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At a target $t$, local polynomial regression fits a polynomial of degree at most $p$ by weighted least squares:Define the response vector $Y=(Y_1,\dots ,Y_n{)}^T$, the design matrix $X$ by $X_{i,j+1}=(x_i-t{)}^j$ for $j=0,\dots ,p$, and the diagonal matrix $W$ by $W_{ii}=K((x_i-t)/h)$. The normal equations for linear least squares giveThis formula assumes the local polynomial Gram matrix $X^{T}WX$ is invertible. It is invertible whenever at least $p+1$ distinct sites have positive weights: a nonzero degree-$p$ polynomial cannot vanish at all of them. With fixed $p$, a nonzero continuous regression kernel, and $nh\to \infty$, this condition holds eventually at the target points considered here. Using weights $h^{-1}K((x_i-t)/h)$ instead gives exactly the same fit.

Use the usual unit-integral normalization ${\mu }_0(K)=1$. A regression kernel need not intrinsically have unit integral, but multiplying all weights by a positive constant leaves local polynomial regression unchanged. With unnormalized raw moments the interior coefficient below is ${\mu }_2(K)/{\mu }_0(K)$; the version written using ${\mu }_2(K)$ alone requires this normalization. A zero regression kernel cannot define an estimator.

Write the local moment sums asThe local constant estimator is ${\hat{m}}_h(t;0)={\sum }_{i}K((x_i-t)/h)Y_i/{\sum }_{i}K((x_i-t)/h)$. Since the errors have zero mean, its bias of an estimator is the corresponding weighted average of $m(x_i)-m(t)$. By the second-order Taylor expansion, uniformly for contributing sites with $|x_i-t|\le h$,where ${\eta }_n\to 0$. This uniform remainder follows from uniform continuity of $m^{\prime \prime }$ on $[0,1]$. Nonnegativity of the regression kernel therefore giveswhenever $A_0$ tends to a positive limit and $A_2=O(h^2)$.

For fixed interior $x$, the moment approximations give $A_0(x)=1+o(1)$, $A_1(x)=O(n^{-1})$, and $A_2(x)=h^2{\mu }_2(K)+o(h^2)$. The hypothesis $n{h}^2\to \infty$ makes $n^{-1}=o(h^2)$, so the odd-moment contribution is negligible at second order. ThusThe symmetry of the regression kernel is what removes the first-order interior bias of an estimator.

At $z_n=\alpha h$, the rescaled observation window is truncated to $[-\alpha ,1]$. Put ${\mu }_{r,\alpha }={\int }_{-\alpha }^1{u}^{r}K(u)du$ and ${\beta }_r={\mu }_{r,\alpha }/{\mu }_{0,\alpha }$. Symmetry, nonnegativity, and unit mass give ${\mu }_{0,\alpha }\ge 1/2$. The supplied moment approximations already imply the leading boundary bias of local constant regression:To retain the second-order term rigorously, we need slightly more precise control of the first moment. The rescaled grid $u_i=i/(nh)-\alpha$ has spacing $1/(nh)$. Each truncated function $u^{r}K(u)1_{[-\alpha ,1]}$ has bounded variation, because $K$ is continuously differentiable on its support. The Riemann sum error is bounded by this variation times the grid spacing: summing the oscillation bounds on the grid cells provesEndpoint cells, including the omitted site at zero, contribute the same error order. ConsequentlyBoth discretization errors are $o(h^2)$ under $n{h}^2\to \infty$. The preceding Taylor expansion therefore proves the second-order boundary expansion for local constant regression:Equivalently, expanding the derivatives about zero givesIn general the truncated first moment is nonzero, so the boundary bias of an estimator has order $h$, although it can be smaller when $m^{\prime }(0)=0$ or the omitted part of the regression kernel carries no mass.

For local linear regression, the polynomial reproduction property of local polynomial regression removes the constant and linear terms exactly. Let $\Delta =A_0{A}_2-A_1^2$. Its effective kernel weights areso ${\sum }_i{l}_i=1$, ${\sum }_i{l}_i(x_i-t)=0$, andAt either an interior or the specified boundary target, $\Delta$ is asymptotic to a positive constant times $h^2$. At the boundary that constant is ${\mu }_{0,\alpha }{\mu }_{2,\alpha }-{\mu }_{1,\alpha }^2>0$, the mass squared times the variance of the continuous normalized truncated regression kernel. Moreover ${\sum }_i|l_i|=O(1)$, as follows by bounding $|A_2-A_1(x_i-t)|=O(h^2)$ in the window. Thus the Taylor remainder still contributes $o(h^2)$, even though some effective kernel weights may be negative. In particular,At an interior point this coefficient reduces to ${\mu }_2(K)$ by symmetry and normalization. Hence local linear regression has bias $O(h^2)$ at both interior and boundary points; its linear reproduction removes the generic order-$h$ boundary term.

At a target $x$, approximate the unknown mean locally by $a+b(X_i-x)$ and solve the local linear regression weighted least-squares problemwhere $K$ is a nonnegative regression kernel and $h>0$ a smoothing bandwidth. A common factor $1/h$ in all weights makes no difference. Estimate $f(x)$ by the fitted intercept ${\hat{a}}_x$, and optionally $f^{\prime }(x)$ by ${\hat{b}}_x$. Repeat at each target rather than fitting a single global straight line.

For an explicit expression put $S_k={\sum }_i{w}_i(X_i-x{)}^k$ and $T_k={\sum }_i{w}_i(X_i-x{)}^k{Y}_i$. The normal equations yieldThe denominator is positive provided there are at least two distinct positively weighted predictor values. The weights sum to one and their first centered moment is zero, so local linear regression exactly reproduces affine functions and reduces boundary bias compared with a local constant fit. Smaller $h$ uses more local detail and increases variance; larger $h$ smooths more and can increase bias. Choose the bandwidth by cross-validation or another justified criterion. A nearest-neighbour span provides an adaptive bandwidth, as in the local fits used later.

Both fits use the Gaussian generalized additive modelwhere $e_i$ is experience and $d_i$ education. Centering makes the additive decomposition identifiable. No experience-education interaction term is included.

The first fit uses local linear regression for each term. span=1 includes all observations in each nearest-neighbour neighbourhood, but distance weights and a moving target still vary: it does not make the fit exactly a global straight line. The second uses cubic smoothing splines, with separate tuning values spar=0.8 and spar=1.2. Increasing spar increases the curvature penalty and smooths more; spar is a monotone tuning parametrization, not the numerical $\lambda$ in part (b). The relation depends on the predictor scaling and design.

The plotted experience effect is increasing in both fits. The local fit gives a broadly smooth, nearly linear rise; the spline fit follows more of the steep low-experience rise and subsequent flattening, especially near the lower boundary. Education has a weaker increasing effect. The local estimate shows some curvature, while the spline fit at spar=1.2 is closer to a straight line. The clearest difference is the amount of smoothing and the experience curvature, rather than opposite effect directions. These are centered partial-effect plots with partial residuals, not scatterplots of wage against each predictor alone.

A smaller local span would allow the first fit to capture more experience curvature. A moderately smaller education spar could reveal nonlinear detail suppressed in the second fit; its small but significant nonparametric component in part (e) supports checking that possibility. Do not tune solely to follow every point, especially at sparse boundaries. Compare candidate spans or separate spline penalties using cross-validation or generalized cross-validation; increase smoothing if flexible fits introduce unsupported wiggles. See the documented monotone spar–penalty convention at stat.ethz.ch/R-manual/R-patched/library/stats/html/smooth.spline.html.