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An integer pattern must survive the stated measurement uncertainty. For an original simulated example, choose a reference electric charge $Q_{\ast }>0$, define $y_i=|q_i|/Q_{\ast }$, and stipulate state meansHere each ${\sigma }_i$ is an independent standard uncertainty of a state mean, before shared calibration effects. The generator constructs repeated noisy observations with these means; the numbers are deliberately chosen, not historical data. There is no conversion of $Q_{\ast }$ to coulombs and no measured value of the elementary charge in this example.

For a candidate spacing $a>0$ in units of $Q_{\ast }$ and proposed integers $n_i$, use weighted least squares:The last two expressions follow by differentiating the quadratic in $a$ and propagating the independent errors. They are conditional on fixed, correct integer assignments and the stated error model. They do not include uncertainty from choosing those assignments. For $\mathbf{n}=(2,3,5,4)$, they give $\hat{a}=1$, $u(\hat{a})\simeq 0.01089$, and ${\chi }^2=0.46875$. Such a small residual is compatible with this chosen error scale; it is not evidence that the errors vanish. A chi-squared distribution interpretation additionally needs the appropriate normal distribution error model and allowance for the fitted parameter; searching many assignments changes that interpretation.

The obstruction is identifiability: replacing $(a,n_i)$ by $(a/m,m{n}_i)$ for any positive integer $m$ leaves every predicted reading unchanged. In particular, an $e$-spaced pattern also fits an $e/2$ lattice using even indices. Even the coarsest well-fitting spacing need not be the physical unit: a dataset containing only even charge states can have twice that unit. Worse, if arbitrarily fine spacings and unrestricted integers are allowed, rounding each $y_i/a$ givesConsequently a finite fit cannot prove a minimal transferable electric charge. Reproducible smallest observed jumps supply additional evidence; interpreting them as single-carrier transfers requires independent physical support. One visibly abrupt transition could contain several unresolved transfers. In the simulation the successive differences are $0.93$, $2.06$, and $-1.03$, each with standard uncertainty $\sqrt{2}(0.08)\simeq 0.1131$. These support changes near $a,2a,-a$ conditional on that unit; they equally support $2,4,-2$ units of $a/2$. Neighboring differences share a state reading and are therefore correlated, even though the state means were stipulated independent.

A shared scale error can preserve every apparent integer relation. To first order, an independent fractional scale standard uncertainty $s$ adds $s^2{a}^2{n}_i{n}_j$ to the covariance matrix of the $y_i$. For the simulation, stipulating $s=0.02$ changes the conditional uncertainty of $\hat{a}$ to approximately $\sqrt{0.01089^2+0.02^2}=0.02277$, despite the same residual pattern. Repeating timings cannot average away this common scale contribution. In the uncorrected continuum model in elementary charge, holding the other measured inputs fixed gives $|q|\propto {\eta }^{3/2}$, so a small fractional error in air dynamic viscosity causes $3/2$ times that fractional error in inferred electric charge. The gas slip correction to Stokes drag requires its own propagation. Pattern agreement, drop-stability controls and absolute measurement calibration together make the inference persuasive within resolution.

A nonnegative regression kernel supplies weights $K((x_i-x)/h)$ in a local weighted least squares fit. Unlike a probability density function, its integral need not equal one, because rescaling every weight does not change the fitted coefficients.

Local linear regression fits $a+b(X_i-x)$ to responses near each target $x$ by weighted least squares, then returns the fitted intercept. With weights $w_i=K((X_i-x)/h)$, put $S_k={\sum }_i{w}_i(X_i-x{)}^k$ and $T_k={\sum }_i{w}_i(X_i-x{)}^k{Y}_i$. If $S_0{S}_2-S_1^2>0$, then $\hat{f}(x)=(S_2{T}_0-S_1{T}_1)/(S_0{S}_2-S_1^2)$. Its polynomial reproduction property of local polynomial regression reproduces affine functions, reducing boundary bias relative to a local constant fit.

Local polynomial regression fits a low-degree polynomial by weighted least squares near each target point.

Choose the diagonal matrix $W=\mathrm{diag}(w_1,\dots ,w_m)$. Multiplying a vector by $W$ scales its $i$th component by $w_i$. ThereforeThe Euclidean norm objective for ordinary linear least squares on $(WA,Wb)$ is exactly the weighted least squares objective, so the sets of minimizing vectors coincide. Zero weights simply discard the corresponding residuals.

