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✓ 5 results · 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. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Delta Method and Asymptotic Normality — Examples

1 · Prerequisites

2 · Summary

These finite-dimensional examples compute a logarithmic derivative and a ratio Jacobian, show the nontrivial quadratic scale at a zero gradient, and exhibit why continuity without differentiability does not support a first-order linear limit.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

A logarithmic transformation

Example

Let $Ω = {- 1, 1}$ with each point having probability $1 / 2$ , let $Z (ω) = ω$ , and set $T_{n} = \max {1 / 2, 1 + Z / \sqrt{n}}$ for $n \geq 1$ . Define the Borel map $g (t) = \log t$ for $t > 0$ and $g (t) = 0$ for $t \leq 0$ . Then $\sqrt{n} {\log T_{n} - \log 1} \Rightarrow Z .$ Thus the log transformation has first-order scale $\sqrt{n}$ and derivative $g^{'} (1) = 1$ in this example.

Facts & Assumptions

Given: The finite two-point law and sequence in the example.

[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: if $r_{n} > 0$ is deterministic with $r_{n} \to \infty$ , $r_{n} (T_{n} - θ) \Rightarrow Z$ , and the Borel map $g$ has derivative $A$ at $θ$ with first-order remainder $o (∥ h ∥)$ , then $r_{n} (g (T_{n}) - g (θ)) \Rightarrow A Z$ .

Verification

Given: The finite two-point law, $r_{n} = \sqrt{n}$ , $θ = 1$ , and the function $g$ in the example.

1.1A1givenalgebra

By [A1], $Z \in {- 1, 1}$ . For $n \geq 4$ , $1 + Z / \sqrt{n} \geq 1 - 1 / 2 = 1 / 2$ , so $T_{n} = 1 + Z / \sqrt{n}$ and $\sqrt{n} (T_{n} - 1) = Z$ exactly. Hence $\sqrt{n} (T_{n} - 1) \Rightarrow Z$ .

2.1F1givenstep 1.1∎

The map $g$ is differentiable at $1$ with derivative $1$ . Apply [F1] to obtain $\sqrt{n} {\log T_{n} - \log 1} \Rightarrow Z$ .

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

A ratio transformation

Example

Give $Ω = {- 1, 1}^{2}$ equal mass $1 / 4$ at each point, and let $Z = (Z_{1}, Z_{2})$ be the coordinate vector. Put $θ = (1, 2)$ , $T_{n} = θ + n^{- 1 / 2} Z$ for $n \geq 1$ , and define $g (x, y) = x / y$ for $y \neq 0$ and $g (x, 0) = 0$ . The denominator of $g (T_{n})$ is positive for every $n$ . The first-order delta method gives $\sqrt{n} (\frac{T_{n, 1}}{T_{n, 2}} - \frac{1}{2}) \Rightarrow \frac{1}{2} Z_{1} - \frac{1}{4} Z_{2} .$ The limit takes the values $1 / 4, 3 / 4, - 3 / 4, - 1 / 4$ with equal probability and has variance $5 / 16$ .

Facts & Assumptions

Given: The four-point probability law and sequence in the example.

[A1]

$Ω = {- 1, 1}^{2}$ has full sigma-field and assigns mass $1 / 4$ to each point; the coordinate variables $Z_{1}, Z_{2}$ are independent Rademacher variables.

[F1]

Verification

Given: The finite four-point law, $r_{n} = \sqrt{n}$ , $θ = (1, 2)$ , and the ratio map $g$ in the example.

1.1A1givenalgebra

By [A1], $Z_{2} \in {- 1, 1}$ , so $T_{n, 2} = 2 + Z_{2} / \sqrt{n} \geq 1$ for every $n \geq 1$ . For $∣ v ∣ < 1$ , the exact remainder after subtracting the linear map is $g (1 + u, 2 + v) - \frac{1}{2} - (u / 2 - v / 4) = - (2 u - v) v / [2 (4 + 2 v)]$ , whose absolute value is bounded by a constant times $u^{2} + v^{2}$ . Thus the derivative at $(1, 2)$ is $(u, v) \mapsto u / 2 - v / 4$ . Also $\sqrt{n} (T_{n} - θ) = Z$ exactly.

