1 / 2$ . In both full-data models, set $P (A = 1 ∣ L = 0) = 0$ and $P (A = 1 ∣ L = 1) = 1 / 2$ , and set $Y^{0} = 0$ everywhere and $Y^{1} = 1$ when $L = 1$ . In model $P_{0}$ , also set $Y^{1} = 0$ when $L = 0$ ; in model $P_{2}$ , set $Y^{1} = 2$ when $L = 0$ . Treatment assignment at $L = 1$ is independent of these deterministic potential outcomes, so conditional exchangeability holds in each model. Define the observed outcome by consistency, $Y = Y^{A}$ .

Facts & Assumptions

Given: The two specified full-data laws, consistency, and the potential-outcome conventions of Potential outcomes and consistency. The target is the population mean $E [Y^{1}]$ .

[A1]

Treatment assignment has zero probability of treatment $1$ at $L = 0$ in both models; the point-treatment setup is as in Potential outcomes and consistency.

[F1]

The standardization and IPW functionals require the conditional probability $e_{1} (L)$ to be positive almost surely. (Standardization for a causal mean, Inverse probability weighting for a causal mean)

[F2]

Conditional treatment-indicator factorization gives $E [1_{{A = 1}} Y^{1} ∣ σ (L)] = e_{1} (L) E [Y^{1} ∣ σ (L)]$ under conditional exchangeability. (Conditional treatment-indicator factorization)

[F3]

Identifiability requires equal observed laws to imply equal target values. (Identifiability of an estimand)

Verification

1.1givenA1construct

In both models the observed law assigns probability $1 / 2$ to $(L, A, Y) = (0, 0, 0)$ , probability $1 / 4$ to $(1, 1, 1)$ , and probability $1 / 4$ to $(1, 0, 0)$ . The unobserved value of $Y^{1}$ at $L = 0$ does not enter any observed outcome because $A = 0$ there.

2.1F3step 1.1algebra

The target means differ: $E_{P_{0}} [Y^{1}] = \frac{1}{2}$ while $E_{P_{2}} [Y^{1}] = \frac{3}{2}$ . Since the observed laws in step 1.1 coincide, [F3] shows that $E [Y^{1}]$ is not identified in this model class.

3.1A1F1F2step 2.1∎

Here $e_{1} (0) = 0$ , so the observed treated conditional mean at $L = 0$ is unavailable and both $q_{1} (0) / e_{1} (0)$ and the treated weight $I_{1} / e_{1} (L)$ are undefined. Also [F2] gives $E [I_{1} Y^{1} ∣ L = 0] = 0 \cdot E [Y^{1} ∣ L = 0] = 0$ in both models, leaving the differing target-stratum means unrecovered.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Extreme IPW weights under poor overlap

Example

For each $ε \in (0, 1 / 2)$ , let $P (L = 0) = P (L = 1) = 1 / 2$ , let $P (A = 1 ∣ L = 0) = ε$ and $P (A = 1 ∣ L = 1) = 1 / 2$ , and set $Y^{1} = 1_{{L = 0}}$ , $Y^{0} = 0$ , and $Y = Y^{A}$ . The potential outcomes are deterministic given $L$ , so conditional exchangeability holds. For every $ε > 0$ , positivity holds; as $ε$ decreases, the treated weight in stratum $L = 0$ grows.

Facts & Assumptions

Given: The finite-stratum family in the Example, consistency, and AC for the conditional-expectation conventions.

[A1]

$Y^{1}$ is integrable in the point-treatment setup of Potential outcomes and consistency, and the true treatment probability is $e_{1} (L) = P (A = 1 ∣ L)$ .

[F1]

The IPW functional uses the population weight $I_{1} / e_{1} (L)$ and its expectation, when integrable. (Inverse probability weighting for a causal mean)

[F2]

Under the stated causal assumptions and positive propensity, the population IPW mean equals $E [Y^{1}]$ . (Inverse probability weighting identifies a causal mean)

Verification

1.1givenA1F1F2algebra

The event ${L = 0, A = 1}$ has probability $ε / 2$ and there the IPW summand $W = I_{1} Y / e_{1} (L)$ equals $1 / ε$ . Everywhere else $W = 0$ . Hence $E [W] = (ε / 2) (1 / ε) = 1 / 2 = E [Y^{1}]$ , consistently with [F2].

2.1step 1.1algebra

The second moment and variance are $E [W^{2}] = \frac{ε}{2} \frac{1}{ε^{2}} = \frac{1}{2 ε}$ and $Var (W) = \frac{1}{2 ε} - \frac{1}{4}$ . As $ε ↓ 0$ , the realized treated weight $1 / ε$ and this population variance diverge, although the exact population identity remains valid at each $ε > 0$ .

3.1givenstep 2.1∎

Thus strict positivity at each model does not impose a uniform weight or variance bound across models; the example makes no claim about finite-sample estimator consistency.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.