This thing appeared in my lecture on autoregressive models (AR1). It talks about how $$y_t=\alpha + \phi y_{t-1} + \varepsilon_t$$ can be written as: $$\Delta y_t=\alpha + \rho y_{t-1}+\varepsilon_t \ where \ \rho=\phi-1 $$

Now I'm okay with this information but then this statement appears:

Stationarity condition $|\rho|<1$ is equivalent to $-2<\rho<0$

In my understanding $\rho$ should be from $-0.\bar{9}$ to $0.\bar{9}$ how is this equivalent to $-2<\rho<0$?

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    $\begingroup$ It's not equivalent. But, based on $\rho = \phi - 1$, $|\rho| < 1$ does imply $0 < \phi < 2$. $\endgroup$ – gammer Apr 1 '17 at 19:35
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    $\begingroup$ The first equation does not lead to the second one. Both alpha and epsilon terms disappear in the second equation. You are right. It is not equivalent to the inequality given in the lecture. $\endgroup$ – Michael Chernick Apr 1 '17 at 19:36
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    $\begingroup$ said AR(1) will be stationary for $ | \phi | < 1 ~ not ~ for ~ | \rho | < 1$. I think there is a mistake in the statement given. you may see equivalency now. $\endgroup$ – Hemant Rupani Apr 1 '17 at 19:58
  • $\begingroup$ not surprising ... the slides are a mess. Thank you! $\endgroup$ – Alex Apr 1 '17 at 20:19
  • $\begingroup$ By the way, $0.\bar{9} = 1$. $\endgroup$ – Kodiologist Apr 6 '17 at 20:07

For the sake of marking this as solved, the answer to this question is that, in effect, they are not equivalent. This is a typo in the slides. The statement should read as:

Stationarity condition $|\phi|<1$ is equivalent to $−2<\rho<0$

To add some value to the answer, the second format (with $\rho$) is quite interesting. First, notice that the long run equilibrium of the system is given by

$$ \bar{y} = \frac{\alpha}{1-\phi} $$

using this result, we can re-write the second equation, by getting rid of $\alpha$. This equation becomes:

$$ \Delta y_t = \rho(y_{t-1} - \bar{y}) + \varepsilon_t$$

This tells us that the adjustment of the process against any shock depends on the direction of the initial deviation. If the shock is positive (thus $y_{t-1} - \bar{y}$ is positive), the right hand side will be negative. Conversely, if the shock is negative, (thus $y_{t-1} - \bar{y}$ is negative), the right hand side is positive. The magnitude constraint of $\rho$ ensures there is no explosiveness. As usual, $\rho < -1 $ means convergence through oscillation. $\rho = -1$ means one-period convergence.


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