Using the paper's named weights $w_i$, takeThe least-squares normal equations for weighted least squares becomeMultiplying by four gives $14{\hat{\beta }}_0+33{\hat{\beta }}_1=59$ and $33{\hat{\beta }}_0+97{\hat{\beta }}_1=143$. Their determinant is $269$, givingThus, to two significant digits, $\boxed{{\hat{\beta }}_0=3.7,{\hat{\beta }}_1=0.20}$, with the slope measured in expression units per hour.

There is a wording convention to make explicit. Because the stated objective squares $w_i$, the choice above gives the final two squared residuals one-quarter of the influence of the others. If “half as much” instead means half the coefficient of each squared residual, use $W=\mathrm{diag}(1,1,1,1/\sqrt{2},1/\sqrt{2})$. Then the least-squares normal equations aregiving ${\hat{\beta }}_0=150/37\approx \boxed{4.1}$ and ${\hat{\beta }}_1=1/37\approx \boxed{0.027}$. Stating which quantity is halved resolves the ambiguity.

The missing output is needed to identify the actual regression formula, response column, transformations and coefficient estimates. A useful possible analysis regresses the maintained-school percentage among acceptances on the corresponding percentage among applications, allowing for year; it must be described as a model of college-level composition, not a direct estimate of individual acceptance probabilities.

For instance, if $a_{jt}$ and $p_{jt}$ are these two percentages, a normal linear model might specify $\mathbb{E}(a_{jt})=\alpha +\beta p_{jt}+\delta 1_{\{t=2000\}}$, with a slope-by-year interaction term if needed. The intercept is the expected response at zero predictor, often an extrapolation; the slope is a percentage-point change in acceptance composition per percentage-point change in application composition. A slope of one is not by itself evidence of equal acceptance chances: equal chances within every college would give the entire relation $a_{jt}=p_{jt}$, including zero intercept and no year shift. A fit with positive slope mainly confirms that applicant composition helps predict acceptance composition.

Regression diagnostics should include residuals versus fitted values and predictors, a residual normal Q-Q plot, checks of heteroscedasticity, regression leverage, studentized residuals and Cook's distance. The aggregate mature-college category merits an influence check and a sensitivity analysis: combining several colleges can conceal different within-college patterns. Repeated observations from the same college across years may have correlated errors, so treating them as independent needs justification. Check nonlinearity, the year interaction and the effects of excluding no observation except with a substantive reason.

Percentages are bounded, and their precision depends on their denominators. A simple weighted least squares or binomial analysis requires the underlying numbers accepted, not merely percentages. For a proportion based on $m$ people, the working variance is $\pi (1-\pi )/m$ under independent sampling, rather than constant across colleges. If the predictor percentage is itself estimated, an ordinary regression also ignores its uncertainty. The supplied summaries cannot recover all four cells of each college contingency table or the actual fitted numerical output.

The published Cambridge percentage table nevertheless allows a descriptive comparison of acceptance rates. Write $p$ for the maintained-school proportion among applicants and $a$ for its proportion among acceptances. The relative risk from group compositions givesThe overall acceptance probability cancels by Bayes' theorem. This risk ratio compares acceptance probabilities; it is not the acceptance odds ratio, which also needs the overall acceptance probability.

For Christ's, the two published application/acceptance pairs give risk ratios $37/77\approx 0.481$ and $147/187\approx 0.786$. If the displayed percentages were rounded to the nearest whole percentage point, the relative risk bounds from rounded group compositions give ranges approximately $[0.461,0.501]$ and $[0.755,0.818]$. Thus both comparisons remain below one under that rounding assumption. These are descriptive comparisons of observed groups, without adjustment for applicant qualifications or a causal effect interpretation. The published compositions determine approximate relative acceptance rates, but not their standard errors or the original regression results.

The response errors have standard deviations proportional to $|{\mu }_i|$. Thus, when fitted means approximate the true means, increasing absolute fitted values tends to accompany increasing residual spread. A residual-versus-fitted plot shows a fan for positive means, or spread increasing away from zero for means of either sign. Standardization by one common residual standard deviation, even with the usual regression leverage correction, does not remove this systematic heteroscedasticity.

Under independent errors the exact covariance matrix after fitting iswhere $k>0$ is the common proportionality constant. Hence the simple fan description is a diagnostic tendency, not an assertion that each fitted residual has exactly variance $k{\mu }_i^2$.

For positive responses, the intended variance-stabilizing transformation is the natural logarithm. The delta method givesUse $\log Y_i$ for positive data with modest relative error, then refit and check the transformed mean and residual structure. In particular, if the original mean was $X_i^T\beta$, the transformed mean is approximately $\log (X_i^T\beta )-k/2$, not automatically a linear function of the original covariates.

