Re: OT: finding the weights used in weighted least squares regression

[James Sherman Jr.]

On Wed, Apr 27, 2011 at 1:19 AM, Kamaraju S Kusumanchi <address@hidden> wrote:

Ben Abbott wrote:
>
> If I understand what you'd like to do, I think the solution is ...
>
> W = (A*(A\b)-b) ./ (A*x-b)
>

No. This is not the solution. Consider the following example

octave:10> n=5, m=2, W = diag(rand(n,1)), A = rand(n,m), b = rand(n,1), x =
(W*A) \ (W*b), (A*(A\b)-b) ./ (A*x-b)
n =  5
m =  2
W =

  0.63718   0.00000   0.00000   0.00000   0.00000
  0.00000   0.75466   0.00000   0.00000   0.00000
  0.00000   0.00000   0.69982   0.00000   0.00000
  0.00000   0.00000   0.00000   0.08992   0.00000
  0.00000   0.00000   0.00000   0.00000   0.94621

A =

  0.833095   0.568790
  0.792483   0.536521
  0.702069   0.588282
  0.036704   0.529085
  0.238762   0.427252

b =

  0.86356
  0.77943
  0.77652
  0.98000
  0.49344

x =

  0.32941
  0.97441

ans =

    2.07510
   -7.79993
    2.55568
    0.23533
  112.87186


Hence W is not same as (A*(A\b)-b) ./ (A*x-b) .

thanks
--
Kamaraju S Kusumanchi
http://malayamaarutham.blogspot.com/

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This seems like an ill-posed problem to me.  (or in other words the answer is no, you can't solve for W or at least not a specific W).  

If you're given A, x, and b, then either

1)   Ax =b, then no matter what W you choose, left multiplying by W on both sides gives you a true statement WAx = Wb

or

2) Ax ~=b, then if you want to solve for a W such that WAx = Wb, equates to W(Ax-b) = 0, which is to say that as long as Ax-b is in the null space of W (or that Ax-b is an eigenvector with eigenvalue 0) you are free to choose any W that satisfies this condition.  This is in essence a linear system with n^2 unknowns (size of W) with only n equations.