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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 correlation function in terms of covariance of increments expressed with reference to Z(x0).

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    See my answer...2011-05-18

1 Answers 1

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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).