Variance and Asymptotic normality of sample variance of normal distribution

Question

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, Commented Apr 22, 2012 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, Commented Apr 22, 2012 at 19:48
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, Commented Apr 22, 2012 at 20:11

mpiktas · Accepted Answer · 2012-04-23 09:12:15Z

1

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.

mpiktas

35.2k6 gold badges88 silver badges145 bronze badges

answered Apr 23, 2012 at 6:14

Questions-Math

1211 silver badge4 bronze badges

1

$\begingroup$ I've cleaned up the language. The general idea is ok, but the answer needs some polish. $\endgroup$
– mpiktas
Commented Apr 23, 2012 at 9:16

Add a comment |

1 Answer 1