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GEOSTATS: stationarity

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  • Donald Myers
    Stationarity is a property of the random function, not the data. If one computes a mean on a set of (spatial) data then you are computing a spatial average not
    Message 1 of 1 , Apr 22, 1997
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      Stationarity is a property of the random function, not the
      data. If one computes a mean on a set of (spatial) data then
      you are computing a spatial average not an ensemble average.
      Stationarity pertains to the invariance of the ensemble
      mean.

      Unfortunately there is no way to compute the ensemble mean
      because there are no replications.

      See my article in Math Geology, circa 1988.

      However there are characteristics that would suggest
      non-stationarity such as the quadratic or higher growth
      rate of the sample variogram. Plotting data values
      against a position coordinate may identify a possible
      non-stationarity.

      Note that there are several forms of stationarity of
      a random function, strong (which depends on the translation
      invariance of joint distributions and this form is not
      usually used in geostatistics), second order which depends
      not only on the mean being constant but also the covariance
      being a function of the separation vector, finally there
      is intrinsic which relates to the variogram (this is weaker
      than second order). If one is working with IRF-k's then
      there are even weaker forms of stationarity.

      Stationarity is not testable, i.e., a statistical test, because
      the test would require stationarity and would require having
      data for more than one realization of the random function.

      With respect to the idea of using different variogram models
      for different directions, that would in general violate the
      requirement of having a valid model. The conditional
      negative definiteness condition is necessary (and sufficient)
      to ensure that the estimation variance is non-negative
      and that the kriging system has a unique solution. That is
      not to say that there might not be some combinations of
      variograms in this manner that are valid but it is very
      hard to prove. See Myers and Journel, Math. Geology
      circa 1990 for examples of how this fails in even a very
      simple case.

      Donald E. Myers
      Department of Mathematics
      University of Arizona
      Tucson, AZ
      myers@...
      http://www.u.arizona.edu/~donaldm
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