Chi-Square Variance Ratio Proof

May 2010
3
0
How would you go about proving that

[ ( n - 1 ) * s^2 ] / σ^2


has a Chi-Square Distribution?
 

Moo

MHF Hall of Honor
Mar 2008
5,618
2,802
P(I'm here)=1/3, P(I'm there)=t+1/3
Hello,

Some information necessary about s^2
Anyway, I can just guess what it is. Note that aX~N(0,a²), where X~N(0,1)
Now think about what happens if you divide a normal distribution N(0,σ²) by σ² !

And finally, remember that a chi-square distribution is the sum of squares of iid random variables following a N(0,1)
 
Oct 2009
340
140
One way to get at this is to assume \(\displaystyle \sigma^2 = 1\), prove it for the case n = 2, then induct by showing that

\(\displaystyle (n - 1) S^2_n = (n - 2) S^2_{n - 1} + \left(\frac{n - 1}{n}\right)(X_n - \bar{X}_{n - 1})^2\),

which will get you where you need since you will have the sum of two independent chi-squares. Some facts that you will use are that the sum of independent chi-squares is chi-square, that the square of an \(\displaystyle N(0, 1)\) is \(\displaystyle \chi^2_1\), and that \(\displaystyle \bar{X}_n \perp S^2_n\). The definitions are:

\(\displaystyle S^2 _n = \frac{\sum_{i = 1} ^ n (X_i - \bar{X}_{n})^2}{n - 1}
\)
\(\displaystyle \bar{X}_n = \sum_{i = 1} ^ n \frac{X_i}{n}\)

Once you have this, it's just a matter of undoing the assumption that \(\displaystyle \sigma^2 = 1\), which isn't really a big deal. To be honest, if this is just for kicks, it's actually more of a pain in the head than it appears, particularly in proving the identity above.
 
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