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For $A(v)=\int L^{\prime \prime }(t)L^{\prime \prime }(t+v)dt$, this U-statistic has expectation $\Vert L_b\ast f^{\prime \prime }{\Vert }_2^2$. It removes the diagonal term $\Vert L^{\prime \prime }{\Vert }_2^2/(n{b}^5)$ from the integral of a squared second derivative kernel density estimator, but leaves smoothing bias of an estimator.

The displayed sum is the first-order projection of a centred U-statistic onto sums of functions of individual observations. Its summands are independent and have mean zero. Subtracting it leaves a remainder with conditional expectation zero in each individual observation, a useful decomposition for variance and normal-limit calculations.

A one-sample U-statistic of fixed degree $m$ has a symmetric U-statistic kernel $h$ and averages it over all distinct subsets of an iid sample:It is unbiased for $\theta =\mathbb{E}h(X_1,\dots ,X_m)$. With finite second moment, define the first Hoeffding projection $h_1(x)=\mathbb{E}[h(x,X_2,\dots ,X_m)]-\theta$. Orthogonality of the higher projections givesIf ${\zeta }_1=\mathrm{Var}(h_1(X))>0$, the central limit theorem therefore proves the U-statistic central limit theoremEquivalently the influence function of a U-statistic is $m{h}_1(x)$, obtained by differentiating its population functional $\int hd{F}^m$ in each distribution argument. If ${\zeta }_1=0$, the root-$n$ limit degenerates and the second or higher projection determines a different, often nonnormal, limit.

For example, $h(x,y)=(x-y{)}^2/2$ gives the unbiased sample variance with denominator $n-1$. Its first projection is $[(x-\mu {)}^2-{\sigma }^2]/2$, so the influence function is $(x-\mu {)}^2-{\sigma }^2$. When the fourth moment is finite, the leading variance of the sample variance is $(\mathbb{E}(X-\mu {)}^4-{\sigma }^4)/n$. This example connects unbiased symmetric estimation to the functional influence calculation.

The empirical distribution function is ${\hat{F}}_n(x)=n^{-1}{\sum }_i{1}_{X_i\le x}$. For continuous F, the probability integral transform gives independent uniform variables $U_i=F(X_i)$. Strict increase of F is unnecessary. More explicitly, generate the observations by $X_i=F^{-1}(U_i)$ with the generalized quantile inverse. Then $1_{X_i\le x}=1_{U_i\le F(x)}$, so ${\hat{F}}_n(x)=G_n(F(x))$, where $G_n$ is the uniform empirical distribution.

Continuity and the limits of F ensure that its range contains $(0,1)$. The endpoint discrepancies for $G_n$ are zero. Hencewhich depends only on n. Equivalently it is ${\max }_{1\le i\le n}\{i/n-U_{(i)},U_{(i)}-(i-1)/n\}$, a function of uniform order statistics. This proves the distribution-free property of the Kolmogorov-Smirnov statistic even when F has flat portions.

Choose $c_{n,\alpha }$ from this common distribution so that $\mathbb{P}(D_n\le c_{n,\alpha })=1-\alpha$. The Kolmogorov-Smirnov confidence band isThe event that all these inequalities hold is exactly the event used to choose the critical value, giving simultaneous coverage $1-\alpha$. The Glivenko-Cantelli theorem states that for an independent identically distributed sample from any distribution function, continuous or not, ${\sup }_x|{\hat{F}}_n(x)-F(x)|\to 0$ almost surely.

For an integrable r-argument U-statistic kernel, replace it by $h_s(x_1,\dots ,x_r)=(r!{)}^{-1}{\sum }_{\pi }h(x_{\pi (1)},\dots ,x_{\pi (r)})$. Every permutation has the same expectation by exchangeability, so this symmetrization of a U-statistic kernel preserves the target functional. For a symmetric U-statistic kernel h and $n\ge r$, the U-statistic isEach summand has expectation theta, hence the statistic is unbiased.

For the population variance, use $h(x,y)=(x-y{)}^2/2$. Independence gives $\mathbb{E}(X_1-X_2{)}^2=2\mathrm{Var}(X_1)$ when the second moment is finite. The corresponding statistic simplifies using ${\sum }_{i<j}(X_i-X_j{)}^2=n{\sum }_i(X_i-\overline{X}{)}^2$:Thus variance as a U-statistic is precisely the usual unbiased sample variance, for $n\ge 2$.

