= Solution
For the <unit-width box kernel>, the <convolution> and <kernel density estimator> are
$$
g_h(x)=(K_h*f)(x)=\int_{\mathbb R}K_h(x-y)f(y)\,dy=\frac1h\int_{x-h/2}^{x+h/2}f(y)\,dy,
$$
and
$$
\widehat f_{n,h}^K(x)=\frac1n\sum_{i=1}^nK_h(x-X_i)=\frac1{nh}\sum_{i=1}^n\mathbf1_{\{|x-X_i|\le h/2\}}.
$$
In particular $\mathbb E\widehat f_{n,h}^K=g_h$. The scaling gives $\|K_h\|_2^2=h^{-1}\|K\|_2^2=h^{-1}$. Even if $f$ is not square-integrable, <Young's convolution inequality> gives $g_h\in L^2$ because $f\in L^1$ and $K_h\in L^2$.
Using independence to eliminate the cross terms in the centred estimator, and <Tonelli theorem> to integrate the nonnegative variance, gives the exact <integrated variance of a kernel density estimator>:
$$
\mathbb E\|\widehat f_{n,h}^K-g_h\|_2^2=\frac1n\left(\int_{\mathbb R}\mathbb E K_h(x-X)^2\,dx-\|g_h\|_2^2\right)=\frac1n\left(\frac1h-\|g_h\|_2^2\right)\le\frac1{nh}.
$$
The <Cauchy-Schwarz inequality> now yields
$$
\boxed{\mathbb E\|\widehat f_{n,h}^K-K_h*f\|_2\le\frac1{\sqrt{nh}},\qquad\kappa=1.}
$$
This constant comes from the unit $L^2$ norm of the unscaled <unit-width box kernel>.
For the piecewise constant <probability density function>, integrability forces the constants on both unbounded outer intervals to be zero. Thus $f$ is bounded, has compact support, and has finitely many jumps. Let $\Delta_r=f(x_r+)-f(x_r-)$ denote the jump at $x_r$. For $h$ smaller than the least gap between consecutive breakpoints, the smoothing regions $[x_r-h/2,x_r+h/2]$ do not overlap. Outside these regions the <convolution> equals $f$.
Within a region, put $v=x-x_r$. For $-h/2<v<0$, the bias is $\Delta_r(1/2+v/h)$; for $0<v<h/2$, it is $-\Delta_r(1/2-v/h)$. Endpoint values do not affect an $L^2$ norm. Integrating these two triangular errors gives
$$
\|K_h*f-f\|_2^2=\sum_{r=0}^k2\Delta_r^2\int_0^{h/2}(1/2-v/h)^2\,dv=\frac h{12}\sum_{r=0}^k\Delta_r^2.
$$
This is the <box-kernel bias of a piecewise constant density>. By the triangle inequality, with $C_f=(\sum_r\Delta_r^2/12)^{1/2}$,
$$
\mathbb E\|\widehat f_{n,h}^K-f\|_2\le\frac1{\sqrt{nh}}+C_f\sqrt h.
$$
Balancing this <bias-variance tradeoff> with $h_n=n^{-1/2}$ gives, for all sufficiently large $n$,
$$
\boxed{\mathbb E\|\widehat f_{n,h_n}^K-f\|_2\le(1+C_f)n^{-1/4}=O(n^{-1/4}).}
$$
Back to article page
