Anisotropy Models for Spatial Data


This work addresses the question of building useful and valid models of anisotropic variograms for spatial data that go beyond classical anisotropy models, such as the geometric and zonal ones. Using the concept of principal irregular term, variograms are considered, in a quite general setting, having regularity and scale parameters that can potentially vary with the direction. It is shown that if the regularity parameter is a continuous function of the direction, it must necessarily be constant. Instead, the scale parameter can vary in a continuous or discontinuous fashion with the direction. A directional mixture representation for anisotropies is derived, in order to build a very large class of models that allow to go beyond classical anisotropies. A turning band algorithm for the simulation of Gaussian anisotropic processes, obtained from the mixture representation, is then presented and illustrated.

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Correspondence to D. Allard.

Appendix A: Proofs

Appendix A: Proofs

Proof of Proposition 1

The “only if” part is proven by building a simple counter-example in \(\mathbb {R}^2\). Let \((r,\theta )\) be the polar coordinates of \(h\). Since \( \beta (\theta )\) is a continuous function of \(\theta \) on a compact set, there exists \(\beta _1\) such that \(0 < \beta _1 = \min _\theta \{ \beta (\theta ) \} \le \beta (\theta ) \). Without loss of generality, the X-axis is a (non necessarily unique) direction for which \(\beta (\theta )\) is minimum, i.e. \(\beta (0)=\beta _1\). Recall that \(\beta (\theta ) = \beta (\theta +\pi )\) because the variogram is an even function. Let us consider the points \(s_0= (0,0)\), \(s_1= (-x,y)\) and \(s_2=(x,y)\), with \(x,y > 0\). The contrast

$$\begin{aligned} Z(0,0) - Z(-x,y)/2 - Z(x,y)/2 = \sum _{i=0}^2 w_i Z(s_i), \end{aligned}$$

corresponds to the error made when predicting the value at \(s_0\) with the average of the values located at \(s_1\) and \(s_2\). Let us examine the sign of the quadratic form

$$\begin{aligned} Q= & {} -\sum _{i=0}^2 \sum _{j=0}^2 w_i w_j \gamma (s_i-s_j) \\= & {} \gamma _\mathrm{AP}\{(-x,y)\} + \gamma _\mathrm{AP}\{(x,y)\} - 0.5\gamma _\mathrm{AP}\{(2x,0)\} \\= & {} a(\pi - \theta ) r^{\beta (\pi - \theta )} + a(\theta ) r^{\beta (\theta )} - 0.5 a(0) (2x)^{\beta _1}, \end{aligned}$$

where \(\theta \) denotes the angle of the vector \(s_2-s_0\) with the X-axis and \(r^2 = x^2 + y^2\). The quadratic form \(Q\) would correspond to the variance of the linear combination if \(\gamma _\mathrm{AP}(h)\) was a valid variogram. Let us denote \(\beta _m = \min \{ \beta (\pi - \theta ), \beta (\theta ) \} \ge \beta _1\). Using \(x = r \cos \theta \) and dividing by \(r^{\beta _m}\) the above expression becomes

$$\begin{aligned} \frac{Q}{r^{\beta _m}} = a(\pi - \theta ) r^{\beta (\pi -\theta ) - \beta _m} + a(\theta ) r^{\beta (\theta ) - \beta _m} - 0.5a(0) (2r\cos \theta )^{\beta _1 - \beta _m}. \end{aligned}$$

As \(r\rightarrow 0\), the first two terms in (18) are bounded by \(a(\pi -\theta ) + a(\theta )\) and the last term is proportional to \(-r^{\beta _1 - \beta _m}\) with \(\beta _1 \le \beta _m\). Hence, \(Q\) becomes negative as \(r\rightarrow 0\) unless \(\beta _m = \beta _1\). Hence, for all \(0 < \theta < \pi /2\)

$$\begin{aligned} Q \ge 0 \ \hbox {as} \ x \rightarrow 0 \Leftrightarrow \min _\theta \{ \beta (\pi - \theta ), \beta (\theta ) \} = \beta _1. \end{aligned}$$

Using topological arguments, one can then show that \(\beta (\theta ) = \beta _1\), for all \(\theta \). The proof is completed by noticing that if a function is not c.d.n. in \(\mathbb {R}^2\), it is not c.d.n. in \(\mathbb {R}^d\), for \(d \ge 2\). \(\square \)

Proof of Theorem 2

For the same construction as in Proposition 1, the quadratic form \(Q\) now reads

$$\begin{aligned} Q = a(\pi - \theta ) r^{\beta (\pi - \theta )} f(r) + a(\theta ) r^{\beta (\theta )} f(r) -0.5 a(\theta ) (2x)^{\beta _1} f(2x), \end{aligned}$$

and thus

$$\begin{aligned} \frac{Q}{f(r)r^{\beta _m}}= & {} a(\pi - \theta ) r^{\beta (\pi -\theta ) - \beta _m} + a(\theta ) r^{\beta (\theta ) - \beta _m}\nonumber \\&- 0.5a(0) (2\cos \theta )^{\beta _1 - \beta _m} \frac{f(2r\cos \theta )}{f(r)}r^{\beta _1 - \beta _m}. \end{aligned}$$

As \(r\rightarrow 0\), the first two terms in (20) are bounded and, since \(f\) is a slowly varying function at \(r=0\), the last term tends to \(-\infty \). Hence, \(Q\) becomes negative as \(r\rightarrow 0\) unless \(\beta _m = \beta _1\). The end of the proof is now exactly similar to that of Proposition 1. \(\square \)

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Allard, D., Senoussi, R. & Porcu, E. Anisotropy Models for Spatial Data. Math Geosci 48, 305–328 (2016).

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  • Anisotropy
  • Covariance
  • Isotropy
  • Spatial statistics
  • Turning band method
  • Variogram