Sufficient and Completeness of the bivariate order statistics Let $(X_1,Y_1),...,(X_n,Y_n)$ be i.i.d. random vectors from a population $P \in \mathcal{P}$ that is the family of all bivariate populations with Lebesgue p.d.f.'s. Show that the set of $n$ pairs $(X_i,Y_i)$ ordered according to the value of their first coordinate constitutes a sufficient and complete statistic for $P \in \mathcal{P}$. Hence, find the UMVUEs of $P(X_i \le Y_i)$ and $P(X_i \le X_j \& Y_i \le Y_j)$.
My approach :
I just learnt about a similar result in the 1D case. That is we can show that $T(\mathbf{X})=(X_{(1)},...,X_{(n)})$ is sufficient and complete for the family of distributions having Lebesgue pdf's. The proof was constructed based on a one-one correspondence between $T(\mathbf{X})$ and $\mathbf{U}=(U_1,...,U_n)$ where $U_j = \sum_{i=1}^{n} X_i ^j$.
My feeling is that we should be able to solve the 2D case by following similar lines but I am not sure!
Finally, for finding the UMVUEs we can exploit the fact that any statistic invariant under the permutation of the $n$ pairs is a function of the order statistics.
Can anyone help me out?
 A: I'll address the the estimation part of the problem.
UMVUE of $P(X_i\le Y_i)$ is fairly easy to guess since this is equivalent to the UMVUE of $P(Z_i\le 0)$ where $Z_i=X_i-Y_i$. Note that the $Z_i$'s are i.i.d univariate with an absolutely continuous distribution.
The symmetric kernel $\phi(Z_1)=I(Z_1\le 0)$ is unbiased for $P(Z_1\le 0)$ and so the UMVUE is given by the corresponding one-sample U-statistic
$$U=\frac1n\sum_{i=1}^n \phi(Z_i)=\frac1n\sum_{i=1}^n I(X_i\le Y_i)$$
Of course, $U=\frac1n\sum\limits_{i=1}^n \phi(Z_{(i)})$ is based on the complete sufficient statistic $(Z_{(i)})_{1\le i\le n}$.

On the other hand, an unbiased estimator of $P(X_1\le X_2,Y_1\le Y_2)$ is
$$\phi^0(\underline{Z_1},\underline{Z_2})=I(X_1\le X_2, Y_1\le Y_2)\,,$$
where $\underline {Z_i}=\begin{pmatrix} X_i \\ Y_i\end{pmatrix}$ for every $1\le i\le n$.
This $\phi^0$ is an asymmetric kernel of degree $2$. The corresponding symmetric kernel is
\begin{align}
\phi(\underline{Z_1},\underline{Z_2})&=\frac12\left(\phi^0(\underline{Z_1},\underline{Z_2})+\phi^0(\underline{Z_2},\underline{Z_1})\right)
\\&=\frac12\left(I(X_1\le X_2, Y_1\le Y_2)+I(X_2\le X_1, Y_2\le Y_1)\right)
\end{align}
Then UMVUE of $P(X_i\le X_j,Y_i\le Y_j)$ for $i\ne j$ is likely to be the U-statistic
$$U=\frac1{\binom{n}{2}}\sum_{1\le i<j\le n}\phi(\underline{Z_i},\underline{Z_j})$$