2.1F1step 1.1given

Apply [F1] with this derivative to obtain the stated limit.

3.1A1givenstep 2.1algebra∎

By [A1], the four equiprobable values of $\frac{1}{2} Z_{1} - \frac{1}{4} Z_{2}$ are $1 / 4, 3 / 4, - 3 / 4, - 1 / 4$ . Their mean is zero and their second moment is $(1 + 9 + 9 + 1) / 64 = 5 / 16$ , which is the stated variance.

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

A zero-gradient quadratic limit

Example

Let $Ω = {- \sqrt{2}, 0, \sqrt{2}}$ with probabilities $1 / 4, 1 / 2, 1 / 4$ , let $Z (ω) = ω$ , and set $T_{n} = Z / \sqrt{n}$ for $n \geq 1$ . For $g (t) = t^{2}$ at $θ = 0$ , the first derivative vanishes and the second derivative is $2$ . The second-order delta method gives $n T_{n}^{2} \Rightarrow Z^{2}$ , which equals $0$ or $2$ with probability $1 / 2$ each.

Facts & Assumptions

Given: The finite three-point law and sequence in the example.

[A1]

$Ω = {- \sqrt{2}, 0, \sqrt{2}}$ has full sigma-field and assigns probabilities $1 / 4, 1 / 2, 1 / 4$ respectively; $Z (ω) = ω$ .

[F1]

First- and second-order delta method for random vectors: if $r_{n} > 0$ is deterministic with $r_{n} \to \infty$ , $r_{n} (T_{n} - θ) \Rightarrow Z$ , the Borel map $g$ has zero first derivative at $θ$ , and $g (θ + h) - g (θ) = \frac{1}{2} H [h, h] + o (∥ h ∥^{2})$ for a continuous symmetric bilinear map $H$ , then $r_{n}^{2} (g (T_{n}) - g (θ)) \Rightarrow \frac{1}{2} H [Z, Z]$ .

Verification

Given: The three-point law, $r_{n} = \sqrt{n}$ , $θ = 0$ , and $g (t) = t^{2}$ .

1.1A1givenalgebra

By [A1], $Z$ has the stated three-point law. We have $\sqrt{n} (T_{n} - 0) = Z$ exactly and $g (h) - g (0) = h^{2} = \frac{1}{2} (2) h^{2}$ for every $h$ , so the first derivative is zero, the second derivative is $2$ , and the second-order remainder vanishes.

2.1A1F1givenstep 1.1algebra∎

Apply [F1] to obtain $n T_{n}^{2} \Rightarrow Z^{2}$ . By [A1], $Z^{2} = 0$ at $ω = 0$ and $Z^{2} = 2$ at $ω = \pm \sqrt{2}$ , so each value has probability $1 / 2$ .

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Sample variance from a multivariate delta method

Example

Assume AC. Let $X_{1}, X_{2}, \dots$ be iid real random variables with $E ∣ X_{1} ∣^{4} < \infty$ . Put $μ_{j} = E [X_{1}^{j}]$ for $1 \leq j \leq 4$ , $σ^{2} = μ_{2} - μ_{1}^{2}$ , ${\overset{ˉ}{X}}_{n} = n^{- 1} \sum_{i} X_{i}$ , $Q_{n} = n^{- 1} \sum_{i} X_{i}^{2}$ , and $S_{n}^{2} = Q_{n} - {\overset{ˉ}{X}}_{n}^{2}$ . Then $\sqrt{n} (S_{n}^{2} - σ^{2}) \Rightarrow N_{1} (0, τ^{2}), τ^{2} = Var ((X_{1} - μ_{1})^{2}) = E [(X_{1} - μ_{1})^{4}] - σ^{4} = μ_{4} - 4 μ_{1} μ_{3} + 8 μ_{1}^{2} μ_{2} - 4 μ_{1}^{4} - μ_{2}^{2} .$ Here $σ^{4} = (σ^{2})^{2}$ and $τ^{2} = Var ((X_{1} - μ_{1})^{2})$ .

