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A Robust Gaussian variogram estimator for cartography of hydrological extreme events

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Abstract

Kriging is used for spatial interpolation of geophysical and hydrological variables. This technique requires the definition of a mathematical function of spatial correlation known as variogram. Extreme value mapping is critical for the analysis of hydrological problems such as multiple risk and disaster damage valuation. However, the formulation of a variogram is complex and the selection of the variogram model is often crucial in the final representation of extreme values. There is some evidence that a variogram estimator may reduce the negative impact of using data samples with outliers. This is an approach that eliminates the term that takes into consideration the squared differences of empirical values in the calculation of a variogram. Known as Cressie–Hawkins Estimator (CH), it is based on considering a Gaussian distribution for the calculation of the variogram. The CH satisfies the main statistical considerations of normality; however, the formulation of kurtosis is not used to guarantee total normality. The present paper modifies the CH, adding the term corresponding to the fourth statistical moment. The numerical example proposed in the literature to evaluate the efficiency of the CH is reproduced, and it is verified that the new variogram (CH-GLo) decreases further the effect of the extreme values in the cartography. CH-GLo is applied to cartography an extreme storm that happened in August 2014 in the city of Queretaro, Mexico. It is concluded that this new variogram formulation is a suitable alternative for mapping extreme values monitored in real-time in urban areas.

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Acknowledgments

The author is grateful to the Risk Management Unit of the UNESCO Regional Office of Science for Latin America and the Caribbean. To Roberto Mercado for the mathematical review. Special thanks to Rafael Porras, Technical Director of RedCIAQ for the information provided, imagery, and meteorological stations database. To Karla Isabel Barrera Gonzalez, for her support in the detailed checking of the equations. To Libna Herrera and Marilu Meza for the exhaustive analysis of rainfall data. He expresses his sincere gratitude to Giselle Galvan Lewis for their detailed style review.

Funding

This work was financially supported by Consejo Nacional de Ciencia y Tecnología (CONACYT). This research received funding from Secretaria de Educacion Publica (SEP-Mexico). Programa para el Desarrollo Profesional Docente (PRODEP) and was supported by the Fund for the Reinforcement of Research UAQ-2018 Research and Postgraduate Division (FOFIUAQ/FIN201911).

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Correspondence to Alfonso Gutierrez-Lopez.

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Appendix A

Appendix A

\({\chi }^{2}\) and Gaussian distribution relationship.

The \({\chi }^{2}\) distribution is related to the Guassian one, in the following sense: If \({E}_{i}, i=1, n,\) are random variables with standard Gaussian distributions, then the random variable \({\rm E}=\Sigma {\rm E}_{I}^{2}\) is a \({\chi }^{2}\) with \(\alpha =n\) degrees of freedom. So when \(\alpha \) is an entire, the distribution is obtained from the sum of standard Gaussian distribution squares.

If a random variable has Gaussian distribution, then let us find out how is the characteristic function of the square of the same random variable; for that, we calculate the average of \({\mathrm{e}}^{{\mathrm{ikx}}^{2}}\) under standard Gaussian.

$$\mathrm{E}\left({\mathrm{e}}^{{\mathrm{ikx}}^{2}}\right)={\int }_{\mathcal{R}}{\mathrm{e}}^{{\mathrm{ikx}}^{2}}\frac{{\mathrm{e}}^{-\frac{{\mathrm{x}}^{2}}{2}}}{\sqrt{2\uppi }}\mathrm{dx}$$

Then,

$$ \begin{aligned} {\text{E}}\left( {{\text{e}}^{{{\text{ikx}}^{2} }} } \right) & = { }\mathop \smallint \limits_{{\mathcal{R}}}^{{}} \frac{1}{{\sqrt {2\pi } }} {\text{e}}^{{ - \frac{{x^{2} }}{2} + ikx^{2} }} {\text{d}}x \\ & = { }\mathop \smallint \limits_{{\mathcal{R}}}^{{}} \frac{1}{{\sqrt {2\pi } }} {\text{e}}^{{ - \frac{{x^{2} }}{2} \left( {1 - 2ik} \right) }} {\text{d}}x \\ \end{aligned} $$

