Birnbaum–Saunders spatial modelling and diagnostics applied to agricultural engineering data

Abstract

Applications of statistical models to describe spatial dependence in geo-referenced data are widespread across many disciplines including the environmental sciences. Most of these applications assume that the data follow a Gaussian distribution. However, in many of them the normality assumption, and even a more general assumption of symmetry, are not appropriate. In non-spatial applications, where the data are uni-modal and positively skewed, the Birnbaum–Saunders (BS) distribution has excelled. This paper proposes a spatial log-linear model based on the BS distribution. Model parameters are estimated using the maximum likelihood method. Local influence diagnostics are derived to assess the sensitivity of the estimators to perturbations in the response variable. As illustration, the proposed model and its diagnostics are used to analyse a real-world agricultural data set, where the spatial variability of phosphorus concentration in the soil is considered—which is extremely important for agricultural management.

Appendices

Appendix 1: The score vector

For the BS log-linear spatial model, the score vector is defined by

$$\begin{aligned} {\varvec{U}}( {\varvec{\theta} }) =\left( \frac{\partial \ell ({\varvec{\theta} })}{\partial \alpha },\,\frac{\partial \ell ({\varvec{\theta} })}{\partial \mu },\,\frac{\partial \ell ({\varvec{\theta} })}{\partial {\varvec{\varphi} }}\right) ^{\top} =\left( U( \alpha) ,\,U( \mu ) ,\, U\left(\varphi _{1}\right),\, U\left(\varphi _{2}\right),\, U\left(\varphi _{3}\right)\right) ^{\top} . \end{aligned}$$

Using the log-likelihood function defined in (20), we have

$$\begin{aligned} U( \alpha)&= \frac{\partial \ell ({\varvec{\theta} })}{\partial \alpha }= \frac{\partial }{\partial \alpha }\left( -n\log (\alpha )-\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}\right)\\& =-\frac{n}{\alpha }+\frac{4}{\alpha ^{3}}{\varvec{V}}^{\top} {\varvec{\Sigma }}^{-1}{\varvec{V}},\\ U( \mu )&= \frac{\partial \ell ({\varvec{\theta} })}{\partial \mu }=\frac{\partial }{\partial \mu }\left( -\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}+ \sum\limits_{i=1}^{n}\log \left( \cosh \left( \frac{y_{i}-\mu }{2}\right) \right) \right) \\&= \frac{2}{\alpha ^{2}}{\varvec{W}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}-\frac{1}{2} \sum\limits_{i=1}^{n}\tanh\left( \frac{y_{i}-\mu }{2}\right) ,\\ U\left( \varphi _{j}\right)&= \frac{\partial \ell ({\varvec{\theta} })}{\partial \varphi _{j}}= \frac{\partial }{\partial \varphi _{j}}\left( -\frac{1}{2}\log ( |{\varvec{\Sigma} }|) -\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}\right) \\&= -\frac{1}{2}{\text {tr}}\left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\varphi _{j}}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\varphi _{j}}{\varvec{\Sigma} }^{-1}{\varvec{V}},\quad j=1,\,2,\,3, \end{aligned}$$

where ${\varvec{W}}=(W_{1},\ldots ,W_{n})^{\top},$ with $W_{i}=\cosh ({(y_{i}-\mu )}/{2}),$ and ${\varvec{V}}=(V_{1},\ldots ,V_{n})^{\top}$ such as given in (19) with $V_{i} =\sinh({(y_{i}-\mu )}/{2}),$ for $i=1,\ldots ,n.$ Thus, considering that ${\varvec{\Sigma} }=\varphi _{1}{\varvec{I}}_{n}+\varphi _{2}{\varvec{R}},$ we have

$$\begin{aligned} \frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{1}}={\varvec{I}}_{n},\quad \frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{2}}={\varvec{R}},\quad \frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{3}}=\varphi _{2}\frac{\partial {\varvec{R}}}{\partial \varphi _{3}}. \end{aligned}$$

Therefore,

$$\begin{aligned} U\left( \varphi _{1}\right)&= -\frac{1}{2}\text {tr}\left( {\varvec{\Sigma} }^{-1}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} \left( {\varvec{\Sigma} }^{-1}\right) ^{2}{\varvec{V}},\\ U\left( \varphi _{2}\right)&= -\frac{1}{2}\text {tr}\left( {\varvec{\Sigma} }^{-1}{\varvec{R}}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{R}}{\varvec{\Sigma} }^{-1}{\varvec{V}},\\ U\left( \varphi _{3}\right)&= -\frac{\varphi _{2}}{2}\text {tr}\left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _{3}}\right) +\frac{2\varphi _{2}}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _{3}}{\varvec{\Sigma} }^{-1}{\varvec{V}}. \end{aligned}$$

