TITLE: If the probability density on a random vector is symmetric, then each variable is identically distributed?
QUESTION [5 upvotes]: Let $X$ be a random vector with joint distribution $F$ and density $f$. If $f$ is symmetric, is this equivalent to each random variable being identically distributed? We say $f$ is symmetric if it is invariant to a permutation in its arguments. 
For $X$ a vector of length $n$ of discrete random variables (say integer valued just for the sake of it) this gives $\Pr(X_i=x)=\sum_{Y\in\mathbb{Z}^{n-1}} f(X_i=x,X_{-i}=Y)$. But with $f$ symmetric, whichever $i$ we choose should be arbitrary, so $\Pr(X_i=x)=\Pr(X_j=x)$ for any choice of $x,i,j$.
Is this correct? Does this change at all for continuous random variables? I would just replace the sum with an integral?

REPLY [3 votes]: Yes you are right and this is the notion of exchangeability, see http://en.wikipedia.org/wiki/Exchangeable_random_variables.