# Complete sufficient statistics for bivariate observations

We have observations $$(X_i,Y_i), 1\le i\le n$$ from a family of distributions $$\mathcal F$$, consisting of all absolutely continuous bivariate distributions. I wish to show that $$(X_{(i)},Y_{\text{corresponding to }X_{(i)}}), 1\le i\le n$$ form a set of complete sufficient statistics for the given family.

My attempt This is intuitively obvious. Afterall, it's just permuting the data, by preserving association. Sufficiency is easy to prove, since the joint density of $$\{(X_i,Y_i)\mid 1\le i\le n\}$$ is just $$\frac1{n!}$$ times the joint density of the bivariate order statistics defined above, so the density is a function of this statistic, and then Neyman-Fisher factorization theorem does the job. But for completeness neither I want to do something like we do for univariate case (constructing exponential family and then using the result for subfamily) nor I know how to do it in this case.

I would appreciate a bivariate analogue of that proof, and would like it better if somehow we can use the completeness of order statistics for univariate case to prove it in this case, without going into the full construction here again.