estimating variance of dependent normal distributed data

Question

Let $X_{ij}$ with $1\leq i<j\leq n$ (that are $X_{12},\dots, X_{1n},\dots,X_{(n-1)n}$) be ${n \choose 2}$ identically normal distributed $N(0,\sigma^2)$ such that $ \text{corr}(X_{ij},X_{rs})=\rho $ if $|\{i,j\}\cap\{r,s\}|=1$ and $0$ if $|\{i,j\}\cap\{r,s\}|=0$.

I'd like to estimate $\sigma^2$ if in the large sample setting. I have proved that $\hat\sigma^2=\frac1{{n\choose 2}}\sum_{i=1}^n\sum_{j=i+1}^nX_{ij}^2$ converges to $\sigma^2$ in probability. But, how can I know that this estimator is the best estimator that we can do? For example, in the efficiency setting, do we have $\sqrt{{n\choose 2}}(\hat\sigma^2-\sigma^2)\to N(0,\tau^2)$ for some $\tau$? Any help will be appreciated.

Answer 1

Not an answer, but no comments for me.

I don't think the problem is fully specified. You've given marginal distributions and second moments, but no joint distribution. For example, if 0 < rho < 1/2, there is a multivariate normal with these properties; X_ij = Y_i + Y_j + e_ij, where the Y's and e's are mutually independent mean 0 normals, with variances producing sigma^2 and rho. Is this what you intended? If so, standard likelihood theory should produce "best" estimators. If not, you should same more about the joint distribution in order to think about what is best.

Answer 2

Your problem is not meaningful in the currently given formulation because the covariance matrix you described is not a valid covariance matrix in general.

Covariance matrices of multivariate Gaussian variables are always positive semidefinite. Let us consider the example of $n=4$. The covariance matrix $C$ of the variables $X_{1,2}, X_{1,3}, X_{1,4}, X_{2,3}, X_{2,4}, X_{3,4},$ is

$$C = \left[ \begin{array}{cccccc} \sigma^2 & \rho & \rho & \rho & \rho & 0 \\ \rho & \sigma^2 & \rho & \rho & 0 & \rho \\ \rho & \rho & \sigma^2 & 0 & \rho & \rho \\ \rho & \rho & 0 & \sigma^2 & \rho & \rho \\ \rho & 0 & \rho & \rho & \sigma^2 & \rho \\ 0 & \rho & \rho & \rho & \rho & \sigma^2 \end{array} \right] $$

If you take $\sigma^2 = 1$ and $\rho = 0.9$, then $\lambda_{1,2} = -0.8$ are two negative eigenvalue.