Facts & Assumptions

Given: AC, iid real observations, and the finite fourth absolute moment.

[A1]

AC is assumed so the library's multivariate iid central limit theorem and multivariate normal law apply.

[A2]

$E ∣ X_{1} ∣^{4} < \infty$ .

[F1]

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

[F2]

Measurable coordinatewise functions preserve independence: measurable coordinatewise maps of an independent family remain independent.

[F3]

Multivariate iid central limit theorem: iid finite-dimensional vectors with finite second moment have a centered Gaussian weak limit with their covariance, under AC.

[F4]

Multivariate normal law, including singular covariance: every scalar projection of a centered multivariate normal law is normal with the projected covariance.

[F5]

First- and second-order delta method for random vectors: a first-order expansion of a map at the target transfers a vector weak limit through its derivative.

[F6]

The Axiom of Choice: AC states that every family of nonempty sets has a choice function.

[F7]

Moments, variance, and covariance on a probability space: for square-integrable $W$ , $Var (W) = E [(W - E W)^{2}]$ ; covariance is defined by the corresponding centered product.

Verification

Given: The data, moments, sample variance, and assumptions in the example.

1.1A1A2F1F2F3F6

Let $Y_{i} = (X_{i}, X_{i}^{2})$ . The map $x \mapsto (x, x^{2})$ is Borel, so [F1]–[F2] show that $(Y_{i})$ is iid. By [A2], $E ∥ Y_{1} ∥^{2} = μ_{2} + μ_{4} < \infty$ , with mean $(μ_{1}, μ_{2})$ and covariance $C = (\begin{array}{cc} μ_{2} - μ_{1}^{2} & μ_{3} - μ_{1} μ_{2} \\ μ_{3} - μ_{1} μ_{2} & μ_{4} - μ_{2}^{2} \end{array}) .$ By [A1] and [F3], $\sqrt{n} {({\overset{ˉ}{X}}_{n}, Q_{n}) - (μ_{1}, μ_{2})} \Rightarrow Z \sim N_{2} (0, C)$ .

2.1F5step 1.1givenalgebra

The map $g (x, y) = y - x^{2}$ has $g (μ_{1}, μ_{2}) = σ^{2}$ and derivative $(u, v) \mapsto - 2 μ_{1} u + v$ . Apply [F5] to step 1.1 to get $\sqrt{n} (S_{n}^{2} - σ^{2}) \Rightarrow - 2 μ_{1} Z_{1} + Z_{2}$ .

3.1F4F7step 1.1step 2.1givenalgebra∎

By [F4], the limit is centered normal with variance $(- 2 μ_{1}, 1) C (- 2 μ_{1}, 1)^{T} = 4 μ_{1}^{2} (μ_{2} - μ_{1}^{2}) - 4 μ_{1} (μ_{3} - μ_{1} μ_{2}) + μ_{4} - μ_{2}^{2} = μ_{4} - 4 μ_{1} μ_{3} + 8 μ_{1}^{2} μ_{2} - 4 μ_{1}^{4} - μ_{2}^{2}$ . Expanding $E [(X_{1} - μ_{1})^{4}] - (μ_{2} - μ_{1}^{2})^{2}$ gives the same expression. By [F7], it is $Var ((X_{1} - μ_{1})^{2})$ , so the displayed $τ^{2}$ is exactly the limit variance.

Remarks

The cited Duchi sample-variance example supplies the setup and covariance matrix, but its printed final variance $μ_{4} - μ_{2}^{2}$ omits the nonzero-mean terms. Step 3.1 derives the corrected variance directly; the printed expression agrees with it when $μ_{1} = 0$ .

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

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].

Delta Method and Asymptotic Normality — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A logarithmic transformation

Example

Facts & Assumptions

Verification

A ratio transformation

Example

Facts & Assumptions

Verification

A zero-gradient quadratic limit

Example

Facts & Assumptions

Verification

Sample variance from a multivariate delta method

Example

Facts & Assumptions

Verification

Remarks

Continuity alone does not give a delta-method limit

Statement refuted

Facts & Assumptions

Counterexample

Sources