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For $0\le t\le \theta$, independence and the uniform distribution give $P(X_{(n)}\le t)=(t/\theta {)}^n$. Consequently, for fixed $x\ge 0$ and $n>x$,For $x<0$ the survival probability is one. Thus the scaled endpoint gap converges in distribution to a unit-rate exponential law; it is not exactly exponential at finite $n$.

The plug-in estimator principle replaces the unknown sampling distribution by its empirical distribution and evaluates its population functional at that fitted distribution. Here the fitted support endpoint is $\hat{\theta }=X_{(n)}$, also the maximum-likelihood estimate: the likelihood is ${\theta }^{-n}{1}_{\{\theta \ge X_{(n)}\}}$ and decreases over its allowed support. For an empirical bootstrap sample $Y_1,\dots ,Y_n$, the corresponding gap isAlmost surely the original maximum is unique. The bootstrap gap equals zero precisely when this observation is selected at least once, soThe true limiting exponential distribution has no atom at zero. This nonvanishing discrepancy proves bootstrap failure for a uniform endpoint: the bootstrap distribution cannot converge to the correct limit, even though the endpoint estimate itself is consistent. It fails to recreate the unsampled population tail beyond the observed maximum. A parametric bootstrap drawing from $U[0,X_{(n)}]$, rather than from the discrete empirical sample, reproduces the scaled-gap law here.

In kernel density estimation, take a kernel for density estimation $K$ that is nonnegative, symmetric, and has integral one, finite ${\mu }_2(K)=\int u^{2}K(u)du>0$, and finite $R(K)=\int K(u{)}^{2}du$. Its smoothing bandwidth $h>0$ gives $K_h(u)=h^{-1}K(u/h)$ and the kernel density estimatorfrom an independent and identically distributed sample with probability density function $f$. Nonnegativity and unit mass make the estimate itself a probability density function. We compare kernel for density estimation shapes by their optimized mean integrated squared error, rather than by their performance at an arbitrarily common numerical smoothing bandwidth.

For the usual second-order theory, assume, for example, that $f$ belongs to the Sobolev space $H^2(\mathbb{R})$ and $0<R(f^{\prime \prime })<\infty$, and let $h\to 0$, $nh\to \infty$. The expectation is the convolution $K_h\ast f$. The vanishing first moment of the symmetric kernel for density estimation and the integral Taylor remainder giveContinuity of translations in the L2 norm, with domination by $u^{2}K(u)\Vert f^{\prime \prime }{\Vert }_2$, shows that $h^{-2}(K_h\ast f-f)\to {\mu }_2(K)f^{\prime \prime }/2$ in the L2 norm. Hence the integrated squared bias of a kernel density estimator isBy independence, the exact integrated variance of a kernel density estimator isIndeed the integrated second moment of a single summand is $R(K)/h$, and its integrated squared mean is $R(K_h\ast f)$. The latter quantity is bounded by $R(f)$, by Young's convolution inequality, so its contribution is $O(n^{-1})$. The bias-variance decomposition of mean squared error therefore yields the leading criterionThis asymptotic mean integrated squared error displays the bias-variance tradeoff: a small smoothing bandwidth increases integrated variance, while a large one increases squared bias of an estimator.

Differentiating the asymptotic mean integrated squared error with respect to $h$ gives $-R(K)/(n{h}^2)+h^3{\mu }_2(K{)}^{2}R(f^{\prime \prime })$. It changes sign once, from negative to positive, so its unique minimum isAt this minimum the integrated variance term is four times the squared bias of an estimator term. Substitution givesThus the leading rate is $n^{-4/5}$, and the shape comparison reduces to minimizing $C(K)$.

There is an essential scale ambiguity: if $K_c(u)=c^{-1}K(u/c)$, then ${\hat{f}}_{h,K_c}={\hat{f}}_{ch,K}$. Changing the scale of the kernel for density estimation merely changes the numerical convention for the smoothing bandwidth. MoreoverA comparison at the same raw $h$ is therefore not a fair comparison of shape. One can normalize the second moment to one, but that does not make the optimal smoothing bandwidth independent of shape.

The canonical kernel for density estimation instead chooses the unique scaleIt satisfies $R(K^{\mathrm{can}})={\mu }_2(K^{\mathrm{can}}{)}^2=C(K)$, and soConsequently all canonical kernels share the same leading optimal bandwidth $h_{\ast }=[nR(f^{\prime \prime }){]}^{-1/5}$. Only the multiplicative constant $C(K)$ depends on shape. This is precisely the separation of shape choice and scale choice that makes the canonical kernel normalization useful. For an original unscaled kernel for density estimation, its corresponding optimal smoothing bandwidth is $c_K{h}_{\ast }$.

To prove optimal shape, scale a candidate kernel for density estimation to have second moment one. Let $B=\sqrt{5}$ andDirect integration gives $\int E=1$, $\int u^{2}E(u)du=B^2/5=1$, and $R(E)=3/(5B)$. Thus $E$ is a rescaled Epanechnikov kernel. For any admissible $K$ with the same mass and second moment,Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, because $K\ge 0$. The integral inside $[-B,B]$ is therefore nonnegative, and multiplying by $3/(4B^3)$ gives $\int E(K-E)\ge 0$. Expanding the square now provesEquality forces $K=E$ almost everywhere. At fixed second moment, minimizing $R(K)$ is equivalent to minimizing $C(K)$. Since $C(K)$ is scale invariant, this proves the optimality of the Epanechnikov kernel over all nonnegative symmetric second-order kernels for density estimation, including the equality case up to rescaling.