Let $X_i=Y_i-{\theta }_0$. Under the symmetric continuous null hypothesis, positive and negative signs have probability $1/2$ and zeros have probability zero. For the sign test, take $N_{+}={\sum }_i{1}_{\{X_i>0\}}$ and reject for small values: $N_{+}\sim \mathrm{Bin}(n,1/2)$ under the null. Choose $k$ with ${\mathbb{P}}_0(N_{+}\le k)\le \alpha$; randomization at the next value attains exact size if desired. The lower tail is appropriate for $\theta <{\theta }_0$.

For the Wilcoxon signed-rank test, rank $|X_i|$ increasingly and put $W_{+}={\sum }_i{R}_i{1}_{\{X_i>0\}}$. Continuous observations have no absolute-value ties almost surely. Conditional on the absolute values, the signs are independent fair signs because symmetry makes the sign independent of magnitude. Hence $W_{+}$ has the distribution of ${\sum }_{r=1}^{n}r{B}_r$ for independent Bernoulli random variables $B_r$ with parameter $1/2$. Again reject in the lower tail, using this exact subset-sum distribution or a normal approximation. Independence gives the requested null moments:For a symmetric kernel $h$ of fixed order $m$, a one-sample U-statistic is $U_n={\left(\genfrac{}{}{0ex}{}{n}{m}\right)}^{-1}{\sum }_{i_1<\cdots <i_m}h(X_{i_1},\dots ,X_{i_m})$. State the U-statistic central limit theorem: if $\mathbb{E}{h}^2<\infty$, $\eta =\mathbb{E}h$, and ${\zeta }_1=\mathrm{Var}(\mathbb{E}[h\mid X_1])>0$, then $\sqrt{n}(U_n-\eta )\Rightarrow N(0,m^2{\zeta }_1)$. The nondegeneracy condition is important.

For $h(x,y)=1_{\{x+y>0\}}$, its order-two U-statistic satisfies the exact Wilcoxon signed-rank statistic identityTo prove it, in an unordered pair the sign of the sum is the sign of whichever observation has larger absolute value. Counting the pairs for which a positive observation is larger contributes its rank minus one; its own positive-sign indicator supplies the remaining one.

Write $F_0$ for the distribution function of $X$ under the null. Symmetry gives $\eta =1/2$ andThe probability integral transform makes $F_0(X_1)$ uniform on $(0,1)$, so ${\zeta }_1=1/12$. Therefore $\sqrt{n}(U_n-1/2)\Rightarrow N(0,1/3)$. The centred $N_{+}$ is $O_p(\sqrt{n})$ and is negligible on the scale $n^{3/2}$ of $W_{+}$. Combining the exact decomposition and its mean givesThe lower-tail approximate level-$\alpha$ rejection rule compares this standardized statistic with ${\Phi }^{-1}(\alpha )$; a half-unit continuity correction can refine discrete-tail calculations.

Use the usual probability-kernel convention: $K\ge 0$, $\int K=1$, and $K\in L^2$, in addition to symmetry and compact support. These conditions are needed for the stated finite bound; symmetry and compact support by themselves do not impose normalization or nonnegativity. Interpret the differentiability assumption in the usual Sobolev space sense $f\in H^2(\mathbb{R})$, or assume classical regularity sufficient for the integral Taylor remainder. The kernel density estimator isThe bias-variance decomposition of mean squared error, followed by Tonelli theorem, yieldsIndependence and a change of variables give the exact integrated variance of a kernel density estimatorAll changes of integration order are justified either by nonnegativity or by the displayed finite integrals.

For the bias of a kernel density estimator, symmetry gives $\int uK(u)du=0$. Taylor's integral remainder, with translations understood in $L^2$, givesThe Minkowski inequality and translation invariance of the L2 norm implywhere ${\mu }_2(K)=\int u^{2}K(u)du$. Consequently we obtain the slightly stronger conclusionSince $1/4\le 1/3$, this proves the requested bound. For signed kernels, the same argument has $\int u^2|K(u)|du$ in place of ${\mu }_2(K)$; the printed bound is not generally justified merely by signed-kernel symmetry. For example, let $U_a=(2a{)}^{-1}{1}_{[-a,a]}$ and $K=(4/3)U_1-(1/3)U_2$. This symmetric compactly supported signed kernel has integral one and ${\mu }_2(K)=0$, but its fourth moment is nonzero. For a normal density its convolution has nonzero bias at any fixed positive bandwidth. Sending $n\to \infty$ would therefore contradict a bound containing only the integrated variance term.

