- 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 - 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

>

>

>

> - On 02-May-05 Simone Sammartino wrote:
> Dear all

As well as Isobel's very practically-oriented explanation,

> 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

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 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 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