Distribution of the sample variance

Question

This is my first post to this great website :) It seems like an excellent place to learn. I have a question however that is bothering me as I cannot figure it out through my textbook.

The sample variance $$ S^2 = \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)^2 $$ Of course, we need to assume that $X_i$ is normal but here is where I am very confused. The next step in my notes shows this:

$$X_i - \bar{X} = \sigma [(X_i - \mu) / \sigma - (\bar{X} - \mu) / \sigma]$$

Where did $\sigma$ come from? And can anyone help me with the proof of this? It appears that the distribution of the sample variance is that of a Chi-Squared R.V. with $n-1$ DoF. Thanks so much!

Answer 1

The statement you've got is true for every $\sigma$ that isn't zero, since you multiplied by $\sigma/\sigma=1$ and that doesn't change the value. $\sigma$ is probably going to end up representing some sort of standard-deviation-like object, but for now it doesn't really matter.

To be a bit clearer, you can simplify the right-hand side into the left-hand side by combining the fractions (which cancels the $\mu$) and then multiplying the $\sigma$ through (which also cancels).

Answer 2

Your second display equation has nothing to do with a normality or any other assumption. As long as $0<\sigma<\infty$ it holds. All you're doing is to multiply by $\sigma$ and to divide by $\sigma$ and to add and subtract $\mu$.

You do need the normality assumption, however, to establish the $\chi_{n-1}^2$ distribution. Use the fact that the sum of the squares of $n-1$ independent normals has a $\chi_{n-1}^2$ distribution.