# Variance and Asymptotic normality of sample variance of normal distribution

Let $X_1,X_2$,....,be a random sample from $N(q,w^2)$; $q,w$ are unknown. Let $S_n$ be the sample standard deviation.

i.e $S_n^2=\frac{1}{n-1}\sum(X_i-\bar{X})^2$

What is $Var(S_n^2)$? and how to show that $S_n^2$ is asymptotically normal?

I tried to do Variance part using moment generating functions but expressions are getting extremely complicated.

• Welcome to the site. I've improved your formatting but you should check that I haven't introduced any errors. If you want your posts to be readable you can use latex code in your questions, inserted between $signs. – Peter Ellis Apr 22 '12 at 19:41 • Is this homework? If so, you should add the "homework" tag. – Peter Ellis Apr 22 '12 at 19:41 • Do you know the finite-sample distribution of$S_n^2$? If so, you can start from there, and see what happens as$n \to \infty$. – jbowman Apr 22 '12 at 19:48 • I donot know distribution of$S_n^2$. – Questions-Math Apr 22 '12 at 20:05 • Also I need to show that$\sqrt(n)(S_n^2-w^2)->N(0,m)$for some m function of w as n->$\infty$in distribution.That is the meaning of asymptotic normality. – Questions-Math Apr 22 '12 at 20:11 ## 1 Answer I got it! I came to know that$(n-1)S_n^2/w^2$is distributed as$\chi^2_{n-1}$. From there I can calculate the variance. Now$S_n^2/w^2$is the sum of$n-1\chi^2_1$random variables divided by$(n-1)$. From central limit theorem,$\sqrt{n-1}(S_n^2/w^2 - 1)\xrightarrow{D} N(0,2)\$. Hence the asymptotic normality.

• I've cleaned up the language. The general idea is ok, but the answer needs some polish. – mpiktas Apr 23 '12 at 9:16