3
$\begingroup$

I saw the top answer on the following post: Stationarity of AR(1) model.

It says that an AR(1) process $X_{t} = \phi X_{t-1} + \epsilon_{t}$ where $\epsilon_{t} \sim WN(0, \sigma^{2})$ has a stationary solution if and only if $|\phi| < 1$. However, it is also commonly known that explosive processes with $|\phi| > 1$ can be represented as $$ X_{t} = \sum_{j=1}^{\infty} -\phi^{-j} w_{t+j} $$ which is a stationary time series. So does it not follow that all AR(1) processes other than when $|\phi| =1 $ have a stationary solution?

$\endgroup$
2
  • $\begingroup$ You have a crucial typo in your prompt; the index on omega should be {t+j} $\endgroup$ Commented Nov 11, 2021 at 22:49
  • $\begingroup$ that post has no answer, but only a comment with an incorrect statement $\endgroup$
    – Aksakal
    Commented Nov 11, 2021 at 23:47

1 Answer 1

5
$\begingroup$

Take this $AR(1)$ process: $X_t-\phi X_{t-1}$=$Z_t$

If $\lvert {\phi} \rvert<1: X_t=\sum_{j=0}^{\infty}\phi^jZ_{t-j}$ is the unique stationary solution of the above $AR(1)$ process.

If $\lvert {\phi} \rvert>1: X_t=-\sum_{j=1}^{\infty}\phi^{-j}Z_{t+j}$ is the unique stationary solution of the above $AR(1)$ process.

So, yes, the $AR(1)$ process is stationary as long as $\lvert {\phi} \rvert\ne1$. Note however that when $\lvert {\phi} \rvert>1$, $X_t$ is expressed in terms of future values of $Z_{t+j}$. This is considered unnatural because $X_t$ is future dependent. In contrast when $\lvert {\phi} \rvert<1$, $X_t$ is future independent, also known as causal. And since every $AR(1)$ process with $\lvert {\phi} \rvert>1$ can be written as an $AR(1)$ process with $\lvert {\phi} \rvert<1$ and a different white noise sequence, not much is lost by focusing on only stationary causal $AR(1)$ models, which is what most textbooks do.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.