Considering the Matérn model to describe the spatial variability given in (11), we have that ${\partial {\varvec{R}}}/{\partial \varphi _{3}}=({\partial r_{ij}}/{\partial \varphi _{3}}),$ where, for $K^{\prime}_{\delta }(u)={\partial K_{\delta} (u)}/{\partial u}=- ({1}/{2})(K_{\delta -1}(u)+ K_{\delta +1}(u)),$

$$\begin{aligned} \frac{\partial r_{ij}}{\partial \varphi _{3}}=-\left( \frac{1}{\varphi _{3}}\right) \left( \delta r_{ij}+\frac{1}{2^{\delta -1}\mu (\delta )}\left( \frac{h_{ij}}{\varphi _{3}}\right) ^{\delta +1}K^{\prime}_{\delta }\left( \frac{h_{ij}}{\varphi _{3}}\right) \right) ,\quad i\ne j,\quad i,\,j=1,\ldots ,n. \end{aligned}$$

Appendix 2: The observed information matrix

The observed Fisher information matrix for the BS log-linear spatial model is defined by $-{\ddot{\varvec{\ell }}}({\varvec{\theta} })$ evaluated at ${\varvec{\theta} }=\widehat{\varvec{\theta }},$ where ${\ddot{\varvec{\ell }}}({\varvec{\theta} })$ is the Hessian matrix given by

$$\begin{aligned} {\varvec{\ddot{{\ell} }}}({\varvec{\theta} })=\left( \begin{array}{ccc} \frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \alpha^2 } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \alpha \partial \mu } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \alpha \partial {\varvec{\varphi} }^{\top} }\\ \frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \mu \partial \alpha } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \mu^2 } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \mu \partial {\varvec{\varphi} }^{\top} }\\ \frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial {\varvec{\varphi} }\partial \alpha } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial {\varvec{\varphi} }\partial \mu } &\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial {\varvec{\varphi} }\partial {\varvec{\varphi} }^{\top} } \end{array} \right) =\left( \begin{array}{ccc} {\ddot{\ell }}_{\alpha \alpha } &{\ddot{\ell }}_{\alpha \mu } &{\varvec{\ddot{{\ell} }}}_{\alpha {\varvec{\varphi} }} \\ {\ddot{\ell }}_{\mu \alpha } &{\ddot{\ell} }_{\mu \mu } &{\varvec{\ddot{{\ell} }}}_{\mu {\varvec{\varphi} }} \\ {\varvec{\ddot{{\ell} }}}_{{\varvec{\varphi} } \alpha } &{\varvec{\ddot{{\ell} }}}_{{\varvec{\varphi} } \mu } &{\varvec{\ddot{{\ell} }}}_{{\varvec{\varphi} } {\varvec{\varphi} }} \end{array} \right) , \end{aligned}$$

with, for ${\ddot{\ell }}_{\alpha \mu }={\ddot{\ell} }_{\mu \alpha },$

$$\begin{aligned} \ddot{\ell }_{\alpha \alpha }= & {} \frac{\partial ^2\ell (\varvec{\theta })}{\partial \alpha ^2}= \frac{\partial }{\partial \alpha }\left( -\frac{n}{\alpha }+\frac{4}{\alpha ^3}\varvec{V}^\top \varvec{\Sigma }^{-1}\varvec{V}\right) = \frac{n}{\alpha ^2}-\frac{12}{\alpha ^4}\varvec{V}^\top \varvec{\Sigma }^{-1}\varvec{V},\\ \ddot{\ell }_{\mu \alpha }= & {} \frac{\partial ^2\ell (\varvec{\theta })}{\partial \mu \partial \alpha }= \frac{\partial }{\partial \mu }\left( \frac{4}{\alpha ^3}\varvec{V}^\top \varvec{\Sigma }^{-1}\varvec{V}\right) = -\frac{4}{\alpha ^3}\varvec{U}^\top \varvec{\Sigma }^{-1}\varvec{V}, \end{aligned}$$

$$\begin{aligned} \ddot{\ell }_{\mu \mu }= & {} \frac{\partial ^2\ell (\varvec{\theta })}{\partial \mu ^2}= \frac{\partial }{\partial \mu }\left( \frac{2}{\alpha ^2}\varvec{U}^\top \varvec{\Sigma }^{-1}\varvec{V}-\frac{1}{2}\displaystyle \sum _{i=1}^{n}\text {tanh}\left( \frac{y_i-\mu }{2}\right) \right) \\= & {} - \frac{1}{\alpha ^2}\left( \varvec{V}^\top \varvec{\Sigma }^{-1}\varvec{V}+\varvec{U}^\top \varvec{\Sigma }^{-1}\varvec{U}\right) +\frac{1}{4} \sum _{i=1}^{n}\text {sech}^2\left( \frac{y_i-\mu }{2}\right) . \end{aligned}$$

