E.g. is a linear process exchangeable? What about strict vs weak stationarity?
EDIT: for clarity, I'm asking if/when the sequence of data points $(x_1, x_2, \ldots, x_n)$ in the time series is exchangeable. I'm mainly interested because I am not sure that I've fully understood the concepts of exchangeability and stationarity, and could not find any other questions that address this directly.
Exchangeability of random variables $X_1,X_2, \dotsc,X_t$ means that all permutations of the random variables do have the same (multivariate) distribution. That is, let $\pi \colon \{1,2,\dotsc,t\} \mapsto \{1,2,\dotsc,t\}$ be a permutation of the index set. Then $$ (X_{\pi 1}, X_{\pi 2}, \dotsc, X_{\pi t}) \stackrel{D}{=} (X_1, X_2, \dotsc, X_t) $$ for all permutations $\pi$. This clearly implies (assuming existence) that means and variances are constant, so implies stationarity, but is much stronger than stationarity.
For a time series this really rules out typical models of serial correlation, none of those will be exchangeable, since for example $X_1$ must have the same correlation with $X_2$ as with $X_t$. The only models consistent with exchangeability are equicorrelation models$^\dagger$: All pairwise correlations must have the same correlation $\rho$. That include independence models with $\rho=0$. But, in reality, with only one realization observed of the time series $X_1,X_2, \dotsc,X_t$, there is no observable difference between an exchangeable model and an independence model (there might be strange counterexamples with $\rho=1$).
See also Exchangeability and IID random variables and Why are words in a document for bag-of-words model exchangeable but not independent?.
$^\dagger$ Assuming correlations exist. Otherwise we can assume all pairwise joints are identical. A multivariate Cauchy distribution can be exchangeable, but without correlations.
