The question I'm trying to solve is shown below - from a practice qualifying exam in probability which has no posted solution (self study). I really don't know where to start (even for part (a)) and am looking for some major help.
Am I supposed to apply one of Kolmogorov's two/three series theorem? Thanks for any ideas you have!
a) is answered in the comments
b) For $\phi \in (0, 1/e)$ you can modify @Michael's hint and note that if
$$ \sum_{n=1}^\infty P(\log^+(Z) > rn) < \infty $$ for any $r > 0$, then $\mathbb{E}[\log^+(Z)] < \infty$, which means you can just modify your calculation in the comment for $\phi \in (0, 1/e)$ by
$$\sum_{n=1}^\infty P(\log^+(Z) > rn) = \sum_{n=1}^\infty P(Z > e^{rn} )$$ and now pick $r$ s.t. $e^r = 1/\phi$.
c) Provided the indexing on $Z_t$s is from $-\infty$ to $\infty$, yes, this sequence is strictly stationary. To see that, note that shifting the indices of $Z_i$s does not change anything about the joint distribution of finitely many $X_{t_1}, X_{t_2}, ..., X_{t_n}$ and the joint distribution does not depend on the absolute values of $t_1, t_2, ..., t_n$, but only their differences.
I think$^*$ sequence, since $\phi$ is a random variable, is not ergodic. To see that, formally, we have
$$\lim_{n \rightarrow \infty} \frac{1}{2n} \sum_{t=-n}^{n} X_n = \lim_{n \rightarrow \infty} \frac{1}{2n} \sum_{t=-n}^{n} \sum_{j=0}^\infty \phi^j Z_{t-j} = \lim_{n\rightarrow \infty} \frac{1}{2n} \sum_{k=-\infty}^{\infty} \sum_{j= \max(|k| - n, 0) }^{|k| + n} \phi^j Z_k \rightarrow \frac{1}{1 - \phi} \mathbb{E} Z $$
where we can swap sums because everything is non-negative. Since this limit is not trivial provided $\phi$ is not, by the ergodic theorem, this sequence is not ergodic.
$*$: I am not fully sure; in particular, I am very unsure about the limit. I am quite sure it isn't ergodic though, since the limit of this running average is going to depend on the value of $\phi$.