A <kernel for density estimation> is an integrable function $K$ with $\int K=1$. With $K_h(u)=h^{-1}K(u/h)$, the <kernel density estimator> is
$$
\widehat f_{n,h,K}(x)=\frac1n\sum_{i=1}^nK_h(x-X_i).
$$
A <kernel of order ell> satisfies
$$
\int u^jK(u)\,du=0\quad(1\leq j<\ell),
\qquad
\int|u|^\ell|K(u)|\,du<\infty.
$$

Split the integrated $L^p$ risk into stochastic and bias terms:
$$
\mathbb E\lVert\widehat f-f\rVert_p^p
\leq2^{p-1}\mathbb E\lVert\widehat f-\mathbb E\widehat f\rVert_p^p
+2^{p-1}\lVert K_h*f-f\rVert_p^p.
$$
For fixed $x$, put $A_i(x)=K_h(x-X_i)$ and $W_i(x)=A_i(x)-\mathbb EA_i(x)$. The supplied <Rosenthal inequality> gives
$$
\mathbb E\left|\frac1n\sum_iW_i(x)\right|^p
\leq C_p\left\{
\frac{\mathbb E|W_1(x)|^p}{n^{p-1}}
+\frac{(\mathbb EW_1(x)^2)^{p/2}}{n^{p/2}}
\right\}.
$$
Now
$$
\int\mathbb E|W_1(x)|^pdx
\leq2^p\int\mathbb E|A_1(x)|^pdx
=2^ph^{1-p}R_p(K).
$$
Also $\mathbb EW_1(x)^2\leq(K_h^2*f)(x)$. The <Young convolution inequality> with exponent $p/2$ gives
$$
\int(K_h^2*f)(x)^{p/2}dx
\leq\left(\frac{R(K)}h\right)^{p/2}
\lVert f\rVert_{p/2}^{p/2}.
$$
After integration and multiplication by the outer factor $2^{p-1}$, the stochastic contribution is at most
$$
\frac{2^{2p-1}C_pR_p(K)}{(nh)^{p-1}}
+\frac{2^{p-1}C_pR(K)^{p/2}\lVert f\rVert_{p/2}^{p/2}}
{(nh)^{p/2}}.
$$

It remains to control the bias. Here $\ell=\lceil\beta\rceil=m+1$. The <Taylor formula with integral remainder>, the vanishing kernel moments, and <Minkowski integral inequality> give
$$
\lVert K_h*f-f\rVert_p
\leq\frac1{m!}\int|K(u)|
\lVert f^{(m)}(\,\cdot-hu)-f^{(m)}\rVert_p
|hu|^mdu.
$$
The defining Nikol'skii smoothness bound is $L|hu|^{\beta-m}$, so
$$
\lVert K_h*f-f\rVert_p
\leq\frac{L}{m!}h^\beta\mu_\beta(K).
$$
Raising to the $p$th power and applying the outer factor $2^{p-1}$ proves
$$
\mathbb EL(\widehat f_{n,h,K},f)
\leq
\frac{2^{2p-1}C_pR_p(K)}{(nh)^{p-1}}
+\frac{2^{p-1}C_pR(K)^{p/2}\lVert f\rVert_{p/2}^{p/2}}{(nh)^{p/2}}
+\frac{2^{p-1}L^p}{(m!)^p}\mu_\beta(K)^ph^{p\beta}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 210 / 2 / Solution

If $f^{(m-1)}$ is locally absolutely continuous, then its Taylor remainder can be written as an integral involving the weak derivative $f^{(m)}$. This form permits norm bounds when the highest derivative is only integrable.


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 Incoming links: Taylor formula with integral remainder