Recall that stationary processes have the following properties

1. no systematic change in the mean or variance
   
2. no systematic trend
   
3. no periodic variations or seasonality

We seek a means for identifying whether our AR(p) models are also stationary

We can write out an AR(p) model using the backshift operator

\[ x_t = \phi_1 x_{t-1} + \phi_2 x_{t-2} + \dots + \phi_p x_{t-p} + w_t \\ \Downarrow \\ \begin{align} x_t - \phi_1 x_{t-1} - \phi_2 x_{t-2} - \dots - \phi_p x_{t-p} &= w_t \\ (1 - \phi_1 \mathbf{B} - \phi_2 \mathbf{B}^2 - \dots - \phi_p \mathbf{B}^p) x_t &= w_t \\ \phi_p (\mathbf{B}) x_t &= w_t \\ \end{align} \]

If we treat \(\mathbf{B}\) as a number (or numbers), we can out write the characteristic equation as

\[ \phi_p (\mathbf{B}) x_t = w_t \\ \Downarrow \\ \phi_p (\mathbf{B}) = 0 \]

To be stationary, all roots of the characteristic equation must exceed 1 in absolute value

For example, consider this AR(1) model

\[ \begin{align} x_t &= 0.5 x_{t-1} + w_t \\ x_t - 0.5 x_{t-1} &= w_t \\ x_t - 0.5 \mathbf{B} x_t &= w_t \\ (1 - 0.5 \mathbf{B})x_t &= w_t \\ \end{align} \]

For example, consider this AR(1) model from earlier

\[ \begin{align} x_t &= 0.5 x_{t-1} + w_t \\ x_t - 0.5 x_{t-1} &= w_t \\ x_t - 0.5 \mathbf{B} x_t &= w_t \\ (1 - 0.5 \mathbf{B})x_t &= w_t \\ \Downarrow \\ 1 - 0.5 \mathbf{B} &= 0 \\ -0.5 \mathbf{B} &= -1 \\ \mathbf{B} &= 2 \\ \end{align} \]

This model is indeed stationary because \(\mathbf{B} > 1\)

What about this AR(2) model from earlier?

\[ \begin{align} x_t &= -0.2 x_{t-1} + 0.4 x_{t-2} + w_t \\ x_t + 0.2 x_{t-1} - 0.4 x_{t-2} &= w_t \\ (1 + 0.2 \mathbf{B} - 0.4 \mathbf{B}^2)x_t &= w_t \\ \end{align} \]

What about this AR(2) model from earlier?

\[ \begin{align} x_t &= -0.2 x_{t-1} + 0.4 x_{t-2} + w_t \\ x_t + 0.2 x_{t-1} - 0.4 x_{t-2} &= w_t \\ (1 + 0.2 \mathbf{B} - 0.4 \mathbf{B}^2)x_t &= w_t \\ \Downarrow \\ 1 + 0.2 \mathbf{B} - 0.4 \mathbf{B}^2 &= 0 \\ \Downarrow \\ \mathbf{B} \approx -1.35 ~ \text{and}& ~ \mathbf{B} \approx 1.85 \end{align} \]

This model is not stationary because only one \(\mathbf{B} > 1\)

Consider our random walk model

\[ \begin{align} x_t &= x_{t-1} + w_t \\ x_t - x_{t-1} &= w_t \\ (1 - 1 \mathbf{B})x_t &= w_t \\ \end{align} \]

Consider our random walk model

\[ \begin{align} x_t &= x_{t-1} + w_t \\ x_t - x_{t-1} &= w_t \\ (1 - 1 \mathbf{B})x_t &= w_t \\ \Downarrow \\ 1 - 1 \mathbf{B} &= 0 \\ -1 \mathbf{B} &= -1 \\ \mathbf{B} &= 1 \\ \end{align} \]

Random walks are not stationary because \(\mathbf{B} = 1 \ngtr 1\)

We can define a space over which all AR(1) models are stationary

For \(x_t = \phi x_{t-1} + w_t\), we have

\[ \begin{align} 1 - \phi \mathbf{B} &= 0 \\ -\phi \mathbf{B} &= -1 \\ \mathbf{B} &= \frac{1}{\phi} > 1 \Rightarrow 0 < \phi < 1\\ \end{align} \]

For \(x_t = \phi x_{t-1} + w_t\), we have

\[ \begin{align} 1 - \phi \mathbf{B} &= 0 \\ -\phi \mathbf{B} &= -1 \\ \mathbf{B} &= \frac{1}{\phi} > 1 \Rightarrow 0 < \phi < 1\\ \end{align} \]

For \(x_t = -\phi x_{t-1} + w_t\), we have

\[ \begin{align} 1 + \phi \mathbf{B} &= 0 \\ \phi \mathbf{B} &= -1 \\ \mathbf{B} &= \frac{-1}{\phi} > 1 \Rightarrow -1 < \phi < 0\\ \end{align} \]

Thus, AR(1) models are stationary if and only if \(\lvert \phi \rvert < 1\)