3
$\begingroup$

I am currently reading Jun Shao's Mathematical Statistics, and in his discussion of U statistics, he proves that

$Var(U_n) = $ $n\choose m$$^{-1} \sum_{k=1}^m $$m \choose k$$n - m \choose m-k$$\zeta_k$

where $\zeta_k$ is the variance of the conditional expectation of the kernel conditioning on $X_1,\ldots,X_k$, n is the sample size, and m is the number of arguments to the kernel function. He then states the following three facts as a corollary:

(i) $\frac{m^2}{n}\zeta_1 \leq Var(U_n) \leq \frac{m}{n}\zeta_m$

(ii) $(n+1)Var(U_{n+1}) \leq nVar(U_n)$

(iii) For any fixed m and $k = 1,\ldots,m$, if $\zeta_j = 0$ for $j < k$ and $\zeta_k > 0$, then

$Var(U_n) = \dfrac{k! {m\choose k}^2\zeta_k}{n^k} + O\left(\frac{1}{n^{k+1}}\right)$

How can we go from the statement of Hoeffding's Theorem to these? I am in particular having trouble working through simplifying the choose operations.

$\endgroup$
1
$\begingroup$

Using variance decomposition ($Var(X) = Var(E(X|Y)) + E(Var(X|Y))$), you can show that $\zeta_1 \le ... \le \zeta_m$.

Note that $n \choose m$$^{-1}$ $m \choose k$ $n-m \choose m-k$ = $k!$$m \choose k$$^2\frac{(n-m)(n-m-1)...(n-2m+k+1)}{n(n-1)...(n-m+1)}$, $k=1...,m-1$. The denominator has $m$ terms and the numerator has $m-k$ terms. If $k=1$, the ratio is $1/n+O(1/n^2)$; if $k=2$, the ratio is $1/n^2+O(1/n^3)$ and so forth. This suffices to prove (iii).

Part (ii) is hard and I think induction might be a good point to start.

The second inequality of Part(i) uses that $nVar(U_n)$ is a decreasing sequence (Part(ii)). The first inequality is by the fact that $nVar(U_n)$ converges to $m^2 \zeta_1$ (Part(iii)) and $nVar(U_n)$ is a decreasing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.