The problem I am dealing with is the following:
"Prove that in an $AR(2)$ process $X_t - \mu = \phi_1(X_{t-1}-\mu)+\phi_2(X_{t-2}-\mu)+Z_t$, $(Z_t \sim N(0,\sigma^2))$, the marginal distribution of the first observation has mean $\mathbb{E}(X_1) = \mu$ and $\mathbb{Var}(X_1)=\frac{\sigma^2}{1-\phi_1^2-\phi^2_2}$ and for the second observation $\mathbb{E}(X_2|X_1) = \mu+ \phi_1(X_1-\mu)$ and $\mathbb{Var}(X_2)=\frac{\sigma^2(1-\phi_1^2)}{1-\phi_1^2-\phi^2_2}$."
Useful results for an $AR(2)$ process are
- $\boldsymbol{X} \sim N(\boldsymbol{\mu},\Gamma_2)$, where $\Gamma_2 = \left(\begin{matrix} \gamma(0) & \gamma(1)\\ \gamma(1) & \gamma(0) \end{matrix}\right)$ is the variance-covariance matrix and
- $X_t|X_{t-1},X_{t-2} \sim N(\mu + \phi_1(x_{t-1}-\mu)+\phi_2(x_{t-2}-\mu), \sigma^2)$.
So, I had no problem in finding the expectations, but I do not understand how can I find the variances.
I started with $\mathbb{Var}(X_1)$:
$\mathbb{Var}(X_1) = \phi_1^2\mathbb{Var}(X_0) + \phi_2^2\mathbb{Var}(X_{-1})+\sigma^2 + 2\phi_1\phi_2\mathbb{Cov}(X_0,X_{-1})$ , and since the process is stationary, I obtain $\mathbb{Var}(X_1) = (\sigma^2 + 2\phi_1\phi_2\mathbb{Cov}(X_0,X_{-1}))/(1-\phi_1^2-\phi_2^2)$, that is not the correct answer.
Any help or suggestions in how to find the variances would be really appreciated. Thanks in advance.