Compositional Bayesian indicator estimation

Abstract

Indicator kriging is widely used for mapping spatial binary variables and for estimating the global and local spatial distributions of variables in geosciences. For continuous random variables, indicator kriging gives an estimate of the cumulative distribution function, for a given threshold, which is then the estimate of a probability. Like any other kriging procedure, indicator kriging provides an estimation variance that, although not often used in applications, should be taken into account as it assesses the uncertainty of the estimate. An alternative approach to indicator estimation is proposed in this paper. In this alternative approach the complete probability density function of the indicator estimate is evaluated. The procedure is described in a Bayesian framework, using a multivariate Gaussian likelihood and an a priori distribution which are both combined according to Bayes theorem in order to obtain a posterior distribution for the indicator estimate. From this posterior distribution, point estimates, interval estimates and uncertainty measures can be obtained. Among the point estimates, the median of the posterior distribution is the maximum entropy estimate because there is a fifty-fifty chance of the unknown value of the estimate being larger or smaller than the median; that is, there is maximum uncertainty in the choice between two alternatives. Thus in some sense, the latter is an indicator estimator, alternative to the kriging estimator, that includes its own uncertainty. On the other hand, the mode of the posterior distribution estimator, assuming a uniform prior, is coincidental with the simple kriging estimator. Additionally, because the indicator estimate can be considered as a two-part composition which domain of definition is the simplex, the method is extended to compositional Bayesian indicator estimation. Bayesian indicator estimation and compositional Bayesian indicator estimation are illustrated with an environmental case study in which the probability of the content of a geochemical element in soil being over a particular threshold is of interest. The computer codes and its user guides are public domain and freely available.

Appendices

Appendix A. Equivalence of the maximum likelihood indicator estimator and simple indicator kriging estimator

The starting point is the likelihood given by Eq. (5) and that is repeated here:

$$ \ell (I({\mathbf{u}}_{0} );{\mathbf{I}},{\tilde{\varvec{\mu }}}_{I} ,\widetilde{{\varvec{\Upsigma}}}_{I} ) = (2\pi )^{{{\frac{ - (n + 1)}{2}}}} \left| {\widetilde{{\varvec{\Upsigma}}}_{I} } \right|^{{ - \frac{1}{2}}} \exp \left\{ { - \frac{1}{2}\left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]} \right\} $$

(17)

The maximum likelihood estimator is obtained by maximizing Eq. 17 or equivalently, by convenience, by minimizing the negative of the loglikelihood:

$$ L(I({\mathbf{u}}_{0} )) = - \ln \left( {\ell (I({\mathbf{u}}_{0} );{\mathbf{I}},{\tilde{\varvec{\mu }}}_{I} ,\widetilde{{\varvec{\Upsigma}}}_{I} )} \right) $$

(18)

$$ L(I({\mathbf{u}}_{0} )) = {\frac{n + 1}{2}}\ln (2\pi ) + \frac{1}{2}\ln \left| {\widetilde{{\varvec{\Upsigma}}}_{I} } \right| + \frac{1}{2}\left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right] $$

(19)

Thus the maximum likelihood estimator is given by the solution to the equation:

$$ {\frac{{\partial L(I({\mathbf{u}}_{0} ))}}{{\partial I({\mathbf{u}}_{0} )}}} = 0 $$

(20)

Thus:

$$ {\frac{{\partial \left[ {\frac{1}{2}\left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]} \right]}}{{\partial I({\mathbf{u}}_{0} )}}} = 0 $$

(21)

Using the previous notation:

$$ \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} = \left[ {I({\mathbf{u}}_{0} ) - m_{I} ,({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} } \right] $$

(22)

$$ \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right] = \left[ {\begin{array}{*{20}c} {I({\mathbf{u}}_{0} ) - m_{I} } \\ {({\mathbf{I}} - {\varvec{\mu}}_{I} )} \\ \end{array} } \right] $$

(23)

$$ \widetilde{{\varvec{\Upsigma}}}_{I} = \left[ {\begin{array}{*{20}c} {\Upsigma_{00} } & {\Upsigma_{0I} } \\ {\Upsigma_{I0} } & {{\varvec{\Upsigma}}_{I} } \\ \end{array} } \right] $$

(24)

Having $\Upsigma_{I0} = \Upsigma_{0I}^{T} :( 1\times n)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{a}}\,{\text{vector}}\,{\text{of}}\,n\,{\text{elements}}.$ $({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} :( 1\times n)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{a}}\,{\text{vector}}\,{\text{of}}\,n\,{\text{elements}}.$ $\Upsigma_{00} :( 1\times 1)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{an}}\,{\text{scalar}}$ ${\varvec{\Upsigma}}_{I} :(n \times n)\,{\text{matrix}}$ $\widetilde{{\varvec{\Upsigma}}}_{I} = \left[ {\begin{array}{*{20}c} {\Upsigma_{00} } & {\Upsigma_{0I} } \\ {\Upsigma_{I0} } & {{\varvec{\Upsigma}}_{I} } \\ \end{array} } \right]$

