2502Re: [ai-geostats] Re: Puzzling question

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• Apr 9, 2006
Hi

You could also express the variogram in this other  form........

2gamma(h) = Var{Z(u)-Z(u+h)} = E{[Z(u)-Z(u + h)]^2}-E{ Z(u)-Z(u+h)}^2

the second element of the right equation goes away when the phenomenon is stationary.

Sebastiano

At 21.15 09/04/2006, Isobel Clark wrote:
Hi. At last one I can answer (i.e. an easy one):

If the phenomenon is non-stationary then the semi-variogram is NOT:

(1)  2\gamma(x,x')=Var[Y(x)-Y(x')]^2

but

(2)  2\gamma(x,x')=Var[(Y(x)-m(x))-(Y(x')-m(x'))]^2

The form (1) is a simplification of the full definition (2) when m(x)=m(x'). This is why, if you do have a non-stationary mean, you get a parabola added to the expected shape of the semi-variogram graph.

Isobel
http://courses.kriging.com

"M.J. Abedini" <abedini@...> wrote:

Dear Colleagues

The following arguement puzzling me. Your clarification will be greatly
appreciated.

When a random funtion Y (x) is not stationary, then one could prove the
relation between variogram and covarinace as:

2\gamma(x,x')=\sigma(x,x)+\sigma(x',x')+[m(x)-m(x')]^2-2\sigma(x,x')

I was hoping to derive the above relationship starting with the following
definition of variogram with no success.

2\gamma(x,x')=Var[Y(x)-Y(x')]^2

I was not able to reproduce the term [m(x)-m(x')]^2.

Am I missing something in this process?

Thanks
MJA

p.s. Most likely, my problem has something to do with misconception of
intrinsic hypothesis

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