Why is $|\rho|<1 \leftrightarrow -2<\rho<0$? 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 $-2<\rho<0$?
 A: 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.
