2

Edited to give the answer... but I still don't understand where it came from!

Suppose we have $$X_1, X_2,..., X_n \overset{i.i.d.}{\sim} N(0, \Omega^{-1})$$ where $\Omega \in \mathbb{R}^{2 \times 2}$ is the precision matrix, the mean is known to be zero, and the covariance matrix $\Sigma = \Omega^{-1}$ is unknown. We estimate the covariance matrix by $$\hat{\Sigma} = \frac{1}{n} \sum_{i = 1}^{n} X_iX_i^T$$ and let $\hat{\Omega} = \hat{\Sigma}^{-1}$. It is the case that $$\sqrt{n}(\hat{\Omega}_{12} - \Omega_{12}) \overset{d}{\to} N(0, \Omega_{11}\Omega_{22} + \Omega_{12}^2)$$ but I have no clue why. This convergence obviously suggests that we use a central limit theorem, but the formula for $\hat{\Omega}_{12}$ is $$\hat{\Omega}_{12} = \frac{-\hat{\Sigma}_{12}}{\det(\hat{\Sigma})}$$ which isn't the sample mean of anything. Any input would be appreciated.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
  • Be careful: your formula for the sample covariance matrix is incorrect. Do you really want to ask about it as defined by you or do you want to ask about the correct one, in which the *sample mean* is first subtracted from each $X_i$? – whuber Feb 17 '20 at 14:15
  • 1
    I should've clarified that we assume the true mean is known to be zero, so the covariance matrix that I wrote isn't technically the sample covariance, but rather an estimator. I'll update the original post –  Feb 17 '20 at 19:27

0 Answers0