= Central limit theorem for the sample variance
{title2=$\sqrt n(S_n^2-v)\Rightarrow N(0,\mu_4-v^2)$}
Let $X_i$ be real <independent and identically distributed random variables>, with <mean> $m$, <variance> $v$ and finite fourth <central moment> $\mu_4=\mathbb E[(X_i-m)^4]$. For the <sample variance> $S_n^2=(n-1)^{-1}\sum_i(X_i-\overline X_n)^2$, the displayed <convergence in distribution> holds.
Indeed $n^{-1}\sum_i(X_i-\overline X_n)^2=n^{-1}\sum_i(X_i-m)^2-(\overline X_n-m)^2$. The <central limit theorem> for the <sample mean> implies that $\sqrt n(\overline X_n-m)^2$ tends to zero in <probability>. Changing the divisor from $n$ to $n-1$ also has negligible effect at this scale, by the <law of large numbers>. Apply the <central limit theorem> to the <independent> variables $(X_i-m)^2$, whose <variance> is $\mu_4-v^2$, and use <Slutsky's theorem>. Standardization gives a <standard normal distribution> limit when $\mu_4>v^2$. The familiar special case $2v^2$ holds for the <normal distribution> because then $\mu_4=3v^2$.
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