Suppose $X_1,...,X_n$ and $Y_1,...,Y_n$ are all independent copies of a standard Pareto random variable. For each of the 'two' random samples we can denote the order statistics $X_{n:n} \geq X_{n-1:n} \geq \ldots \geq X_{1:n}$, and, $Y_{n:n} \geq Y_{n-1:n} \geq \ldots \geq Y_{1:n}$.
Suppose we take the top 3 order statistics from the first sample $X_{n:n}, X_{n-1:n}, X_{n-2:n}$ and just $Y_{n-2:n}$ from the other sample. If we are given that $Y_{n-2:n}>X_{n-3:n}$, is there any way to characterize the joint distribution of the order statistics of $(X_{n:n}, X_{n-1:n}, X_{n-2:n}, Y_{n-2:n})=(\xi_1, \xi_2, \xi_3, \xi_4)$? It is probably not true, but somehow I am wondering if we have the equality in distribution $$(\xi_{4:4},\xi_{3:4}, \xi_{2,4}, \xi_{1:4})\overset{d}{=}(Z_{n+1:n+1}, Z_{n:n+1}, Z_{n-1:n+1}, Z_{n-2:n+1}) \qquad \mid Y_{n-2:n}>X_{n-3:n} $$ for the order statistics $Z_{n+1:n+1} \geq \ldots \geq Z_{1:n+1}$ of another random sample of the Pareto distribution.