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[ai-geostats] A banal question...

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  • Simone Sammartino
    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
    Message 1 of 6 , May 2, 2005
      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!
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    • Pierre Goovaerts
      Hi Simone... one way to look at it is to call Z(x) the tail variable and Z(x+h) the head variable. These are two variables describing the relative position in
      Message 2 of 6 , May 2, 2005
        Hi Simone... one way to look at it is to call Z(x) the tail variable and Z(x+h) the head variable.
        These are two variables describing the relative position in space of values of the same physical
        attribute. Then, you have your 2 variables to compute the covariance fuction for a given vector h...
        By pooling together pairs of data from different parts of the study area, you are ignoring the
        location x and simply using the length (and possibly orientation) of the vector joining these 2 data...
        which requires the assumption of stationarity of the covariance..

        Hope it helps,

        Pierre

        -----Original Message-----
        From: Simone Sammartino [mailto:marenostrum@...]
        Sent: Mon 5/2/2005 10:14 AM
        To: Geostat newsgroup
        Cc:
        Subject: [ai-geostats] A banal question...



        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
      • 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 3 of 6 , May 2, 2005
          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
          >
          >
          >
          >
        • 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 4 of 6 , May 2, 2005
            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 5 of 6 , May 7, 2005
              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 6 of 6 , May 10, 2005
                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@...
                -----------------------------



                ____________________________________________________________
                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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