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[ai-geostats] Puzzling question

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  • M.J. Abedini
    Dear Colleagues The following arguement puzzling me. Your clarification will be greatly appreciated. When a random funtion Y(x) is not stationary, then one
    Message 1 of 3 , Apr 9, 2006
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      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
    • Isobel Clark
      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
      Message 2 of 3 , Apr 9, 2006
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        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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      • 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 3 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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          ( see http://www.ai-geostats.org/help_ai-geostats.htm )

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          Signoff ai-geostats


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