Consider an AR(1) model with coefficient $\rho =1$, i.e. a random walk, $Y_t=Y_{t-1}+\epsilon_t$. We usually say that since the AR polynomial has a unit root, it cannot be inverted and therefore the AR(1) cannot be written as an MA($\infty$).

Putting this into numbers, my understanding is that if we rewrite the model using the lag operator,

$(1-L)Y_t=\epsilon_t \to Y_t=(1-L)^{-1} \epsilon_t$,

then the second expression is not defined, since $(1-L)^{-1}$ does not exist. However, Hayashi (2000, p.372) does actually define this as

$ (1-L)^{-1}=1+L+L^2+L^3+...$

Hence I understand my original question is actually mathematically correct. But then:

1) Are all lag polynomials invertible?

2) Why do we refer to a stationary AR(1) as having an invertible AR polynomial and to a non-stationary AR(1) as having a non-invertible AR polynomial?


2 Answers 2


Regarding your questions:

1.) No, only stationary time series have invertible AR polynomials, and

2.) the AR(1) model is stationary iff the polynomial $\phi(z) = 1-\phi z$ doesn't have a root equal to plus or minus $1$. For higher-order AR(p) models, with $p > 1$, this $z$ can be a complex number. But in the case of an AR(1), this condition is the same as $\phi$ not being equal to plus or minus $1$. As long as this isn't true, there exists a $\delta > 0$ such that $\frac{1}{\phi(z)} = \sum_{j=-\infty}^{\infty} \phi^j z^j$ for all $1 - \delta < |z| < 1+\delta$, and $$ \sum_{j=-\infty}^{\infty} |\phi|^j < \infty \tag{*}. $$ This condition is called absolute summability, and it is needed to even define linear processes.

Why? Intuitively, you can see that if $\phi = \pm 1$, then the above sum wouldn't be finite, and it doesn't make sense to write $$ \lim_{n \to \infty} X_t^n = \lim_{n \to \infty}\sum_{j=-n}^{n} \phi^j \epsilon_{t-j} = \sum_{j=-\infty}^{\infty} \phi^j \epsilon_{t-j}. $$ The limit would not exist in either the mean square sense or the almost-surely sense.

  • $\begingroup$ Hi Taylor, thank you for your answer. Just to clarify: given 1.) and 2.), how can I reconcile that with my reference to Hayashi (and with the title of my question), i.e. with a unit root $(1-\phi L)^{-1}=(1-L)^{-1}=1+L+L^2+...$ (the polynomial is inverted even though the series is not stationary)? $\endgroup$
    – econ86
    Feb 17, 2019 at 11:54
  • $\begingroup$ @econ86 I can’t say with any certainty because I haven’t read the paper. If he/she didn’t assume stationarity and defined that anyway, then he/she made a mistake, because you need stationarity to define that. If there’s a unit root, you can difference your data, though. That’s the “I” in “ARIMA.” $\endgroup$
    – Taylor
    Feb 17, 2019 at 20:24

There is a lot of information missing, e.g. no-one will know what your variables are.

In general, with $\rho=1$, you have to ensure that the operator $L$ has no eigenvalue of 1. Otherwise the sequence on the left hand side does not converge in the direction of that eigenvector/eigenfunction.

  • $\begingroup$ I reframed the question, hope it is now understandable. $\endgroup$
    – econ86
    Feb 16, 2019 at 16:59
  • $\begingroup$ Hi: In that particular case, you can invert it that way because the difference is stationary. But a ( truly ) non-stationary AR(1) can't be inverted for greater than 1 AR(1) coefficient . you happened to use the example of a random walk which, when differenced, is not explosive so it can be written as an infinite sum of epslon's. ( but that's not really an MA). $\endgroup$
    – mlofton
    Feb 16, 2019 at 17:36
  • $\begingroup$ well, it's an MA but it's an MA where all the coefficients are (1) so really non stationary itself in that the errors are all permanent. Usually, the MA has transitory error terms that die out in the long run. $\endgroup$
    – mlofton
    Feb 16, 2019 at 17:39
  • $\begingroup$ mlofton: so you are saying that an AR(1) with rho=1 is nonstat and its AR polynomial can be inverted; AR(1) with rho>1 is nonstat and its AR polynomial cannot be inverted. Is that correct? Then 1/(1-rhoL) in the latter case is not defined? $\endgroup$
    – econ86
    Feb 16, 2019 at 18:53
  • $\begingroup$ Hi: $Y_t = Y_{t-1} + \epsilon_{t}$ isn't really an AR(1). It's more correct to say that the first difference is white noise. This pseudo- "AR(1)" is not stationary nor is it really AR(1) because it has a unit root. By unit root, I just mean that the AR(1) coefficient equals one and for the AR(1) model to be stationary, the absolute value of the AR(1) coefficient should really be strictly less than one. I think this is what Taylor is saying also. $\endgroup$
    – mlofton
    Feb 17, 2019 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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