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.

✓ 1 result · all verified · 1 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; all 1 also cleared it.

Missingness weights and sensitivity — Examples

1 · Prerequisites

2 · Summary

Finite laws show that strict response positivity can coexist with arbitrarily large weighted-contribution variance as an observation probability approaches zero. At a positivity failure, two full-data laws can give the same observed records but different means. A location-shift sensitivity curve likewise keeps the observed law fixed while changing the full-data mean; the resulting range depends on the stated sensitivity model and is not a confidence interval.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Positivity failure, near-positivity weights, and an MNAR sensitivity curve

Example

All laws below are finite. Encode an observed record by its response mask and the available coordinates; when convenient write the missing outcome slot as a placeholder zero, with the mask distinguishing it from an observed zero.

First consider $X \in {0, 1}$ with $P (X = 0) = P (X = 1) = 1 / 2$ , full outcome $Y = X$ , and response probabilities $π (0) = 1, π (1) = ε, 0 < ε \leq 1.$ For the response-weighted contribution $V = R Y / π (X)$ , positivity holds for every fixed $ε > 0$ , and $E [V] = E [Y] = 1 / 2$ . However, $V = 1 / ε$ with probability $ε / 2$ and is zero otherwise, so $Var (V) = \frac{1}{2 ε} - \frac{1}{4} ⟶ \infty as ε ↓ 0.$ Thus strict positivity gives the population identity but no uniform finite-variance bound as positivity deteriorates.

At the boundary $ε = 0$ , let $X$ still be fair, let $R = 1$ exactly when $X = 0$ , and set $Y = 0$ on $X = 0$ . Consider two full-data laws indexed by $θ \in {0, 1}$ that set $Y = θ$ on $X = 1$ . Their observed record $O = (X, R, R Y)$ has the same law under both: mass $1 / 2$ at $(0, 1, 0)$ and mass $1 / 2$ at $(1, 0, 0)$ . The full-data mean is $ψ (θ) = E_{θ} [Y] = θ / 2$ , so it differs between observationally equivalent laws. The zero response probability at $X = 1$ prevents identification of the mean.

Finally let $P (R = 1) = P (R = 0) = 1 / 2$ , take $Y ∣ R = 1 \sim Bernoulli (0.4)$ , and, for a sensitivity value $Δ \in [0, 1 / 2]$ , specify $Y ∣ R = 0 \overset{d}{=} B + Δ$ for $B \sim Bernoulli (0.4)$ . This is the location-shift pattern-mixture family of Location-shift pattern-mixture sensitivity function with constant $X$ . Every member has the same observed law: $P (R = 1, R Y = 0) = 0.3, P (R = 1, R Y = 1) = 0.2, P (R = 0, R Y = 0) = 0.5.$ Its full-data mean is $μ (Δ) = \frac{1}{2} (0.4) + \frac{1}{2} (0.4 + Δ) = 0.4 + 0.5 Δ,$ so the sensitivity range $Δ \in [0, 1 / 2]$ yields means in $[0.4, 0.65]$ despite an unchanged observed law. This is a model-indexed range of target values, not a confidence interval or a coverage statement.

Facts & Assumptions

Given: The finite laws and parameter ranges specified in the Example.

[F1]

On a finite probability space, expectation is the probability-weighted finite sum. (Expectation of a real random variable on a finite probability space)

[F2]

A finite probability law has nonnegative atom weights summing to one, and event probabilities are sums of those weights. (Finite probability spaces, outcome weights, events, and event probabilities)

[F3]

Variance is the expectation of the squared deviation from the mean. (Moments, variance, and covariance on a probability space)

[F4]

An estimand is a specified function of the parameter in a parametric model. (Estimands on a statistical model)

[F5]

Two parameter values are observationally equivalent exactly when their sampling laws agree. (Observational equivalence)

[F6]

An estimand is identifiable only if observationally equivalent parameters have the same target value. (Identifiability of an estimand)

[F7]

The location-shift sensitivity family specifies the nonrespondent conditional law by translating the respondent law by $Δ (X)$ . (Location-shift pattern-mixture sensitivity function)

[F8]

In the one-outcome pattern, MAR requires the response probability for the missing pattern to depend only on that pattern's observed coordinates. (MCAR, MAR, and MNAR relative to an observation pattern)

[F9]

Conditional independence given a sigma-algebra is the conditional product identity for bounded measurable functions. (Conditional independence given a sigma-algebra)

[A1]

$ε \in (0, 1]$ in the near-positivity law.

