I have a question that I cannot answer myself. I am close and seem to know the answer, but I do not know how to get there. My question is closely related to this question. That is, let's call the sampled values from a multivariate normal distribution $X_i$. What is an estimator of the population variance of the sampled values $X_i$? I assume that all the means $\mu$ are the same, there is one common variance $\sigma_i^2$, and all covariances $\sigma_{ij}$ are the same.
The linked question states that you can take the expectation of the sampling variance, which seems to make sense to me. That is, we need to solve
$$\begin{align*} \mathbb{E} \left( \frac{1}{k} \sum_{i=1}^k \left( X_i - \frac{1}{k} \sum_{j=1}^k X_j \right)^2 \right) &= \frac{1}{k} \sum_{i=1}^k \mathbb{E}(X_i^2) - \frac{2}{k^2} \sum_{i=1}^k \sum_{j=1}^k \mathbb{E}(X_i X_j) + \frac{1}{k^2} \sum_{j=1}^k \sum_{\ell=1}^k\mathbb{E}(X_j X_{\ell}). \\ &= \frac{1}{k} \sum_{i=1}^k \mathbb{E}(X_i^2) - \frac{1}{k^2} \sum_{i=1}^k \sum_{j=1}^{k} \mathbb{E}(X_i X_j). \end{align*}$$ where we sum over $i=1, ..., k$.
However, I do not know how to take the expectation in case of a multivariate normal distribution. I know how to do this for a (univariate) normal distribution but not for a multivariate. I think the answer is $\sigma_i^2 - \sigma_{ij}$, but I am not totally sure.
Thank you in advance for your help!