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I am trying to get a understanding of how good the estimation is of my covariance matrix. Suppose I have random variables X1,..., X10 and they are all iid $N(\mu, \Sigma)$ where $\Sigma$ is a NxN matrix and $\mu$ is a Nx1 vector.

I collect a sample of size K for each $X_i$, i.e. I have a Kx10 matrix of observations. I use this Kx10 data matrix to compute the sample estimator of the covariance $\hat{\Sigma}$ (using the plain vanilla sample covariance estimator).

What is the formula that I can use to calculate the confidence interval for the covariance? Is the distribution for each cov(xi,xj) Chi-squared distributed?

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    $\begingroup$ Very closely related: stats.stackexchange.com/questions/59478. The issue here is whether you want individual confidence intervals for each of the $N(N-1)/2=45$ covariances or a single, simultaneous confidence region for them all. $\endgroup$
    – whuber ♦
    Commented Jul 17, 2023 at 20:11
  • $\begingroup$ Please also tell us the context of this problem (by editing the post, not only as a comment). A confidence region for a $10 \times 10$ covariance matrix is unlikely to be helpful! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Jul 17, 2023 at 21:21
  • $\begingroup$ Thanks for linking to 59478. I will check it out. $\endgroup$
    – user2337857
    Commented Jul 18, 2023 at 7:11
  • $\begingroup$ @whuber In the context of t-test on many mean, it there exists test that allow us to test many different means, i.e. ANOVA. When it comes to the correlation matrix, is it best practise to test all N(N-1)/2 individually? or is there a signle simultaneous test available similar to ANOVA but for covariances? $\endgroup$
    – user2337857
    Commented Jul 18, 2023 at 13:48
  • $\begingroup$ I have never worked out the result because the assumption that ten variables are multivariate Normal is both crucial and, in most situations, implausible. But in general, it's difficult to test all the values individually since (a) you need to apply a multiple testing adjustment but (b) it's hard to determine what it should be due to the strong associations among the covariances. One approach is to apply an omnibus multiple testing correction, like Bonferroni, and if the resulting (highly conservative) intervals are acceptable, then you don't have to work any harder. $\endgroup$
    – whuber ♦
    Commented Jul 18, 2023 at 14:34

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