Precision matrix of AR(2) matrix I am trying to construct the inverse covariance matrix of an AR(2) process of the form $X_t=\theta_1 X_{t-1}+\theta_2 X_{t-2} + \epsilon_t$ with i.i.d. $\epsilon_i$, $\mathbb{E} \epsilon_i =0$, $\mathbb{E}\epsilon_i\epsilon_j=\sigma^2 \delta_{i,j}$. We assume the process to be stationary.
I have already calculated the variance  $Var(X_t)=\frac{(1-\theta_2)\sigma^2}{(1+\theta_2)((1-\theta_2)^2-\theta_1^2)}$ and the autocorrelation coefficients $\rho_1=\frac{\mathbb{E}X_t x_{t-1}}{Var(X_t)}=\frac{\theta_1}{1-\theta_2}$, $\rho_2=\frac{\mathbb{E}X_t x_{t-2}}{Var(X_t)}=\theta_2+\frac{\theta_1^2}{1-\theta_2}$ and $\rho_s=\theta_1 \rho_{s-1}+\theta_2 \rho_{s-2}$ for $s>2$.
How do I now proceed to construct the inverse covariance matrix? I tried to calculate it analytically but it did not work out.
EDIT: For fix $T$ I am looking for the inverse matrix of the covariance matrix $\Phi$ with entries $\Phi_{ij}=Cov(X_i,X_j)$, $i, j = 1,\dots,T$. The entries of this covariance matrix are already given by the autocorrelation coefficients above since the process is assumed to be stationary.
Any help would be very appreciated!
 A: Ok, so this is not the full solution, but I will present how to get the explicit form for $\Phi$, maybe somebody figures out how to invert it. So, according to the above notation, $\Phi_{ij} = \rho_{|i-j|}$, where $\rho_0 = 1$ and $\rho_1$ and $\rho_2$ are specified in the question
We will now solve the recurrence relation $\rho_s = \theta_1 \rho_{s-1} + \theta_2 \rho_{s-2}$ with help of generating functions
Define $G(x) = \sum_{i=1}^{\infty}\rho_ix^i$, then use recurrence relation
$G(x)
= \sum_{i=1}^{\infty}\rho_{i+1} x^i 
= \rho_1 + x \rho_2 + \sum_{i=2}^{\infty}x^i \theta_1\rho_{i} + \sum_{i=2}^{\infty}x^i \theta_2 \rho_{i-1}
= \rho_1 + x(\rho_2-\theta_1\rho_1) + x \theta_1 G(x) + x^2 \theta_2 G(x)
$
Let $c_1 = \rho_2-\theta_1\rho_1 = \theta_2$. Solving for $G(x)$, we get
$$G(x)
= \frac{\rho_1 + x c_1}{1 - \theta_1x-\theta_2x^2}
= \frac{\rho_1 + x c_1}{-\theta_2 (x-k_1)(x-k_2)}
= \frac{\rho_1 + x c_1}{\theta_2 (k_1 - k_2)}(\frac{1}{x-k_2} - \frac{1}{x-k_1})
= \frac{\rho_1 + x c_1}{\theta_2 (k_1 - k_2)}(\frac{1}{k_1}\frac{1}{1-\frac{x}{k_1}} -\frac{1}{k_2}\frac{1}{1-\frac{x}{k_2}})
$$
Here we have factored the quadratic equation, resulting in solutions $k_{1,2} = \frac{1}{2 \theta_2}(-\theta_1 \pm \sqrt{\theta_1^2 + 4 \theta_2})$
Now we substitute the series $\frac{1}{1-x}=1+x+x^2+...$
$$G(x)
= \frac{\rho_1 + x c_1}{\theta_2 (k_1 - k_2)} (\frac{1}{k_1}\sum_{i=0}^{\infty} \frac{x^i}{k_1^i} - \frac{1}{k_2}\sum_{i=0}^{\infty} \frac{x^i}{k_2^i})
= (\rho_1 + c_1 x) \sum_{i=0}^{\infty}a_ix^i
$$
Where
$$a_i = \frac{1}{\theta_2(k_1-k_2)}(\frac{1}{k_1^{i+1}} - \frac{1}{k_2^{i+1}})$$
Finally,
$$\rho_{i+1} = \rho_1 a_i + c_1 a_{i-1} = \rho_1 a_i + \theta_2 a_{i-1}$$
So the next step is to try to simplify the explicit form of $\Phi_{ij}$ to see if it is somehow invertible. I'm sure there probably is a faster way to do this, if one somehow exploits some known symmetry of the matrix, but that's beyond me atm
