# Confidence interval for $\sigma^2$

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

• Well you mess up the range! You should use $N^2(0,1)\sim \chi^2(1)$ – TPArrow Dec 3 '18 at 13:17
• 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. – UCLstudent420 Dec 3 '18 at 14:01

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}} .

• 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 – UCLstudent420 Dec 3 '18 at 15:03
• 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. – user158565 Dec 3 '18 at 15:06
• That is helpful, but do you know how to calculate $\mu_4$? – UCLstudent420 Dec 3 '18 at 15:10
• $\frac{1}{n} \sum_{i=1}^n (X_i - \bar X)^4$ – user158565 Dec 3 '18 at 15:13