There is a domain qualification in the literal printed model: a nondegenerate normal distribution has negative support. For positive ${\mu }_i$, the probability of $Y_i\le 0$ is $\Phi (-1/\sqrt{k})$, so a real $\log Y_i$ is not defined throughout the stated sampling model. When $k$ is small this is a negligible-tail approximation, not an exact transformation to normal errors. If signed values must be retained and all ${\mu }_i\ne 0$, write $Y_i={\mu }_i(1+\sqrt{k}{Z}_i)$ with standard normal $Z_i$. ThenThe second term has a common distribution with finite variance, so log absolute value stabilization of a normal scale family is exact, but loses the sign of the mean and does not produce normal errors. Zeros and a zero mean are excluded from this transform. Alternatively retain the response scale and fit the stated variance model with weighted least squares weights proportional to $1/{\hat{\mu }}_i^2$ where the means stay away from zero. No arbitrary addition of a constant before taking logarithms exactly solves the given variance relation.

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.

For ordinary homoscedastic fixed-design nonparametric regression, take observations $Y_i=g(x_i)+{ϵ}_i$ at known distinct sites, with independent mean-zero errors of common variance. A degree-$p$ local polynomial estimator at a target $u$ solves the finite-dimensional weighted least squares problemHere $K$ is a nonnegative regression kernel, $h>0$ is the smoothing bandwidth, and sufficient local design rank ensures uniqueness. The estimate of the regression function is $\hat{g}(u)={\hat{\beta }}_0(u)$. Each target has its own locally fitted polynomial; this is not one global polynomial fit. The bandwidth controls the localization and the bias-variance tradeoff.

A natural cubic smoothing spline instead solves the global penalized problemThe Sobolev space here consists of functions with square-integrable weak derivatives through order two. The minimizer is a natural cubic spline with knots at the observation sites and linear tails outside the extreme sites. Equivalently one may perform the minimization on that finite-dimensional spline space: the minimum roughness property of the natural cubic spline interpolant shows that replacing a candidate by the natural spline with the same site values preserves the data loss and decreases the second-derivative penalty. The smoothing parameter $\lambda$ controls curvature; it differs from the local bandwidth $h$.

For the interval-average observation problem, the printed upper limit in the loss is an indexing error: there are only $n-1$ responses and intervals, so a term with $i=n$ would involve the undefined $Y_n$ and $x_{n+1}$. The coherent loss uses the upper limit $n-1$. Put ${\Delta }_i=x_{i+1}-x_i$ and define the linear observation functionalsThe intended criterion on $C^1[a,b]$ is thereforeUnlike the ordinary cubic smoothing spline, this uses interval averages as data and a first-derivative penalty; the resulting spline degree will be two.

Fix any candidate $v\in C^1[a,b]$. By the supplied interpolation property, there is a unique quadratic spline $q\in C^1[a,b]$ with knots $x_1,\dots ,x_n$, constant on both exterior intervals, such that $L_{i}q=L_{i}v$ for every $i$. Put $d=v-q$. ThenOn each interior knot interval $q^{\prime \prime }$ is a constant $c_i$, and on the two exterior intervals $q^{\prime }=0$. Integrate $q^{\prime }{d}^{\prime }$ by parts separately on the intervals. At every knot the boundary contributions cancel because both $q^{\prime }$ and $d$ are continuous; the contributions at $a$ and $b$ vanish because $q^{\prime }$ is zero there. ThusThis orthogonality is the key to the quadratic smoothing spline for interval averages. Expanding the roughness penalty givesThe data-fit terms are equal because the interval averages are equal. Consequently,Equality can occur only if $d^{\prime }=0$ everywhere: the derivative is continuous and its squared integral is zero. Then $d$ is constant, and any of its zero interval averages forces that constant to be zero. Therefore every minimizer must equal its matching quadratic spline $q$.

We conclude that the minimizer is a continuously differentiable quadratic spline with knots at the specified sites and constant tails on both exterior intervals. Existence and uniqueness also follow from the supplied interpolation property. The interval-average map identifies the finite-dimensional space of these quadratic splines with ${\mathbb{R}}^{n-1}$, and the criterion becomes a strictly convex quadratic function of that average vector: its data-fit term is $\Vert Y-y{\Vert }^2$, while its penalty is nonnegative and quadratic. It has a unique minimizer, and the strict equality argument above proves uniqueness in the full function class.

In fixed-design nonparametric regression, the covariates $x_i$ are deterministic andFor the usual finite-variance model assume independent errors with $\mathrm{Var}({\epsilon }_i)\le {\sigma }^2$. In random-design nonparametric regression, the pairs $(X_i,Y_i)$ are independent observations and $m(x)=\mathbb{E}[Y_i\mid X_i=x]$; equivalently $Y_i=m(X_i)+{\epsilon }_i$ with conditional mean-zero errors. Conditional variances are generally assumed bounded for risk bounds.

