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Consider an intrinsic RF Z(x) that is not second order stationary.

Considering an arbitrary reference RV Z(x0), how to express traditional variogram in terms of covariance of increments expressed with reference to Z(x0).

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In that link, see the equation after "For a non-stationary process the square of the difference between expected values at both points must be added:". Does it help? – Shai Covo May 17 '11 at 22:46
hmm..i'll see..thanks Shai Covo! – Pupil May 17 '11 at 23:02
@Shai: Can you figure out from that expression why the difference between expected values at both points should be added? – Pupil May 18 '11 at 0:49
See my answer... – Shai Covo May 18 '11 at 10:03
up vote 1 down vote accepted

Using the notation of the Wikipedia article, it is easy to show that $$ 2\gamma (x,y) = C(x,x) + C(y,y) - 2C(x,y) + [{\rm E}(Z(x)) - {\rm E}(Z(y))]^2 . $$ Indeed, this follows straightforwardly from $$ 2\gamma (x,y): = {\rm E}(|Z(x) - Z(y)|^2 ) = {\rm E}(Z(x)^2 ) + {\rm E}(Z(y)^2 ) - 2{\rm E}(Z(x)Z(y)), $$ $$ C(x,x): = {\rm Cov}(Z(x),Z(x)) = {\rm Var}(Z(x)), $$ $$ C(y,y): = {\rm Cov}(Z(y),Z(y)) = {\rm Var}(Z(y)), $$ and $$ C(x,y):={\rm Cov}(Z(x),Z(y)) = {\rm E}(Z(x)Z(y)) - {\rm E}(Z(x)){\rm E}(Z(y)). $$ In case of a weak stationary $Z$ (hence, in particular, in case of a second-order/strict stationary $Z$ with finite covariance), the expectation does not vary with respect to location, that is ${\rm E}(Z(x)) = {\rm E}(Z(y))$; hence the Wikipedia formula $$ 2\gamma (x,y) = C(x,x) + C(y,y) - 2C(x,y) $$ for stationary $Z$ (with finite covariance function), and, in turn, the Wikipedia claim that for a non-stationary process the square of the difference between expected values at both points must be added (however, in the equation following that claim, that term was mistakenly subtracted, instead of being added).

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