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.
$(X_n)_{n=1}^N$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