Dear Dr. Abedini,

As far as I understand there are two different definitions of the variogram 
around:

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

and 

2\gamma(x,x)=E[(Y(x)-Y(x'))^2]=VAR[Y(x)-m(x)-Y(x')+m(x')]+ (m(x')-m(x))^2

and furthermore the second definition is often applied to 

Y(x)=Z(x)-m(x)

which is than again numerically equal to the first definition (in any case).

All these definitions are equal as long as the field is intrinsically 
stationary. And as most Textbooks are discussing the stationary case, both 
definitions are used without contradiction.

However in the instationary case you get the difference you observed. Cressie 
insisted in some publications (e.g. in his 1993 book) that the second 
definition is the correct definition for several reasons. 

Best regards,
Gerald



Am Samstag, 5. Mai 2007 10:20 schrieb M.J. Abedini:
> Dear Colleagues
>
> I would greatly appreciate it if someone could shed some light on the
> following issue:
>
> For a nonstationary random function, variogram is defined as:
>
> 2*\gamma(x,x')=VAR[Y(x)-m(x)-Y(x')+m(x')]
>
> If one simplifies the right hand side, the whole equation becomes:
>
> 2*\gamma(x,x')=R(x,x)+R(x',x')-2*R(x,x')
>
> In a number of textbook, relation between covariance and variogram is
> written as:
>
> 2*\gamma(x,x')=R(x,x)+R(x',x')-2*R(x,x')+[m(x)-m(x')]^2
>
> Am I missing something from this mathematical manipulation?
>
> Thanks
> Abedini
> +
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-- 
-------------------------------------------------
Prof. Dr. K. Gerald v.d. Boogaart
Professor als Juniorprofessor fuer Statistik
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