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2043Re: [ai-geostats] A quasi-stationary framework...

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  • Digby Millikan
    May 7 12:13 AM
    • 0 Attachment
      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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