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Re: [ai-geostats] A quasi-stationary framework...

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  • Digby Millikan
    The variogram is only the difference between the absolute values of the grade, so if you subtract the mean from every z(x), you will still have the same
    Message 1 of 3 , May 7, 2005
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      The variogram is only the difference between the absolute values of the
      grade, so if you
      subtract the mean from every z(x), you will still have the same variogram?

      ----- Original Message -----
      From: "Simone Sammartino" <marenostrum@...>
      To: "Geostat newsgroup" <ai-geostats@...>
      Sent: Wednesday, May 04, 2005 11:36 PM
      Subject: [ai-geostats] A quasi-stationary framework...


      It's me again!...:-))
      My problem now is:
      about a quasi-stationary framework...
      Assume Z(x) is not exactly stationary but its mean varies weakly in the
      space...
      Thus E[Z(x)]=m(x)...let's consider a new variable, said residual,
      Y(x)=Z(x)-m(x), with zero mean.
      Variogram for Z(x) is
      (1) 2*Gamma(x)=E{[Z(x)-Z(x+h)]^2}-[m(x)-m(x+h)]^2
      At this point the book says "...and it's easy to realize how variogram of
      Y(x) is exactly the same of (1)..." How??!?!?!?
      I tried everything but I did not manage to obtain the same result....
      Anyone helping me?
      Thanks as always
      Simone


      -----------------------------
      Dr. Simone Sammartino
      PhD student
      - Geostatistical analyst
      - G.I.S. mapping
      I.A.M.C. - C.N.R.
      Geomare-Sud section
      Port of Naples - Naples
      marenostrum@...
      -----------------------------



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