= Solution
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:
$$
U_n=\binom nm^{-1}\sum_{i_1<\cdots<i_m}h(X_{i_1},\ldots,X_{i_m}).
$$
It is unbiased for $\theta=\mathbb E h(X_1,\ldots,X_m)$. With finite second moment, define the first <Hoeffding projection> $h_1(x)=\mathbb E[h(x,X_2,\ldots,X_m)]-\theta$. Orthogonality of the higher projections gives
$$
U_n-\theta=\frac mn\sum_i h_1(X_i)+R_n,\qquad\mathbb E R_n^2=O(n^{-2}).
$$
If $\zeta_1=\operatorname{Var}(h_1(X))>0$, the <central limit theorem> therefore proves the <U-statistic central limit theorem>
$$
\boxed{\sqrt n(U_n-\theta)\Rightarrow N(0,m^2\zeta_1),\quad\operatorname{Var}(U_n)=\frac{m^2\zeta_1}{n}+O(n^{-2}).}
$$
Equivalently the <influence function of a U-statistic> is $m h_1(x)$, obtained by differentiating its population functional $\int h\,dF^m$ 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.
Back to article page
