We say that $(x_1,x_2,\dots)$is an infinitely exchangeable sequence of random variables iff for any permutation $\pi$, $p(x_1,\dots,x_n)=p(x_{\pi(1)},\dots,x_{\pi(n)})$.
Let $(x_1,x_2,\dots)$ be an infinite sequence of iid r.v, and $x_0$ an independent r.v. from the sequence.
Then why is $(x_1+x_0,x_2+x_0,\dots)$ infinitely exchangeable but not iid?