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Nadaraya–Watson bias from design gradients and boundaries

Statement refuted

Symmetry of the kernel alone makes the population Nadaraya–Watson smoother reproduce every affine regression function exactly in the interior and keeps its bias of order h2 at a support boundary.

Facts & Assumptions

Given: The two models below and K(u)=121{∣u∣≤1}, which is symmetric, nonnegative, normalized, and has μ2=1/3.

[F1]

Under its stated interior hypotheses, Pointwise population bias of the Nadaraya–Watson smoother identifies the population smoother by the kernel-weighted ratio and gives the interior expansion used in step 1.1.

[F2]

The regression target is a chosen version of E[Y∣σ(X)]. Regression target and chosen version.

[F3]

A density f defines the measure f dλ by (f dλ)(A)=∫Af dλ. The measure with density f relative to μ.

[A1]

In the interior model, f(t)=(1+t)/2 on [−1,1] and Y=X. Choose the regular conditional law Q(t,dy)=δt(dy) for every t∈R, giving the pointwise version m(t)=t; also f′(0)/f(0)=1.

[A2]

In the boundary model, X is uniform on [0,1], Y=X, and the prediction point is the support boundary x=0. Choose the regular conditional law Q(t,dy)=δt(dy) for every t∈R, giving the pointwise version m(t)=t.

[A3]

AC is inherited through the chosen versions of the regression target The Axiom of Choice; the finite integrations use no choice.

Counterexample

1.1givenF1F2F3A1algebra

In the interior model, f(−hu)=(1−hu)/2 and m(−hu)=−hu. Evenness gives ∫K(u)u du=0, while ∫K(u)u2 du=1/3. Hence the denominator is 1/2 and the numerator is h2/6, so their ratio is h2/3. The planned interior formula gives the same nonzero design-gradient term 12h2(1/3)(2m′(0)f′(0)/f(0))=h2/3.

1.2givenF2F3A2algebra

In the boundary model, define the population kernel-weighted response mean by m~h(0)=E[K(−X/h)Y]/E[K(−X/h)]. By [A2] and [F3], it equals ∫01K(−t/h)t dt/∫01K(−t/h) dt. Here K(−t/h)=1/2 exactly for 0≤t≤h and is zero for h<t≤1; hence the denominator is h/2 and the numerator is h2/4, yielding m~h(0)=h/2.

2.1givenstep 1.1step 1.2A3algebra∎

The first ratio is not m(0) despite affine m, and the second has a nonzero coefficient of h, not h2. Both contradict the refuted assertion.

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