Intuitive and mathematical explanation of stationarity for AR(p)? I'd be grateful if someone could explain me what's the link between stationarity conditions of AR(p) processes in theory and in the practical sense.
I was given the following short definition about weakly stationarity: a process is weakly stationary if the fist and second moments of the distribution are finite and time-invariant. 
As a practical criterion to check for stationarity, I've understood that we have to rely on the analysis of the roots of the characteristic equation associated to the lag polynomial, i.e all the roots have to lie outside the unit circle (greater than 1 in absolute value). 
As I mentioned earlier, I don't understand why the condition on the roots should imply finite first and second moments. Is this linked to the convergence of some lag polynomial, or to complex numbers? This seems to suggest a criterion for the ratio of some series...
Moreover, what's the relation between stationarity and invertibility condition? They seem to me as being the same thing for AR(p). What's the statistical explanation? How does it work in general? (I.e can I say that stationarity implies invertibility of the lag polynomial? Is it necessary condition or both necessary and sufficient?) 
Thank you very much. 
 A: Note: I will not go through all the details in this answer, but merely lay out the intuition and ideas. Some of the arguments are a bit handwavy and taken as truths when indeed they require (non-trivial) proofs.

Let us consider an $\operatorname{AR}(p)$ model given by $X_t = Z_t + \phi_1X_{t-1} + \phi_2X_{t-2} + \ldots + \phi_pX_{t-p}$
where $\{Z_t\}_t$ is a white noise process. 
Define the backshift operator $B$ such that $BX_t = X_{t-1}$. Consider the polynomial 
$$
\Phi(z) = 1-\phi_1z - \phi_2z^2 - \ldots - \phi_pz^p, \quad z\in \mathbb C.
$$
Then we can rewrite $X_t$ with the polynomial in the backshift operator, i.e. 
$$
\Phi(B) X_t = Z_t.
$$
The question is when this representation is stationary, which is not obvious. So we take a step back and consider $p = 1$ (i.e. an $\operatorname{AR}(1)$ model). Then the polynomial is $\Phi(z) = 1-\phi z$. This constitutes to
$$
X_t = Z_t +\phi X_{t-1}. \qquad \qquad (1)
$$
If we iterate (1) a couple of times by substituting $X_{t-1}$, $X_{t-2}$, etc. we get 
\begin{align*}
X_t &= 
Z_t + \phi(Z_{t-1} + X_{t-2})
\\&=
Z_t + \phi Z_{t-1} + \phi(\phi Z_{t-2} + X_{t-3})
\\&= \ldots \\&=
Z_t + \phi Z_{t-1} + \phi^2 Z_{t-2} + \ldots + \phi^{n}Z_{t-n}\phi^n + \phi^{n+1}X_{t-n}.
\end{align*}
This suggests that $X_t$ be given by the representation
$$
X_t = \sum_{k = 0}^\infty \phi^kZ_{t-k}.
$$
If this is true, then we can rule out $\lvert \phi \rvert = 1$ immediately; otherwise the sum would not converge. 
Thus we consider $\lvert \phi\rvert < 1$. (*) One would now have to show that this representation gives a stationary time series by satisfying the two moment requirements (which indeed it does). 

Now, what does that have to do with the polynomial $\Phi$? Well, indeed $\lvert \phi\rvert < 1$ means that $\Phi(z) = 1-\phi z$ does not have any roots for $\lvert z\rvert \leq 1$. Or equivalently, the inverse $\Phi(z)^{-1}$ is a convergent power series in the one-dimensional "circle" $\lvert z\rvert \leq 1$:
$$
\sum_{k = 0}^\infty \phi^k z^k =  \frac1{1-\phi z } = \Phi(z)^{-1}  , \qquad \lvert \phi\rvert < 1, \lvert z\rvert \leq 1.
$$
For the general setting where $\Phi(z) = 1- \sum_{k = 0}^p \phi_k z^k$ we want a similar property of the inverse. That means no roots for the polynomial, i.e. $\Phi(z)\neq 0$ for $\lvert z\rvert = 1$, $z\in \mathbb C$. Note how we changed to complex numbers: That happened because the higher order polynomials may have complex roots. (**)

(*) Note that for $\lvert \phi \rvert >1$ we would iterate by substituting $Z_t$, $Z_{t+1}$, etc., suggesting the representation
$$
X_t = -\sum_{k = 1}^\infty \phi^{-k}Z_{t+k}.
$$
However, this means the time series depends on its future, which in practice is nonsense. This is why you often see the requirement of causality to avoid these cases.
(**) Causality here constitutes to no roots in the circle $\lvert z\rvert \leq 1$. 
