# Is there any quick way to derive the maximum likelihood estimator in the following example?

Let's say we are given a sample: $Z_1,Z_2,\cdots,Z_n$. We know their joint distribution is $$\begin{bmatrix} Z_1\\ Z_2\\ \vdots\\ Z_n \end{bmatrix} \sim N\left(0, \begin{bmatrix} \sigma^2+2\omega^2 & -\omega^2 & 0 & 0 &\cdots & 0\\ -\omega^2 & \sigma^2+2\omega^2 & -\omega^2 & 0 &\cdots & 0\\ 0 & -\omega^2 & \sigma^2+2\omega^2 & -\omega^2 & \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots\\ 0 & \cdots &\cdots &\cdots & -\omega^2 & \sigma^2 + 2\omega^2 \\ \end{bmatrix} \right)$$ and want to derive the maximum likelihood estimator for both $\sigma^2$ and $\omega^2$.

If we put $\Sigma$ as the covariance matrix of the sample, we may get a closed-form solution for $|\Sigma|$ and $\Sigma^{-1}$. Since it's a tridiagonal matrix, what I usually do is to use the formulas for determinants and inverse of block matrix inductively. Then we should be able to get the first order conditions and then solve for the MLE.

However, I doubt if such method is too laboursome. At the very first glance, I guess that the MLE should be the same as moment estimator. That is, $$\begin{bmatrix} \hat{\sigma}^2_{ML}\\ \hat{\omega}^2_{ML}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{n} \sum_{j=1}^n Z_j^2 + \frac{2}{n-1} \sum_{j=2}^n Z_j Z_{j-1} \\ -\frac{2}{n-1} \sum_{j=2}^n Z_j Z_{j-1} \end{bmatrix}$$ Now my question is: if my guess is correct, is there any quick way to derive the maximum likelihood estimator than to do the tedious derivation? Or maybe I'm wrong, that is, $|\Sigma|$ and $\Sigma^{-1}$ can be easily solved?

Any help will be greatly appreciated!