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GEOSTATS: Gaussian variograms

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  • Donald Myers
    It is likely that you have encountered a common problem with the Gaussian model, a simple fix is to include a small nugget term. Assuming you are using
    Message 1 of 1 , Oct 14, 1998
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      It is likely that you have encountered a common problem with the Gaussian
      model, a simple fix is to include a small nugget term.

      Assuming you are using ordinary kriging the coefficient matrix is in
      general not positive definite but still invertible if a valid variogram is
      used. The problem is the zero in the last row/last column.

      However the problem with the Gaussian is a further complication. If you
      examine the graph of a Gaussian you see that it has values that are almost
      zero (assuming no nugget) for a considerable distance (relative to the
      range) hence when one uses the usual kind of a search neighborhood, i.e.,
      only the nearest data locations, this can result in lot of variogram values
      in the kriging matrix that are nearly zero. This likely results in an
      ill-conditionned matrix and depending on how they attempt to solve the
      system of equations it may not have a unique solution.

      If the kriging system is written in terms of covariances rather than
      variograms then the same kind of a problem occurs but rather than a lot of
      values that are close to zero, the values are all close to the sill value.

      If you are using simple kriging (as opposed to ordinary) then the same
      problem can still occur with a Gaussian model although theoretically the
      kriging matrix is postive definite.

      I haven't looked to see which equation solver they are using in S+ and it
      may be their choice is unusually sensitive to ill-conditionned matrices.

      Donald E. Myers
      Department of Mathematics
      University of Arizona
      Tucson, AZ 85721

      (520) 621-6859
      myers@...

      http://www.u.arizona.edu/~donaldm

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