Alphabeta Math
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Gaussian-kernel moments and pointwise constants

Example

For d≥1, take the standard Gaussian kernel Kd(u)=(2π)−d/2e−∥u∥2/2. Its second-moment matrix is μjk=δjk and ∥Kd∥22=(4π)−d/2, where δjk is 1 for j=k and 0 otherwise. Thus, under the pointwise theorem hypotheses, Ef^h(x)−f(x)=12h2Δf(x)+o(h2),Var⁡(f^h(x))=f(x)nhd(4π)−d/2+o((nhd)−1). Here Δf=∑j=1d∂jjf.

Facts & Assumptions

Given: d≥1, the kernel Kd above, and a density/sample satisfying the hypotheses of Pointwise second-order bias of the KDE and Pointwise variance of the KDE.

[A1]

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

[F1]

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

[F2]

Tonelli factors nonnegative integrals of products on Euclidean product spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F3]

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

[F4]

Fubini factors absolutely integrable products on sigma-finite product spaces (Fubini's theorem for L^1 functions on a sigma-finite product).

[F7]

The exponential dominates each fixed polynomial at infinity (The exponential dominates every fixed nonnegative integer power at +∞).

[F8]

The pointwise bias and variance expansions hold with their stated regularity and bandwidth conditions (Pointwise second-order bias of the KDE, Pointwise variance of the KDE).

[F9]

Integrals of nonnegative functions pass to increasing-domain limits (Monotone convergence for the integral).

Verification

1.1givenA1F1F2F3F5F6F7F9

Put I=∫Re−t2dt. Tonelli [F2] and [F1] give I2=∫R2e−(x2+y2)d(x,y)=π. Since I≥0, I=π. Let ϕ(t)=(2π)−1/2e−t2/2. The substitution s=t/2 in [F3] then gives ∫ϕ=1. By [F5], ϕ′(t)=−tϕ(t). Integration by parts [F6] gives ∫0Rtϕ(t)dt=ϕ(0)−ϕ(R). By [F7], ϕ(R)→0 and Rϕ(R)→0, so ∫∣t∣ϕ(t)dt=2ϕ(0)<∞. Oddness then gives ∫tϕ(t)dt=0. Integration by parts [F6] on [−R,R] also yields ∫−RRt2ϕ(t)dt=∫−RRϕ(t)dt−[tϕ(t)]−RR. As R→∞, Rϕ(R)→0 by [F7] (put s=R2/4 and use R≤2s for R≥2), and the right side tends to 1, so ∫t2ϕ(t)dt=1 by [F9].

2.1step 1.1F1F2F4algebra

Since Kd(u)=∏j=1dϕ(uj), repeated use of Tonelli [F2] shows ∫Kd=1. Fubini [F4], the one-dimensional moments from step 1.1, and absolute integrability give μjj=1 and μjk=0 for j≠k. Also ∫Kd(u)2du=(2π)−d∏j=1d∫Re−uj2duj=(2π)−dπd/2=(4π)−d/2.

3.1step 2.1F8algebra∎

Substituting μjk=δjk and ∥Kd∥22=(4π)−d/2 into [F8] yields the two displayed leading terms. The Gaussian kernel is nonnegative and integrates to one by step 2.1, so A nonnegative normalized kernel estimate integrates to one also confirms that each finite estimate is a density.

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