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Suppose we have the following model:

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ADF would test if $\gamma$ equals 0 (null hypothesis), and rejects it if the statistic is sufficiently negative in favor of $\gamma < 0.$ But what if, say, gamma equals -5? Wouldn't this mean that yes, there's no unit root, but since $\gamma = -5$ the process still won't be stationary, it would explode? If so, is it really correct to use ADF to conclude that the process is stationary when we get a low enough negative value (which is what I was taught) as opposed to just concluding that the process has no unit root? I must be missing something. Thanks!

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To understand more clearly, lets exclude the lags of dependent variable and focus on the following (which is the Dickey Fuller model) :

\begin{equation} \Delta y_{t}=\alpha+\beta t+\gamma y_{t-1} (1) \end{equation}

This can be also written as:

\begin{equation} y_{t}=\alpha+\beta t+ (1+\gamma) y_{t-1} (2) \end{equation}

Replacing $(1+\gamma)$ with $\rho$ gives \begin{equation} y_{t}=\alpha+\beta t+ \rho y_{t-1} (3) \end{equation} The fact that null is unit root (non stationary) comes from third equation and can be stated in terms of null hypothesis as: \begin{equation} H_{0} :\rho =1 \end{equation}

So, if

\begin{equation} |\rho|\le 1 \end{equation}

then, the process is stationary. In your case this is 4 , so the process is non-stationary.

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  • $\begingroup$ Right, so going back to my question, I think $\rho$ would equal -4. But, an Augmented Dickey Fuller test would leat me to reject the null that $\gamma=0$ in favor of $\gamma<0,$ meaning that there is no unit root (en.wikipedia.org/wiki/…). But in this case, I would have no unit root, but the process would still be non-stationary, right? $\endgroup$
    – dfgkjsldf
    Commented Sep 27, 2013 at 14:50
  • $\begingroup$ Yes $\rho$ is -4 but for the convergence (which is stationary) in eq 3 (this is the first difference equation), absolute of $\rho$ should be less than or equal to one. With gamma=-5 this convergence is not achieved (you are, however, correct to say that literature just says null is rejected when gamma<0) (unit root and non stationary are same thing). You can try doing simulation and see the plots. $\endgroup$
    – Metrics
    Commented Sep 27, 2013 at 15:33
  • $\begingroup$ I don't think I'm following. It seems to me like under the usual specification of the test, the alternative hypothesis can be true (e.g., the -5 value) yet the series won't be stationary (as in the case with -5). How exactly are these reconciled? $\endgroup$
    – dfgkjsldf
    Commented Sep 28, 2013 at 8:24

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