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Suppose you have a general distribution, and you have a sample of iid random variable $(X_1,X_2,...X_n)$.

We all know that the unbiased variance estimator is $$\hat {\sigma^2} = \frac{\sum_{i=1}^n (X_i - \hat \mu)^2}{n-1} $$ Where $\hat \mu = \frac{\sum_{i=1}^n X_i}{n}$

But how to calculate the variance of $\hat {\sigma^2}$?

Thank you so much!

Dorian
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    Maybe [this Q&A](https://math.stackexchange.com/questions/2476527/variance-of-sample-variance) will help. – BruceET Oct 15 '19 at 08:22

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You have to use the chisquare distribution for that. $$(n-1)S^2/\sigma^2\sim\chi^2(n-1)$$

BruceET
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October15
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    But this is only valid for the normal distribution random variable, I think the general distribution random variable is much more complicated – Dorian Oct 15 '19 at 18:02