# When is an AR(1) process strictly stationary?

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 $$a$$?

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

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.