You could also express the variogram in this other form........
2gamma(h) = Var{Z(u)-Z(u+h)} = E{[Z(u)-Z(u + h)]^2}-E{ Z(u)-Z(u+h)}^2
the second element of the right equation goes away when the phenomenon is stationary.
Sebastiano
At 21.15 09/04/2006, Isobel Clark wrote:
Hi. At last one I can answer (i.e. an easy one):
If the phenomenon is non-stationary then the semi-variogram is NOT:
(1) 2\gamma(x,x')=Var[Y(x)-Y(x')]^2
but
(2) 2\gamma(x,x')=Var[(Y(x)-m(x))-(Y(x')-m(x'))]^2
The form (1) is a simplification of the full definition (2) when m(x)=m(x'). This is why, if you do have a non-stationary mean, you get a parabola added to the expected shape of the semi-variogram graph.
Isobel
http://courses.kriging.com
"M.J. Abedini" <[EMAIL PROTECTED]> wrote:
- Dear Colleagues
- The following arguement puzzling me. Your clarification will be greatly
- appreciated.
- When a random funtion Y (x) is not stationary, then one could prove the
- relation between variogram and covarinace as:
- 2\gamma(x,x')=\sigma(x,x)+\sigma(x',x')+[m(x)-m(x')]^2-2\sigma(x,x')
- I was hoping to derive the above relationship starting with the following
- definition of variogram with no success.
- 2\gamma(x,x')=Var[Y(x)-Y(x')]^2
- I was not able to reproduce the term [m(x)-m(x')]^2.
- Am I missing something in this process?
- Thanks
- MJA
- p.s. Most likely, my problem has something to do with misconception of
- intrinsic hypothesis
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