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Re: [ai-geostats] Re: Puzzling question

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  • seba
    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
    Message 1 of 3 , Apr 9, 2006
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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 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" <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')]^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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