An very well known relation that is needed is the inverse of a partitioned matrix (Graybill 1976):

$$ \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} = \left[ {\begin{array}{*{20}c} {\Upsigma_{00} } & {\Upsigma_{0I} } \\ {\Upsigma_{0I}^{T} } & {{\varvec{\Upsigma}}_{I} } \\ \end{array} } \right]^{ - 1} = \left[ {\begin{array}{*{20}c} {\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} } & { - \left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} } \\ { - \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} } & {\left( {\Upsigma_{I} - \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} } \\ \end{array} } \right] $$

(25)

Then

$$ \frac{1}{2}\left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right] $$

(26)

can be expanded as:

$$ \frac{1}{2}\left[ {I({\mathbf{u}}_{0} ) - m_{I} ,({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} } \right]\left[ {\begin{array}{*{20}c} {\Upsigma_{00} } & {\Upsigma_{0I} } \\ {\Upsigma_{0I}^{T} } & {{\varvec{\Upsigma}}_{I} } \\ \end{array} } \right]^{ - 1} \left[ {\begin{array}{*{20}c} {I({\mathbf{u}}_{0} ) - m_{I} } \\ {({\mathbf{I}} - {\varvec{\mu}}_{I} )} \\ \end{array} } \right] = \frac{1}{2}\left[ {A,B^{T} } \right]\left[ {\begin{array}{*{20}c} C & D \\ E & F \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} A \\ B \\ \end{array} } \right] = \frac{1}{2}\left[ {A^{2} C + AB^{T} E + ADB + B^{T} FB} \right] $$

(27)

With $A = I({\mathbf{u}}_{0} ) - m_{I} \quad ( 1\times 1)\quad {\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{an}}\,{\text{scalar}}$ $B = ({\mathbf{I}} - {\varvec{\mu}}_{I} )\quad (n \times 1)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{a}}\,{\text{vector}}\,{\text{of}}\,n\,{\text{elements}}$ $C = \left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} \quad ( 1\times 1)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{an}}\,{\text{scalar}}$ $D = - \left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \quad ( 1\times n){\text{ matrix}},\,{\text{i}}.{\text{e}}.\,{\text{a}}\,{\text{vector}}\,{\text{of}}\,n\,{\text{elements}}$ $E = - \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} \quad (n \times 1)\,{\text{matrix}},\,{\text{i}}.{\text{e}}.\,{\text{a}}\,{\text{vector}}\,{\text{of}}\,n\,{\text{elements}}$ $F = \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} {\varvec{\Upsigma}}_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \quad (n \times n){\text{ matrix}}$

Thus

$$ {\frac{{\partial \left[ {\frac{1}{2}\left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]^{T} \widetilde{{\varvec{\Upsigma}}}_{I}^{ - 1} \left[ {\widetilde{{\mathbf{I}}} - {\tilde{\varvec{\mu }}}_{I} } \right]} \right]}}{{\partial I({\mathbf{u}}_{0} )}}} = AC + \frac{1}{2}B^{T} E + \frac{1}{2}DB = 0 $$

(28)

$$ \begin{aligned} AC + \frac{1}{2}B^{T} E + \frac{1}{2}DB & = (I(u_{0} ) - m_{I} )\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} - \frac{1}{2}({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} \\ & \quad - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} \Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(29)

Pre-multiplying Eq. 29 by $\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right),{\text{a}}(1 \times 1)\,{\text{matrix,}}\,{\text{i}} . {\text{e}} .\,{\text{a}}\,{\text{scalar:}}$

$$ (\Upsigma_{00} - \Upsigma_{0I} \Upsigma_{I}^{ - 1} \Upsigma_{0I}^{T} )(I(u_{0} ) - m_{I} )\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)^{ - 1} - \frac{1}{2}(\Upsigma_{00} - \Upsigma_{0I} \Upsigma_{I}^{ - 1} \Upsigma_{0I}^{T} )({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}(\Upsigma_{00} - \Upsigma_{0I} \Upsigma_{I}^{ - 1} \Upsigma_{0I}^{T} ) \left( {\Upsigma _{{00}} - \Upsigma _{{0I}} {\mathbf{\Upsigma }}_{I}^{{ - 1}} \Upsigma _{{0I}}^{T} } \right)^{{ - 1}} \Upsigma _{{0I}} {\mathbf{\Upsigma }}_{I}^{{ - 1}} ({\mathbf{I}} - {\mathbf{\mu }}_{I} ) = 0$$

