LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

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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.

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A nonnegative normalized kernel estimate integrates to one

Statement

Assume the Axiom of Countable Choice. If $K \geq 0$ pointwise and $\int K d λ_{d} = 1$ , then for every sample $z_{1}, \dots, z_{n} \in R^{d}$ and bandwidth $h > 0$ , the function $x \mapsto \frac{1}{n h^{d}} \sum_{i = 1}^{n} K ⁣ (\frac{x - z_{i}}{h})$ is nonnegative and has Lebesgue integral one.

Facts & Assumptions

Given: Countable Choice; a pointwise nonnegative Borel kernel $K$ of integral one; $n \geq 1$ ; $h > 0$ ; and a sample $z_{1}, \dots, z_{n}$ .

[A1]

The Axiom of Countable Choice, written ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The estimate is the finite sum $(n h^{d})^{- 1} \sum_{i} K ((x - z_{i}) / h)$ and $\int K = 1$ (Density targets and the kernel density estimator).

[F2]

Under ${A C}_{ω}$ , a $C^{1}$ diffeomorphism obeys the change-of-variables formula for every nonnegative measurable integrand (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

[F3]

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

Proof

1.1givenA1F1F2

Every summand is nonnegative because $K \geq 0$ . For fixed $z_{i}$ , the map $T_{i} (u) = z_{i} + h u$ is a $C^{1}$ diffeomorphism of $R^{d}$ with $∣ \det D T_{i} ∣ = h^{d}$ . Applying [F2] to $K$ gives $\int_{R^{d}} h^{- d} K ⁣ (\frac{x - z_{i}}{h}) d x = \int_{R^{d}} h^{- d} K (u) h^{d} d u = 1.$

2.1step 1.1F3algebra∎

Each scaled translate is in $L^{1}$ by step 1.1. Linearity [F3] therefore gives $\int_{R^{d}} \frac{1}{n h^{d}} \sum_{i = 1}^{n} K ⁣ (\frac{x - z_{i}}{h}) d x = \frac{1}{n} \sum_{i = 1}^{n} 1 = 1.$ This proves the claim.

Depends on

Used by

Gaussian-kernel moments and pointwise constants Example

Cited to discharge well-definedness by Density targets and the kernel density estimator.

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Sources

Whitney K. Newey, Nonparametric and Semiparametric Estimation, MIT OpenCourseWare 14.385, Fall 2007, PDF pp. 1–3 (standard reference, not scraped)
Adrian Bowman and Ludger Evers, Nonparametric Smoothing, APTS 2013–14 course notes, §1.2, PDF pp. 6–7 (standard reference, not scraped)