Partition theorem for the chi-square distribution

Hi all,

So I've reading several textbooks this morning and they all have a very
complicated proof of the "partition theorem for the chi-square
distribution". It's when one doesn't know the exact mean of a normal
variable and estimates the variance with the estimate of the mean.

So if u_i are N(0,1) then we want to know the distribution of
Q = sum [(u_i - u)^2]

where u = 1/n sum [u_i]

So Q is chi-square with n-1 degrees of freedom, right?

By putting the u_i's in a vector U, it can be seen that Q = U' A' A U,
where A is a (n,n) matrix equal to I - 1/n F (where I is the identity
matrix, 'eye(n)' in Matlab and F is a matrix full of ones, 'ones(n)' in
Matlab). If we diagonalize A'A, we get A'A = V'DV where V is an
orthogonal matrix and D contains a single zero and n-1 ones. Then Q =
U' V' D V U = W' D W. The vector W still contains iid N(0,1) variables,
since V is an orthogonal transform, and therefore Q is chi-square(n-1).

Well, this "proof" is much shorter than what I read, so I am suspicious
about it and I would like you guys to give me feedback. I posted a pdf
with a little more detail (and it's easier to read too).

Thanks in advance,
Tony Bruguier

http://antoine.caltech.edu/chi-square.pdf

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