ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

An empty risk set prevents root-n tightness

Example

Assume AC. Let $T_{1}, T_{2}, \dots$ be IID uniform on $[0, 1]$ event times and set $X_{i} = 1$ for every subject. For each $n \geq 2$ , use the moving terminal time $t_{n} = 2^{- 1 / n}$ and risk indicators $Y_{n i} = 1 {T_{i} \geq t_{n}}$ . Write $K_{n} = \sum_{i = 1}^{n} Y_{n i}$ . The target mean among subjects at risk is the fixed value $θ = 1$ . Define the risk-set ratio

${\hat{θ}}_{n} = {\begin{cases} \frac{\sum_{i = 1}^{n} X_{i} Y_{n i}}{K_{n}}, & K_{n} > 0, \\ 0, & K_{n} = 0. \end{cases}$

The convention makes the estimator defined even when the risk set is empty.

Facts & Assumptions

Given: The IID uniform event times and ratio estimator in the Example. Show that its root- $n$ error is not tight, so the terminal influence theorem cannot be applied without a separate expansion.

[A1]

The event times are IID, their common law satisfies $P (T_{i} < t) = t$ for $0 \leq t \leq 1$ , and subjects are mutually independent.

[A2]

Assume AC. The Axiom of Choice

[F1]

Identical distribution and IID families: IID means a common law and mutual independence of the whole family.

[F2]

Independent families of event classes: For finite selections of distinct independent indices, the probability of an intersection is the product of the event probabilities.

[F3]

Stochastic order and convergence notation for finite-dimensional random vectors: $Z_{n} = O_{P} (1)$ requires that its tail probabilities can be made uniformly small in the limit by choosing a sufficiently large bound.

[F4]

Finite-dimensional stochastic-order calculus and tightness: Any finite-dimensional sequence converging in distribution is bounded in probability.

[F5]

IID terminal influence limit from a separately established expansion: If this estimator had a centered finite-variance IID influence expansion, its root- $n$ error would converge in distribution under AC.

Verification

1.1A1F1F2algebra

For each $n \geq 2$ , [A1] gives $P (Y_{n i} = 0) = t_{n} = 2^{- 1 / n}$ . The events ${Y_{n i} = 0}$ are independent by the finite event-product identity in [F2], hence $P (K_{n} = 0) = (2^{- 1 / n})^{n} = 1 / 2$ . Since $X_{i} = 1$ , the ratio is $1$ when $K_{n} > 0$ and is $0$ otherwise; therefore $\sqrt{n} ({\hat{θ}}_{n} - 1)$ is $0$ when $K_{n} > 0$ and $- \sqrt{n}$ when $K_{n} = 0$ .

2.1A2F3F4F5step 1.1∎

For every fixed $M > 0$ and every integer $n > M^{2}$ , $P (∣ \sqrt{n} ({\hat{θ}}_{n} - 1) ∣ > M) = P (K_{n} = 0) = 1 / 2$ . Taking $η = 1 / 4$ in [F3] shows that the root- $n$ errors are not $O_{P} (1)$ . If they converged in distribution, [F4] would make them $O_{P} (1)$ , a contradiction. Under [A2], the expansion required by [F5] would imply a weak limit, so it cannot hold for this estimator. The example shows why a consumer must control its own random denominator before claiming a terminal influence limit.

Depends on

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