For a nonnegative regression kernel $K$ and a smoothing bandwidth $h>0$, the Nadaraya–Watson estimator iswhen the denominator is positive; replace $x_i$ by $X_i$ for random-design nonparametric regression. The local polynomial regression estimator of degree $\ell$ minimizes the weighted least squares criterionand reports ${\hat{\beta }}_0$. If the weighted Gram matrix is invertible, this minimizer is unique. The Nadaraya-Watson estimator is exactly local polynomial regression of degree zero. Higher degrees reproduce nonconstant polynomials and reduce boundary bias of an estimator.

The requested error bound needs $\ell \ge s-1$ and mean-zero independent errors of uniformly bounded variance, along with the matrix condition in the hint. The degree requirement is missing from the printed question. To see the problem, take $\ell =0$, $K=1_{[-1,1]}$, $m(t)=t$, zero errors, $x=0$, and $h=n^{-1/4}$. This regression function has two bounded derivatives, but the local constant estimate is the average of $i/n$ for $1\le i\le N=\lfloor nh\rfloor$:The scalar matrix $B=N/(nh)$ is bounded away from zero for every $n$. Nevertheless $(nh{)}^{-1/2}+h^2=n^{-3/8}+n^{-1/2}=o(h)$, so the printed conclusion with $s=2$ fails. A bounded-variance error hypothesis is also essential; otherwise an expectation bound need not even be finite.

Under the qualified hypotheses, use $U(t)=(1,t,\dots ,t^{\ell }/\ell !{)}^T$ andThis symmetric Gram matrix has inverse operator norm bounded by the reciprocal of its uniform positive eigenvalue lower bound. If $K$ is supported in $[-R,R]$, only $|x_i-x|\le Rh$ contribute. The equally spaced design and $nh\ge 1$ imply that there are at most $2Rnh+1\le Cnh$ such points. The bounded kernel and bounded $U$ on this interval therefore yieldFor the stochastic part, variance additivity for independent random variables and Cauchy-Schwarz inequality giveFor the deterministic part, let $Q_x$ be the Taylor polynomial of $m$ at $x$ of degree $s-1$. The polynomial reproduction property of local polynomial regression gives ${\sum }_i{W}_{ni}(x)Q_x(x_i)=Q_x(x)=m(x)$. By Taylor's theorem, $|m(x_i)-Q_x(x_i)|\le C|x_i-x{|}^s\le C{h}^s$ on the kernel window. ThusCombining the two parts provesThe argument also applies at the endpoints when the assumed matrix bound holds. With arbitrary degree, the generally valid smoothness term is $h^{\min (s,\ell +1)}$; extra symmetry can improve some interior biases but does not fix the general boundary counterexample.

For fixed-design nonparametric regression, let $x_i=i/n$ and $K_h(u)=h^{-1}K(u/h)$. The degree-$p$ local polynomial estimator takes the fitted intercept ${\hat{m}}_h(x;p)={\hat{\beta }}_0$, whereUse a nonnegative regression kernel and an invertible local polynomial Gram matrix to obtain a unique fit. For $p=0$, solving the single weighted least squares normal equation gives the local constant estimator:At the interior point $x_0$, use the standard unit-integral regression kernel convention ${\mu }_0(K)=1$. Otherwise the leading variance below is $R(K)/(nh{\mu }_0(K{)}^2)$; unlike a kernel density estimator, the local constant estimator itself is unchanged by rescaling $K$.

The local constant estimator is a linear estimator in nonparametric regression with weights $W_i=K_h(x_i-x_0)/(n{s}_{0,h})$. The independent unit-variance errors implyTo control the bias of an estimator, differentiability at this one point is enough. WriteSymmetry makes ${\mu }_1(K)=0$, and the supplied moment approximation gives $s_{1,h}=o(h)$. Therefore the linear contribution to the bias of an estimator is $m^{\prime }(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta (h)h$, because only $|x_i-x_0|\le h$ receive positive weight. This is the interior first-order bias cancellation for local constant regression, and gives squared bias of an estimator $o(h^2)$. Combining with the variance provesFor fixed $\epsilon >0$, take $h_{\epsilon }={\epsilon }^{-1/3}{n}^{-1/3}$. It satisfies both asymptotic smoothing bandwidth conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\epsilon$. ThusLetting $\epsilon \downarrow 0$ proves the optimized scaled pointwise error tends to zero. The order of these limits matters: no remainder uniform in $\epsilon$ has been assumed.

