Given $AR(2)$ stationary process
$$ y_t = 2 + 0.6y_{t-1} - 0.08y_{t-2} +u_t$$
where $u_t$ white noise from $N(0,4)$
Find $Var(y_t)$
My problem:
When I take the variances of left and right I have a kind of recursive covariance of $Cov(y_{t-1}, y_{t-2})$
Indeed, you will have two unknown variables, so you need to write two equations.
Let $Cov(y_t, y_{t+k}) = \gamma_k$.
Then $$Var(y_t) = \gamma_0 = 0.6^2 Var(y_{t-1}) + 0.08^2 Var(y_{t-2}) + 2\cdot 0.6 \cdot 0.08 Cov(y_{t-1}, y_{t-2}) = $$ $$ = 0.6^2 \gamma_0 + 0.08^2 \gamma_0 + 2\cdot 0.6 \cdot 0.08 \gamma_1 $$ $$ Cov(y_t, y_{t+1}) = \gamma_1 = 0.021\gamma_0 - 0.0018 \gamma_1 - 0.014 \gamma_1 - 0.00012 \gamma_0 $$ After solving the system of two equations you should obtain $\gamma_0, \gamma_1$.
Note that $\gamma_0 = Cov(y_t, y_{t+0}) = Var(y_t)$