I've been searching in bibliography about this test applied to an AR(p) model. $$Q(L)(Y_{t})=c+\epsilon_{t}$$

Where L represent the Lag Operator and $Q=1-\phi_{1}x-.....-\phi_{p}x^{p}$ is the polynomial expression associated to the model.

I know that if $Q(r)=0$ implies $|r|>1$, then the process is stationary (at least in weak sense).

My question is: Why the Null Hypothesis of Augmented Dickey-Fuller test is stated as: "$r=1$ is a root of the polynomial"? Rejecting that hypothesis implies that every single root of Q lies outside the unit circle??

I'm new at this area so every recommendation or suggestion will be useful. Thanks.

  • 4
    $\begingroup$ This is a purely statistical question, you might get the best answers here: stackexchange.com $\endgroup$
    – Richi Wa
    Jan 14, 2016 at 7:11
  • 1
    $\begingroup$ You may also try here: stats.stackexchange.com. $\endgroup$
    – Gordon
    Jan 14, 2016 at 16:28
  • $\begingroup$ I'm voting to close this question as off-topic because it belongs on stats.stackexchange.com $\endgroup$
    – chollida
    Jan 14, 2016 at 21:11

1 Answer 1



Your interpretation is wrong. If r>1 (not in absolute value) the series follows and explosive and therefore is not stationary. If you reject the unit root it means that the series does not have a unit root, because as it is stated in the comments the hypothesis is h0: r=1 H1: r<1. so rejecting means that all every single root of Q lies inside the unit circle.

Thanks Richard Hardy for the correction

  • 2
    $\begingroup$ The augmented Dickey Fuller test rejects the null hypothesis of presence of a unit root only when the test statistic is sufficiently low, which happens when the process has a root smaller than 1. It does not reject when the test statistic is high where the root is above 1. Thus your answer is incorrect. $\endgroup$ Feb 26, 2016 at 16:34
  • $\begingroup$ the usual hypothesis test is H0: r=1, which means that the series has a unit root, or H1: r<1 which means that the series does not have a unit root and therefore is stationary. if r is > 1 means that the series has an explosive behavior. so my interpretation is not wrong $\endgroup$ Feb 28, 2016 at 18:17
  • 1
    $\begingroup$ The point is that you do not reject in the tail where $r>1$ but only in the tail where $r<1$. Your statement If r>1 (not in absolute value) the series follows and explosive and therefore is not stationary is correct by itself but it is redundant in the context of the augmented Dickey-Fuller test. Therefore rejecting the null is only compatible with stationarity (and not with explosivity). $\endgroup$ Feb 28, 2016 at 22:25
  • $\begingroup$ Yeah you are right thanks for the correction $\endgroup$ Feb 29, 2016 at 16:09
  • $\begingroup$ Would you mind editing your post accordingly? It is better to have accurate information not only in the comments but in the answer itself (so that it would attract upvotes rather than downvotes, for example). $\endgroup$ Feb 29, 2016 at 19:12

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.