Well, it looks like my "feeling" might have been wrong this time. The way
you may want to go through the "tedious" proof is the following.
1) You construct the Aij matrix (also sometimes called the Hessian matrix)
Aij=D^2chi^2/DxiDxj (D is the partial derivative sign. E-mail is
still too primitive)
In the simple case of a single crystal this is often approximated as
Aij=Sum Wh*DIh/Dxi*DIh/Dxj
2) You invert the matrix: Bij=(Aij)^-1
3) Assuming chi^2=1, the ESD are the square roots of the diagonal elements
of Bij.
You can do this for very simple cases, say with 3 reflections and 2
parameters:
I1=x1+x2; I2=x1+2*x2, I3=x2, and assuming unitary weights.
Taking into account all 3 reflections, you get:
2 -1
Bij=
-1 2/3
Excluding I3 (which is insensitive to x1) you get
5 -3
Bij=
-3 2
Clearly, the ESD on x1 have worsened in this simple case. This does not
prove the general case, but there might be a proof for that as well.
Sorry about the confusion.
Paolo
