How would you go about proving that

[ ( n - 1 ) * s^2 ] / σ^2

has a Chi-Square Distribution?

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- May 13th 2010, 06:17 AMNintendo64Chi-Square Variance Ratio Proof
How would you go about proving that

[ ( n - 1 ) * s^2 ] / σ^2

has a Chi-Square Distribution? - May 13th 2010, 01:55 PMMoo
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) - May 14th 2010, 03:14 PMtheodds
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.