I can accept the fact that $Z^2 = \dfrac{\left(X-\mu\right)^2}{\sigma^2} \sim \chi^2(1)$ without knowing too much about this mysterious $\chi$-function, but I'm wondering how I can show that $(n-1)\dfrac{S^2}{\sigma^2} \sim \chi^2(n-1)$ from this?
Let $X_1 + X_2+\cdots+X_n$ be independent normally distributed variables:
$S = \dfrac{1}{n-1}\sum_{i=1}^n \left(X_i-\bar X\right)^2$$W^2 = \dfrac{(\sum_{i=0}^n X_i -\bar X)^2}{\sigma^2} = \;? = \dfrac{n-1}{n-1}\dfrac{\sum\left(X_i-\bar X\right)^2}{\sigma^2} = \dfrac{(n-1)S^2}{\sigma^2} \overset{?}{\sim} \chi^2(n-1)$
I'm assuming some of this is wrong, and I don't really understand the argument of $\chi$ (why isn't it $\chi(n)$?), but I'm mostly interested in seeing a better derivation of the last result, if there is one.
Any help is appreciated!
