 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  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')]^2but(2) 2\gamma(x,x')=Var[(Y(x)m(x))(Y(x')m(x'))]^2The 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.Isobelhttp://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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 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
 * By using the aigeostats mailing list you agree to follow its rules
 ( see
http://www.aigeostats.org/help_aigeostats.htm )
 * To unsubscribe to aigeostats, send the following in the subject or
in the body (plain text format) of an email message to
sympa@...
 Signoff aigeostats
* By using the aigeostats mailing list you agree to follow its rules
( see http://www.aigeostats.org/help_aigeostats.htm )
* To unsubscribe to aigeostats, send the following in the subject or in the body (plain text format) of an email message to sympa@...
Signoff aigeostats  Dear Colleagues