= Solution
For a normalized <kernel for density estimation> $K$, with $\int K=1$, the <kernel density estimator> is
$$
\boxed{\widehat f_{n,h}(x)=\frac1{nh}\sum_{i=1}^nK\!\left(\frac{x-X_i}{h}\right).}
$$
It replaces each atom of the <empirical measure> by a translated, scaled copy of $K$: averaging these small probability-density bumps gives a smooth approximation to the sampling density. Integral one is part of the probability-kernel convention; without it, the estimator generally targets $f(x)\int K$.
\b[The printed bandwidth assumption is insufficient for the claimed normal limit.] The additional condition $nh_n\to\infty$ is needed at points with positive density. For a counterexample, take a <normal distribution> density $f$, $x=0$, the <unit-width box kernel> and $h_n=n^{-2}$. Then $nh_n^3\to0$, but the probability that any observation lies in the window is at most $n\|f\|_\infty h_n\to0$. On the complementary event the estimator is zero, and $\sqrt{nh_n}(\widehat f_{n,h_n}(0)-f(0))=-f(0)/\sqrt n\to0$. The limit in probability is therefore zero, whereas the stated normal limit has positive variance.
We now prove the intended <pointwise central limit theorem for a kernel density estimator> under $h=h_n\to0$, $nh\to\infty$ and $nh^3\to0$. The first condition also follows from the last one. A <mean value theorem> bound, with $L=\|f'\|_\infty$, gives the <bias of a kernel density estimator>
$$
|\mathbb E\widehat f_{n,h}(x)-f(x)|
=\left|\int K(u)(f(x-hu)-f(x))\,du\right|
\le Lh\int|u|K(u)\,du.
$$
Thus the bias multiplied by $\sqrt{nh}$ tends to zero.
The <Lindeberg-Feller central limit theorem> says that independent centered variables in each row of a triangular array have sum converging in distribution to $N(0,\sigma^2)$ if the sum of their variances tends to $\sigma^2>0$ and, for every $\eta>0$, the sum of their truncated second moments $\mathbb E[Z_{n,i}^2\mathbf1_{\{|Z_{n,i}|>\eta\}}]$ tends to zero. Put $Y_{n,i}=K((x-X_i)/h)$ and $Z_{n,i}=(Y_{n,i}-\mathbb EY_{n,i})/\sqrt{nh}$. The <independence> hypothesis holds within each row. A change of variable gives
$$
\sum_{i=1}^n\operatorname{Var}(Z_{n,i})
=\frac{\operatorname{Var}(Y_{n,1})}{h}
=\int K(u)^2f(x-hu)\,du-h\left(\int K(u)f(x-hu)\,du\right)^2
\longrightarrow f(x)\|K\|_2^2.
$$
Continuity of $f$ at $x$ and the compact support and boundedness of $K$ justify the limit. Also $|Z_{n,i}|\le2\|K\|_\infty/\sqrt{nh}\to0$, so the <Lindeberg condition> is eventually identically zero for every $\eta$. The sum is exactly $\sqrt{nh}(\widehat f_{n,h}(x)-\mathbb E\widehat f_{n,h}(x))$. Combining the centered limit with the negligible bias by the <Slutsky theorem> proves
$$
\boxed{\sqrt{nh_n}(\widehat f_{n,h_n}(x)-f(x))\xrightarrow{d}N(0,f(x)\|K\|_2^2).}
$$
If $f(x)=0$, the centered variance tends to zero, and the <Chebyshev inequality> proves the corresponding degenerate limit directly.
For $f(x)>0$, let $q_\beta$ be the supplied $\beta$-quantile of this limiting <normal distribution>. Inverting the limiting central probability gives the <confidence interval>
$$
\boxed{\left[\widehat f_{n,h_n}(x)-\frac{q_{1-\alpha/2}}{\sqrt{nh_n}},\quad
\widehat f_{n,h_n}(x)-\frac{q_{\alpha/2}}{\sqrt{nh_n}}\right].}
$$
Continuity of the limiting distribution gives asymptotic coverage $1-\alpha$. Equivalently, its half-width is $z_{1-\alpha/2}\sqrt{f(x)\|K\|_2^2/(nh_n)}$ under the stipulated oracle quantiles. A degenerate limit at $f(x)=0$ does not by itself justify this nominal-coverage normal interval.
Back to article page
