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Re: [ai-geostats] A banal question...

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  • M.J. Abedini
    Simone Well, first of all one has to acknowledge that stationarity is NOT the property of data, it is a property of a conceptualized model. Generally speaking,
    Message 1 of 6 , May 2 9:49 AM
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      Simone

      Well, first of all one has to acknowledge that stationarity is NOT the
      property of data, it is a property of a conceptualized model. Generally
      speaking, one cannot verify whether a set of spatial data are stationary
      of not. For a set of data to be stationary, one has to have:

      a. m=E[Y(x)], i.e. mean function independent of spatial location
      b. Cov[Y(x), Y(x+h)]=cov(h), i.e., Covariance function is a function of
      separation vector and
      c. Variance is finite

      Unfortunately, one cannot verify the above conditions for a snapshot of
      data. Only for the case when the closed form of random function Y(x) is
      given, one could verify these conditions.

      Generally speaking, covariance function is a function of absolute
      location. But that is a dead end for estimation purposes. What we are
      really doing is that we are imposing stationarity on our data to exit from
      this dead end. Even to make life simpler, we assume that covariance
      function is a function of separation distance which has something to do
      with isotropy. See the level of complexity is like this:

      covariance function is a function of absolute location
      covariance function is a function of separation vector, i.e., stationarity
      covariance function is a function of separation distance, i.e, isotropy

      I would suggest to take a look at Kitanidis book on "Introduction to
      Geostatistics: Application to Hydrogeology" for an example on this topic
      in ch. 3, sec 3.2.

      Hope this helps

      Thanks
      Abedini

      On Mon, 2 May 2005, Simone Sammartino wrote:

      > Dear all
      > a banal question...
      > I'm not able to understand the stationarity of covariance in second order stationarity theory...
      > On any book or article I can read:
      > ....covariance between Z(x) e Z(x+h) exist and does not depend on x, but only on h; in fact
      > Cov[Z(x),Z(x+h)]=Cov(h)....
      > It is considered so banal that in any text I consulted this part is described with the same sentence...but it is not explicated via mathematical formalism....
      > Why should E[Z(x)Z(x+h)]-m^2 be so logically reduced to Cov(h)
      > You'll laugh for my request, but I'm not able to understand why it should be so logical....
      > In some text I found also...=Cov(x1-x2)=Cov(h) where distance between x1 and x2 is exactly h, but it does not help me to understand it....
      > I can't realize how to calculate Cov(h) that is a variable (it is in reality at least a vector of constant), when usually covariance is calculated between two variables....
      > Please have the patience to help me to solve this trick
      > Thanks
      > 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@...
      > -----------------------------
      >
      >
      >
      > ____________________________________________________________
      > 6X velocizzare la tua navigazione a 56k? 6X Web Accelerator di Libero!
      > Scaricalo su INTERNET GRATIS 6X http://www.libero.it
      >
      >
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    • Ted Harding
      ... As well as Isobel s very practically-oriented explanation, I think it would be useful to look at it from a theoretical point of view, since this is in fact
      Message 2 of 6 , May 2 10:12 AM
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        On 02-May-05 Simone Sammartino wrote:
        > Dear all
        > a banal question...
        > I'm not able to understand the stationarity of covariance in second
        > order stationarity theory...
        > On any book or article I can read:
        > ....covariance between Z(x) e Z(x+h) exist and does not depend
        > on x, but only on h; in fact Cov[Z(x),Z(x+h)]=Cov(h)....
        > It is considered so banal that in any text I consulted this part
        > is described with the same sentence...but it is not explicated
        > via mathematical formalism....
        > Why should E[Z(x)Z(x+h)]-m^2 be so logically reduced to Cov(h)
        > You'll laugh for my request, but I'm not able to understand why
        > it should be so logical....
        > In some text I found also...=Cov(x1-x2)=Cov(h) where distance
        > between x1 and x2 is exactly h, but it does not help me to
        > understand it....
        > I can't realize how to calculate Cov(h) that is a variable (it
        > is in reality at least a vector of constant), when usually
        > covariance is calculated between two variables....
        > Please have the patience to help me to solve this trick
        > Thanks
        > Simone

        As well as Isobel's very practically-oriented explanation,
        I think it would be useful to look at it from a theoretical
        point of view, since this is in fact important!

        For instance, in the part of the world where I live
        (approx 0deg 23.5min E, 52deg 28.6min N), the height of
        the land above sea level varies rather little -- from
        about -2m to +2m for several miles in all directions,
        with occasional exceptions. It is, overall, very flat!

        Indeed, you can sum it up by saying that it looks the
        same in all places and in all directions.

        Now first let me work within a particular square, 2km by 2km,
        in the region, say centred on the above geographical
        coordinates.

        Suppose I pick a random point in this square, say A,
        and then, in a random direction, another point B at a
        distance of h metres from A (for some value of h up to
        say 500m). I then determine the heights X at A, and Y at B,
        above sea level.

        I then have two random variables X and Y and, with respect
        to the random mechanism by which I have selected them,
        X has an expectation E(X), Y has an expectation E(Y),
        and X*Y has an expectation E(X*Y). In the usual way, the
        covariance between X and Y is defined as E(X*Y) - E(X)*E(Y).