With the change of variable \({\mathrm{x}}^{2}\left(1-2\mathrm{ik}\right)={\mathrm{y}}^{2}\) or. \(\mathrm{x }{\left(1-2\mathrm{ik}\right)}^{1/2}=\mathrm{y},\mathrm{ dy}= {\left(1-2\mathrm{ik}\right)}^{1/2}=\mathrm{dx}\),

$$ \begin{aligned} {\text{E}}\left( {{\text{e}}^{{{\text{ikx}}^{2} }} } \right) & = { }\mathop \int \limits_{{\mathcal{R}}}^{{}} \frac{1}{{\sqrt {2\pi } }} {\text{e}}^{{ - \frac{{y^{2} }}{2} }} \frac{{{\text{d}}y}}{{\left( {1 - 2ik} \right)^{1/2} }} \\ & = \frac{1}{{\left( {1 - 2ik} \right)^{1/2} }}{ }\mathop \smallint \limits_{{\mathcal{R}}}^{{}} \frac{1}{{\sqrt {2\pi } }} {\text{e}}^{{ - \frac{{y^{2} }}{2} }} {\text{d}}y \\ \end{aligned} $$

Then,

$$\mathrm{E}\left({e}^{{\mathrm{ikx}}^{2}}\right)= \frac{1}{{\left(1-2ik\right)}^{1/2}}$$

Then, the sum of the squares of random variables distributed according to the normal produce

$$ \begin{aligned} {\text{E }}\left( {{\text{e}}^{ik} \mathop \sum \limits_{j = 1}^{n} x_{j}^{2} } \right) & = {\text{E }}\left( {\mathop \prod \limits_{j = 1}^{n} {\text{e}}_{j}^{{ikx_{j}^{2} }} } \right) \\ & = \mathop \prod \limits_{j = 1}^{n} E \left( {{\text{e}}^{{ikx_{j}^{2} }} } \right) \\ & = \mathop \prod \limits_{j = 1}^{n} \frac{1}{{\left( {1 - 2ik} \right)^{1/2} }} \\ & = \frac{1}{{\left( {1 - 2ik} \right)^{1/2} }} \\ \end{aligned} $$

which is the characteristic function of the \({\chi }^{2}\) distribution.

In particular, if a random variable follows the \({\chi }^{2},\) we can obtain its relationship with the Gaussian one. Let x be the variable x that follows \({\chi }^{2}\); then, the measure of probability is:

$$= \frac{1}{{2}^{\propto /2}} \frac{{x}^{\frac{\propto }{2}-1}}{\Gamma \left(\frac{\propto }{2}\right)}{\mathrm{e}}^{-\frac{x}{2}} \mathrm{d}x.$$

and, with the change of variable, \(x={\left(\frac{y}{\surd 2\sigma }\right)}^{2}, 0<x< \infty , dx=2\frac{y}{\surd 2\sigma }d \left(\frac{y}{\surd 2\sigma }\right), {(*)}\),

there is:

