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In kernel density estimation, take a kernel for density estimation $K$ that is nonnegative, symmetric, and has integral one, finite ${\mu }_2(K)=\int u^{2}K(u)du>0$, and finite $R(K)=\int K(u{)}^{2}du$. Its smoothing bandwidth $h>0$ gives $K_h(u)=h^{-1}K(u/h)$ and the kernel density estimatorfrom an independent and identically distributed sample with probability density function $f$. Nonnegativity and unit mass make the estimate itself a probability density function. We compare kernel for density estimation shapes by their optimized mean integrated squared error, rather than by their performance at an arbitrarily common numerical smoothing bandwidth.

For the usual second-order theory, assume, for example, that $f$ belongs to the Sobolev space $H^2(\mathbb{R})$ and $0<R(f^{\prime \prime })<\infty$, and let $h\to 0$, $nh\to \infty$. The expectation is the convolution $K_h\ast f$. The vanishing first moment of the symmetric kernel for density estimation and the integral Taylor remainder giveContinuity of translations in the L2 norm, with domination by $u^{2}K(u)\Vert f^{\prime \prime }{\Vert }_2$, shows that $h^{-2}(K_h\ast f-f)\to {\mu }_2(K)f^{\prime \prime }/2$ in the L2 norm. Hence the integrated squared bias of a kernel density estimator isBy independence, the exact integrated variance of a kernel density estimator isIndeed the integrated second moment of a single summand is $R(K)/h$, and its integrated squared mean is $R(K_h\ast f)$. The latter quantity is bounded by $R(f)$, by Young's convolution inequality, so its contribution is $O(n^{-1})$. The bias-variance decomposition of mean squared error therefore yields the leading criterionThis asymptotic mean integrated squared error displays the bias-variance tradeoff: a small smoothing bandwidth increases integrated variance, while a large one increases squared bias of an estimator.

Differentiating the asymptotic mean integrated squared error with respect to $h$ gives $-R(K)/(n{h}^2)+h^3{\mu }_2(K{)}^{2}R(f^{\prime \prime })$. It changes sign once, from negative to positive, so its unique minimum isAt this minimum the integrated variance term is four times the squared bias of an estimator term. Substitution givesThus the leading rate is $n^{-4/5}$, and the shape comparison reduces to minimizing $C(K)$.

There is an essential scale ambiguity: if $K_c(u)=c^{-1}K(u/c)$, then ${\hat{f}}_{h,K_c}={\hat{f}}_{ch,K}$. Changing the scale of the kernel for density estimation merely changes the numerical convention for the smoothing bandwidth. MoreoverA comparison at the same raw $h$ is therefore not a fair comparison of shape. One can normalize the second moment to one, but that does not make the optimal smoothing bandwidth independent of shape.

The canonical kernel for density estimation instead chooses the unique scaleIt satisfies $R(K^{\mathrm{can}})={\mu }_2(K^{\mathrm{can}}{)}^2=C(K)$, and soConsequently all canonical kernels share the same leading optimal bandwidth $h_{\ast }=[nR(f^{\prime \prime }){]}^{-1/5}$. Only the multiplicative constant $C(K)$ depends on shape. This is precisely the separation of shape choice and scale choice that makes the canonical kernel normalization useful. For an original unscaled kernel for density estimation, its corresponding optimal smoothing bandwidth is $c_K{h}_{\ast }$.

To prove optimal shape, scale a candidate kernel for density estimation to have second moment one. Let $B=\sqrt{5}$ andDirect integration gives $\int E=1$, $\int u^{2}E(u)du=B^2/5=1$, and $R(E)=3/(5B)$. Thus $E$ is a rescaled Epanechnikov kernel. For any admissible $K$ with the same mass and second moment,Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, because $K\ge 0$. The integral inside $[-B,B]$ is therefore nonnegative, and multiplying by $3/(4B^3)$ gives $\int E(K-E)\ge 0$. Expanding the square now provesEquality forces $K=E$ almost everywhere. At fixed second moment, minimizing $R(K)$ is equivalent to minimizing $C(K)$. Since $C(K)$ is scale invariant, this proves the optimality of the Epanechnikov kernel over all nonnegative symmetric second-order kernels for density estimation, including the equality case up to rescaling.

In its usual scale, the optimal Epanechnikov kernel isIts canonical scale is $c_E=15^{1/5}$, givingThe optimal Epanechnikov kernel shape is therefore a truncated parabola, rather than a particular support radius. The canonical kernel and unit-second-moment version represent the same shape but different scale conventions.

