> To be a valid covariance function, it must be

I hate to sound ignorant here, but aren't most of the

> positive definite (as a function). In particular

> this implies that the function is bounded

> (hence no polynomials)

standard semi-variogram models polynomials of one kind

or another?

I remember seeing a paper a few years ago by a coupl

eof blokes from Pretoria University on a generalised

polynomial fit which would be positive definite. I

don't have it to hand but can probably track it down

if given sufficient motivation ;-)

Isobel Clark

http://geoecosse.bizland.com/news.html

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* Support to the list is provided at http://www.ai-geostats.org- I think that Pierre and Don Myers gave the correct response. For beginners

the lesson must be that you cannot use any old function (and in particular a

polynomial) as a covariance function. It will lead to singular kriging

matrices for some configurations of data points. That is why we stick to a

few tried and tested covariance functions in most computer packages. In fact

there are lots of 'esoteric' covariance functions known. You don't have to

just stick to the spherical, exponential and Gaussian ( it might be an idea

to have a list of covariance functions somewhere on the AI geostats site)

To get after the point that Isobel said about polynomials - the facts are

best stated in terms of the type of random function

If the random function is stationary - then a covariance function C(h)

exists and is bounded - C(h) is a positive definite function so absolutely

no polynomials allowed whatsoever!

If the data is from an intrinsic function of order 0 (IRF-0) then the

variogram must have growth less than h**2, so the only polynomial is of

the form "gamma(h) = a+b*h where b>=0" (Of course you can have "gamma(h)

= h**c where c<2 but the only polynomial is with c=1)

If the data has more general nonstationarity such as the IRF-k, then the

generalised covariance can be a polynomial. However

1) the polynomial is not arbitrary (there are constraints on the

coefficients)

2) it is definitely not found by fitting a curve to the experimental

variogram - you need special fitting techniques which are not found in many

packages

In other words, if your variogram appears to grow without bounds, you have

got non staionarity. You can either try a linear model (if appropriate) or

try to tackle the non-stationarity head on by removing the trend from your

data and then using ordinary variogram fitting techniques to the stationary

residuals (or by applying an IRF-k model if you have the software)

For more details, see the book by Chiles and Delfiner for example (or for

the real purist you can go to Matheron's original publication. "The

intrinsic random functions and their applications Adv. App. Prob., 5, pp

439-468" but beware the maths is not simple in Matheron's paper)

Regards

Colin Daly

----- Original Message -----

From: "Isobel Clark" <drisobelclark@...>

To: <ai-geostats@...>

Sent: Thursday, November 21, 2002 10:23 AM

Subject: Re: AI-GEOSTATS: curve fitting summary

> > To be a valid covariance function, it must be

> > positive definite (as a function). In particular

> > this implies that the function is bounded

> > (hence no polynomials)

> I hate to sound ignorant here, but aren't most of the

> standard semi-variogram models polynomials of one kind

> or another?

>

> I remember seeing a paper a few years ago by a coupl

> eof blokes from Pretoria University on a generalised

> polynomial fit which would be positive definite. I

> don't have it to hand but can probably track it down

> if given sufficient motivation ;-)

>

> Isobel Clark

> http://geoecosse.bizland.com/news.html

>

> __________________________________________________

> Do You Yahoo!?

> Everything you'll ever need on one web page

> from News and Sport to Email and Music Charts

> http://uk.my.yahoo.com

>

> --

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"unsubscribe ai-geostats" followed by "end" on the next line in the message

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