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Subtract the diagonal variance contribution from the derivative estimate's squared norm, then insert the result into asymptotic mean integrated squared error. The expected curvature estimate equals $R(f^{\prime \prime })+O(h^2)$, apart from a negligible finite-n term. This makes the resulting criterion asymptotically correct even though it is not an exactly unbiased estimate of integrated risk. Under standard conditions its selector is ratio-consistent with the optimal bandwidth and has relative fluctuation order $n^{-1/10}$.

For a nonnegative symmetric unit-integral kernel for density estimation with $0<{\mu }_2(K)=\int u^{2}K(u)du<\infty$ and $R(K)=\int K^2<\infty$, the canonical scaling satisfies $R(K_c)={\mu }_2(K_c{)}^2$. Since $R(K_c)=R(K)/c$ and ${\mu }_2(K_c)=c^2{\mu }_2(K)$, the unique scale is $c=[R(K)/{\mu }_2(K{)}^2{]}^{1/5}$. Consequently its asymptotic mean integrated squared error is $C(K)[(nh{)}^{-1}+h^{4}R(f^{\prime \prime })/4]$, where $C(K)=R(K{)}^{4/5}{\mu }_2(K{)}^{2/5}$ is invariant under kernel rescaling. The leading optimal smoothing bandwidth becomes $[nR(f^{\prime \prime }){]}^{-1/5}$, independent of the kernel shape. Unit second moment is a different normalization and does not have this separation property.

The Epanechnikov kernel for density estimation is a nonnegative symmetric truncated parabola with integral one, ${\mu }_2(K_E)=1/5$, and $R(K_E)=3/5$. It has the smallest optimized asymptotic mean integrated squared error among nonnegative second-order density kernels, up to rescaling. Its canonical kernel has scale $15^{1/5}$ and support $[-15^{1/5},15^{1/5}]$.

The integrated squared second derivative measures a smooth density's curvature in L2. It supplies the leading integrated squared bias constant of a second-order kernel density estimator. For nonzero second moment ${\mu }_2(K)$, the asymptotic mean integrated squared error has bias term $h^4{\mu }_2(K{)}^{2}R(f^{\prime \prime })/4$, so greater curvature calls for a smaller smoothing bandwidth.

Fix the second moment $v>0$. Put $B=\sqrt{5v}$ and $E(u)=3(B^2-u^2{)}_{+}/(4B^3)$. For every nonnegative unit-integral kernel for density estimation $K$ with the same second moment and finite $R(K)$, the mass and second-moment constraints give $\int (B^2-u^2)(K-E)=0$. Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, so $\int E(K-E)\ge 0$. Expanding the square gives $R(K)-R(E)=\int (K-E{)}^2+2\int E(K-E)\ge 0$, with equality only when $K=E$ almost everywhere. Since $R(K{)}^2{\mu }_2(K)$ is scale invariant, rescaled Epanechnikov kernels minimize the optimized asymptotic mean integrated squared error over all such second moments. This conclusion concerns nonnegative second-order kernels; signed higher-order kernels obey a different bias expansion.

A kernel for density estimation is an integrable function K of unit integral, ordinarily nonnegative and centered. Square integrability is needed for integrated variance; a symmetric second-order kernel for density estimation also has $\int uK(u)du=0$ and finite nonzero second moment ${\mu }_2(K)$. Its scaled form is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator isHigher-order signed kernels are possible, but need not give a nonnegative probability density function estimate.

The mean integrated squared error is $\mathbb{E}\int ({\hat{f}}_h(x)-f(x){)}^{2}dx$. Put $g_h=K_h\ast f$. Independence gives the pointwise mean $g_h$ and variance $n^{-1}[(K_h^2\ast f)-g_h^2]$. Integrating, use $\int f=1$ and Fubini to obtain $\int (K_h^2\ast f)=R(K_h)$. The bias-variance decomposition proves the exact formulaEquivalently this is $R(f)-2\int f(K_h\ast f)+(1-n^{-1})R(K_h\ast f)+R(K)/(nh)$. These identities assume the displayed integrals are finite; they are exact, rather than a small-smoothing bandwidth expansion.

For a symmetric second-order kernel for density estimation, Taylor expansion gives leading bias $h^2{\mu }_2(K)f^{\prime \prime }/2$, and the leading integrated variance is $R(K)/(nh)$. Thus the asymptotic mean integrated squared error balances inverse smoothing bandwidth against fourth-power bias, with optimal smoothing bandwidth $[R(K)/(n{\mu }_2(K{)}^{2}R(f^{\prime \prime })){]}^{1/5}$. Its appearance of the integrated squared density curvature motivates the scale comparison.

