AR(1) - Stationarity condition Consider the well-known AR(1) model:
$$x_t = \phi X_{t-1} + \epsilon_t$$
where, as usual, $\epsilon$ is an independent white noise process.
I have read many sources. All of them get away saying that the process is (covariance/weakly)-stationary iff $|\phi| < 1$.
I believe them but I can't really find the proof of this theorem (and I am really interested).
I then started with the definition of weekly stationary:

*

*$\mathbb{E}[X_t] = \mu < \infty$ for every $t$

*$\gamma(t, k) = \gamma_k < \infty$ for every $t$
In my notation $\gamma(t, k)$ is the auto-covariance function.
Showing 1. yields no problem as I found $\mathbb{E}[X_t] = X_0$ (the only assumption needed here is that $X_0 < \infty$ but this doesn't seem to be a big problem as often $X_0$ is assumed to be zero, at least in my book).
Far more complicated (at least for me) is showing 2.
After few trials I arrived to a convincing result:
$$\gamma(t, k) = \phi^{k} \cdot \mathbb{V} \left[ X_{t-k}\right]$$
My aim would now be to show that $\mathbb{V} \left[ X_{t-k}\right] < \infty$ iff $|\phi| < 1$. This seems to me the only place in which that assumption is needed. Am I mistaken? How should I proceed? Any hint/remark/suggestion is greatly appreciated!
 A: I consider a more general case. The AR(1) process is given by
\begin{align}
X_t=\phi_0+\phi_1X_{t-1}+\epsilon_t \quad \epsilon_t \sim WN(0,\sigma^2)
\end{align}
First you calculate the mean:
\begin{align}
E(X_t)=E(\phi_0+\phi_1X_{t-1}+\epsilon_t)=\phi_0+\phi_1E(X_{t-1})+E(\epsilon_t)
\end{align}
Since $\epsilon_t$ is a white noise process, $E(\epsilon_t)=0$. In order for the process to be stationary, it must hold that $E(X_t)=E(X_{t-1})$. Therefore
\begin{align}
E(X_t)=\phi_0+\phi_1E(X_t) \Leftrightarrow E(X_t)=\frac{\phi_0}{1-\phi_1}
\end{align}
You see that $E(X_t)<\infty$ if $\phi_1\neq1$. Now look at the variance.
\begin{align}
V(X_t)=V(\phi_0+\phi_1X_{t-1}+\epsilon_t)=\phi_2^2V(X_{t-1})+\underbrace{2Cov(X_{t-1},\epsilon_t)}_{=0}+\underbrace{V(\epsilon_t)}_{\sigma^2}
\end{align}
If the process is stationary, we have $V(X_t)=V(X_{t-1})$ and therefore:
\begin{align}
V(X_t)=\phi^2V(X_t)+\sigma^2 \Leftrightarrow V(X_t)=\frac{\sigma^2}{1-\phi_1^2}
\end{align}
The variance is positive an finite if $\phi_1^2<1 \Leftrightarrow \vert \phi \vert <1$.
Since $\phi_0=(1-\phi_1)E(X_t)$, we can center the process around $E(X_t)$ and get:
\begin{align}
X_t-E(X_t)=\phi_1(X_{t-1}-E(X_t))+\epsilon_t
\end{align}
We can use this result in order to derive the autocovariance function:
\begin{align}
\gamma_k&=E[(X_t-E(X_t))(X_{t-k}-E(X_t))] \\
&=E[(\phi_1(X_{t-1}-E(X_t))+\epsilon_t)(X_{t-k}-E(X_t))] \\
&=E[\phi_1(X_{t-1}-E(X_t))(X_{t-k}-E(X_t))+\epsilon_t(X_{t-k}-E(X_t))] \\
&=E[\phi_1(X_{t-1}-E(X_t))(X_{t-k}-E(X_t))]+E[\epsilon_t(X_{t-k}-E(X_t))] \\
&=\phi_1E[(X_{t-1}-E(X_t))(X_{t-k}-E(X_t))]+\underbrace{E[\epsilon_t]}_{=0}E[(X_{t-k}-E(X_t))] \\
&=\phi_1E[(X_{t-1}-E(X_t))(X_{t-k}-E(X_t))] \\
&=\phi_1\gamma_{k-1}
\end{align}
And by repeated substitution:
\begin{align}
\gamma_k=\phi_1^k\gamma_0=\phi_1^kV(X_t)=\phi_1^2\frac{\sigma^2}{1-\phi_1^2}<\infty
\end{align}
If and only if $\vert \phi \vert <1$. So you see that if $\vert \phi_1\vert<1$ the mean of the process and $\gamma_k$ are finite (also $\gamma_k$ does not depent on $t$). Therefore the process is weakly stationary.
