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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?

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Exchangeability is a generalization of IID, so IID implies Exchangeability but not the other way around. As a trivial example take $(X_1,X_2)$ where $X_1$ is bernoulli (-1,1) and $X_2=-X_1$, which is exchangeable but not iid.

Assuming each $x_i$ is independent in your example and non-constant, then it's not IID because $\mbox{Cov}(x_1+x_0,x_2+x_0)=\mbox{Var}(x_0,x_0)\neq 0$ (recall IID would imply 0 covariance).

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    $\begingroup$ Perhaps you should change the first sentence to "All iid sequences are exchangeable, but the reverse is not true." $\endgroup$ – jaradniemi Apr 13 '16 at 18:50
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    $\begingroup$ Note that the example given here, while it is exchangeable, is not infinitely exchangeable! $\endgroup$ – kjetil b halvorsen Apr 13 '16 at 19:45

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