‘Fully’ Exchangeable Random Sequences

Recall that an infinite sequence $(X_i)_{i \in I}$ is exchangeable if its distribution is invariant under any permutation $\sigma : I \rightarrow I$ which is finite, i.e. $\# \{ i \colon \sigma(i) \neq i \} < \infty$.

Clearly i.i.d. sequences are exchangeable. My question is the following, what can be said about sequences that are invariant under any permutation $\sigma : I \rightarrow I$ (i.e. fully exchangeable)?

It may be nice to believe that a sequence has this property if and only if it is i.i.d. but I am not certain this is true, so I would be interested in knowing (what may be a very obvious and elementary) counterexample.

Happy thinking.

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With probability of $1/2$, all values equal $1$, with a probability of $1/2$, all values equal $0$. That is certainly invariant under all permutations and not i.i.d. – Michael Greinecker Apr 10 2012 at 12:08
Thank you; truly a trivial example! – Owen Daniel Apr 10 2012 at 13:12
I do not understand your question. Two infinite sequences $(X_n)$ and $(Y_n)$ have the same joint distribution iff for all $N$ the $N$-tuples $(X_n)_{n=1}^N$ and $(Y_n)_{n=1}^N$ have the same joint distribution. So fully exchangeable is the same as exchangeable. Google de Finetti's Theorem to find a characterization of exchangeable families of random variables. – Bill Johnson Apr 10 2012 at 13:14
Hi Bill. You are quite right! I'm forgetting the elementary stuff! – Owen Daniel Apr 10 2012 at 17:14