Hi
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 nonstationary then the semivariogram 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 nonstationary mean, you get a
parabola added to the expected shape of the semivariogram graph.
Isobel
http://courses.kriging.com
"M.J. Abedini" <abedini@...>
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')]^22\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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