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Let $\Sigma$ be the population covariance matrix and $\hat{\Sigma}$ be the sample covariance matrix. It is well known that $\hat{\Sigma} \rightarrow \Sigma$ in the large sample limit.

I have also heard that $\hat{\Sigma}^{-1} \rightarrow \Sigma^{-1}$ also holds, but I am not sure how to prove it. Does anyone know how?

To me, this seems similar to arguments involving continuous functions, i.e., if $x_n \rightarrow x$ then $f(x_n) \rightarrow f(x)$.

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  • $\begingroup$ I think, since covariance matrix is invertible, thus we will have a new covariance matrix which preserve specifications of a covariance matrix. Thus, You can consider a new population with a new covariance matrix $\endgroup$ – Cardinal Jul 29 '15 at 22:27
  • $\begingroup$ Notice that the set of nonsingular matrices is open and inversion is smooth on that set. So if $\Sigma$ is positive definite, apply a variant of continuous mapping theorem according to the convergence mode of $\hat\Sigma\to\Sigma$. $\endgroup$ – user251257 Jul 30 '15 at 1:10

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