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AI-GEOSTATS: summary variogram modelling with nested structures

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  • Dobler, Lorenz
    this was my question(s) I am dealing with heavy metals in upper soil layers of forests. Some of my (indicator) variograms behave like a spherical model at the
    Message 1 of 1 , Feb 21, 2001
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      this was my question(s)

      I am dealing with heavy metals in upper soil layers of forests. Some
      of my (indicator) variograms behave like a spherical model at the
      beginning but in stead of reaching the sill they behave like a
      power-model for greater distances. I tried
      several power models but they didn't fit the variogram very well.
      Just two questions:

      1. What does this special behavior of the experimental variogram tell
      me about the spatial structure or stationarity?
      2. Can I „mix" a spherical and and a power model to build up one
      nested model?

      and here are the answers:

      Isobel Clark [drisobelclark@...] wrote:


      It sounds like you have a trend in your values.

      You should have received your copy of the book by now.
      Check out Chapter 7 for full information on fitting
      trend surfaces, Chapter 9 on diagnosing trend in the
      semi-variogram (Wolfcamp example) and Chapter 12 for
      Universal Kriging.

      second answer of isobel:

      Let me clear up the indicator thing for you first.
      An indicator transform changes all values below
      the 'cutoff' or discriminator value to 0 and all
      values above it to 1. At least, that's what ours
      does, other packages do it the other way round --
      change all values below cutoff to 1 and all values
      above to 0.

      You only need to do multiple indicators if you have
      (a) a very strange distribution or (b) a mixture of
      different populations. See Chapter 12 on indicators.

      I would recommend you look at the distribution
      (histogram,
      probability plot, etc) before resorting to something
      as
      complicated as a multi-indicator approach. If you are
      getting decent semi-variograms from your ordinary
      values
      you don't need indicator kriging!

      As far as your semi-variogram goes, if your 'power
      model'
      looks like a power less than 1 you probably just have
      a
      nested Spherical structure which hasn't reached the
      later
      range of influence yet. Try modelling a second
      component
      with a range longer than yoru maximum lag distance.

      Re-reading your e-mail carefully (sorry, I tend to
      skim)
      you say your first 4 deciles have this unbounded
      nature
      and the others are bounded. What this is saying is
      that
      for low cutoffs your ranges are much longer than for
      higher cutoffs. Your high values tend to occur in
      smaller more coherent 'blobs' than the lower values
      which are diffuse and related to much greater
      distances.

      This could be multiple populations with the high
      values
      coming from a more localised concentrated source and
      the
      rest forming a 'background' phenomenon. Best thing to
      do
      is simply postplot the indicators for the higher
      cutoffs
      and see WHERE they congregate. Then try to relate that
      to
      what you know about the origins of your variable. For
      example, are these high human concentrations?
      Industrial
      areas? Forest? Water? Landfill? and so on. You know
      your
      data better than I do.

      Visual assessment of patterns is still a very valuable
      tool!

      answers of
      Donald E. Myers
      Department of Mathematics
      University of Arizona
      Tucson, AZ 85721
      http://www.u.arizona.edu/~donaldm

      The indicator variogram has to be bounded if the random function is
      stationary, note not just second order stationary but stationary. An
      unbounded sample indicator variogram suggests that the stationarity
      condition is not satisfied.

      I mentioned before that "stationarity" in the case of the indicator
      transform is not the same as second order stationarity which is the
      common
      assumption underlying geostatistics. "Stationarity" in this case means
      that
      for any choice of points s1,.....sn and any vector h, the joint
      probability distribution of Z(s1),....,Z(sn) is the same as the joint
      probability distribution of Z(s1 +h),...., Z(sn + h). That is, all the
      joint probability distributions are translation invariant. Unfortunately
      one
      will never have the right kind of data to test this condition. This
      condition also implies that the probability distribution of Z(s) is the
      same
      as the probability distribution for Z(s+h) for any point s and any
      vector h,
      this common distribution is usually called the marginal and might be
      denoted
      as F(z) (no subscript on the F). Then one can show that the indicator
      variogram is actually F(z)[1-F(z)] which is always bounded by 0.25.
      This
      bound applies for any choice of the cutoff "z". Hence it would not be
      theoretically possible for the sample indicator variogram to be bounded
      for
      some cutoffs and unbounded for others, IF the stationarity condition is
      satisfied. Unfortunately all of indicator geostatistics is based on this
      assumption and unlike the problem with a non-constant mean it would not
      be
      sufficient to compute some kind of residuals and then do the analysis on
      the
      residuals.

      A point I should have mentioned to be sure there is no confusion, you
      made
      reference to water quality "indicators". That use of the term indicator
      is
      not the same as is referred to in discussing indicator variograms.

      Dr. Heinz Burger
      Freie Universitaet Berlin
      - Geoinformatik -
      Malteserstr. 74-100
      12249 BERLIN, Germany
      Tel. (49) 30-838-70561 Fax: (49) 30-775-2075
      e-mail: hburger@...-berlin.de
      Web-Seite: http://userpage.fu-berlin.de/~hburger/hb

      Linear combinations of admissible models produce admissible models.
      The problem is: what will you do? For kriging you only need a good model
      within the search radius. For an interpretation of the spatial structure
      of
      your phenomenon you don't need a model-function in every case (except
      for comparison). A detailed trend analysis and resiual analysis (incl.
      variogram) will give more information about local and regional spatial
      structures. The variogram is only one tool among others.

      Edzer J. Pebesma [e.pebesma@...]

      >
      > Hello list,
      >
      > I am dealing with heavy metals in upper soil layers of forests. Some
      > of my (indicator) variograms behave like a spherical model at the
      > beginning but in stead of reaching the sill they behave like a
      > power-model for greater distances. I tried
      > several power models but they didn't fit the variogram very well.
      > Just two questions:
      >
      > 1. What does this special behavior of the experimental variogram tell
      > me about the spatial structure or stationarity?

      It may be non-stationary.

      > 2. Can I ?mix" a spherical and and a power model to build up one
      > nested model?
      >
      Yes, most common software, like GSLIB, Variowin, gstat, ...

      Yetta Jager
      Environmental Sciences Division
      Oak Ridge National Laboratory
      P.O. Box 2008, MS 6036
      Oak Ridge, TN 37831-6036
      U.S.A.

      OFFICE: 865/574-8143
      FAX: 865/576-8543
      Work email: jagerhi@...
      In my opinion, it suggests that one might be better off finding a
      covariate
      to explain the drift and removing it before modeling the autocorrelation
      of
      residuals. See our TM report on my website with Pattern-Plus in the
      title.


      Marco Alfaro [malfaro@...]

      You cannot use the power model in the case of indicator variograms.

      The variogram g(h) of a random set (indicator 0 or 1) has the following
      relationships (not accomplished by the power variogram):

      g(h) <= 0.5
      g(2h) <= 2g(h)


      thanks to all
      Lenz




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