The numerical gains over other reasonable kernels for density estimation are modest. Define asymptotic relative efficiency against the optimal Epanechnikov kernel by the sample-size fraction giving the same optimized leading mean integrated squared error:For the Gaussian density kernel $K_N(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$, $R(K_N)=1/(2\sqrt{\pi })$ and ${\mu }_2(K_N)=1$, so its efficiency is $6\sqrt{\pi }/(5\sqrt{5})\simeq 0.9512$. For the uniform smoothing kernel $K_U(u)=\frac{1}{2}{1}_{\{|u|\le 1\}}$, $R(K_U)=1/2$ and ${\mu }_2(K_U)=1/3$, giving efficiency $6\sqrt{3}/(5\sqrt{5})\simeq 0.9295$. The Gaussian density kernel is infinitely smooth, whereas the optimal Epanechnikov kernel has compact support and allows each observation to affect only a finite window. Such computational and smoothness considerations may outweigh a small leading-efficiency difference.

In practice $R(f^{\prime \prime })$ is unknown. A plug-in estimator estimates this roughness functional from a pilot estimate of the second derivative and substitutes it into the optimal smoothing bandwidth formula. Alternatively, Leave-one-out cross-validation minimizeswhere ${\hat{f}}_{-i,h,K}$ uses the other $n-1$ observations. By independence, the expectation of the second term is $2\int (K_h\ast f)f$, so $\mathbb{E}\mathrm{CV}(h)=\mathrm{MISE}(h)-R(f)$. Thus it targets integrated error without knowing $f$. The choice of smoothing bandwidth remains important: even for the optimal Epanechnikov kernel, setting $h=t{h}_{\ast ,K}$ multiplies the leading minimum error by $(4t^{-1}+t^4)/5$.

The preceding optimum concerns nonnegative second-order kernels for density estimation and twice-smooth densities on the real line. Signed kernels of order ell can cancel higher moments for smoother densities and yield different rates, but may produce negative density estimates. A support boundary or a density jump also requires separate bias of a kernel density estimator analysis. These settings do not contradict the proved second-order optimality of the Epanechnikov kernel.

Iterating the recurrence yieldsThus every finite collection of the $X_j$ is a linear transformation of independent standard normal random variables, and has a multivariate Gaussian distribution. The expected values are zero. Independence of the $Y_k$ givesFor $i<j$, the recurrence writes $X_j={\rho }^{j-i}{X}_i$ plus a sum of future innovations independent of $X_i$. Hence $\mathrm{Cov}(X_i,X_j)={\rho }^{j-i}$. The required vector law isThe covariance matrix is positive definite: the lower-triangular transformation from $(Y_1,\dots ,Y_n)$ has nonzero diagonal entries $1,\sqrt{1-{\rho }^2},\dots ,\sqrt{1-{\rho }^2}$. This also identifies a stationary autoregressive process of order one, with marginal density $f=\varphi$.

With the standard normal density as the kernel for density estimation, the two characteristic functions are $\psi (t)=e^{-t^2/2}$ and ${\psi }_K(ht)=e^{-h^2{t}^2/2}$. The Gaussian random variable $X_1-X_{j+1}$ has mean zero and variance $2(1-{\rho }^j)$, soThis expression is real. Substitution in the given integrated covariance expression, followed by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-a{t}^2}dt=\sqrt{\pi /a}$ for $a>0$, gives

To bound the total Gaussian autoregressive kernel variance correction, apply the mean value theorem to $u\mapsto u^{-1/2}$. For $j\ge 1$, the whole interval between $1+h^2-{\rho }^j$ and $1+h^2$ is bounded below by $1-\rho >0$. ThereforeSumming the geometric series gives the uniform boundIt follows from the supplied variance identity thatThe constant depends on the fixed $\rho$, but not on $h$ or $n$. In particular this applies to the requested shrinking-bandwidth regime; the covariance bound itself does not need $nh\to \infty$.

All observations have the same marginal density $f$, so dependence changes the variance but not the bias of a kernel density estimator: $\mathbb{E}{\hat{f}}_h=K_h\ast f$. Write $R(q)={\int }_{\mathbb{R}}{q}^2$ and ${\mu }_2(K)=\int u^{2}K(u)du$. For a smooth density with $f^{\prime \prime }\in L^2$, the integrated variance of a kernel density estimator in the independent case is $R(K)/(nh)+O(n^{-1})$. A second-order Taylor expansion of the convolution gives $K_h\ast f-f=(h^2{\mu }_2(K)/2)f^{\prime \prime }+o(h^2)$ in the L2 norm. Thus the bias-variance decomposition of mean squared error givesThe first two terms are the asymptotic mean integrated squared error. The additional dependence term is smaller than $n^{-4/5}$, so the leading optimal bandwidth and optimal rate are the same as for independent data:Here both $K$ and $f$ are the standard normal density. Since $f^{\prime \prime }(x)=(x^2-1)\varphi (x)$, direct Gaussian integrals giveConsequently the explicit leading bandwidth isAt this bandwidth the leading integrated variance is four times the leading integrated squared bias, which is also a check on the minimization constant.

