<p>Consider an AR(q) process, <span class="math-container">$u_t$</span>. If the roots of a characteristic polynomial are outside of the unit circle, the AR(q) process is weakly stationary.</p>
<p>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. <span class="math-container">$Cov(u_t,u_{t-k})$</span> only depends on <span class="math-container">$k$</span>.</p>
<p>Is this condition on the roots also sufficient for strongly stationary? If not, what condition is needed?</p>

CC BY-SA 4.0
2021-07-15T17:24:58.820Z
No, I mean outside.
Michael Gmeiner
CC BY-SA 4.0
2021-07-15T17:28:18.647Z
Maybe take a look at this https://stats.stackexchange.com/questions/297991/when-is-armap-q-strictly-strongly-stationary-and-what-is-its-unconditional-d
manifold
CC BY-SA 4.0
2021-07-16T08:30:43.667Z
According to  [this](http://www.phdeconomics.sssup.it/documents/Lesson4.pdf) 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](https://dsp.stackexchange.com/questions/481/does-the-autocorrelation-function-completely-describe-a-stochastic-process/488#488).
tdm
CC BY-SA 4.0
2021-07-16T11:43:37.050Z
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?
Michael Gmeiner
CC BY-SA 4.0
2021-07-17T13:36:23.110Z
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 $x_t$.
mark leeds
<econometrics><time-series><autoregressive>
AR(q) Strongly Stationary
Michael Gmeiner
# AR(q) Strongly Stationary

(Asked by: username_0 on 2021-07-15)

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?

**Tags:** econometrics, time-series, autoregressive

> **Comments:**
> **username_0**: No, I mean outside.  
> **username_1**: Maybe take a look at this https://stats.stackexchange.com/questions/297991/when-is-armap-q-strictly-strongly-stationary-and-what-is-its-unconditional-d  
> **username_2**: According to  [this](http://www.phdeconomics.sssup.it/documents/Lesson4.pdf) 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](https://dsp.stackexchange.com/questions/481/does-the-autocorrelation-function-completely-describe-a-stochastic-process/488#488).  
> **username_0**: 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?  
> **username_3**: 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 $x_t$.  

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