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Let $X_{1}, ...,X_{n}$ be i.i.d random variables and let $\sigma$ denote a parameter of their common density function. It is known that $\operatorname{E}|X_i|=\sigma$ and $\operatorname{Var}|X_i|=\sigma^2$. The MLE is $\hat{\sigma} =1/n\sum_{i=1}^{n}|X_{i}|$. My question: why is $\sqrt{n}(\hat{\sigma} - \sigma)$ asymptoticly normal distributed with mean $0$ and variance $\sigma^2$? According to wikipedia that's only true when $\hat{\sigma} =1/n\sum_{i=1}^{n}X_{i}$? Will the central limit theorem still hold if you apply an arbitrary function $g(X_i)$ like in this case the absolute value function?

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    $\begingroup$ Indeed the sequence $(g(X_i)$ is i.i.d. hence the CLT says that $$\frac1{\sqrt n}\sum_{i=1}^n(g(X_i)-E(g(X)))$$ converges in distribution to a centered normal distribution. $\endgroup$ – Did Oct 4 '17 at 11:36
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Define a new sequence of random variables by $Y_n := |X_n|$ -- or more generally, $Y_n = g(X_n)$. These are now perfectly valid new random variables, so if they satisfy the hypotheses of the Central Limit Theorem, it can be applied to them.

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