[R] help computing a covariance
spencer.graves at pdf.com
Fri Jul 2 19:23:54 CEST 2004
Another thought: Do you really need A to be an arbitrary k-vector?
Might it have other structure that you can use? spencer graves
Have you tried simulating several special cases? The right set of
simulations should suggest an answer, which might then lead you to a
hope this helps. spencer graves
Eugene Salinas (R) wrote:
> 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
>>> 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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