For $f_a(x)=af(ax)$, $a>0$, two differentiations give $f_a^{\prime \prime }(x)=a^3{f}^{\prime \prime }(ax)$. A change of variables therefore gives $R(f_a^{\prime \prime })=a^{5}R(f^{\prime \prime })$, which tends to zero as $a\downarrow 0$ if the original curvature integral is finite. The distribution with probability density function $f_a$ is that of X divided by a, so $\sigma (f_a)=\sigma (f)/a$. ConsequentlyThis scale-invariant density curvature measures shape without dependence on the units used to measure X.

For the final minimization, write $c=35/32$ and use the triweight kernel $f_0=c(1-x^2{)}^3$ on $(-1,1)$, zero outside. Integration gives total mass one, mean zero and variance $1/9$. Its value and its first two derivatives vanish at both endpoints, so its extension is twice continuously differentiable. On the interior,In particular $f_0^{\prime \prime \prime }(1)=-48c$ and $f_0^{\prime \prime \prime }(-1)=48c$.

Put $e=h-f_0$. The mass and moment constraints give $\int e=\int x^{2}e=0$, since both means vanish and their variances match. If $R(h^{\prime \prime })=\infty$ the claim is immediate; otherwise expand the squared norm and integrate its cross term twice by parts over $[-1,1]$. The first endpoint term vanishes because $f_0^{\prime \prime }(\pm 1)=0$, but the next one must be retained:For the second equality use the zero mass and second moment of e and $e=h\ge 0$ outside the support. All terms in the last expression are nonnegative, including the endpoint values of a probability density function. It follows thatThis proves that the triweight density minimizes integrated squared curvature under the specified constraints. Equality forces $e^{\prime \prime }=0$; a globally affine integrable difference of probability density functions is zero, so equality occurs only at $h=f_0$.

Write $R(v)={\int }_{\mathbb{R}}v(u{)}^{2}du$. The kernel density estimator with bandwidth $h>0$ isSufficient conditions are that $K$ be a nonnegative symmetric probability density function with $R(K)<\infty$ and finite nonzero second moment ${\mu }_2(K)=\int u^{2}K(u)du$, and that $h\to 0$, $nh\to \infty$. Symmetry and the second-moment assumption give $\int uK(u)du=0$. The expectation is a convolution:Taylor's theorem with integral remainder givesThe zeroth-order term integrates to $f(x)$, and the first-order term vanishes. The second derivative is bounded and continuous, so dominated convergence, with dominating function proportional to $u^{2}K(u)$, yieldsCombining this with the supplied leading variance gives the asymptotic mean squared errorWhen $f(x)>0$ and $f^{\prime \prime }(x)\ne 0$, its derivative is $-R(K)f(x)/(n{h}^2)+{\mu }_2(K{)}^2{f}^{\prime \prime }(x{)}^2{h}^3$. It changes sign exactly once, giving the pointwise optimal kernel bandwidthPositivity of $f(x)$ is needed for this positive interior optimum. The printed condition $f^{\prime \prime }(x)\ne 0$ alone does not ensure it. For instance, $f(x)=x^2\varphi (x)$ is a normalized bounded probability density function with bounded continuous square-integrable second derivative, but $f(0)=0$ and $f^{\prime \prime }(0)=2\varphi (0)>0$. At this point the displayed leading variance vanishes and the two-term AMSE has no positive interior minimizer; finer variance terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.

For integrated error, the same expansion holds in $L^2$: translation continuity of $f^{\prime \prime }\in L^2$ applied to the integral remainder givesIndeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^{2}K(u){\int }_0^1(1-s)\Vert f^{\prime \prime }(\cdot -shu)-f^{\prime \prime }{\Vert }_{2}dsdu$; each translated difference tends to zero and is bounded by $2\Vert f^{\prime \prime }{\Vert }_2$. This proves the claimed integrated bias expansion without assuming a uniform pointwise remainder over the whole real line.

If $K_h(x)=h^{-1}K(x/h)$, direct integration of the variance givesThe last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and convolution with a probability density function does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the asymptotic mean integrated squared error and its minimizer areHere $R(f^{\prime \prime })>0$: otherwise the continuous second derivative vanishes everywhere, making $f$ affine, impossible for a probability density function on the whole real line.

Now take the standard normal density $f=\varphi$. Differentiation gives ${\varphi }^{\prime \prime }(x)=(x^2-1)\varphi (x)$. Gaussian integration givesThe kernel constants cancel from the bandwidth ratio. At $x\ne \pm 1$,Put $t=x^2$. The logarithmic derivative of $e^{t/2}/(t-1{)}^2$ is $1/2-2/(t-1)$. On $0\le t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the global minimizers are $x=\pm \sqrt{5}$, and the normal-density local-to-global bandwidth ratio isThe ratio diverges at the two inflection points $\pm 1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.

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.