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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 $(U, V)$?

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up vote 2 down vote accepted

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 $(u,v)$.

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