$$ \begin{aligned} {\text{d}}P\left( {x;\alpha } \right) & = { }\frac{1}{{2^{\alpha /2} }} \frac{{x^{{\frac{\alpha }{2} - 1}} }}{{{\Gamma } \left( {\frac{\alpha }{2}} \right)}}{\text{e}}^{{ - \frac{x}{2}}} {\text{d}}x \\ & = { }\frac{2}{{2^{\alpha /2} }} \frac{{\left( {\frac{y}{\surd 2\sigma }} \right)^{\alpha - 1} }}{{{\Gamma } \left( {\frac{\alpha }{2}} \right)}}{\text{e}}^{{ - \frac{1}{2} \left( {\frac{y}{\surd 2\sigma }} \right)^{2} }} {\text{d}} \left( {\frac{y}{\surd 2\sigma }} \right) \\ & =^{\alpha = 1} { }\frac{2}{{2^{1/2} }} \frac{1}{{\surd {\uppi }}}{\text{e}}^{{ - \frac{1}{2} \left( {\frac{y}{\surd 2\sigma }} \right)^{2} }} {\text{d}} \left( {\frac{y}{\surd 2\sigma }} \right) \\ & = 2\frac{1}{{\surd 2{\uppi }}}{\text{e}}^{{ - \frac{1}{2} \left( {\frac{y}{\surd 2\sigma }} \right)^{2} }} {\text{d}} \left( {\frac{y}{\surd 2\sigma }} \right) \\ \end{aligned} $$
$$\left(*\right)\frac{1}{{2}^{\frac{\alpha }{2}}} {\left(\frac{{\left(\frac{y}{\sqrt{2\alpha }}\right)}^{2}}{\Gamma \left(\frac{\alpha }{2}\right)}\right)}^{\frac{\alpha }{2}-1}{\mathrm{e}}^{-\frac{1}{2}}{\left(\frac{y}{\sqrt{2\sigma }}\right)}^{2}2\frac{y}{\sqrt{2\sigma }}\mathrm{d} \left(\frac{2}{\sqrt{2\sigma }}\right)=\frac{2}{{2}^{\frac{\alpha }{2}}}{\left(\frac{\frac{y}{\sqrt{2\sigma }}}{\Gamma \frac{\alpha }{2}}\right)}^{\alpha -2+1}{\mathrm{e}}^{-\frac{1}{2}}{\left(\frac{y}{\sqrt{2\sigma }}\right)}^{2}\mathrm{d}\left(\frac{2}{\sqrt{2\sigma }}\right)=\frac{2}{{2}^{\alpha /2 }}{\left(\frac{{\left(\frac{y}{\sqrt{2\alpha }}\right)}^{2}}{\Gamma \left(\frac{\alpha }{2}\right)}\right)}^{\frac{\alpha }{2}-1}{\mathrm{e}}^{-\frac{1}{2}}{\left(\frac{y}{\sqrt{2\sigma }}\right)}^{2}\mathrm{d} \left(\frac{2}{\sqrt{2\sigma }}\right)=\frac{2}{{2}^{1/2}}\frac{1}{\surd \pi }{\mathrm{e}}^{-\frac{1}{2} \frac{y}{\surd 2\sigma }}\mathrm{d}\left(\frac{2}{\sqrt{2\sigma }}\right)$$

Then the variable y follows a Gaussian, null mean and variance \(2\sigma \).

$$\frac{1}{{2}^{\alpha /2 }}{\left(\frac{\frac{{y}^{2}}{{\sigma }^{2}}}{\Gamma \left(\frac{\alpha }{2}\right)}\right)}^{\frac{\alpha }{2}-1}{\mathrm{e}}^{-\frac{{y}^{2}}{2{\sigma }^{2}}}\frac{2}{{\sigma }^{2}}y\mathrm{d}y = \frac{2}{{2}^{\alpha /2 }} \frac{1}{{\sigma }^{\alpha -2+2}} \frac{{\left(y\right)}^{\alpha -2+1}}{\Gamma \left(\frac{\alpha }{2}\right)} {\mathrm{e}}^{-\frac{{y}^{2}}{2{\sigma }^{2}}} \mathrm{d}y = \frac{2}{{2}^{\alpha /2 }}\frac{1}{{\sigma }^{\alpha }}\frac{{\left(y\right)}^{\alpha -2+1}}{\Gamma \left(\frac{\alpha }{2}\right)}{\mathrm{e}}^{-\frac{{y}^{2}}{2{\sigma }^{2}}} \mathrm{d}y =2\frac{1}{\surd 2\pi \sigma }{\mathrm{e}}^{-\frac{{y}^{2}}{2{\sigma }^{2}}} \mathrm{d}y$$

