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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 $h=f_0$.

The triweight kernel is a compactly supported symmetric probability density with variance $1/9$. Its value and first two derivatives vanish at the support endpoints. Its integrated squared second derivative is $35$, and it minimizes that curvature among twice continuously differentiable densities with zero mean and the same variance, by the triweight density minimizes integrated squared curvature inequality.