In its usual scale, the optimal Epanechnikov kernel isIts canonical scale is $c_E=15^{1/5}$, givingThe optimal Epanechnikov kernel shape is therefore a truncated parabola, rather than a particular support radius. The canonical kernel and unit-second-moment version represent the same shape but different scale conventions.

The numerical gains over other reasonable kernels for density estimation are modest. Define asymptotic relative efficiency against the optimal Epanechnikov kernel by the sample-size fraction giving the same optimized leading mean integrated squared error:For the Gaussian density kernel $K_N(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$, $R(K_N)=1/(2\sqrt{\pi })$ and ${\mu }_2(K_N)=1$, so its efficiency is $6\sqrt{\pi }/(5\sqrt{5})\simeq 0.9512$. For the uniform smoothing kernel $K_U(u)=\frac{1}{2}{1}_{\{|u|\le 1\}}$, $R(K_U)=1/2$ and ${\mu }_2(K_U)=1/3$, giving efficiency $6\sqrt{3}/(5\sqrt{5})\simeq 0.9295$. The Gaussian density kernel is infinitely smooth, whereas the optimal Epanechnikov kernel has compact support and allows each observation to affect only a finite window. Such computational and smoothness considerations may outweigh a small leading-efficiency difference.

In practice $R(f^{\prime \prime })$ is unknown. A plug-in estimator estimates this roughness functional from a pilot estimate of the second derivative and substitutes it into the optimal smoothing bandwidth formula. Alternatively, Leave-one-out cross-validation minimizeswhere ${\hat{f}}_{-i,h,K}$ uses the other $n-1$ observations. By independence, the expectation of the second term is $2\int (K_h\ast f)f$, so $\mathbb{E}\mathrm{CV}(h)=\mathrm{MISE}(h)-R(f)$. Thus it targets integrated error without knowing $f$. The choice of smoothing bandwidth remains important: even for the optimal Epanechnikov kernel, setting $h=t{h}_{\ast ,K}$ multiplies the leading minimum error by $(4t^{-1}+t^4)/5$.

The preceding optimum concerns nonnegative second-order kernels for density estimation and twice-smooth densities on the real line. Signed kernels of order ell can cancel higher moments for smoother densities and yield different rates, but may produce negative density estimates. A support boundary or a density jump also requires separate bias of a kernel density estimator analysis. These settings do not contradict the proved second-order optimality of the Epanechnikov kernel.

Condition on the independent external sample, and abbreviate ${\delta }_{\pi }={\hat{\pi }}_m-\pi$ and ${\delta }_{\mu }={\hat{\mu }}_m-\mu$. Expanding the plug-in estimator around the true nuisance functions giveswhere $\sqrt{n}{R}_{n,m}=o_p(1)$ because the two nuisance mean squared errors vanish, the evaluation sample is independent of the nuisance fits, $A$ is bounded, and $Y$ has bounded support. The linear term is the empirical average of the influence function, so the central limit theorem givesThe remaining bias obeys the Cauchy-Schwarz inequalityThus the claimed conclusion follows under the standard product-rate conditionequivalently $\sqrt{n}\sqrt{\mathrm{MSE}({\hat{\pi }}_m)operatornameMSE({\hat{\mu }}_m)}\to 0$. Slutsky theorem then yields

As printed, the paper instead assumes only $\sqrt{n}\mathrm{MSE}({\hat{\pi }}_m)\mathrm{MSE}({\hat{\mu }}_m)\to 0$, which is insufficient with its stated definition of MSE. For example, deterministic nuisance errors of size $n^{-3/16}$ have both MSEs equal to $n^{-3/8}$ and satisfy the printed condition, while the scaled product bias is $\sqrt{n}{n}^{-3/8}=n^{1/8}$. The result therefore requires the stronger condition above, or “MSE” in the printed rate must be read as root mean squared error.

Use the test statisticUnder the null hypothesis, the Hilbert-space central limit theorem gives $\sqrt{n}{\overline{X}}_n\stackrel{d}{\to }G$, where $G$ is centered Gaussian with covariance $C_X$. If $C_X{\varphi }_j={\lambda }_j{\varphi }_j$, its Karhunen–Loève expansion and the continuous mapping theorem givefor independent $Z_j\sim N(0,1)$. Reject for $T_n$ above the $(1-\alpha )$ quantile of this weighted chi-squared law; replacing the ${\lambda }_j$ by empirical covariance eigenvalues gives a plug-in estimator of the critical value.

Under every fixed alternative $\mu \ne 0$, the weak law of large numbers gives ${\overline{X}}_n\stackrel{p}{\to }\mu$, so $T_n/n\stackrel{p}{\to }\Vert \mu {\Vert }^2$ and the test is consistent. Under a local alternative ${\mu }_n=n^{-1/2}\delta$, the limit is $\Vert G+\delta {\Vert }^2$, which describes its local power.