mse of non-normal variance estimator

Question

Suppose you have a general distribution, and you have a sample of iid random variable $(X_1,X_2,...X_n)$.

We all know that the unbiased variance estimator is $$\hat {\sigma^2} = \frac{\sum_{i=1}^n (X_i - \hat \mu)^2}{n-1} $$ Where $\hat \mu = \frac{\sum_{i=1}^n X_i}{n}$

But how to calculate the variance of $\hat {\sigma^2}$?

Thank you so much!

1

Maybe this Q&A will help. – BruceET Oct 15 '19 at 08:22

score 0 · Answer 1 · edited Oct 15 '19 at 08:19

0

You have to use the chisquare distribution for that. $$(n-1)S^2/\sigma^2\sim\chi^2(n-1)$$

edited Oct 15 '19 at 08:19

BruceET

51,500

answered Oct 15 '19 at 07:26

October15

232

1

But this is only valid for the normal distribution random variable, I think the general distribution random variable is much more complicated – Dorian Oct 15 '19 at 18:02

mse of non-normal variance estimator

1 Answers1