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Regression Estimation — Examples

1 · Prerequisites

2 · Summary

Local-linear weighted least squares reproduces an affine response exactly whenever its moment matrix is nonsingular. For Nadaraya–Watson smoothing, a nonconstant design density creates a second-order interior bias even for a linear regression function, and a support boundary can create first-order bias. A separate finite-sample counterexample shows that the local-linear estimator's expectation need not exist under local smoothness, the displayed bandwidth conditions, and singular-only fallback. This is compatible with the proved conditional response-law bias expansion in probability; it prevents interpreting that expansion as an unconditional expectation formula.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Local-linear regression reproduces affine responses

Example

Assume AC as in Regression target and chosen version. Let $X, Y$ be real random variables with $E ∣ Y ∣ < \infty$ and $Y = a + b X$ almost surely. Choose the regular conditional law $Q (t, d y) = δ_{a + b t} (d y)$ for every $t \in R$ ; then the chosen regression-function version is $m (t) = a + b t$ at every point. For any observed sample and any $x, h$ whose degree-one local moment matrix is nonsingular, the local-linear estimate equals $a + b x$ , even when the design points are asymmetric around $x$ .

Facts & Assumptions

Given: The integrable affine-response model, its specified pointwise regression-function version, and a nonsingular degree-one local moment matrix.

[F1]

For degree one, the estimator is the intercept from the weighted least-squares coefficient vector $(Z^{T} W Z)^{- 1} Z^{T} W Y$ . Local-polynomial estimator.

[F2]

The chosen version $m (t)$ represents $E [Y ∣ X = t]$ in the almost-sure regression-target sense. Regression target and chosen version.

[A1]

AC is inherited only through the selected version in [F2]; the finite-dimensional calculation is choice-free.

[A2]

The specified conditional law $Q (t, d y) = δ_{a + b t} (d y)$ gives $m (t) = a + b t$ for every $t$ , not just almost surely under the law of $X$ .

Verification

Given: The data satisfy $Y_{i} = a + b X_{i}$ , the selected conditional law is as in the Example, and the sample moment matrix is nonsingular.

1.1givenF1A1algebra

With $u_{i} = (x - X_{i}) / h$ , the design row is $(1, u_{i})$ and $X_{i} = x - h u_{i}$ , so $Y_{i} = (a + b x) - b h u_{i}$ . Thus the response vector is exactly $Z γ$ for $γ = (a + b x, - b h)^{T}$ .

2.1givenF1F2A2step 1.1algebra∎

The weighted least-squares formula gives $\hat{β} = (Z^{T} W Z)^{- 1} Z^{T} W Z γ = γ$ . By the chosen-version convention in [F2], the specified conditional law in [A2] fixes $m (x) = a + b x$ at this evaluation point. Thus its intercept is $a + b x = m (x)$ , proving exact reproduction of that chosen version.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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 $h^{2}$ at a support boundary.

Facts & Assumptions

Given: The two models below and $K (u) = \frac{1}{2} 1_{{∣ u ∣ \leq 1}}$ , which is symmetric, nonnegative, normalized, and has $μ_{2} = 1 / 3$ .

[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 $f d λ$ by $(f d λ) (A) = \int_{A} f d λ$ . 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 $Q (t, d y) = δ_{t} (d y)$ for every $t \in R$ , 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 $Q (t, d y) = δ_{t} (d y)$ for every $t \in R$ , 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, $f (- h u) = (1 - h u) / 2$ and $m (- h u) = - h u$ . Evenness gives $\int K (u) u d u = 0$ , while $\int K (u) u^{2} d u = 1 / 3$ . Hence the denominator is $1 / 2$ and the numerator is $h^{2} / 6$ , so their ratio is $h^{2} / 3$ . The planned interior formula gives the same nonzero design-gradient term $\frac{1}{2} h^{2} (1 / 3) (2 m^{'} (0) f^{'} (0) / f (0)) = h^{2} / 3$ .

1.2givenF2F3A2algebra

In the boundary model, define the population kernel-weighted response mean by ${\tilde{m}}_{h} (0) = E [K (- X / h) Y] / E [K (- X / h)]$ . By [A2] and [F3], it equals $\int_{0}^{1} K (- t / h) t d t / \int_{0}^{1} K (- t / h) d t$ . Here $K (- t / h) = 1 / 2$ exactly for $0 \leq t \leq h$ and is zero for $h < t \leq 1$ ; hence the denominator is $h / 2$ and the numerator is $h^{2} / 4$ , yielding ${\tilde{m}}_{h} (0) = h / 2$ .

