CounterexampleConstruction: 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.

Uniform moving support invalidates the naive score bound

Statement refuted

For every differentiable one-parameter density family, the formal score $S_{θ} = \partial_{θ} \log p_{θ} (X)$ and its finite positive second moment $I (θ) = E_{θ} S_{θ}^{2}$ automatically give the Cramér–Rao inequality ${Var}_{θ} (T) \geq ψ^{'} (θ)^{2} / I (θ)$ for every unbiased finite-variance estimator $T$ , even when the support depends on $θ$ .

Facts & Assumptions

Given: AC; $θ > 0$ ; the density family and estimator stated above.

[A1]

The Axiom of Choice: AC includes countable choice and supplies the selected representatives used by the score definition.

[F1]

Statistics and estimators: an estimator is a measurable statistic with a specified target.

[F2]

The measure with density $f$ relative to $μ$ and Integrating against a density agrees with integrating the product: the law with density $p_{θ}$ integrates a nonnegative $g$ as $\int g p_{θ} d λ$ .

[F3]

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included: $λ ((0, θ)) = θ$ and $λ ({0}) = λ ({θ}) = 0$ for $θ > 0$ .

[F4]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , and For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term: continuous functions on $[0, θ]$ are Riemann integrable, their Riemann and Lebesgue integrals agree under countable choice, and antiderivatives evaluate the required polynomial integrals.

[F5]

Score and Fisher information: on the positive-density support, the score is the parameter derivative of the log density and information is the expected squared score.

[F6]

Moments, variance, and covariance on a probability space: variance and covariance are centered second moments.

[F7]

Scalar Cramér–Rao lower bound: the bound requires a mean-zero score and the score-covariance derivative identity; the fixed-support route is sufficient, not necessary.

[F8]

The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set: the derivative is the limit of the difference quotient $(f (x) - f (c)) / (x - c)$ at $c$ .

Counterexample

Assume AC. Let $θ > 0$ and $X \sim U n i f [0, θ]$ . Use the equivalent Lebesgue-density version $p_{θ} (x) = θ^{- 1} 1_{(0, θ)} (x)$ with respect to Lebesgue measure, and set $T = 2 X$ , estimating $ψ (θ) = θ$ . The positive-density set $(0, θ)$ changes with $θ$ .

Proof technique: direct.

1.1A1F1F2F3F4F6givenalgebra

By [F3], the endpoint singletons have measure zero and $λ ((0, θ)) = θ$ , so this density version gives the law $U n i f [0, θ]$ and has total mass one. Using [F2] for the nonnegative functions $2 x$ and $(2 x - θ)^{2}$ , then [F4] to evaluate their integrals, gives $E_{θ} T = θ^{- 1} \int_{0}^{θ} 2 x d x = θ$ and ${Var}_{θ} (T) = E_{θ} [(2 X - θ)^{2}] = θ^{- 1} \int_{0}^{θ} (2 x - θ)^{2} d x = θ^{- 1} {[\frac{4}{3} x^{3} - 2 θ x^{2} + θ^{2} x]}_{0}^{θ} = \frac{θ^{2}}{3}$ . Thus $T$ is unbiased for $ψ (θ) = θ$ and has variance $θ^{2} / 3$ .

2.1F5F6F7F8step 1.1givenalgebra

For $0 < x < θ$ , $ℓ_{θ} (x) = - \log θ$ , so $S_{θ} (x) = - 1 / θ$ . Therefore $I (θ) = E_{θ} S_{θ}^{2} = 1 / θ^{2}$ but $E_{θ} S_{θ} = - 1 / θ \neq 0$ . Since the score is constant $P_{θ}$ -almost surely, ${Cov}_{θ} (T, S_{θ}) = 0$ . On $Θ = (0, \infty)$ , $ψ (θ) = θ$ has difference quotient $1$ at every $x \neq θ$ , so [F8] gives $ψ^{'} (θ) = 1$ . Both score identities needed for [F7] fail here.

3.1F7step 1.1step 2.1givenalgebra∎

The formula obtained by ignoring those failures would give $ψ^{'} (θ)^{2} / I (θ) = θ^{2}$ , contradicting the actual ${Var}_{θ} (T) = θ^{2} / 3$ from step 1.1. This does not contradict [F7]: the moving support violates its regularity route, and the score-mean and covariance-derivative hypotheses fail explicitly in step 2.1.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

94 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

George Deligiannidis, SB2.1 Foundations of Statistical Inference, §3.1, support regularity discussion, PDF pp. 21–24 (standard reference, not scraped)
Anna Mikusheva, MIT 14.381 Lecture 6: Efficient Estimators, Rao-Cramer Bound, PDF pp. 2, 5–6 (standard reference, not scraped)