[A2]

$Δ \in [0, 1 / 2]$ in the sensitivity family.

Proof

1.1A1F2F9construct

The near-positivity law has atom weights $1 / 2$ at $(X, R, Y) = (0, 1, 0)$ , $ε / 2$ at $(1, 1, 1)$ , and $(1 - ε) / 2$ at $(1, 0, 1)$ . By [A1] and [F2], these are nonnegative and sum to one. Given either value $X = x$ , $Y = x$ is constant, so for bounded measurable $f (R)$ and $g (Y)$ , $E [f (R) g (Y) ∣ X = x] = g (x) E [f (R) ∣ X = x] = E [f (R) ∣ X = x] E [g (Y) ∣ X = x]$ . Thus this construction is MAR under the assigned conditional-independence formulation [F9]. Hence $V$ equals $1 / ε$ on the responding $X = 1$ atom, of probability $ε / 2$ , and is zero on the other atoms.

1.2F1F2F4construct

For each $θ \in {0, 1}$ , the full-data law assigns probability $1 / 2$ to $(X, R, Y) = (0, 1, 0)$ and $1 / 2$ to $(1, 0, θ)$ . Mapping these atoms to $O = (X, R, R Y)$ gives the same two observed atoms for both values of $θ$ , while [F1] gives $ψ (θ) = E_{θ} [Y] = θ / 2$ . Thus the observed sampling laws agree but the estimand values are $0$ and $1 / 2$ .

1.3A2F2F7construct

For each $Δ \in [0, 1 / 2]$ , assign observed-pattern masses $0.3$ to $(R, Y) = (1, 0)$ , $0.2$ to $(1, 1)$ , and missing-pattern masses $0.3$ to $(0, Δ)$ , $0.2$ to $(0, 1 + Δ)$ . These masses sum to one by [F2]. Conditional on $R = 1$ , the outcome is Bernoulli $(0.4)$ ; conditional on $R = 0$ , it has the law of that Bernoulli variable translated by $Δ$ , as required by [F7]. The observed record $(R, R Y)$ has masses $0.3, 0.2, 0.5$ at $(1, 0), (1, 1), (0, 0)$ for every $Δ$ .

2.1A1F1F3step 1.1algebra

By [F1], $E [V] = (ε / 2) (1 / ε) = 1 / 2$ and $E [(V - E V)^{2}] = (ε / 2) (1 / ε - 1 / 2)^{2} + (1 - ε / 2) (1 / 2)^{2} = 1 / (2 ε) - 1 / 4$ . This diverges as $ε ↓ 0$ , although $π (X) > 0$ for each fixed $ε > 0$ . At $ε = 0$ , $π (1) = 0$ , so positivity fails on an $X$ stratum of probability $1 / 2$ .

2.2F5F6step 1.2

By [F5], the two parameters are observationally equivalent; by [F6], the different values of $ψ$ therefore show that the full-data mean is not identifiable from this observed-data model. The response probability is zero at $X = 1$ , exactly the stratum on which the two full-data laws differ.

2.3A2F1F4F5F6step 1.3algebra

By [F1], the target mean is $0.2 (1) + 0.3 Δ + 0.2 (1 + Δ) = 0.4 + 0.5 Δ$ . At $Δ = 0$ it is $0.4$ , and at $Δ = 1 / 2$ it is $0.65$ , while step 1.3 shows that the observed sampling law is unchanged. Thus the target is not identified across this sensitivity family; the interval $[0.4, 0.65]$ is only the set of means induced by the stipulated $Δ$ range and carries no repeated-sampling coverage assertion.

3.1A2F8step 1.3givenalgebra∎

If $Δ > 0$ , then $P (R = 0 ∣ Y = 0) = 0$ whereas $P (R = 0 ∣ Y = Δ) = 1$ : both conditioning events have positive probability, and $Δ \in (0, 1 / 2]$ keeps $Δ$ distinct from respondent support ${0, 1}$ . With no always-observed covariate, the missing pattern has an empty observed-coordinate vector, so [F8] requires this response probability to be constant in $Y$ for MAR; it is not, hence this member is MNAR. At $Δ = 0$ , both response groups have the same Bernoulli $(0.4)$ law and $R$ is independent of $Y$ , so the response probability is constant and the member is MAR.