Thanks. This is sort of what I have been trying to do... but I keep
ending up with products of non-independent chi-squares where I am sort
of getting stuck. I knew this Theorem just never knew it was called
Cochran's Thm. Btw, do you know of a good book that deals with
multivariate statistics using vector notation etc. All the books I have
seem to be focused on scalar random variables and they don't even
mention it.)
thanks, eugene.
Spencer Graves wrote:
Have you considered Cochran's theorem? (A Google search just
produce 387 hits for this, the second of which
"http://mcs.une.edu.au/~stat354/notes/node37.html"; provided details
that might help.) By construction, P is n x n, idempotent of rank k,
so y'Py is chi-square(k). Also, xA is an n-vector in the (rank k)
column space of x; indeed, PxA = [x*inv(x'x)*x]xA = xA. I can't see
the details now but I believe you can write (A'x'y)^2 = y'xAA'x'y as a
weighted sum of k independent chi-squares each with one degree of
freedom (since x and P have rank k), and then get what you want from
the sum of the weights. Then check your result using Monte Carlo.
hope this helps. spencer graves
Eugene Salinas (R) wrote:
Hi everyone,
(This is related to my posting on chi-squared from a day ago. I have
tried simulating this but I am still unable to calculate it
analytically.)
Let y be an n times 1 vector of random normal variables mean zero
variance 1 and x be an n times k vector of random normal variables
mean zero variance 1. x and y are independent.
Then P is the projection matrix P=x*inv(x'*x)*x'
I need to figure out the covariance
Cov ( y'*P*y , (A'*x'*y)^2 ) where A is a constant of dimension k
times 1.
thanks, eugene.
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