# Another Chi square - s^2 defintion

• December 12th 2009, 10:57 PM
yvonnehr
Another Chi square - s^2 defintion
Dear Matheagle,
You still there? Here's another one that I've been spending too much time on. Can you please help?

Let $X_1, X_2,..., X_n$ be iid r.v.s from $Normal (\mu, \sigma)$. Given the fact that the rv

$X^2_n = \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2$

has a chi square distribution with degrees of freedom n.

Show that

$\frac{(n-1)s^2}{\sigma^2} = \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2 - \left(\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}\right)^2$.

Thanks a bunch!
• December 12th 2009, 11:00 PM
matheagle
Dear Poisson Ivy,

Use the definition of $s^2$

${(n-1)s^2\over \sigma^2} = {\sum_{i=1}^n(x_i-\bar x)^2\over \sigma^2}$

$= {\sum_{i=1}^n((x_i-\mu)+(\mu-\bar x))^2\over \sigma^2}$

$= {\sum_{i=1}^n (x_i-\mu)^2+2(\mu-\bar x) \sum_{i=1}^n(x_i-\mu)+n(\mu-\bar x)^2\over \sigma^2}$

$= {\sum_{i=1}^n (x_i-\mu)^2+2(\mu-\bar x) n(\bar x-\mu)+n(\mu-\bar x)^2\over \sigma^2}$

$= {\sum_{i=1}^n (x_i-\mu)^2-2n(\bar x-\mu)^2+n(\bar x-\mu)^2\over \sigma^2}$

$= {\sum_{i=1}^n (x_i-\mu)^2-n(\bar x-\mu)^2\over \sigma^2}$

$= {\sum_{i=1}^n (x_i-\mu)^2\over \sigma^2} -{n(\bar x-\mu)^2\over \sigma^2}$

$= \sum_{i=1}^n {(x_i-\mu)^2\over \sigma^2} -{(\bar x-\mu)^2\over \sigma^2/n}$

$= \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2 - \left(\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}\right)^2$.