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Assume that the non-diagonal elements of the covariance matrix are not zero. Please provide a closed form formula. I'm interested in the bivariate case in particular. How does the formula simplify in the bivariate case?

More precisely, consider the decomposition of covariance matrix $\Sigma$ into correlation matrix $R$ and diagonal matrix $S$ that lists standard deviation: $\Sigma= S R S$. I'm interested in $S$ and $R$, when $\Sigma$ is the variance of the estimate of mean of multivariate normal distribution.

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  • $\begingroup$ The estimator will be a vector (since the mean of a multivariate normal distribution is a vector), so the variance of the estimator will be a matrix. When you say standard error of this estimator, do you mean the vector composed of square roots of the diagonal of this matrix? $\endgroup$ Aug 10, 2015 at 14:07
  • $\begingroup$ Covariance matrix of the mean estimate will be always semi-positive definite symmetric matrix, no? I would like to know the full matrix. Just knowing the formula for the diagonal elements would also help. $\endgroup$
    – matus
    Aug 10, 2015 at 14:29
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    $\begingroup$ So more precisely, you need the variance-covariance matrix of the estimator, not the standard error? Then consider editing the post accordingly. $\endgroup$ Aug 10, 2015 at 14:37

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If you have an i.i.d. sample of any random vector $\{ (X_{1t}, \ldots, X_{pt}) \}_{t=1}^{n}$ where $\text{Var}(X_i) = \sigma^2_i$ and $\text{Cov}(X_i, X_j) = \sigma_{ij}$ then the covariance matrix of the mean vector $\bar{X}$ has diagonal elements $\text{Var}(\bar{X}_i) = \sigma^2_i / n$ and off-diagonals \begin{align} \text{Cov}(\bar{X}_i, \bar{X}_j) &= \frac{1}{n^2} \sum_{k=1}^{n} \sum_{l=1}^{n} \text{Cov}(X_{ik}, X_{jl}) \\ &= \frac{\sigma_{ij}}{n} \end{align} for $i \neq j$. So the covariance matrix of $\bar{X}$ is just the covariance matrix of the original vector divided by $n$.

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