Transform an AR(2) to MA($\infty$)

Question

So I have the following AR(2) process:

$z_t = \delta + \phi_2z_{t-2} + \epsilon_t$

where $\epsilon_t$ is white noise $(0, \sigma^2)$

How can I transform this process to an MA($\infty$)?

My thought process is to recursively substitute as follows:

$z_t = \delta + \phi_2(\delta + \phi_2z_{t-4} + \epsilon_{t-2}) + \epsilon_t$
$z_t = \delta +\phi_2\delta + \phi_2^2z_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_t$

Then substitute in for $z_{t-4}$ as follows:

$z_t = \delta +\phi_2\delta + \phi_2^2(\delta + \phi_2z_{t-6} + \epsilon_{t-4}) + \phi_2\epsilon_{t-2} + \epsilon_t$
$z_t = \delta + \phi_2\delta + \phi_2^2\delta + \phi_2^3z_{t-6} + \phi_2^2\epsilon_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_t$

And then keep doing the same thing over and over again. Hopefully you see the pattern of what I am doing.

So this is kind of where I am stuck. If this is correct, how would I write the MA($\infty$) process in sigma notation?

Answer 1

Your $\delta + \phi_2\delta + \phi_2^2\delta + \cdots$ is just a geometric series so let's call this $\mu$ and we have $\mu= \frac\delta{1-\phi_2}$, at least when $|\phi_2|<1$

Meanwhile your $\cdots +\phi_2^3z_{t-6} + \phi_2^2\epsilon_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_{t}$ can be written as $\sum\limits_{i=0}^\infty \psi_i \epsilon_{t-i}$ where $\psi_{2n}=\phi_2^n$ and $\psi_{2n+1}=0$, noting $\psi_{0}=1$. You will have $\sum\limits_{i=0}^\infty |\psi_i| = \frac\delta{1-|\phi_2|} < \infty$ when $|\phi_2|<1$.

So, providing that $|\phi_2|<1$, you can write $$z_t=\mu + \sum\limits_{i=0}^\infty \psi_i \epsilon_{t-i} = \frac\delta{1-\phi_2} + \sum\limits_{n=0}^\infty \phi_2^n \epsilon_{t-2n} $$ which is the form of an $\mathrm{MA}(\infty)$ process