Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 34 / 3 / Solution

For fixed-design nonparametric regression, let $x_i=i/n$ and $K_h(u)=h^{-1}K(u/h)$. The degree-$p$ local polynomial estimator takes the fitted intercept ${\hat{m}}_h(x;p)={\hat{\beta }}_0$, whereUse a nonnegative regression kernel and an invertible local polynomial Gram matrix to obtain a unique fit. For $p=0$, solving the single weighted least squares normal equation gives the local constant estimator:At the interior point $x_0$, use the standard unit-integral regression kernel convention ${\mu }_0(K)=1$. Otherwise the leading variance below is $R(K)/(nh{\mu }_0(K{)}^2)$; unlike a kernel density estimator, the local constant estimator itself is unchanged by rescaling $K$.

The local constant estimator is a linear estimator in nonparametric regression with weights $W_i=K_h(x_i-x_0)/(n{s}_{0,h})$. The independent unit-variance errors implyTo control the bias of an estimator, differentiability at this one point is enough. WriteSymmetry makes ${\mu }_1(K)=0$, and the supplied moment approximation gives $s_{1,h}=o(h)$. Therefore the linear contribution to the bias of an estimator is $m^{\prime }(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta (h)h$, because only $|x_i-x_0|\le h$ receive positive weight. This is the interior first-order bias cancellation for local constant regression, and gives squared bias of an estimator $o(h^2)$. Combining with the variance provesFor fixed $\epsilon >0$, take $h_{\epsilon }={\epsilon }^{-1/3}{n}^{-1/3}$. It satisfies both asymptotic smoothing bandwidth conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\epsilon$. ThusLetting $\epsilon \downarrow 0$ proves the optimized scaled pointwise error tends to zero. The order of these limits matters: no remainder uniform in $\epsilon$ has been assumed.

For the lower bound over the whole function class, the absence of a uniform bound on derivatives permits a particularly direct argument. It also avoids assuming a normal distribution for errors when the model only supplies their first two moments. Fix $n$ and $x_0$. Choose a smooth bump function $b$ on the interval, with $b(x_0)=1$, whose support contains no design point except possibly $x_0$. Such a bump exists because the design is finite; at an endpoint use the restriction of a smooth bump on the real line. Every $m_a=ab$, $a\in \mathbb{R}$, belongs to the differentiability class.

If $x_0$ is not a design point, all observation means are zero for every $a$. The Le Cam two-point lemma applied to $a=\pm A$ has total variation distance zero and gives a lower bound tending to infinity with $A$. In fact, the supremum of the mean squared error is infinite at every such $x_0$.

If $x_0$ is a design point, only its response depends on $a$. The resulting location family is the single observation $Y=a+ϵ$; all remaining observations are independent of $a$ and give no extra information. The single-observation location minimax bound isHere is a proof covering discrete as well as continuous errors. Give $a$ the prior distribution $U[-A,A]$, independently of the noise. Put $p_M=\mathbb{P}(|ϵ|\le M)$ and $v_M=\mathrm{Var}(ϵ\mid |ϵ|\le M)$. Consider an easier experiment in which an oracle also reveals whether $|ϵ|\le M$, and reveals $a$ itself on the complementary event. On the event $|ϵ|\le M$ and $|Y|\le A-M$, every possible truncated noise value places $a=Y-ϵ$ within $[-A,A]$. The flat prior distribution then makes the Bayesian posterior noise law exactly its truncated original law. The smallest conditional squared-error loss is its conditional variance $v_M$, attained by the posterior mean.

The probability of that event is at least $p_M(1-2M/A)$ for $A>2M$, by restricting further to $|a|\le A-2M$. Thus every rule's supremum risk function is at least the original Bayes risk, which is at least the easier experiment's Bayes risk, and therefore at least $p_M(1-2M/A)v_M$. Let $A\to \infty$ and then $M\to \infty$. Finite second moment and zero mean give $p_M{v}_M\to 1$. The rule $T(Y)=Y$ has constant mean squared error one, proving the equality.

Consequently, at every $x_0\in [0,1]$, the unrestricted differentiability class satisfies the stronger conclusionThus $c=1$ works for every $n\ge 1$. There is no conflict with the preceding fixed-function limit: the worst functions, including arbitrarily narrow bumps, can vary with $n$. This is the pointwise versus uniform risk distinction. A smoothness ball with a common derivative bound would pose a different minimax risk problem.