For a normalized kernel for density estimation $K$, with $\int K=1$, the kernel density estimator isIt replaces each atom of the empirical measure by a translated, scaled copy of $K$: averaging these small probability-density bumps gives a smooth approximation to the sampling density. Integral one is part of the probability-kernel convention; without it, the estimator generally targets $f(x)\int K$.

The printed bandwidth assumption is insufficient for the claimed normal limit. The additional condition $n{h}_n\to \infty$ is needed at points with positive density. For a counterexample, take a normal distribution density $f$, $x=0$, the unit-width box kernel and $h_n=n^{-2}$. Then $n{h}_n^3\to 0$, but the probability that any observation lies in the window is at most $n\Vert f{\Vert }_{\infty }{h}_n\to 0$. On the complementary event the estimator is zero, and $\sqrt{n{h}_n}({\hat{f}}_{n,h_n}(0)-f(0))=-f(0)/\sqrt{n}\to 0$. The limit in probability is therefore zero, whereas the stated normal limit has positive variance.

We now prove the intended pointwise central limit theorem for a kernel density estimator under $h=h_n\to 0$, $nh\to \infty$ and $n{h}^3\to 0$. The first condition also follows from the last one. A mean value theorem bound, with $L=\Vert f^{\prime }{\Vert }_{\infty }$, gives the bias of a kernel density estimatorThus the bias multiplied by $\sqrt{nh}$ tends to zero.

The Lindeberg-Feller central limit theorem says that independent centered variables in each row of a triangular array have sum converging in distribution to $N(0,{\sigma }^2)$ if the sum of their variances tends to ${\sigma }^2>0$ and, for every $\eta >0$, the sum of their truncated second moments $\mathbb{E}[Z_{n,i}^2{1}_{\{|Z_{n,i}|>\eta \}}]$ tends to zero. Put $Y_{n,i}=K((x-X_i)/h)$ and $Z_{n,i}=(Y_{n,i}-\mathbb{E}{Y}_{n,i})/\sqrt{nh}$. The independence hypothesis holds within each row. A change of variable givesContinuity of $f$ at $x$ and the compact support and boundedness of $K$ justify the limit. Also $|Z_{n,i}|\le 2\Vert K{\Vert }_{\infty }/\sqrt{nh}\to 0$, so the Lindeberg condition is eventually identically zero for every $\eta$. The sum is exactly $\sqrt{nh}({\hat{f}}_{n,h}(x)-\mathbb{E}{\hat{f}}_{n,h}(x))$. Combining the centered limit with the negligible bias by the Slutsky theorem provesIf $f(x)=0$, the centered variance tends to zero, and the Chebyshev inequality proves the corresponding degenerate limit directly.

For $f(x)>0$, let $q_{\beta }$ be the supplied $\beta$-quantile of this limiting normal distribution. Inverting the limiting central probability gives the confidence intervalContinuity of the limiting distribution gives asymptotic coverage $1-\alpha$. Equivalently, its half-width is $z_{1-\alpha /2}\sqrt{f(x)\Vert K{\Vert }_2^2/(n{h}_n)}$ under the stipulated oracle quantiles. A degenerate limit at $f(x)=0$ does not by itself justify this nominal-coverage normal interval.

For a kernel for density estimation $K$ and smoothing bandwidth $h>0$, the kernel density estimator isWe use the convention that a kernel of order ell hasThe kernel for density estimation is integrable and can have negative values; requiring nonnegativity would prevent its second moment from vanishing. Some conventions additionally require a nonzero moment of order $\ell$. The construction below covers that convention too.

