Chi squared distribution
By David Parkin on Thursday, December 05, 2002 - 10:16 pm:

Can anybody help us with a CLEAR, ACCESSIBLE discussion of why sample variances (s2 ) have a Chi Squared distribution. Some "A level" textbooks avoid the topic or talk gobledegook!

Nailsea School Further Maths Group

By William Hall on Friday, December 06, 2002 - 12:40 am:

David,

Are you aware of the fact that a Chi Squared distribution with n degrees of freedom is defined by the distribution of the sum

Z1 2 + Z2 2 + ... + Zn 2

where each of the Zi are N(0,1) random variables and are independent of each other? This should help you in getting to your result.

Bill

By William Hall on Friday, December 06, 2002 - 11:27 am:
Just to add to this (sorry should have added this last night!) that proving the result properly is actually rather complicated - the full proof involves making a transformation of the random sample variables X1 , X2 , ., using a matrix. If you know about matrices, I'll try and go through it; if not it's probably best you just accept it for now (unless there is another way of proving it which I don't know!) The above gives you an idea as to why it might be that the sample variance is chi squared; but if you look carefully you should see that the random variables in the sum are not independent.

The full result is that if your population variance is σ2 , then S2 / σ2 is Chi squared with n-1 degrees of freedom, if you wanted to know.

Bill

By Kerwin Hui on Friday, December 06, 2002 - 12:08 pm:

One further point: the sample variance is chi-square when the underlying variable is normally distributed. The result is, in general, not true otherwise, e.g. a random variable which takes a constant value will always have sample variance 0.

Kerwin

By William Hall on Friday, December 06, 2002 - 12:12 pm:

Thanks Kerwin, that's totally necessary!