Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 210 / 2 / Solution

A kernel for density estimation is an integrable function $K:\mathbb{R}\to \mathbb{R}$ with $\int K=1$. For bandwidth $h>0$, the kernel density estimator isA kernel of order ell has

Let ${\mathcal{H}}_{n,\delta }$ and $\Gamma$ be as in the question, put $h_{\max }=\max {\mathcal{H}}_{n,\delta }$, and define the positive pilot boundFor every $h\in {\mathcal{H}}_{n,\delta }$, formThe Lepski bandwidth selection method takesSmall bandwidths have little bias but wide intervals; the rule increases the bandwidth until the estimates cease to be mutually compatible.

The grid is geometric with ratio two. Its lower endpoint has order $\log (\log n/\delta )/n$, while its upper endpoint is of orderFor fixed $\beta \le \ell$, the oracle bandwidtheventually lies between these endpoints. Its largest grid predecessor ${\tilde{h}}_{\mathrm{opt}}$ therefore exists, and the dyadic spacing gives

On ${\Omega }_0$, the true value $f(x)$ belongs to every interval with $h\le {\tilde{h}}_{\mathrm{opt}}$. Their intersection is therefore nonempty, soThe intervals at ${\hat{h}}_{\delta }$ and ${\tilde{h}}_{\mathrm{opt}}$ have a common point. Since ${\hat{\sigma }}_{n,h,\delta }$ decreases with $h$,Using ${\hat{f}}_{\infty }\le A$ and ${\tilde{h}}_{\mathrm{opt}}\ge h_{\mathrm{opt}}/2$ givesBecause $|{\mathcal{H}}_{n,\delta }|=O(\log n)$, enlarging the constant gives the required

For the final claim take ${\delta }_n=n^{-3}$ for sufficiently large $n$ and use $\hat{h}={\hat{h}}_{{\delta }_n}$. The squared error on ${\Omega }_0$ is at most a constant timesOff ${\Omega }_0$, boundedness of $K$ and the lower endpoint of the bandwidth grid give a deterministic polynomial bound on $|{\hat{f}}_{n,\hat{h},K}(x)|$; multiplying its square by $\mathbb{P}({\Omega }_0^c)\le n^{-3}$ is $O(n^{-1})$. Since $f(x)\le A$ and $N$ is independent of ${\delta }_n$, this term is no larger than the target rate after changing the constant. Enlarging it once more covers the finitely many $n<N$, proving the claimed adaptive mean squared error bound.