Determine whether an ARMA(p,q) process is stationary and invertible such that $y_t = \sum_{i=1}^{p} \phi_i y_{t-i} + \sum_{i=1}^{p} \theta_{i} \epsilon_{t-i}$ with the restriction that $\theta_{0} = 1$
I'm not familiar with determining this, here's my best shot using the knowledge I have:
Set lag operators: $\epsilon_{t-i} = L_1^i \epsilon_{t}$, and $y_{t-i} = L^{i}_{2} y_{t}$
$y_t - y_t \sum_{i=1}^{p} \phi_i L_1^i = y_t(1 - \sum_{i=1}^{p} \phi_i L_1^i) = \epsilon_{t} \sum_{i=1}^{p} \theta_{i} L_2^i$
I am guessing that this is stationary, because the roots of $1 - \sum_{i=1}^{p} \phi_i L_1^i$ can be outside unit circle. What about $\sum_{i=1}^{p} \theta_{i} L_2^i$ ? Does that mean it's invertible? I would guess and say yes, because if $\sum_{i=1}^{p} \theta_i$ is less than zero then the roots should lie outside of the unit circle as well. Is this correct?
It would also be nice to have more information about this in general - does anyone know of a resource that goes over determining whether a ARMA/AR/MA process is stationary or invertible?
If you set $\Phi(z)=1-\phi_1z-\phi_2z^2-\cdots-\phi_pz^p$ and $\Theta(z)=1+\theta_1z+\cdots+\theta_qz^q$, then your process can be written as $$ \Phi(L)y_t=\Theta(L)\epsilon_t.\tag{$*$} $$ Note I'm assuming you have a typo in that the sum $\sum_{i=1}^p\theta_i\epsilon_{t-i}$, as you have written, should really begin with $i=0$ and should end with $q$. And $L$ here denotes the lag operator with which you can at least raise to any integral powers but, in this simple case, shouldn't be added with a subscript like you have done.
In general, $X_t$ is stationary if $\Phi(z)$ has its roots outside the unit circle and $X_t$ is invertible if $\Theta(z)$ has its roots outside the unit circle.
A standard reference for general results like this is Hamilton (1994).