Alphabeta Math
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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Gaussian KDE boundary bias for a uniform density

Example

In one dimension, choose the Borel density version f=1[0,1] and the standard Gaussian kernel K(u)=(2π)−1/2e−u2/2. At the support boundary x=0, Ef^h(0)=∫−1/h0K(u)du⟶12(h↓0), while the chosen target value is f(0)=1. Hence the uncorrected estimator has limiting bias −1/2 at the boundary. The smoothness hypotheses of the pointwise second-order bias theorem fail at x=0.

Facts & Assumptions

Given: d=1, iid observations with chosen density version f=1[0,1], kernel K(u)=(2π)−1/2e−u2/2, bandwidth h>0, and evaluation point x=0.

[A1]

The Axiom of Countable Choice, written ACω (The Axiom of Countable Choice (ACω)).

[F1]

The estimator is f^h(x)=(nh)−1∑iK((x−Xi)/h) (Density targets and the kernel density estimator).

[F2]

Expectation is integration against the sample law, which has density f (Change of variables for expectation, Integrating against a density agrees with integrating the product).

[F3]

Expectation is linear on integrable variables (The Lebesgue integral is linear on L1(μ)).

[F4]

The nonnegative plane Gaussian integral is ∫R2e−(x2+y2)d(x,y)=π (The plane Gaussian integral equals π by polar coordinates).

[F6]

Nonnegative integrals pass through increasing limits (Monotone convergence for the integral).

[F7]

Under ACω, affine C1 substitutions obey nonnegative change of variables (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

Verification

1.1givenA1F4F5F7F8algebra

The function f=1[0,1] is a Borel density version: [F8] gives λ1([0,1])=1. Put I=∫Re−t2dt. Tonelli [F5] and [F4] give I2=π, so I=π. Scaling with [F7] shows ∫R(2π)−1/2e−t2/2dt=1. Since this kernel is even, ∫−∞0K(u)du=1/2.

2.1givenA1F1F2F3F7step 1.1

By [F1]–[F3] and the density formula, each summand is integrable because K is bounded and the sample law is a probability measure. Thus Ef^h(0)=h−1∫01K(−t/h)dt. Apply [F7] on R to T(u)=−hu and g(t)=1[0,1](t)K(−t/h). It gives ∫01K(−t/h)dt=h∫−1/h0K(u)du, so the displayed expectation equals ∫−1/h0K(u)du.

3.1step 1.1F6step 2.1

Along any sequence hm↓0, the intervals [−1/hm,0] increase to (−∞,0]. Monotone convergence [F6] and the half-mass calculation in step 1.1 give ∫−1/hm0K(u)du⟶∫−∞0K(u)du=12.

4.1givenstep 3.1algebra∎

The density version was chosen with f(0)=1, so step 3.1 gives Ef^h(0)−f(0)→−1/2. The selected density is discontinuous at 0, so this calculation does not satisfy the smooth interior-point theorem's hypotheses.

Depends on

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