        This covariance depends on h and on the place where the
        2km square is centred. By using different values for h
        but still working within the same square, I would get
        the covariance for differet values of h, and so would
        get a function C(h) which is specific to the particular
        2km square I am working in. In view of the extremely flat
        terrain, C(h) would be high for relatively small values
        of h (say up to 100m), but would then diminish and
        would probably be small for h > 500m.

        Now suppose I shift the 2km square by (say) 3km to the East,
        and look at the same procedure within the new square.

        I can similarly get C(h) for this square. I am rather
        confident that the two functions C(h), for the two different
        squares, would be very similar (if not identical).
        Indeed, I could probably position the square anywhere within
        a range up to 9km to the North, 12kn to the NW, 7km to the E,
        10kn to the South, 7km to the SW, 6km to the West, and at
        least 50km to the NW, without this situation changing.
        (Beyond these limits, the terrain changes, becoming more
        hilly, and I would expect C(h) to behave differently in
        such places).

        Given that I expect C(h) to be much the same function of h
        wherever I position the square within that region, I would
        then say (by definition) that "height is second-order
        stationary within that region" -- it doesn't matter where
        my "base" (origin) for the measurements is placed.

        Indeed, from the fact that it looks the same in all directions,
        I would also expect that I would get the same C(h) if, instead
        of choosing the direction from A to B at random, I simply
        chose a constant direction (and it would not matter which
        constant direction I chose). In other words, the spatial
        process I am observing is isotropic as well as second-order
        stationary. [** see at end]

        I could estimate C(h) by performing tha above for several
        different points A, followed by B at distance h, and computing
        the covariance between the X series and the Y series.

        Next, rather than determine C(h) as above (first choose A
        randomly, then B, as described) I could measure the height
        at many different points (Z, say), and then use a variogram
        technique to estimate C(h) from all these (which of course
        will be at allsorts of different pairwise distances from
        each other).

        But in doing so, I would be somewhat relying on stationarity
        and isotropy to validate the variogram technique -- quite
        apart from relying on these to validate the concept of
        C(h) as informative about the process anyway -- you could
        apply the same measurement procedures to the South flank
        of Mt Everest if you wanted to, but I don't think that
        C(h) would tell you much about Mt Everest! On the other
        hand, given the overall featureless terrain I'm describing
        here, C(h) would be capable of giving you quite a lot
        about the detailed behaviour of the terrain. In particular
        it could probably be applied to help determine the overall
        hydrography of the region -- e.g. what complexity of
        drainage systems would you need.

        However, as well as the fact that the informativeness of C(h)
        depends on properties like stationarity and isotropy, also
        a lot of theory about analysing measurements on spatial
        processes depends on assuming these properties in order
        to make progress. The fundamental issue that depends on
        these assumtions is the question: whether the expectation
        of a random variable at an arbitrary point will be the
        same as the average over several fixed points. In the
        mathematical theory, it would be assumed that these held
        to an indefinite distance in all directions. In practice,
        it is often adequate that they should hold for a sufficient
        distance, which is beyind the range at which C(h) falls
        to small values. So if I were only concerned to draw
        conclusions about what happens within 5km of where I live,
        I would not worry about the fact that it all fell apart
        20kn away!

        I hope this contributes further to clarifying your query!

        best wishes,
        Ted.

        [**] There is a potential source of anisotopy: The region
        is intersected by a number of watercourses -- on the size-scale
        of rivers (which some of them are) -- which are contained
        within raised banks (2-3m high), each of which tends to
        run in a straight line for several km. Therefore at certain
        points, given that the height is 2m or more, it will
        remain so for a considerable distance in a particular
        direction. Therefore the assumption of stationarity and
        isotropy do not hold strictly everywhere. However, the
        proportion of the area over which they do not hold is
        a very small fraction of the whole.



        --------------------------------------------------------------------
        E-Mail: (Ted Harding) <Ted.Harding@...>
        Fax-to-email: +44 (0)870 094 0861
        Date: 02-May-05 Time: 18:08:16
        ------------------------------ XFMail ------------------------------
      • Digby Millikan
        Simone, I was curious about the averaging of samples to produce each variogram value, which immediately is resulting in some smoothing, you may be interested
        Message 3 of 6 , May 7 12:06 AM
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          Simone,

          I was curious about the averaging of samples to produce each variogram
          value,
          which immediately is resulting in some smoothing, you may be interested that
          there
          is a simulation method based on uncertainty of the variogram, which is a
          practical
          attempt to account for these assumptions.

          Regard Digby
        • Simone Sammartino
          It seems interesting... Unfortunately I have to write my PhD thesis now and I m proceeding methodically from the beginning... I ll contact you again when I ll
          Message 4 of 6 , May 10 1:48 AM
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            It seems interesting...
            Unfortunately I have to write my PhD thesis now and I'm proceeding methodically from the beginning...
            I'll contact you again when I'll speak about simulations...:-)
            Thank you
            Simone

            > Simone,
            >
            > I was curious about the averaging of samples to produce each variogram
            > value,
            > which immediately is resulting in some smoothing, you may be interested that
            > there
            > is a simulation method based on uncertainty of the variogram, which is a
            > practical
            > attempt to account for these assumptions.
            >
            > Regard Digby
            >
            >
            >
            >

            -----------------------------
            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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