For the lower bound over the whole function class, the absence of a uniform bound on derivatives permits a particularly direct argument. It also avoids assuming a normal distribution for errors when the model only supplies their first two moments. Fix $n$ and $x_0$. Choose a smooth bump function $b$ on the interval, with $b(x_0)=1$, whose support contains no design point except possibly $x_0$. Such a bump exists because the design is finite; at an endpoint use the restriction of a smooth bump on the real line. Every $m_a=ab$, $a\in \mathbb{R}$, belongs to the differentiability class.

If $x_0$ is not a design point, all observation means are zero for every $a$. The Le Cam two-point lemma applied to $a=\pm A$ has total variation distance zero and gives a lower bound tending to infinity with $A$. In fact, the supremum of the mean squared error is infinite at every such $x_0$.

If $x_0$ is a design point, only its response depends on $a$. The resulting location family is the single observation $Y=a+ϵ$; all remaining observations are independent of $a$ and give no extra information. The single-observation location minimax bound isHere is a proof covering discrete as well as continuous errors. Give $a$ the prior distribution $U[-A,A]$, independently of the noise. Put $p_M=\mathbb{P}(|ϵ|\le M)$ and $v_M=\mathrm{Var}(ϵ\mid |ϵ|\le M)$. Consider an easier experiment in which an oracle also reveals whether $|ϵ|\le M$, and reveals $a$ itself on the complementary event. On the event $|ϵ|\le M$ and $|Y|\le A-M$, every possible truncated noise value places $a=Y-ϵ$ within $[-A,A]$. The flat prior distribution then makes the Bayesian posterior noise law exactly its truncated original law. The smallest conditional squared-error loss is its conditional variance $v_M$, attained by the posterior mean.

The probability of that event is at least $p_M(1-2M/A)$ for $A>2M$, by restricting further to $|a|\le A-2M$. Thus every rule's supremum risk function is at least the original Bayes risk, which is at least the easier experiment's Bayes risk, and therefore at least $p_M(1-2M/A)v_M$. Let $A\to \infty$ and then $M\to \infty$. Finite second moment and zero mean give $p_M{v}_M\to 1$. The rule $T(Y)=Y$ has constant mean squared error one, proving the equality.

Consequently, at every $x_0\in [0,1]$, the unrestricted differentiability class satisfies the stronger conclusionThus $c=1$ works for every $n\ge 1$. There is no conflict with the preceding fixed-function limit: the worst functions, including arbitrarily narrow bumps, can vary with $n$. This is the pointwise versus uniform risk distinction. A smoothness ball with a common derivative bound would pose a different minimax risk problem.

Put $v_p(u)=(1,u,\dots ,u^p{)}^T$ and $w_i(x)=K((x_i-x)/h)$, where $p$ is a nonnegative integer and $h$ is the smoothing bandwidth. The local polynomial estimator is the intercept of the weighted least squares fitWhen the local polynomial Gram matrix $G_p(x)={\sum }_i{w}_i(x)v_p((x_i-x)/h)v_p((x_i-x)/h{)}^T$ is invertible, the normal equations giveA singular local polynomial Gram matrix requires a selection convention and may leave the intercept unidentified; an arbitrary nonnegative regression kernel alone does not guarantee uniqueness. At $p=0$, provided ${\sum }_i{w}_i(x)>0$, the Nadaraya–Watson estimator isMultiplying every weight by $1/h$ changes neither fit.

For the uniform kernel, let $I_x=\{i:|i/n-x|\le h\}$ and $N_x=|I_x|$. Then ${\hat{m}}_h(x)=N_x^{-1}{\sum }_{i\in I_x}{Y}_i$. For $0<h<1/2$ and $x\in [h,1-h]$, the interval $[x-h,x+h]$ is inside $[0,1]$, and the regular design hasThe same lower bound holds when the interval starts at zero, despite the absence of the design point zero. In particular the Nadaraya–Watson estimator is defined throughout the integration interval.

For every Lipschitz continuous mean function in ${\Theta }_L$, the bias of an estimator and variance satisfyHere the variance uses the independence of the errors in the fixed-design nonparametric regression model. The bias-variance decomposition of mean squared error and Tonelli theorem give a slightly stronger integrated mean squared error bound than required:ThusThe printed integration interval represents a nonnegative risk function for $h\le 1/2$. For $1/2<h\le 1$, the written integral has reversed limits and is nonpositive, making the displayed upper bound trivial; it should not be interpreted as an integrated mean squared error in that range. For $h>1$, evaluation outside $[0,1]$ would additionally require a specified extension of the mean function. The subsequent optimization restricts to $h<1/3$.