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Suppose I have an AR(1) process $X_t=aX_{t-1}+e_t$, where $e_t$ is a white noise with zero mean and finite variance. Under what conditions do I have $\{X_t\}$ being strictly stationary in the sense that the joint distribution of $[X_{t_1},...,X_{t_k}]$ and $[X_{t_1+a},...,X_{t_k+a}]$ are the same for any set of integers $t_1,...,t_k$ and any integer $a$?

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    $\begingroup$ A implicit assumption seems to be that $e_t$ is independent of the r.vs $X_s$ for $s \leq t-1$? $\endgroup$
    – Yves
    Commented Apr 24, 2023 at 7:13
  • $\begingroup$ @Yves Thanks! Do you mean the independence between $e_t$ and past $X_s$ suffices for strict stationarity? or do you mean this independence assumption is usually maintained for an AR(1) model, regardless of its stationarity? $\endgroup$ Commented Apr 24, 2023 at 7:56
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    $\begingroup$ Without independence I believe that you do not specify the distribution of the process $X_t$. $\endgroup$
    – Yves
    Commented Apr 24, 2023 at 9:02
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    $\begingroup$ I think independence assumption is usually maintained for an AR(1) model, regardless of its stationarity. $\endgroup$ Commented Apr 24, 2023 at 9:12
  • $\begingroup$ Should it not be a sufficient condition that the variance is stationary, $|a|<1$? $\endgroup$ Commented Apr 24, 2023 at 9:29

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Assume that the definition of the AR(1) process includes a specification of $X_0$ (absent which $X_1 = aX_0+ \varepsilon_1$ is difficult to interpret).

The simplest choice is $X_0 = 0$ which makes \begin{array} {}X_1 &= \varepsilon_1\\ X_2 &= \varepsilon_2 + a\varepsilon_1\\ X_3 &= \varepsilon_3 + a\varepsilon_2 + a^2\varepsilon_1\\ \vdots &= \ddots \end{array} For the process to be strictly stationary, it is necessary that $X_1$ and $X_2$ be identically distributed. But, since $\varepsilon_t$ is a white noise process, we have that \begin{array} {}\operatorname{var}(X_1) &= \sigma^2\\ {}\operatorname{var}(X_2) &= \operatorname{var}(\varepsilon_2) +a^2 \operatorname{var}(\varepsilon_1) + 2a\operatorname{cov}(\varepsilon_2,\varepsilon_1)\\ &= \sigma^2 + a^2 \sigma^2 + 0\\ &= \sigma^2(1 + a^2)\\ &\neq \sigma^2 = \operatorname{var}(X_1) \end{array} and so the process cannot possibly be a strictly stationary process.

If $|a| < 1,$ then the variance of $X_n$ converges to $\dfrac{\sigma^2}{1-a^2}$. So, what happens if we set $X_0$ to have variance $\dfrac{\sigma^2}{1-a^2}$ and to be independent of the white noise process? Well, then \begin{align} \operatorname{var}(X_1) &= \dfrac{a^2\sigma^2}{1-a^2}+ \sigma^2\\ &= \dfrac{\sigma^2}{1-a^2}\\ &= \operatorname{var}(X_0). \end{align} Inductively, we get that all the $X_i$ have the same variance $\dfrac{\sigma^2}{1-a^2}$. However, equality of variances is not the same as equality of distributions, and there is no guarantee that the process is stationary even to order $1$, let alone be strictly stationary.. I will leave it as an exercise for the OP to determine whether this process is a weakly stationary process (also called wide-sense-stationary process, or, in the time-series literature, stationary process).

Finally, if the white noise process is a Gaussian white noise process (which requires, among other things, that all the random variables are jointly Gaussian), then for $|a|< 1$, and $X_0 \sim N\left(0,\frac{\sigma^2}{1-a^2}\right)$, the process $\left\{X_n\colon n \geq 0\right\}$ is a strictly stationary (Gaussian) process.

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  • $\begingroup$ Thank you very much! This is very helpful. $\endgroup$ Commented Apr 27, 2023 at 8:21

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