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I'd like to find the asymptotic distribution of

$$\sqrt{n}\left(\log|\mathbf{S}| - \log|\boldsymbol{\Sigma}|\right), ~~~~~n \rightarrow \infty$$

where $\mathbf{S} \sim W_j\left(n, \frac{\boldsymbol{\Sigma}}{n}\right)$ is a multivariate Wishart random variable, and $\boldsymbol{\Sigma}$ is the covariance matrix.

What I have so far is that

$$\sqrt{n}\left(\log|\mathbf{S}| - \log|\boldsymbol{\Sigma}|\right) = \sqrt{n}\log\frac{|\mathbf{S}|}{|\boldsymbol{\Sigma}|}$$

A similar result is that if $\mathbf{X} \sim W_j\left(n, \boldsymbol{\Sigma}\right)$ then $\frac{|\mathbf{X}|}{|\boldsymbol{\Sigma}|} = Y_1Y_2\cdots Y_m$, where $Y_i \sim \chi_{n - i + 1}^2$ and independent (I omit the details).

So, using this result

$$\sqrt{n}\left(\log|\mathbf{S}| - \log|\boldsymbol{\Sigma}|\right) = \sqrt{n}\log\left(\prod_{i=1}^m Y_i\right) = \sqrt{n}\sum_{i=1}^m\log Y_i.$$

For a Chi-squared distribution, say $H \sim \chi_k^2$, by the CLT

$$\frac{H - k}{\sqrt{2k}} \rightarrow N(0,1),~~~~~k \rightarrow \infty$$

I believe that, $\log H$, would also tend towards a normal distribution, but I'm not quite sure how to actually show this via CLT, and then apply it to my problem above.

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    $\begingroup$ Note that $log|S|$ is the sum of the eigenvalues of a Wishart. $\endgroup$ – Xi'an Nov 21 '19 at 9:51

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