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I understand that if $X \sim N(\mu, \sigma^{2}$), then the sample mean for a sample size of N is $\overline{x} \sim N(\mu, \frac{\sigma^{2}}{N})$ but I am not sure how was this derived $ \overline{\sigma}^{2} \sim N(\sigma^{2}, \sigma^{4}\frac{2}{N-1}) $.

Furthermore, the standard error (standard deviation of the parameter) for $\overline{\sigma}$ is $SE(\overline{\sigma}) = \overline{\sigma} \sqrt{\frac{1}{2N}}$, and I didn't find derivation for this either. Most of what I found is on the standard deviation of the sample mean, or $\frac{s}{\sqrt{N}}$ for unknown population standard deviation.

Standard error of $\overline{\sigma}$ and standard error of $\overline{\sigma^{2}}$ are completely different and they do not connect right? (Is there an intuition behind this?)

Could you show me the derivation for those two quantities? Also, should it be $\overline{\sigma}$ or $\hat{\sigma}$?

Thanks!

user101998
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    The sample variance cannot be Normally distributed, as the sample variance must be nonnegative by construction but the Normal distribution has a positive probability of negative numbers. You might want to check your source... – jbowman Jun 04 '18 at 16:07
  • @jbowman I might be reading it wrong, but at the bottom of page 2 of this PDF https://hannig.cloudapps.unc.edu//STOR655/handouts/Handout-asymptotics.pdf it has that the sample standard deviation is following a normal distribution. Is there a more complete source you know that I can read? – user101998 Jun 04 '18 at 16:20
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    That's the limiting distribution as the sample size $\rightarrow \infty$; the finite sample distribution isn't Normal, though. What are you not understanding about the derivation that's given in the PDF? – jbowman Jun 04 '18 at 16:27
  • But for large sample size shouldn't the distribution be close to the limiting distribution? – user101998 Jun 04 '18 at 16:42

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