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I started with any distribution and underwent the CLT on $\sqrt{n}(\widehat{\sigma}^2 - \sigma^2)$ where

$$ \widehat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2 $$ is a sample mean of $\sigma^2$. This leads to a normal distribution $\sqrt{n}(\widehat{\sigma}^2 - \sigma^2) \sim N(0, \mu_4 - \sigma^4)$. Where $\mu_4 = \mathbb{E}((X_i - \mu)^4)$. I know that this can be "rearranged" to be $\frac{\sqrt{n}(\widehat{\sigma}^2 - \sigma^2)}{\sqrt{\mu_4 - \sigma^4}} \sim N(0,1)$ and I used this to try and calculate a 95% confidence interval for $\sigma^2$ by $\mathbb{P}(\widehat{\sigma}^2 - \frac{1.96 \sqrt{\mu_4 - \sigma^4}}{\sqrt{n}} < \sigma^2 < \widehat{\sigma}^2 - \frac{1.96 \sqrt{\mu_4 - \sigma^4}}{\sqrt{n}}) = 0.95$ but when calculating $\sqrt{\mu_4 - \sigma^4}$ on R by estimating $\mu_4$ with a modified version of the 4th sample moment $= \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^4$ as I know $\mu$, it sometimes came about that the estimate of $\mu_4 < \sigma^4$ so the program was trying to square root a negative number.

Any help on how to properly calculate $\mu_4$ or how to manipulate the above so that I am not rooting a negative number?

Thanks

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  • $\begingroup$ Well you mess up the range! You should use $N^2(0,1)\sim \chi^2(1)$ $\endgroup$ – TPArrow Dec 3 '18 at 13:17
  • $\begingroup$ Thanks for your response @TPArrow, I tried using this and when working out the confidence interval still ended up with a $\sqrt{\mu_4 - \sigma^4}$ being calculated in the pivot. $\endgroup$ – UCLstudent420 Dec 3 '18 at 14:01
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From $\frac{\sqrt{n}(\widehat{\sigma}^2 - \sigma^2)}{\sqrt{\mu_4 - \sigma^4}} \sim N(0,1)$.

$$-Z\lt \frac{\sqrt{n}(\widehat{\sigma}^2 - \sigma^2)}{\sqrt{\mu_4 - \sigma^4}} $$

$$Z^2({\mu_4 - (\sigma^2)^2)}\lt {n(\widehat{\sigma}^2 - \sigma^2)^2} $$ $$........$$ $$[(n+Z^2)](\sigma^2)^2 -[2n\hat\sigma^2]\sigma^2+[n(\hat\sigma^2)^2-Z^2\mu_4] > 0$$

Then following $ax^2 + bx +c =0$ to get the solution.

Finish $\frac{\sqrt{n}(\widehat{\sigma}^2 - \sigma^2)}{\sqrt{\mu_4 - \sigma^4}} <Z$.

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  • $\begingroup$ Thank you for your answer, but I can't see how this is useful, would you be able to add a bit more explanation to your answer, for example what $Z$ is? Many thanks $\endgroup$ – UCLstudent420 Dec 3 '18 at 15:03
  • $\begingroup$ Z = 1.96 if alpha = 0.05. In your original approach, $\widehat{\sigma}^2 - \frac{1.96 \sqrt{\mu_4 - \sigma^4}}{\sqrt{n}} < \sigma^2$ does not work, because $\sigma^2$ appears in both sidez. $\endgroup$ – user158565 Dec 3 '18 at 15:06
  • $\begingroup$ That is helpful, but do you know how to calculate $\mu_4$? $\endgroup$ – UCLstudent420 Dec 3 '18 at 15:10
  • $\begingroup$ $\frac{1}{n} \sum_{i=1}^n (X_i - \bar X)^4$ $\endgroup$ – user158565 Dec 3 '18 at 15:13

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