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.
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html