CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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Continuity alone does not give a delta-method limit

Statement refuted

Continuity of $g$ at $θ$ alone suffices for a first-order delta-method conclusion: whenever $r_{n} (T_{n} - θ) \Rightarrow Z$ , there is a linear map $A$ for which $r_{n} {g (T_{n}) - g (θ)} \Rightarrow A Z$ .

Facts & Assumptions

Given: The stated finite two-point law, sequence, and absolute-value function.

[A1]

$Ω = {- 1, 1}$ has full sigma-field and assigns mass $1 / 2$ to each point; $Z (ω) = ω$ .

[F1]

First- and second-order delta method for random vectors: a linear first-order expansion with remainder $o (∥ h ∥)$ at the target is sufficient for the first-order delta conclusion under the theorem's other hypotheses.

Counterexample

Let $Ω = {- 1, 1}$ have equal atom probabilities, let $Z (ω) = ω$ , put $T_{n} = Z / \sqrt{n}$ for $n \geq 1$ , and take $g (t) = ∣ t ∣$ at $θ = 0$ . This $g$ is continuous but not differentiable at $0$ .

Given: The false statement and the counterexample above.

1.1A1givenalgebra

By [A1], we have $\sqrt{n} (T_{n} - 0) = Z$ and $\sqrt{n} {∣ T_{n} ∣ - ∣ 0 ∣} = ∣ Z ∣ = 1$ for every $n$ . Thus the transformed sequence has the constant limit $1$ .

2.1A1F1step 1.1givenalgebra∎

By [A1], $Z$ takes both signs. For any linear map $A : R \to R$ , $A Z = a Z$ for some $a \in R$ ; if $a = 0$ its law is concentrated at $0$ , and if $a \neq 0$ its law has the two distinct values $\pm a$ . Neither law is the constant $1$ . Also the right and left difference quotients of $g$ at zero are $1$ and $- 1$ , so no derivative exists. Therefore the proposed conclusion fails, consistently with the differentiability hypothesis in [F1].

Depends on

First- and second-order delta method for random vectors

Used by

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