ARMA (1,2) model - Auto covariance function I am struggling with finding the Autocovariance function $\gamma(k)$, of the following ARMA(1,2) model:
$x_t-0.9x_{t-1}=e_t+2e_{i-1}+0.5e_{t-2}$.
I have already found this model to be stationary, hence the auto-cov. will only depend on the difference in time, $k$.
where $e_t$ is a regular white noise process with variance $\sigma_e^2=0.2^2$.
I have started out with first finding the Auto-cov for $k>2$, which gives that
$\gamma(k)=0.9\gamma(k-1)$, for $k>2$.
I am however struggling with finding the variance (k=0) and the auto. cov. for $k=1,2$.
Thanks in advance.
 A: So this is one way to compute the equilibrium variance of the process. Note that the equilibrium mean of the process is zero, so that $\mu_x=0$. 
Iterating the finite difference equation you find: 
$x_t=\epsilon_t+2 \epsilon_{t-1}+0.5 \epsilon_{t-2}+\\
+0.9(\epsilon_{t-1}+2 \epsilon_{t-2}+0.5 \epsilon_{t-3})+\\
+0.9^2(\epsilon_{t-2}+2 \epsilon_{t-3}+0.5 \epsilon_{t-4})+\\
+0.9^3(\epsilon_{t-3}+2 \epsilon_{t-4}+0.5 \epsilon_{t-5})+...$
And now you can collect terms, looking for a pattern:
$x_t=\epsilon_t+(2+0.9) \epsilon_{t-1}+\\
+(0.5+0.9*2+0.9^2)\epsilon_{t-2}+\\
+0.9*(0.5+0.9*2+0.9^2)\epsilon_{t-3}+\\
+0.9^2*(0.5+0.9*2+0.9^2)\epsilon_{t-4}+..\\$
Note that this is a MA representation of the process.
Starting from here it is easy to compute the variance of $x_t$, using that the shocks are all independent and summing the resulting geometric sequence. The variance of the sum of independent terms is just the sum of the variances):
$E[x_t^2]=\sigma^2_x=\sigma^2_\epsilon(1+2.9^2+\frac{(0.5+0.9*2+0.9^2)^2}{1-0.9^2})$
, where I leave out the final calculation on the calculator.
Once you have the equilibrium variance of the process, $\sigma^2_x$, it is easy to compute the time-lag 1 covariance just by squaring and taking expectations of both members in the following equation:
$x_t-0.9x_{t-1}=\epsilon_t+2 \epsilon_{t-1}+0.5 \epsilon_{t-2}$
, which gives:
$\sigma^2_x(1+0.9^2)-1.8E[x_t x_{t-1}]=\sigma^2_\epsilon(1+4+0.25)$
so that:
$\gamma(1)=\frac{1.81\sigma^2_x-5.25\sigma^2_\epsilon}{1.8}$
The same trick applies for getting the time-lag 2 covariance, writing $x_t-0.9^2x_{t-2}$ as a sum of shocks, squaring and taking expectations. After that, one can iterate with the already written recurrence relation.
Hope this was helpful!  
