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

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?

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