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An AR($p$) process is any causal and weakly stationary solution to the equations $$ X_t = \beta_1 X_{t-1} + \dotsc + \beta_p X_{t-p} + \epsilon_t, \qquad t \in \mathbb{Z} $$ where the polynomial $B(z) = 1 - \beta_1z - \dotsc - \beta_p z^p$ must not have roots on the unit disk. Here, I assume the noise $\epsilon_t$ to be iid with $E\epsilon_t = 0, \text{Var}(\epsilon_t) = \sigma^2$.

There is a unique solution to these equations which fulfills weak stationarity and causality. It can be written as a MA($\infty$): $$ X_t = \sum_{k = 0}^\infty \chi_k \epsilon_{t-k} $$ with appropiate coefficients $\chi_k$.

I believe that $X_t$ is a strictly stationary process due to the $\epsilon_t$ being iid, but in the literature I can't find a proof for this claim (all proofs that I saw are about weak stationarity).

How does the proof work?

EDIT: Note that I am not fixing a "starting distribution" for $X_0$. This is unnecessary because the process starts at time $t = -\infty$.

L D
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    Does this answer your question? [Must a time series be stationary if it has no unit root?](https://stats.stackexchange.com/questions/330714/must-a-time-series-be-stationary-if-it-has-no-unit-root) – Ben Aug 23 '20 at 18:53
  • Are you assuming the $\epsilon_t$ have Normal distributions? For if not, typically the distributions of the $X_t$ will all be different, regardless of whether they have common expectations or variances. – whuber Aug 23 '20 at 19:46
  • @whuber: No, the $\varepsilon_t$ do not need to have normal distribution. However, they are i.i.d. How come the $X_t$ will be differently distributed, even if they all are of the same form $\sum_{k=0}^\infty \chi_k \epsilon_{t-k}$ ? – L D Aug 23 '20 at 21:21
  • @Ben: I think that it does not answer my question. In the linked question, they fix a "starting" distribution for $X_0$. Here I assume that the process starts at time $t = -\infty$, so no starting distribution has to be established. – L D Aug 24 '20 at 10:01
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    To avoid confusion, I have changed the terminology of the other linked answer to refer to an "anchoring distribution" instead of a "starting distribution". The term was not intended to mean that it is restricted to a process starting at $X_0$. It is merely recognising that we start our analysis by specifying the distribution of an arbitrary element, and then the recursive equation for the AR model and the error distribution do the rest --- this works perfectly well for a process defined for all $t \in \mathbb{Z}$. – Ben Aug 24 '20 at 10:11
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    To see that your process is not necessarily stationary (not even necessarily weakly stationary), consider the case of an AR(1) process where $\sigma=0$ and $\mu_0 \neq 0$ and take $X_t = a^t \mu_0$. That gives a deterministic process that satisfies the recursive equation for the AR(1) but it is not mean stationary. (The reason being that the "anchoring distribution" is not the stationary distribution.) – Ben Aug 24 '20 at 10:14
  • @Ben: You do have a point there, I should have mentioned in my question that I only am interested in weakly stationary solutions to the equations. So let me reformulate: If I have a weakly stationary solution, it is of the form $X_t = \sum_{k=0}^\infty \chi_k \epsilon_{t-k}$, and it is unique -- right? If the noise is iid as mentioned above, can I conclude that the solution is stricly stationary? – L D Aug 24 '20 at 12:09
  • ... I have edited the question to include these details. – L D Aug 24 '20 at 12:17
  • ... I have reformulated the question in https://stats.stackexchange.com/questions/484534/is-the-ma-infty-process-with-i-i-d-noise-strictly-stationary in a much shorter manner. – L D Aug 25 '20 at 15:25

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