The Legendre polynomial kernel construction uses the polynomial reproducing kernel at zero:Every polynomial $p$ of degree at most $\ell -1$ has the expansion $p={\sum }_{m=0}^{\ell -1}⟨p,{\varphi }_m⟩{\varphi }_m$, since the displayed Legendre polynomials have those successive degrees and form an orthonormal basis of this polynomial space. ThereforeTaking $p=1,u,\dots ,u^{\ell -1}$ proves all the required moment identities. Boundedness and compact support give every absolute moment and a finite L2 norm. If a nonzero moment of order $\ell$ is required, replace this kernel byThe lower moments are unchanged by orthogonality. If $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩$, then $a_{\ell }\ne 0$ because otherwise $u^{\ell }$ would lie in the polynomial space of degree at most $\ell -1$. Evaluating its full Legendre expansion at zero gives $\int u^{\ell }{K}_{\ell }=-{\varphi }_{\ell }(0)a_{\ell }$, so $\int u^{\ell }{\tilde{K}}_{\ell }=a_{\ell }\ne 0$.

For instance, a simple kernel satisfying the moment conditions needed for three derivatives isIts negative wings cancel its central contribution to the second moment.

Here is the third-order pointwise kernel error bound. Write $B=\Vert f{\Vert }_{\infty }$ and $M_3=\Vert f^{(3)}{\Vert }_{\infty }$, and use any square-integrable kernel with unit integral, first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$. A change of variables givesBy the Taylor theorem with Lagrange remainder,The moment cancellations consequently give the uniform bias of a kernel density estimator boundFor the variance, independence and another change of variables giveThe triangle inequality and Cauchy-Schwarz inequality now yieldBalancing $h^3$ against $(nh{)}^{-1/2}$ givesThe constant is independent of both $n$ and $x$. The original PDF has exponent $3/7$; the converted TeX's $3/2$ is a transcription error.

To obtain the final parametric rate, we choose a flat-top kernel for density estimation. This kernel also satisfies all the preceding moment conditions, so a single kernel works for both constructions, with different bandwidth choices. With the Fourier transform convention $\hat{g}(t)=\int e^{-itx}g(x)dx$, choose a real even $C^{\infty }$ function $\eta$ equal to one on $[-1,1]$ and zero outside $[-2,2]$, and putSuch a cutoff is explicit: if $\rho (s)=e^{-1/s}$ for $s>0$ and zero otherwise, take $\eta (t)=\rho (2-|t|)/[\rho (2-|t|)+\rho (|t|-1)]$. It is constant near zero, and its endpoint transitions are smooth. Repeated integration by parts shows that $K$ and all its derivatives decay faster than every power, so $K$ lies in the Schwartz space. The Fourier inversion and moment differentiation of the Fourier transform giveThus this kernel can be used with $h_n=n^{-1/7}$ for the preceding general bound.

For the sharper example choose the probability density functionTo check this is a density and identify its Fourier transform, direct integration givesThe continuous value at zero is as displayed. This function is nonnegative and integrable, being bounded near zero and $O(x^{-2})$ at infinity. Fourier inversion gives $\hat{f}(t)=(1-|t|{)}_{+}$, and in particular $\int f=\hat{f}(0)=1$. Differentiation of this compact frequency integral also shows that $f$ and $f^{(3)}$ are bounded. Since $\eta =1$ on the support of $\hat{f}$, the convolution theorem givesBoth sides are continuous, so the equality holds at every $x$. There is therefore exactly zero bias of a kernel density estimator at bandwidth one, and the variance bound above givesThis is parametric-rate kernel estimation of a bandlimited density. A fixed bandwidth is appropriate because it reproduces this particular density exactly; a generic density would retain a nonzero smoothing bias.

For independent and identically distributed random variables sampled from a probability density function $f$, a kernel density estimator with kernel for density estimation $K$ and smoothing bandwidth $h>0$ isThe usual kernel for density estimation is nonnegative; finite $\int K^2$ makes the integrated variance finite. The mean integrated squared error and its bias-variance decomposition of mean squared error areThe interchange is justified by the Tonelli theorem.

For the exponential distribution and the specified unit-width uniform kernel, put $g_h(x)=\mathbb{E}{\hat{f}}_h(x)$. The expected value is $h^{-1}\mathbb{P}\{x-h/2\le X\le x+h/2\}$. Intersecting this interval with the positive half-line givesOn $[-h/4,0)$ the true probability density function vanishes, whereasHere $1-e^{-z}\ge z/2$ for $0\le z\le 1/4$. This leakage across the support boundary proves the integrated squared bias from a density jump:In particular, $h_0=1$ and $c=1/256$ suffice. The phenomenon differs from a smooth interior bias of a kernel density estimator: the order-one boundary error persists over a region of width proportional to $h$.

