# Equivalence of the completeness of the order statistics and the uniqueness of symmetric unbiased estimators

I am reading A.J. Lee's 1990 book "U-statistics: Theory and Practice". There is an equation on page 6 that I cannot explain why it holds, and I hope somebody could help me. Here is the detail.

Let $$X_1,...,X_k$$ be iid random variables with distribution F. A statistic $$T(X_1,...,X_k)$$ is complete with respect to a family of distributions $$\mathcal{F}$$ if $$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k) = 0$$
for all $$F\in\mathcal{F}$$ implies $$h=0$$. Then it is said that in case $$T(x_1,...,x_k)=(x_{(1)},...,x_{(k)})$$ is the order statistics then $$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k) = \int h^{[n]}(x_1,...,x_k)dF(x_1)\cdots dF(x_k).$$ I don't understand why this equality is true. As a side note, it appears there is a typo (even after the typo is corrected, I still don't see why the equality holds): $$h^{[n]}(x_1,...,x_k)$$ should be $$h^{[k]}(x_1,...,x_k)$$, which is defined as $$h^{[k]}(x_1,...,x_k) = \frac{1}{k!}\sum h(x_{i_1},...,x_{i_k}),$$ where the sum is over all permutations $$(x_{i_1},...,x_{i_k})$$ of $$\{1,2,...,k\}$$. Using this equality, it is concluded that the completeness of the order statistics relative to $$\mathcal{F}$$ is exactly equivalent to the uniqueness of symmetric unbiased estimators for all $$F\in\mathcal{F}$$.

• @Xi'an: No, even if the typo is corrected, I still don't know why the equality holds. – legon Apr 17 at 15:42

## 1 Answer

The equality$$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k) = \int h^{[n]}(x_1,...,x_k)dF(x_1)\cdots dF(x_k)$$does not hold for any function $$h$$. For instance, if$$h(x_1,...,x_k)=x_1$$ $$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k)=\mathbb E^F[X_{(1)}]$$ while$$\int h^{[n]}(x_1,...,x_k)dF(x_1)\cdots dF(x_k)=\mathbb E^F[X_1]$$ Hence, I presume there must be an additional constraint on the function $$h$$ to be found in the proof.

• Thank you for pointing this out - I also suspected that it was not true for all function $h$. However, I couldn't find any condition on $h$ in the book. – legon Apr 18 at 17:52