(30)

one reach the expression:

$$ (I(u_{0} ) - m_{I} ) - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} \left( {\Upsigma_{I} - \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\,=\,0 $$

(31)

A needed result is the matrix inversion lemma (Minamide 1985):

$$ \left( {{\varvec{\Upsigma}}_{I} - \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} \Upsigma_{0I} } \right)^{ - 1} = {\varvec{\Upsigma}}_{I}^{ - 1} + {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} (\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} )^{ - 1} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} $$

(32)

Inserting (32) into (31) one obtains:

$$ \begin{aligned} & (I(u_{0} ) - m_{I} ) - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} \\ & \left( {{\varvec{\Upsigma}}_{I}^{ - 1} + {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} (\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} )^{ - 1} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} } \right)\Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(33)

Next, expanding the second term of the left hand side of Eq. 33:

$$ \begin{aligned} & (I(u_{0} ) - m_{I} ) - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} \left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right) \\ & {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} (\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} )^{ - 1} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} {\varvec{\Upsigma}}_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(34)

And after expanding the third term of the left hand side of Eq. 34 and simplification:

$$ \begin{aligned} (I(u_{0} ) - m_{I} ) & - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} {\varvec{\Upsigma}}_{0I}^{T} \Upsigma_{00}^{ - 1} \\ & \quad - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(35)

And because $\frac{1}{2}({\mathbf{I}} - {\varvec{\mu}}_{I} )^{T} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} = \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )$, as $\Upsigma_{I}^{ - 1}$is symmetric, (35) can be expressed as:

$$ \begin{aligned} (I(u_{0} ) - m_{I} ) & - \frac{1}{2}\left( {\Upsigma_{00} - \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} } \right)\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} {\varvec{\Upsigma}}_{0I}^{T} \Upsigma_{00}^{ - 1} \\ & \quad - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(36)

And expanding the second term of (36) one reach the expression:

$$ \begin{aligned} & (I(u_{0} ) - m_{I} ) - \frac{1}{2}\Upsigma_{00} \Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\Upsigma_{00}^{ - 1} + \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) \\ & \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} {\varvec{\Upsigma}}_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(37)

Which after simplification and because $\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T}$ is a (1 × 1) matrix, i.e. an scalar and it has changed places accordingly.one has:

$$ \begin{aligned} & (I(u_{0} ) - m_{I} ) - \frac{1}{2}\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) + \frac{1}{2}\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} )\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} \Upsigma_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} \Upsigma_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) \\ & \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} {\varvec{\Upsigma}}_{0I}^{T} \Upsigma_{00}^{ - 1} - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 \\ \end{aligned} $$

(38)

The third and fourth terms of the left hand side cancel to obtain:

$$ (I(u_{0} ) - m_{I} ) - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) - \frac{1}{2}\Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) = 0 $$

(39)

one finally has

$$ I(u_{0} ) = m_{I} + \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} ({\mathbf{I}} - {\varvec{\mu}}_{I} ) $$

(40)

It may be seen else were (e.g. Goovaerts 1997) how Eq. 32 is exactly the simple indicator kriging estimator.

$$ I(u_{0} ) = m_{I} + \sum\limits_{i = 1}^{n} {\lambda_{i} } (I({\mathbf{u}}_{i} ) - m_{I} ) $$

(41)

were the set of weights is given, in matrix form, by

$$ {\varvec{\uplambda}} = \Upsigma_{0I} {\varvec{\Upsigma}}_{I}^{ - 1} $$

(42)

Appendix B. Trapezoidal prior distribution

A flexible prior distribution for coding prior information at each unsampled location from secondary information is given by the trapezoidal pdf prior. The trapezoidal prior has been implemented in the provided software and it can be defined with four parameters which may be seen in Fig. 11. These four parameters (a, b, c, d) are but four indicator values and thus must be in the interval [0,1] and in non decreasing order $0 \le a \le b \le c \le d \le 1$. With this four parameters, the trapezoidal pdf (Fig. 11a) may be transformed to the uniform in a given interval (Fig. 11e), triangular (Fig. 11b), left rectangular triangular (Fig. 11c), etc. This is an efficient way of coding fuzzy prior information into a prior distribution. Additionaly do not dependent on parametric parameters (mean, standard deviation, shape parameter, …) of parametrized distribution. For example the trapezoidal pdf of Fig. 11a says that the prior value of the indicator is most likely in the interval [b,c], with smaller likelihood is in the interval [a,b] and [c,d] and no likelihood in the interval [0,a] or [d,1]. The uniform prior will be specified with $a = b = 0$ and $c = d = 1$ (Fig. 11e).

Fig. 11

Different forms of prior using the trapezoidal pdf. a trapezoidal, b triangular, c left rectangular triangular, d right rectangular triangular, e uniform, f trapezoidal