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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.

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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.

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  • $\begingroup$ First I'd like to thank for your answer. I don't know what the joint density is. But, we can assume it is also normal. In this case, do you have an advice? $\endgroup$ – Jlamprong Mar 18 '14 at 8:54
  • $\begingroup$ If they are multivariate normal, the answer to your asymptotic normality question is no. sigma^2 is a linear combination of var(Y) and var(e) above. From the representation above, information about var(Y) accumulates only at rate sqrt(n), So you'll probably see asymptotic normality with a fourth root of n normalization. $\endgroup$ – guest Mar 18 '14 at 9:57
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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.

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  • $\begingroup$ Thanks @kolixx for your answer. So, do you have an advice or suggestion about my problem above? $\endgroup$ – Jlamprong Mar 18 '14 at 8:52
  • $\begingroup$ I don't know the situation where your problem formulation comes from. This problem, as you propose it here is just not meaningful as it is. If you can describe the source of the problem, it might help. It might be that there are extra entries in the covariance matrix, or there are restrictions on the value of $\rho$. This should be cleared up. $\endgroup$ – kolixx Mar 18 '14 at 9:41
  • $\begingroup$ It is a problem from myself. OK, thanks for your time $\endgroup$ – Jlamprong Mar 18 '14 at 14:44

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