Therefore, the measure of probability is:

$$\mathrm{d}\stackrel{\sim }{P}\left(y;\sigma \right)=\frac{1}{\surd 2\pi }{\mathrm{e}}^{-\frac{1}{2}{\left(\frac{y}{\surd 2\sigma }\right)}^{2}} \mathrm{d}\left(\frac{y}{\surd 2\sigma }\right), - \infty <y< \infty .$$

On the opposite, if \(\frac{y}{\surd 2\sigma }\) follows a Gaussian

$$\mathrm{d}\stackrel{\sim }{P}=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{1}{2}{\left(\frac{y}{\surd 2\sigma }\right)}^{2}} \mathrm{d}\left(\frac{y}{\surd 2\sigma }\right), - \infty <y< \infty $$

Then, \(x={\left(\frac{y}{\surd 2\sigma }\right)}^{2}, 0<x<\infty , \mathrm{d}x=2\frac{y}{\surd 2\sigma }\mathrm{d}\left(\frac{y}{\surd 2\sigma }\right)\) and, we obtain that x follows a \({\chi }^{2}\) variable of 1 degree of freedom because there is only one standard Gaussian variable.

On the other hand, if the variable x, follows a \({\chi }^{2}\) with a parameter \(\alpha \), then the average square root of x is:

$${\langle {x}^\frac{1}{2}\rangle }_{{\mathrm{Chi}}^{2}}={\int }_{0}^{\infty }{x}^\frac{1}{2} . \frac{1}{{2}^{\frac{\alpha }{2}}} \frac{{x}^{\frac{\alpha }{2}-1}}{\Gamma \left(\frac{\alpha }{2}\right)}{\mathrm{e}}^{-\frac{x}{2}}\mathrm{d}x$$
$$= {\int }_{0}^{\infty }\frac{1}{{2}^{\frac{\alpha }{2}}} \frac{{x}^{\frac{\alpha }{2}-1}}{\Gamma \left(\frac{\alpha }{2}\right)} {\mathrm{e}}^{-\frac{x}{2}}\mathrm{d}x$$
$$= \frac{{2}^\frac{1}{2}\Gamma \left(\frac{\alpha +1}{2}\right)}{\Gamma \left(\frac{\alpha }{2}\right)} {\int }_{0}^{\infty }\frac{1}{{2}^{\frac{\alpha +1}{2}}} \frac{{x}^{\frac{\alpha +1}{2}-1}}{\Gamma \left(\frac{\alpha +1}{2}\right)} {\mathrm{e}}^{-\frac{x}{2}}\mathrm{d}x$$

with \(=\frac{1}{2}\),

$$\left\langle {x\frac{1}{2}} \right\rangle {\text{Chi}}^{2} = \frac{{2\frac{1}{2}\Gamma \left( {\frac{{\frac{1}{2} + 1}}{2}} \right)}}{{\Gamma \left( {\frac{{\frac{1}{2}}}{2}} \right)}} = \frac{1}{{2\sqrt \pi }}\Gamma \left( {\frac{3}{4}} \right)$$

In consequence, \(x={\left(\frac{y}{\sqrt{\sigma }}\right)}^{2}, 0<x<\infty \to {x}^\frac{1}{4}={\left(\frac{y}{\sqrt{\sigma }}\right)}^\frac{1}{2}= \frac{{y}^\frac{1}{2}}{{\left(\sigma \right)}^\frac{1}{4}} , 0<x<\infty , {y}^\frac{1}{2}={\left(\sigma \right)}^\frac{1}{4}{ x}^\frac{1}{4},\)

$${\langle {y}^\frac{1}{2}\rangle }_{\mathrm{G}}={\langle {\left(\sigma \right)}^\frac{1}{4}{x}^\frac{1}{4}\rangle }_{{\mathrm{Chi}}^{2}}= {\left(\sigma \right)}^\frac{1}{4}{\langle { x}^\frac{1}{4}\rangle }_{{\mathrm{Chi}}^{2}}$$

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Gutierrez-Lopez, A. A Robust Gaussian variogram estimator for cartography of hydrological extreme events. Nat Hazards 107, 1469–1488 (2021). https://doi.org/10.1007/s11069-021-04641-9

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