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Consider an AR(q) process, $u_t$. If the roots of a characteristic polynomial are outside of the unit circle, the AR(q) process is weakly stationary.

I've seen this proof that proceeds by showing the mean and variance are constant, and covariance terms only depend on the number of time periods in between, i.e. $Cov(u_t,u_{t-k})$ only depends on $k$.

Is this condition on the roots also sufficient for strongly stationary? If not, what condition is needed?

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    $\begingroup$ No, I mean outside. $\endgroup$ Jul 15 '21 at 17:24
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    $\begingroup$ Maybe take a look at this stats.stackexchange.com/questions/297991/… $\endgroup$
    – manifold
    Jul 15 '21 at 17:28
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    $\begingroup$ According to this a sufficient condition is for weak stationarity to imply strong stationarity is that the variables are Gaussian as in this case the mean and covariance are sufficient to determine the joint distribution. See also here. $\endgroup$
    – tdm
    Jul 16 '21 at 8:30
  • $\begingroup$ Thank you both for these comments. I see that if the process is Guassian, then it is strictly stationary. That's fine, but I'm curious if there are other straightforward conditions under which weakly stationary results in strong stationary. Are there none? $\endgroup$ Jul 16 '21 at 11:43
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    $\begingroup$ Miichael: Aside from the gaussian case, I don't think so because strong says something about the whole distribution ( of any subset of the $x_t}$ ) while weak only says something about the mean and the variance of the indvidual $x_t$. $\endgroup$
    – mark leeds
    Jul 17 '21 at 13:36

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