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 $h^{2}$ . Both contradict the refuted assertion.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Local-linear estimator can lack a finite-sample expectation

Statement refuted

At a fixed one-dimensional interior point, suppose the design density is positive and $C^{2}$ locally, the chosen regression function is $C^{2}$ locally, and the conditional response variance is finite and continuous. Suppose $K$ is nonnegative, symmetric, positive near zero, normalized, with finite positive second moment and finite squared integral. If $h_{n} \to 0$ , $n h_{n} \to \infty$ and $n h_{n}^{3} \to \infty$ , then the empirical local-linear fit with a fixed fallback only on singular designs has a Lebesgue expectation (and hence an unconditional second-order mean-bias expansion).

Facts & Assumptions

Given: $X_{i}$ are independent uniform variables on $[- 1, 1]$ ; $Y_{i}$ are independent Rademacher variables independent of all predictors; $K (u) = \frac{1}{2} 1_{{∣ u ∣ \leq 1}}$ ; $h_{n} = n^{- 1 / 4}$ ; and the degree-one estimator has any fixed fallback.

[F1]

A degree-one local-polynomial estimator is the intercept of the weighted least-squares coefficient vector when its moment matrix is nonsingular, and uses its fallback only when that matrix is singular. Local-polynomial estimator.

[F2]

The chosen regression version represents $E [Y ∣ σ (X)]$ almost surely. Regression target and chosen version.

[F3]

Independent and identically distributed variables have a common law and are mutually independent. Identical distribution and IID families.

[F4]

A density $f$ defines the measure $f d λ$ by $(f d λ) (A) = \int_{A} f d λ$ . The measure with density $f$ relative to $μ$ .

[A1]

AC is used only to specify the conditional-law version The Axiom of Choice; all sample-space and divergence calculations are choice-free.

[A2]

Choose the regular conditional law $Q (t, \cdot) = \frac{1}{2} δ_{- 1} + \frac{1}{2} δ_{1}$ for every $t \in R$ . This is a version of the response law given $X$ and gives the selected version $m (t) = 0$ and conditional variance $1$ at every $t$ .

Counterexample

1.1givenF1F2F3F4A1A2construct

Put $h = h_{n}$ . By [A2], the chosen regression function is $m (t) = 0$ and the conditional response variance is $1$ at every $t$ ; the uniform design density is $1 / 2$ near $0$ . The uniform kernel has $μ_{2} = 1 / 3$ and $R (K) = 1 / 2$ . For each $n \geq 2$ , the event $E_{n} = {X_{1} \in [h / 4, h / 3], X_{2} = X_{1} + δ for 0 < δ < h / 12, ∣ X_{j} ∣ > h for j \geq 3}$ has positive design density; on it exactly two kernel weights are positive. Since $X_{1} \neq X_{2}$ , the two-column degree-one design matrix is invertible, so [F1] gives its intercept $(X_{2} Y_{1} - X_{1} Y_{2}) / (X_{2} - X_{1})$ .

2.1givenstep 1.1F2algebra

Conditional on $E_{n}$ , each sign pattern $(Y_{1}, Y_{2}) = (1, - 1)$ and $(- 1, 1)$ has probability $1 / 4$ , and the intercepts are respectively $(X_{1} + X_{2}) / δ$ and $- (X_{1} + X_{2}) / δ$ . On the chosen $X_{1}$ interval, $X_{1} + X_{2} \geq h / 2$ , while the joint design density in $(X_{1}, δ)$ is a positive constant. Integrating either signed tail therefore includes a positive multiple of $\int_{0}^{h / 12} (h / 2) δ^{- 1} d δ = \infty$ . Both positive and negative parts have infinite expectation, so the estimator has no Lebesgue expectation.

3.1givenF1step 2.1algebra∎

The bandwidth satisfies $h_{n} \to 0$ , $n h_{n} = n^{3 / 4} \to \infty$ , and $n h_{n}^{3} = n^{1 / 4} \to \infty$ , and the matrix is nonsingular on $E_{n}$ . Thus the counterexample satisfies these local smoothness, kernel and bandwidth hypotheses while the finite-sample expectation does not exist; the singular-matrix fallback is not used on these samples.

Regression Estimation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Local-linear regression reproduces affine responses

Example

Facts & Assumptions

Verification

Nadaraya–Watson bias from design gradients and boundaries

Statement refuted

Facts & Assumptions

Counterexample

Local-linear estimator can lack a finite-sample expectation

Statement refuted

Facts & Assumptions

Counterexample

Sources