# Link between stationarity of AR(2) and stability condition of corresponding difference equation

In the standard textbooks, such as Hamilton (1994), it is stated that the conditions for stationarity of a process

$y_t = \theta_0 + \theta_1 y_{t-1} + \theta_1 y_{t-2} + \epsilon_t$

follow from the stability condition of the the characteristic equation of the corresponding difference equation

$(1-\theta_1 z - \theta_2 z^2) = 0$,

which are that its roots lie outside the unit circle.

I don't understand how the stability conditions of the corresponding difference equation relates to stationarity of an AR(2) process. Does anyone know a book or give me a hint on their relation?

Many thanks!

## 1 Answer

this relationship comes from the fact that converting and AR model to an MA model must be stationary. Then assume you are going to convert the problem into an MA problem. As a result you must keep the response, y, on the left and on the right just keep error term (saying white noise).

\begin{align} (1-\theta_1L-\theta_2L^2) y_t=e_t \end{align} where $L$ is the backward shift operator $(Ly_t=y_{t-1})$. Now convert the problem into an MA problem. \begin{align} y_t=\frac{1}{(1-\theta_1L-\theta_2L^2)} e_t \end{align} on the other hand assume that the polynomial, $(1-\theta_1L-\theta_2L^2)$ has two roots saying, $\alpha$ and $\beta$. Then you can write the denominator as, \begin{align} y_t=\frac{1}{(L-\alpha)(L-\beta)} e_t \end{align}

now rewrite each term in denominator as a power series, $\sum_{i=0}^{\infty}(L/\alpha)^i = \frac{-\alpha}{L-\alpha}$ and similarly for the second term. Now you can let the backward shift operator to operate on $e_t$ (before that L was in the denominator and there is no definition about how L can operate). But the point is that $\alpha$ and $\beta$ must be greater than one otherwise the power series diverges, meaning for example variance tends to infinity!

You should notice that $L$ is an operator not like a variable or parameter and it just operates linearly on the index of its next term.