By independence, $\mathrm{Var}({\hat{f}}_h(x))=n^{-1}\{\mathbb{E}{K}_h(x-X{)}^2-g_h(x{)}^2\}$. Integrating the first term and using $\int K^2=1$ yieldsAlso $I(h)\le \int f^2=1/2$: the kernel density estimator mean is an average of translates of $f$, so the Jensen inequality followed by the Tonelli theorem bounds its squared integral.

To optimize over every $h>0$, we must establish that useful smoothing bandwidths approach zero, rather than optimize only a formal small-$h$ expression. An exact calculation supplies that justification. The autocorrelation integral of the exponential distribution density isAveraging over the two uniform windows, or integrating the piecewise expression for $g_h$, givesThese formulas show that $B$ is continuous on $(0,\infty )$, is strictly positive there because $g_h>0$ on $(-h/2,0)$, and tends to $1/2$ as $h\to \infty$. Consequently, $B$ is bounded away from zero on $[\delta ,\infty )$ for every $\delta >0$. Its expansion agrees with the supplied $B(h)=h/12+o(h)$.

The choice $h_n=\sqrt{12/n}$ givesChoose smoothing bandwidths within $1/n$ of the infimum. Their mean integrated squared error tends to zero, and nonnegativity of the integrated variance gives $B(h_n)\to 0$, hence $h_n\to 0$. For any $\epsilon \in (0,1/12)$, eventuallyThe last step is the arithmetic-geometric mean inequality. Taking the lower limit and then $\epsilon \downarrow 0$ matches the upper bound. ThusThe original PDF has this constant; the TeX aid's $1/\sqrt{2}$ is a transcription error.

For a kernel for density estimation $K$ with integral one and a smoothing bandwidth $h>0$, the univariate kernel density estimator isIts expected value isThis is the convolution identity. It holds wherever the integral is defined, in particular everywhere for the bounded kernel used below. For a nonnegative kernel for density estimation, Tonelli theorem also gives the identity as an extended nonnegative integral. A signed kernel for density estimation instead requires absolute integrability for this interchange.

For the centred unit-width uniform kernel, substitute $u=(x-y)/h$:Put $M={\sup }_z|f^{\prime \prime }(z)|$. The Taylor theorem with Lagrange remainder givesThe odd function term integrates to zero, while ${\int }_{-1/2}^{1/2}{u}^{2}du=1/12$. Hence the bias of a kernel density estimator obeysThis is a second-order pointwise bias bound for kernel density estimation. Existence and boundedness of the second derivative suffice for the stated remainder; continuity of that derivative is not an extra assumption.

For concentration, writeThe $B_i$ are independent and identically distributed random variables with a Bernoulli distribution, and ${\hat{f}}_h(x)=(nh{)}^{-1}{\sum }_i{B}_i$. The required concentration bound follows from the Hoeffding lemma. We prove its Bernoulli case directly. Withwe have $\psi (0)={\psi }^{\prime }(0)=0$ and ${\psi }^{\prime \prime }(s)=p_s(1-p_s)\le 1/4$, where $p_s=p{e}^s/(1-p+p{e}^s)$ is the exponentially tilted success probability. Therefore $\psi (s)\le s^2/8$ for every real $s$. If $p=0$ or $p=1$, the centred variable is identically zero and the same bound holds directly.

Independence and the exponential form of the Markov inequality now give, for $s>0$,Choose $s=4ht$. Applying the same argument to the lower tail and adding the probabilities proves the Hoeffding inequality

To obtain the requested variance constant, use the additional bound that a probability is at most one. Set $Z={\hat{f}}_h(x)-\mathbb{E}{\hat{f}}_h(x)$ and $a=2n{h}^2$. The tail integral formula for moments, justified by Tonelli theorem, yieldsThusThe clipping at one is essential for this integration argument to give that constant. It is an instance of a second moment bound from a clipped Gaussian tail. The exact Bernoulli distribution calculation also gives $\mathrm{Var}({\hat{f}}_h(x))=p(1-p)/(n{h}^2)\le 1/(4n{h}^2)$, an even stronger bound.

A kernel for density estimation is an integrable function $K$ with $\int K=1$. The kernel density estimator isIt has order $\ell$ when $\int u^{j}K(u)du=0$ for $1\le j<\ell$ and $\int |u{|}^{\ell }|K(u)|du<\infty$.

Differentiation gives the derivative kernel density estimatorDropping the negative square of its mean from the variance and using Tonelli and the substitution $u=(x-X_1)/h$,Thus $\alpha =3$ and $C_1(K)=\Vert K^{\prime }{\Vert }_2^2$.