ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Mean functional and the sample-mean influence function

Example

Let $(X, A)$ be a measurable observation space and let $X : X \to [0, \infty)$ be measurable. Let $M_{X}$ be the model of probability laws $Q$ on this space satisfying $E_{Q} X < \infty$ , and define $ψ : M_{X} \to R$ by $ψ (Q) = E_{Q} X$ . Take $P \in M_{X}$ with $X \in L^{2} (P)$ , and write $μ = E_{P} X$ . Specify the allowed one-sided paths through $P$ to be $d P_{t, s} = (1 + t s) d P$ for bounded real $s$ with $E_{P} s = 0$ and sufficiently small $t \geq 0$ . Their scores are the bounded mean-zero functions, whose closure is $T = L_{0}^{2} (P)$ . The right path derivative is represented by $X - μ$ . It is the unique representer in $T$ , hence the canonical gradient in this example. For iid observations $X_{1}, \dots, X_{n} \sim P$ in $X$ , define the sample mean of the observable by ${\overline{X}}_{n} = n^{- 1} \sum_{i = 1}^{n} X (X_{i})$ . It has the exact expansion

${\overline{X}}_{n} - μ = (P_{n} - P) (X - μ),$

so $X - μ$ is also its estimator influence function. Only the specified bounded-score path class is asserted here.

Facts & Assumptions

Given: A probability law $P$ , a nonnegative real $X \in L^{2} (P)$ , the bounded-score path class in the example, and an iid sample $X_{1}, \dots, X_{n} \sim P$ .

[F1]

Mean-zero square-integrable functions: $L_{0}^{2} (P)$ is the set of square-integrable real functions modulo almost-sure equality with mean zero.

[F2]

Asymptotically linear estimators and influence functions: an estimator is asymptotically linear when its error is $(P_{n} - P) φ + o_{P} (n^{- 1 / 2})$ with $φ \in L_{0}^{2} (P)$ .

[F3]

Probability measures and probability spaces: $P$ has total mass one, hence $∥ 1 ∥_{L^{2} (P)} = 1$ .

[F4]

Expectation of a nonnegative or integrable random variable: expectations of integrable real functions are their Lebesgue integrals.

[F5]

For measurable representatives $f, g$ of real $L^{2} (P)$ classes, Cauchy-Schwarz inequality for $L^{2}$ gives $\int ∣ f g ∣ d P \leq ∥ f ∥_{2} ∥ g ∥_{2}$ , and The modulus of an integral is bounded by the integral of the modulus then gives $∣ E_{P} [f g] ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ .

[F6]

Dominated convergence: pointwise convergence under one integrable majorant gives convergence of integrals.

[F7]

The Lebesgue integral is linear on $L^{1} (μ)$ : the integral is linear on integrable functions.

[F8]

Integrating against a density agrees with integrating the product: for nonnegative measurable $f$ and $g$ , $\int g d (f P) = \int g f d P$ .

[F9]

The measure with density $f$ relative to $μ$ : the measure with density $q$ relative to $P$ assigns $A$ the mass $\int_{A} q d P$ .

[F10]

Pathwise differentiability and parameter influence curves: DQM is defined by a common-measure square-root $L^{2}$ expansion; the tangent space is the closed span of allowed scores, and a pathwise derivative's unique representer in that tangent space is its canonical gradient.

Verification

Given: The data and path class in the example.

1.1givenF3F4F5F7F9F10algebra

Since $X \in L^{2} (P)$ and $P$ is a probability, Cauchy--Schwarz [F5] gives $μ = E_{P} X < \infty$ . For bounded centered $s$ and sufficiently small $t \geq 0$ , $1 + t s \geq 0$ and $E_{P} (1 + t s) = 1 + t E_{P} s = 1$ , so $P_{t, s}$ is a probability law with density $1 + t s$ by [F9]. Direct rationalization gives $\sqrt{1 + u} - 1 - u / 2 = - u^{2} / [2 (1 + \sqrt{1 + u})^{2}]$ ; for $u = t s$ the $L^{2} (P)$ norm of the DQM remainder in [F10] is at most $t^{2} ∥ s^{2} ∥_{2} / 2 = o (t)$ because $s$ is bounded. Thus the score is $s$ .

1.2givenF1F3F5F6F10

Every bounded mean-zero $s$ is an allowed score. If $g \in L_{0}^{2} (P)$ , truncate it to $g_{m} = \max (- m, \min (g, m))$ . Then $g_{m} \to g$ in $L^{2} (P)$ by [F6], since $∣ g_{m} - g ∣^{2} \leq g^{2}$ and the difference tends pointwise to zero. Also $E_{P} g_{m} \to E_{P} g = 0$ by Cauchy–Schwarz and [F3], [F5]. Hence $s_{m} = g_{m} - E_{P} g_{m}$ is bounded and mean-zero, and $s_{m} \to g$ in $L^{2} (P)$ . Thus the score closure is exactly $L_{0}^{2} (P)$ by the tangent-space convention [F10].

1.3givenF4F7F8F9algebra

Since $P \in M_{X}$ , $E_{P} X < \infty$ ; with bounded $s$ , [F8] gives the finite expectation $E_{P_{t, s}} X = E_{P} [X (1 + t s)]$ , so every designated path lies in $M_{X}$ . Linearity [F7] then yields $ψ (P_{t, s}) = E_{P_{t, s}} X = μ + t E_{P} [X s]$ . Since $E_{P} s = 0$ , the right derivative is $E_{P} [(X - μ) s]$ .

2.1givenstep 1.2step 1.3F1F3F5F10

Cauchy–Schwarz gives $∣ E_{P} [(X - μ) s] ∣ \leq ∥ X - μ ∥_{2} ∥ s ∥_{2}$ , so the derivative extends continuously to $T$ . The vector $X - μ$ is mean-zero and square-integrable, hence belongs to $T$ . If another $\tilde{φ} \in T$ represents the same derivative, then $h = \tilde{φ} - (X - μ) \in T$ is orthogonal to every bounded mean-zero score; by density and continuity it is orthogonal to itself, so $E_{P} h^{2} = 0$ and $h = 0$ . By [F10], $X - μ$ is the unique representer in $T$ and therefore the canonical gradient.

3.1givenstep 2.1F2algebra∎

By the empirical-measure convention in [F2] and $E_{P} (X - μ) = 0$ , $(P_{n} - P) (X - μ) = n^{- 1} \sum_{i = 1}^{n} (X (X_{i}) - μ) = {\overline{X}}_{n} - μ$ exactly. Step 2.1 gives $X - μ \in L_{0}^{2} (P)$ . The remainder is zero, so the sample mean is asymptotically linear with estimator influence function $X - μ$ under [F2]. This algebraic identity invokes no central limit theorem.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 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.