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The loss $|a-\theta |$ for estimating a real parameter $\theta$ by $a$. Its risk of a decision rule is the corresponding expected value. Unlike squared-error loss, its pointwise size is a metric, so the triangle inequality gives a direct overlap version of the Le Cam two-point lemma.

The minimax risk is the infimum over estimators of the supremum of their risk functions over a parameter class. An upper bound supplies an estimator; a lower bound applies to every estimator, often using the Le Cam two-point lemma.

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.

Let $p=dP/d\nu$ and $q=dQ/d\nu$ be Radon-Nikodym derivatives. Use the following normalizations of the Kullback-Leibler divergence, total variation distance and unnormalized Hellinger distance:We use $0\log (0/q)=0$ and assign infinite Kullback-Leibler divergence if $Pot\ll Q$, where $\ll$ denotes absolute continuity of measures. Common domination alone also suffices for the inequalities with the extended-value convention. The negative part of the logarithmic integral is integrable: on $p<q$, $p\log (q/p)\le q-p\le q$. Thus the extended-value integral defining the Kullback-Leibler divergence is well-defined. These expressions do not depend on which common dominating measure is chosen. The Kullback-Leibler divergence is generally asymmetric and therefore fails a defining property of a metric.

For completeness, the integral formula for total variation distance follows by taking $A=\{p\ge q\}$: since $\int (p-q)d\nu =0$, the positive and negative parts of $p-q$ have the same integral, and this set attains the supremum.

Put $a=\int \sqrt{pq}d\nu \le 1$, by the Cauchy-Schwarz inequality. A second application givesThe logarithmic inequality in the hint implies $\log t\le t-1$ for $t>0$. Apply it to $t=\sqrt{q/p}$ on $\{p>0\}$:Multiplying by $p$ and integrating provesIf $q=0<p$ on a set of positive $P$-measure, the right side is finite but the Kullback-Leibler divergence is infinite, so the inequality is immediate. ThusThis total variation–Hellinger–relative entropy inequality requires the stated Hellinger distance normalization for its first constant. With the alternative convention ${\bar{h}}^2=\frac{1}{2}\int (\sqrt{p}-\sqrt{q}{)}^2$, the correct comparison is $\mathrm{TV}\le \sqrt{2}\bar{h}$ and $\bar{h}\le \sqrt{\mathrm{KL}/2}$. In particular, $P=\mathrm{Bernoulli}(0.6)$ and $Q=\mathrm{Bernoulli}(0.4)$ have $\mathrm{TV}=0.2>\bar{h}\approx 0.1421$.

Here is a testing form of the Le Cam two-point lemma. Suppose two parameter values ${\theta }_0,{\theta }_1$ in a metric space have separation $\Delta =d({\theta }_0,{\theta }_1)>0$, and let $P_0,P_1$ be the corresponding observation laws. Every estimator $T$ satisfiesTo prove it, classify $T$ by the closer of the two parameters, breaking a tie in favour of one. Let $A=\{d(T,{\theta }_1)\le d(T,{\theta }_0)\}$. On $A$, the triangle inequality implies $d(T,{\theta }_0)\ge \Delta /2$, while on $A^c$ it implies $d(T,{\theta }_1)>\Delta /2$. Hence the sum of the two probabilities in the display is at leastAt least one is at least half the sum, proving the claim. It also gives a lower bound $\Delta (1-\mathrm{TV})/4$ for the maximum expected metric loss.

For absolute-error loss, or any loss given by a metric, there is a stronger expectation form. Take densities $p_0,p_1$ relative to $m=P_0+P_1$ and use the triangle inequality under their overlap:Therefore the Le Cam lower bound under absolute-error loss isBoth arguments are valid for randomized estimators as well, by adjoining the independent randomization to the observation; doing so does not change total variation distance.

For the normal distribution model, assume the usual nondegenerate convention $\sigma >0$ and use the full product observation laws. ChooseFor one observation, its Gaussian log likelihood ratio isUnder the first law, ${\mathbb{E}}_0(X-{\mu }_1{)}^2={\sigma }^2+({\mu }_1-{\mu }_0{)}^2$ and ${\mathbb{E}}_0(X-{\mu }_0{)}^2={\sigma }^2$. Thus the Kullback-Leibler divergence between normal distributions isFor the independent and identically distributed random variables, the joint log likelihood ratio is the sum of the individual ones, so the Kullback-Leibler divergence of the two full statistical samples is $n({\mu }_1-{\mu }_0{)}^2/(2{\sigma }^2)=1/4$. The comparison just proved implies $\mathrm{TV}(P_0,P_1)\le 1/2$. Applying the expectation form of the Le Cam two-point lemma givesThis is a Gaussian location minimax lower bound under absolute-error loss. The constant may depend on the fixed known $\sigma$, but neither on the estimator nor on $n$. If a signed nonzero scale were used, the same result holds with $|\sigma |$ in place of $\sigma$. If zero noise were allowed, $X_1=\mu$ would estimate $\mu$ exactly, so a positive lower bound would be false; positive variance is necessary.

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 $h<1/3$.

Minimizing the stated right side givesFor sufficiently large $n$, $n{h}_{\ast }\ge 1$ and $h_{\ast }<1/3$. For example take $n_0=1+\lceil \max \{1,2L/(\sqrt{3}\sigma ),81{\sigma }^2/(4L^2)\}\rceil$. Therefore the upper bound holds with

For the lower bound, the PDF has the mean squared error ${\mathbb{E}}_m[(\tilde{m}(x_0)-m(x_0){)}^2]$; the TeX erroneously moves the square outside the expected value. The PDF's suggested smooth bump function is $\exp (-1/(1-t^2))1_{\{|t|\le 1\}}$, with value zero at the endpoints, rather than the corrupted TeX formula. A triangular alternative suffices because only Lipschitz continuity is needed.

