2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Can you add a bit more information: are you asking about the exchangeability of the sequence of data points in one time-series? Or something else? (It might help to also briefly explain why you are asking this) $\endgroup$ – Juho Kokkala Jul 2 '18 at 13:28
  • 1
    $\begingroup$ Here is a similar question with answer: stats.stackexchange.com/questions/353722/… $\endgroup$ – kjetil b halvorsen Jul 2 '18 at 17:09
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.