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How do you find the variance of sample variance, without using moment generating functions but using the fact that variance of a chi squared distribution with n degrees of freedom, is equal to 2n?

Sample variance is defined as $$S^2=\frac1{n-1}\sum_{i=1}^n\left(X_i-\overline{X}\right)^2.$$

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Use Fisher's Lemma: $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$

So $\mathbb{D} \left( \frac{(n-1)s^2}{\sigma^2} \right) = \mathbb{D} \chi^2_{n-1} = 2(n-1)$

$\frac{(n-1)^2 \mathbb{D}(s^2)}{\sigma^4} = 2(n-1) \Rightarrow \boxed{\mathbb{D}(s^2) = \frac{2\sigma^4}{n-1}}$

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