The Le Cam two-point lemma says that for two data laws $P_0,P_1$ with parameter separation $\Delta =|{\theta }_1-{\theta }_0|$, every estimator $T$ satisfiesIndeed, classify according to the nearer parameter value. Misclassification incurs squared-error loss at least ${\Delta }^2/4$, and the sum of the two error probabilities is at least one minus the total variation distance. Averaging the two risk functions proves the stated form.

Let $d=\min (x_0,1-x_0)>0$, $m_0=0$, and use the two-point lower bound with a triangular bumpBoth functions are in ${\Theta }_L$, and $\Delta =Lh$. Assume first that $h\le d$ and $nh\ge 1$. At most $2nh+1$ design points meet the bump's support. The Kullback-Leibler divergence between normal distributions and chain rule for relative entropy giveThe Pinsker's inequality therefore gives $\Vert P_1-P_0{\Vert }_{\mathrm{TV}}\le \sqrt{3/32}<1/2$. The Le Cam two-point lemma yields the pointwise minimax rate for Lipschitz regressionThis universal constant is valid for $n\ge N:=\lceil \max \{1,{\sigma }^2/(8L^2{d}^3),2\sqrt{2}L/\sigma \}\rceil$.

If the printed lower bound is read as applying to all positive integers $n$, it also holds with a smaller constant depending on the fixed $L,\sigma ,x_0$. For $n<N$, use the constant mean functions $m_0=0$ and $m_1=\sigma /\sqrt{n}$. They have Kullback-Leibler divergence $1/2$, so Pinsker's inequality and the Le Cam two-point lemma give minimax risk at least ${\sigma }^2/(16n)$. Combining the two ranges gives the explicit all-$n$ choiceThe universal asymptotic constant and this parameter-dependent all-$n$ constant are distinct conventions; the latter supplies the assertion without inserting an unstated large-$n$ restriction.

The total variation distance isIf $P\ll Q$, the Kullback-Leibler divergence isPinsker's inequality states

The squared-loss Le Cam two-point lemma says that for two experiments $P_0,P_1$ with scalar parameters ${\theta }_0,{\theta }_1$,Indeed, classify the data as $0$ when $\hat{\theta }$ is closer to ${\theta }_0$ and as $1$ otherwise. On a classification error, the estimation error is at least $|{\theta }_1-{\theta }_0|/2$. The sum of the two testing error probabilities is at least $1-\mathrm{TV}(P_0,P_1)$. Averaging the two risks and then bounding their maximum proves the result.

Let $r=\lfloor \beta \rfloor$. The Hölder class $\mathcal{H}(\beta ,L)$ on $[0,1]$ consists of functions with derivatives through order $r$ whose $r$th derivative is Hölder of order $\beta -r$ with constant $L$, with the standard integer-order convention.

Fix any $x_0\in [0,1]$. First compare the constant regression functions $m_0=0$ and $m_1=a$ with $a=c_0{n}^{-1/2}$. Both belong to every Hölder class under the seminorm convention, and the normal-product divergence isPinsker and Le Cam therefore give a lower bound $c/n$.

For the smoothness-dependent term, use the supplied smooth bump $\psi (u)=e^{-1/(1-u^2)}{1}_{\{|u|\le 1\}}$, translated one-sidedly near a boundary when necessary, and compareChoose its fixed normalization so that $m_1\in \mathcal{H}(\beta ,L)$ whenever $a\le c_{1}L{h}^{\beta }$. The Gaussian divergence satisfiesTakewith the bandwidth and amplitude truncated at constants when this expression leaves $[0,1]$. Then the divergence remains bounded andThe same construction can be placed at every $x_0$, with a one-sided bump at the endpoints. Pinsker and Le Cam, combined with the constant alternatives, provewhere $c>0$ depends only on $\beta$.

With densities $p,q$,By Cauchy-Schwarz inequality,Also $(\sqrt{p}-\sqrt{q}{)}^2\le |p-q|$, so $H^2\le 2\mathrm{TV}$.

The Hellinger affinity is $\rho (P,Q)=\int \sqrt{pq}d\mu =1-H^2/2$. Product densities and Fubini's theorem give $\rho (P^n,Q^n)=\rho (P,Q{)}^n$, hence

Le Cam two-point lemma states, for squared-error estimation at parameter points ${\theta }_0,{\theta }_1$, thatTake ${\theta }_0=0$, ${\theta }_1=\delta =1/(4n)$. The one-observation uniform densities overlap on length $1-\delta$, so $H^2(P_0,P_1)=2\delta$. ThereforeThe first distance inequality gives $\mathrm{TV}(P_0^n,P_1^n)\le 1/\sqrt{2}$. Le Cam's lemma now yieldsThis proves the claim with the displayed universal positive constant.

For fixed-design nonparametric regression, compare $m_0=0$ with $m_1(x)=L(h-|x-x_0|{)}_{+}$. This pair has parameter separation $Lh$ at $x_0$, Lipschitz bound $L$, and Kullback-Leibler divergence at most $L^2{h}^2(2nh+1)/(2{\sigma }^2)$. Choosing $h\asymp ({\sigma }^2/(n{L}^2){)}^{1/3}$ controls total variation distance by Pinsker's inequality and yields the pointwise minimax rate for Lipschitz regression through the Le Cam two-point lemma.