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## RE: [ai-geostats] A banal question...

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• 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 1 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@...
-----------------------------

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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,
Message 2 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
>
>
>
>
• ... 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 3 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 ------------------------------
• 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 4 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
• 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 5 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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