So I have the following AR(2) process:
$z_t = \delta + \psi_2z_{t-2} + \epsilon_t$
which has the following MA($\infty$) representation:
$\frac{\delta}{1 - \psi_2} + \sum_{n=0}^\infty \psi_2^n \epsilon_{t-2n}$
Now I am looking for $E[z_t|z_1, z_2, . . . ,z_{t-2}, z_{t-1}]$
This is where I am kind of confused.
Would it just be $\frac{\delta}{1 - \psi_2} + \sum_{n=1}^\infty \psi_2^n \epsilon_{t-2n}$
Therefore, the only difference would be the sum begining at $1$ instead of $0$?
Why do you need the $MA(\infty)$ representation here? If you're conditioning on the previous values, they can be taken out of the conditional expectation. So we have
$$\mathbb{E}[z_t|z_1,\dots,z_{t-1}]=\mathbb{E}[\delta +\psi_2 z_{t-2}+\epsilon_t|z_1,\dots,z_{t-1}] = \delta + \psi_2z_{t-2}+\mathbb{E}[\epsilon_t|z_1,\dots,z_{t-1}]$$
You mention the process is AR(2), so presumably the white noise terms are i.i.d. and independent of the history of the process, i.e., $\mathbb{E}[\epsilon_t|z_1,\dots,z_{t-1}]=\mathbb{E}[\epsilon_t]=0$. So we would have
$$\mathbb{E}[z_t|z_1,\dots,z_{t-1}]= \delta + \psi_2z_{t-2}$$