Stationarity of ARMA model Let be $$y_{t} = 2 + 1.3y_{t-1}-0.4y_{t-2}+z_t+z_{t-1}$$ an ARMA process where $z_t$ is modeled as a white noise $(0,\sigma^2)$ and $y_t$ is the output. I have to verify if this ARMA process is covariance-stationary. So, I know how to calculate the mean of this process, that is simply $$\frac{2}{1-1.3+0.4}=20$$ But how can I calculate the autocovariance of this process? Do you know if exists a general formula that provides the autocovariance coefficients?
 A: We have $$(1-1.3B+0.4B^2)y_t=2+(1+B)z_t$$ where $B$ is the backward shift operator. So for this process to be stationary, the roots of the equation $$1-1.3x+0.4x^2=0$$ must lie outside the unit circle. They are $x=2,1.25$ so indeed it is causal and hence stationary.
Thus the autocovariance is $$\begin{align}\gamma_h&=\text{Cov}(y_t,y_{t+h})=\text{Cov}(y_t,2+1.3y_{t+h-1}-0.4y_{t+h-2}+z_{t+h}+z_{t+h-1})\\&=\text{Cov}(2,y_t)+1.3\text{Cov}(y_t,y_{t+h-1})-0.4\text{Cov}(y_t,y_{t+h-2})+\text{Cov}(y_t,z_{t+h})+\text{Cov}(y_t,z_{t+h-1})\\&=0+1.3\gamma_{h-1}-0.4\gamma_{h-2}+0+0\quad\color{red}{(*)}\\&=1.3\gamma_{h-1}-0.4\gamma_{h-2}\end{align}$$ for $h\ge2$ which does not depend on $t$. 
$\color{red}{(*)}:$ Since $y_{t-i}$ is a linear combination of $z_{t-i}$, $z_{t-i-1}$, ... and since $\mathbb{E}(z_{t-i-j}z_t)=\mathbb{E}(z_{t-i-j})\mathbb{E}(z_t)=0$ due to independence of white noise, we must have that $$\mathbb{E}(y_{t-i}z_t)=\mathbb{E}\left(\sum_{k=0}^\infty\delta_kz_{t-k}z_t\right)=0$$ 
Can you now finish it by calculating $\gamma_1$?
