# proof of a property of the Chi-Squared distribution

• Jun 4th 2011, 08:07 AM
cooltowns
proof of a property of the Chi-Squared distribution
Hi

I was wondering if someone could please provide me with the proof of the following property of a Chi-Squared distribution.

If $X_{1},X_{2},..........,X_{n}$are random observations from a normal distribution with parameters $\mu$ and $\sigma$, then the statistic

$\frac{1}{\sigma^{2}}\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}\approx\chi_{n-1}^{2}$where $\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$is the sample mean

Thanks
• Jun 4th 2011, 08:34 AM
theodds
There are a couple of proofs of this result. The quickest uses a decent amount of linear algebra.

A less efficient proof first establishes the result for $n = 2$ and then proceeds by induction, using the two identities:

$\displaystyle \bar X_n = \frac{X_{n} + (n-1)\bar X_{n - 1}}{n}$

and

$\displaystyle (n - 1) S^2 _n = (n - 2)S_{n-1} ^2 + \frac{n-1}{n}(X_n - \bar X_{n - 1})^2$

where $\bar X_k = \sum_{i = 1} ^ k k^{-1} X_i$ and $S^2 _k = \frac 1 {k - 1}\sum_{i = 1} ^ k (X_i - \bar X_k)^2$. The result follows from the second identity after division by $\sigma^2$ because has been expressed as the sum of a $\chi^2_{n - 2}$ and an independent $\chi^2_1$. I think I've given enough for you to fill in the gaps in terms of what needs you actually need to show to complete this argument (that the relevant terms are independent as well as that the second term on the right hand side of the second identity is, indeed, and $\chi^2_1$).

The second identity is quite a pain to derive. Lots of sticky algebra. The first identity is just used to help prove the second one.