Consider the confidence interval for the population variance $\sigma^2$ constructed under normality: $$\left[A_n,B_n\right]=\left[\frac{(n-1)s^2}{\chi^2_{n-1,\alpha/2}},\frac{(n-1)s^2}{\chi^2_{n-1,1-\alpha/2}}\right] $$

Supposing that our data it's non-normal, what would be the limit coverage of said interval? i.e, what's the limit: $$\lim_{n\rightarrow\infty}\mathbb{P}\left(A_n<\sigma ^2 <B_n \right)?$$

I know that: $$\chi^2_{n-1,\alpha/2}=\sqrt{2n-2}~z_{\alpha/2} + n-1 + o(1),$$

so if you substitute on the C.I expression (after a little manipulation) you end up with: $$ \mathbb{P}\left(-\sqrt{2n-2}~z_{\alpha/2} + n-1 + o(1)< \frac{(n-1)s^2}{\sigma^2}<\sqrt{2n-2}~z_{\alpha/2} + n-1 + o(1)\right).$$ Now,i also know that the limit distribution of the sample variance $s^2$ is: $$N\left(\sigma^2,\frac{\mu_4-\sigma^4}{n}\right)$$, but I fail to see how to approach the problem with this.

Edit: We can assume that that $\mu_4$ is finite

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    $\begingroup$ The sample variance doesn't necessarily have a limit distribution because generally $\mu_4$ could be infinite. The problem is that "non-normal" is not sufficiently specific to do any useful analysis. $\endgroup$ – whuber Jan 1 at 17:15
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    $\begingroup$ Yes, i forgot to mention that we assume that $\mu_4$ is finite. Knowing that $s^2$ converges into a normal there is any way to conclude anything about the subject? $\endgroup$ – Hendrik Jan 1 at 17:33
  • $\begingroup$ Yes, but wouldn't that be tautological? The only way the coverage could fail to equal the stated value would be for $s^2/n$ not to have Normal asymptotic behavior in distribution. $\endgroup$ – whuber Jan 3 at 16:01

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