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May I use Fisher information for constructing confidence interval for MLE variance estimator?

Fisher information should be: $I(\theta) = E[I_0(\theta,Y)] = E\left[ \frac{m}{2(\sigma^2)^2} \right] = \frac{m}{2(\sigma^2)^2}$

Variance of estimated parametr is : $V(\hat{\theta}) = I(\theta)^{-1} = \frac{2(\sigma^2)^2}{m}$

Supposing the sample size is large enough, may I use following equation for 95% CI? $$ \sigma^2 \in \hat{\theta} \pm z_{\frac{\alpha}{2}} \sqrt{V(\hat{\theta})} $$

Durden
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casstel
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    The sampling distribution of $\sigma^2$ follows a $\chi^2$ distribution, not a standard normal. Note that the $\chi^2$ distribution is not symmetric (which means the confidence interval cannot be written $\pm$ some value). The 95% confidence interval for your estimate of $\sigma^2$ should look something like this. – Durden Mar 21 '24 at 15:25
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    You may, but you'll need a pretty substantial sample size - in the hundreds - for that to be (reasonably, IMO) accurate. As @Durden says, we know the exact sampling distribution of $\sigma^2$ in this case, so you'd do better to use that. – jbowman Mar 21 '24 at 15:27

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