Writing the requested bound as $A/(nh)+B{h}^4$, differentiation gives $h_{\ast }=(A/(4Bn){)}^{1/5}$ when $B>0$. Thus $h\asymp n^{-1/5}$ gives mean integrated squared error of order $n^{-4/5}$. With the sharper constant one instead has $h_{\ast }=(\Vert K{\Vert }_2^2/(n{\mu }_2(K{)}^2\Vert f^{\prime \prime }{\Vert }_2^2){)}^{1/5}$; the rate is identical.

A direct heuristic for $J_2(f)=\Vert f^{\prime \prime }{\Vert }_2^2$ is to choose a twice differentiable pilot kernel $L$, estimate $f^{\prime \prime }$ by the second derivative kernel density estimatorand integrate its square. Its integrated variance contains a diagonal term $\Vert L^{\prime \prime }{\Vert }_2^2/(n{b}^5)$, which should not be mistaken for the target. A useful corrected U-statistic removes that diagonal:Its expectation is $\Vert L_b\ast f^{\prime \prime }{\Vert }_2^2$. Thus smoothing bias of an estimator, rather than the diagonal contribution, remains. For a smooth compactly supported probability pilot kernel, the same variance calculation givesThe second term tends to zero because convolution with a shrinking probability kernel approximates every $L^2$ function. Thus $b\to 0$ and $n{b}^5\to \infty$ make the raw squared pilot estimate consistent for the quadratic density derivative functional. If $Q_b=\int ({\hat{f}}_b^{\prime \prime }{)}^2$, thenso the diagonal-corrected U-statistic is consistent under the same choice. Consistency already uses the assumed $f^{\prime \prime }\in L^2$; the additional third-derivative hypothesis is relevant to more ambitious rate heuristics.

Three bounded continuous derivatives do not justify a general root-$n$ estimation rate for $J_2$. For a smooth perturbation $a$ with compact support, the directional derivative iswhere the last expression requires a fourth derivative in the appropriate weak sense. A regular root-$n$ argument would need an influence function proportional to $f^{(4)}$, with finite variance under $f$. Three derivatives do not supply that condition.

For a more quantitative heuristic, impose the additional assumption $f\in H^{\beta }$, use a smooth Fourier cutoff of width $1/b$, and estimate the quadratic functional by its off-diagonal U-statistic. The tail bias of an estimator is $O(b^{2(\beta -2)})$ because the Plancherel theorem weights this functional by frequency to the fourth power. The degenerate part of its standard deviation is $O(n^{-1}{b}^{-9/2})$: the squared $L^2$ norm of the quadratic kernel scales as $b^{-9}$. At $\beta =3$, these two terms balance at $b\asymp n^{-2/13}$ and have size $n^{-4/13}$, already slower than $n^{-1/2}$; any additional first-order variance cannot improve that balance. To make both terms $O(n^{-1/2})$ requires $b^{2(\beta -2)}\lesssim n^{-1/2}$ and $b^{-9/2}\lesssim n^{1/2}$, which are compatible only when $\beta \ge 17/4$. This is an illustrative smoothness calculation under extra Sobolev and moment assumptions, not a rate theorem asserted solely from bounded third derivatives. Particular much smoother probability density functions may of course permit root-$n$ estimation.

Exchangeability of an independent identically distributed sample makes every permuted kernel have the same expectation. Averaging over permutations therefore leaves the target functional unchanged and gives a symmetric kernel. The U-statistic then averages this kernel over the unordered r-element sample subsets.

A U-statistic averages a symmetric function of $m$ distinct sample observations over all unordered $m$-tuples. For independent and identically distributed observations it is an unbiased estimator of $\mathbb{E}H(X_1,\dots ,X_m)$. Omitting repeated indices can remove diagonal contributions in quadratic plug-in estimators.

The symmetric kernel $h(x,y)=(x-y{)}^2/2$ has expectation equal to the population variance for independent identically distributed observations with finite second moment. Its U-statistic reduces to the displayed unbiased sample variance, using ${\sum }_{i<j}(X_i-X_j{)}^2=n{\sum }_i(X_i-\overline{X}{)}^2$.