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Let $S_n^2\sim\mathcal{\chi_n}^2$ and let $\hat\Sigma_n$ be an estimator of $\Sigma\in\mathbb{R^p}$ such that $(n-1)\hat{\Sigma}_n\sim W(\Sigma,n-1)$ (Wishart distribution) for some $n>p $.

a)Show that $\mathcal{L}(\sqrt nlog(\frac{S_n^2}{n}))$ onverges weakly to $\mathcal{N}(0,2)$.

I know that $$S_n^2=X_1^2+...+X_n^2$$ where $X_i\sim \mathcal{N}(0,1)$. After that I don´t what argument apply to proof a). I don´t know any result to analyze the convergence of $\sqrt{n}log(\frac{S_n}{n})$ Maybe I can use the central limit theorem but I don´t know how to apply in this case.

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    $\begingroup$ The information about $\hat{\Sigma}_n$ and $\Sigma$ seems redundant and disturbing $\endgroup$ – Zhanxiong Oct 30 '17 at 1:01
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If you apply the Central Limit Theorem to $$S_n^2=X_1^2+...+X_n^2$$ you get that $$\frac{1}{\sqrt{n}}(X_1^2+...+X_n^2-n)\stackrel{\text{dist}}{\longrightarrow}\mathcal{N}(0,2)$$ since $\mathbb{E}[X_1^4]=2$. Meaning that $S_n^2/n$ is asymptotically $\mathcal{N}(1,2/n)$.

Now if you apply the continuous mapping theorem, $\log\{S_n^2/n\}$ is asymptotically $\log\{\mathcal{N}(1,2/n)\}$, which is equivalent to a $$\mathcal{N}\left(\log\{1\},\left[\frac{d\log(\mu)}{d\mu}\right]_{\mu=1}^2 2/n\right)$$ by a Taylor expansion.

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