It is possible to write an AR(p) model as an MA(\(\infty\)) model

For example, consider an AR(1) model

\[ \begin{align} x_t &= \phi x_{t-1} + w_t \\ x_t &= \phi (\phi x_{t-2} + w_{t-1}) + w_t \\ x_t &= \phi^2 x_{t-2} + \phi w_{t-1} + w_t \\ x_t &= \phi^3 x_{t-3} + \phi^2 w_{t-2} + \phi w_{t-1} + w_t \\ & \Downarrow \\ x_t &= w_t + \phi w_{t-1}+ \phi^2 w_{t-2} + \dots + \phi^k w_{t-k} + \phi^{k+1} x_{t-k-1} \end{align} \]

If our AR(1) model is stationary, then

\[ \lvert \phi \rvert < 1 ~ \Rightarrow ~ \lim_{k \to \infty} \phi^{k+1} = 0 \]

so

\[ \begin{align} x_t &= w_t + \phi w_{t-1}+ \phi^2 w_{t-2} + \dots + \phi^k w_{t-k} + \phi^{k+1} x_{t-k-1} \\ & \Downarrow \\ x_t &= w_t + \phi w_{t-1}+ \phi^2 w_{t-2} + \dots + \phi^k w_{t-k} \end{align} \] \[ \]

An MA(q) process is invertible if it can be written as a stationary autoregressive process of infinite order without an error term

For example, consider an MA(1) model

\[ \begin{align} x_t &= w_t + \theta w_{t-1} \\ & \Downarrow \\ w_t &= x_t - \theta w_{t-1} \\ w_t &= x_t - \theta (x_{t-1} - \theta w_{t-2}) \\ w_t &= x_t - \theta x_{t-1} - \theta^2 w_{t-2} \\ & ~~\vdots \\ w_t &= x_t - \theta x_{t-1} + \dots + (-\theta)^k x_{t-k} + (-\theta)^{k+1} w_{t-k-1} \\ \end{align} \]

If we constrain \(\lvert \theta \rvert < 1\), then

\[ \lim_{k \to \infty} (-\theta)^{k+1} w_{t-k-1} = 0 \]

and

\[ \begin{align} w_t &= x_t - \theta x_{t-1} + \dots + (-\theta)^k x_{t-k} + (-\theta)^{k+1} w_{t-k-1} \\ & \Downarrow \\ w_t &= x_t - \theta x_{t-1} + \dots + (-\theta)^k x_{t-k} \\ w_t &= x_t + \sum_{k=1}^\infty(-\theta)^k x_{t-k} \end{align} \]

Q: Why do we care if an MA(q) model is invertible?

A: It helps us identify the model's parameters

For example, these MA(1) models are equivalent

\[ x_t = w_t + \frac{1}{5} w_{t-1}, ~\text{with} ~w_t \sim ~\text{N}(0,25) \]

\[ x_t = w_t + 5 w_{t-1}, ~\text{with} ~w_t \sim ~\text{N}(0,1) \]