Find joint density from transformation I'm trying to learn some tricks for how to find joint densities from transformations, but this problem looks really hard. Any help?
Suppose the joint density of continuous random variables $X,Y,Z$ is $f(x, y, z)$. 
Suppose moreover that: $$f(x; y; z) = f(x; z; y) = f(y; x; z) = f(y; z; x) = f(z; x; y) = f(z; y; x)$$
Let: $$U = max(X, Y, Z), V = min(X, Y, Z)$$
How can I find the joint density of $(U, V)$?
 A: As usual with maxima/minima, the event $A_{u,v}=[v\leqslant V,U\leqslant u]$ for $v\leqslant u$ is well suited to the computation of the distribution of $(U,V)$ since $A_{u,v}=[v\leqslant X,Y,Z\leqslant u]$, hence
$$
\mathbb P(A_{u,v})=\int\!\!\!\iint f(x,y,z)\,\mathbf 1_{v\leqslant x,y,z\leqslant u}\,\mathrm dx\mathrm dy\mathrm dz.
$$
Due to the symmetries of $f$, this is
$$
\mathbb P(A_{u,v})=6\int_v^u\int_v^x\int_v^yf(x,y,z)\,\mathrm dz\mathrm dy\mathrm dx,
$$
hence
$$
\partial_u\mathbb P(A_{u,v})=6\int_v^u\int_v^yf(u,y,z)\,\mathrm dz\mathrm dy=6\int_v^u\int_z^uf(u,y,z)\,\mathrm dy\mathrm dz,
$$
and
$$
\partial_v\partial_u\mathbb P(A_{u,v})=-6\int_v^uf(u,y,v)\,\mathrm dy.
$$
This yields the density $f_{U,V}$ of $(U,V)$ as, for example,
$$
f_{U,V}(u,v)=6\int_v^uf(u,v,w)\,\mathrm dw\cdot\mathbf 1_{u\leqslant v},
$$
which has the following infinitesimal interpretation: to get $U\approx u$ and $V\approx v$, order the sample ($6$ choices), put the lowest value of the sample at $u$, the highest value at $v$ and the middle value at $w$ somewhere in $(u,v)$.
