I've seen somewhat similar questions, but nothing that is quite like what I am concerned with currently.
Given that $ X_t=0.3X_{t-1}+e_{1t}+0.3e_{1,t-1} $ and $z_t=X_t+X_{t-1}$ ; is $z_t$ (weakly) stationary and/or invertible?
So, I've tried to do this using the stationarity conditions, starting with a mean independent of time.
Note: $e_{1t}$~$WN(0,1)$
$$z_t=0.3X_{t-1}+0.3X_{t-2}+e_{1t}+1.3e_{1,t-1}+0.3e_{1,t-2}$$
Take expectation:
\begin{align*}E(z_t)&=E(0.3X_{t-1}+0.3X_{t-2}+e_{1t}+1.3e_{1,t-1}+0.3e_{1,t-2}) % \\&=0.3E(X_{t-1})+0.3E(X_{t-2}) \\&= 0.3E(z_{t-1}) \end{align*} And what I initially though is that because $z_t$ is dependent on time (due to the lags) then it is non-stationary. But I realised that this means that any model with an AR lag would be non-stationary, so my interpretation here is clearly incorrect.
Using the Unit root method:
Given that $z_t=X_t+X_{t-1}$,
\begin{align*} z_t&=0.3X_{t-1}+0.3X_{t-2}+e_{1t}+1.3e_{1,t-1}+0.3e_{1,t-2} \\ &=0.3(X_{t-1}+X_{t-2} )+e_{1t}+1.3e_{1,t-1}+0.3e_{1,t-2} \\ &=0.3z_{t-1}+e_{1t}+1.3e_{1,t-1}+0.3e_{1,t-2} \end{align*} From here I can find the AR and MA polynomials. However, I am unsure if this method will provide correct results given the unchanged error terms..
My two questions
Is the working out I have done for the stationarity conditions methods correct? Because, it implies that any model with any ar lags will be non-stationary.
Can I simply use the unit root method after writing the equation in terms of $z_t$ to find stationarity/invertability?
1.
Your work is correct but intermediate. You do not know anything about the expectation of $z_{t-1}$ (yet). However, if you reiterate a couple of times, you may realize that $z_t$ can be written on the form
$$ z_t = e_t + 2(\phi+1) e_{t-1} + 2(\phi + 1) \sum_{j = 1}^\infty \phi^j e_{t-1-j} $$ for $\phi = 0.3 (<1)$. To show that the process is a mean-zero process, it suffices to show that $$ 2(\phi + 1) E\left[ \sum_{j = 1}^\infty \phi^j e_{t-1-j} \right] = 0. $$ By a dominated convergence argument, using that $\phi<1$, we may interchange sum and expectation, giving the result.
To show that the autocorrelation function is independent of time, follow similar arguments.
2.
The problem with using the unit root method in this case is that the MA polynomial has a root on the unit disc $S = \{z \in \mathbb C : \lvert z \rvert = 1\}$, so the method is inconclusive.