Note that ${\ddot{\ell} }_{\varphi \alpha }={\ddot{\ell} }_{\alpha \varphi} ^{\top}$ is a $3\times 1$ vector with elements given by

$$\begin{aligned} {\ddot{\ell }}_{\alpha \varphi _{j}}= \frac{\partial ^{2}\ell ({\varvec{\theta} })}{\partial \alpha \partial \varphi _{j}}= \frac{\partial }{\partial \alpha }\left( -\frac{1}{2}{\text {tr}}\left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial\varphi _{j}}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial\varphi _{j}}{\varvec{\Sigma} }^{-1}{\varvec{V}}\right) = -\frac{4}{\alpha ^{3}}{\varvec{V}}^{\top} \left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\right) {\varvec{V}}, \quad j = 1, 2, 3\end{aligned}$$

where ${\partial {\varvec{\Sigma} }}/{\partial \varphi _{i}}$ is given in Appendix 1. Then, with ${\ddot{\ell} }_{{\alpha }{\varphi }_{1}}={\ddot{\ell} }_{{\varphi }_{1}{\alpha }},$ ${\ddot{\ell} }_{{\alpha }{\varphi }_{2}}={\ddot{\ell} }_{{\varphi }_{2}{\alpha }}$ and ${\ddot{\ell} }_{{\alpha }{\varphi }_{3}}={\ddot{\ell} }_{{\varphi }_{3}{\alpha }},$ we have

$$\begin{aligned} {\ddot{\ell} }_{\alpha \varphi _{1}} =-\frac{4}{\alpha ^{3}}{\varvec{V}}^{\top} \left( {\varvec{\Sigma} }^{-1}\right) ^{2}{\varvec{V}},\quad {\ddot{\ell} }_{\alpha \varphi _{2}} =-\frac{4}{\alpha ^{3}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{R}}{\varvec{\Sigma} }^{-1}{\varvec{V}},\quad {\ddot{\ell} }_{\alpha \varphi _{3}} =-\frac{4\varphi _{2}}{\alpha ^{3}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _{3}}{\varvec{\Sigma} }^{-1}{\varvec{V}}. \end{aligned}$$

Furthermore, ${\ddot{\ell} }_{\varphi \mu }={\ddot{\ell} }_{\mu \varphi} ^{\top}$ is a $3\times 1$ vector with elements

$$\begin{aligned} {\ddot{\ell} }_{\mu \varphi _{j}}=\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \mu \partial \varphi _{j}}=\frac{\partial }{\partial \mu }\left( -\frac{1}{2}{\text {tr}}\left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial\varphi _{j}}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial\varphi _{j}}{\varvec{\Sigma} }^{-1}{\varvec{V}}\right) =- \frac{2}{\alpha ^{2}}{\varvec{U}}^{\top} \left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\right) {\varvec{V}}, \quad j = 1, 2, 3 \end{aligned}$$

Then, with ${\ddot{\ell} }_{{\mu }{\varphi }_{1}}={\ddot{\ell} }_{{\varphi }_{1}{\mu }},\, {\ddot{\ell} }_{{\mu }{\varphi }_{2}}={\ddot{\ell} }_{{\varphi }_{2}{\mu }}$ and ${\ddot{\ell} }_{{\mu }{\varphi }_{3}}={\ddot{\ell} }_{{\varphi }_{3}{\mu }},$ we have

$$\begin{aligned} {\ddot{\ell} }_{\mu \varphi _{1}}=-\frac{2}{\alpha ^{2}}{\varvec{U}}^{\top} \left( {\varvec{\Sigma} }^{-1}\right) ^{2}{\varvec{V}},\quad {\ddot{\ell} }_{\mu \varphi _{2}}=-\frac{2}{\alpha ^{2}}{\varvec{U}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{R}}{\varvec{\Sigma} }^{-1}{\varvec{V}},\quad {\ddot{\ell} }_{\mu \varphi _{3}}=-\frac{2\varphi _{2}}{\alpha ^{2}}{\varvec{U}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _{3}}{\varvec{\Sigma} }^{-1}{\varvec{V}}. \end{aligned}$$

Moreover, ${\varvec{\ddot{\ell }}}_{{\varvec{\varphi} }{\varvec{\varphi }}} =({\ddot{\ell }}_{\varphi _{j}\varphi _{k}})$ is a $3\times 3$ symmetric matrix with elements given by

$$\begin{aligned} {\ddot{\ell }}_{{\varphi }_{j}{\varphi }_{k}}=\frac{\partial ^{2} \ell ({\varvec{\theta} })}{\partial \varphi _{j}\varphi _{k}}&= -\frac{1}{2}{\text {tr}}\left( -{\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{k}}+{\varvec{\Sigma} }^{-1}\frac{\partial ^{2}{\varvec{\Sigma} }}{\partial \varphi _{j}\partial \varphi _{k}}\right) +\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} \left( -{\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\varphi _{k}}{\varvec{\Sigma} }^{-1}\right) {\varvec{V}}\\&\quad+\frac{2}{\alpha ^{2}}{\varvec{V}}^{\top} \left( {\varvec{\Sigma} }^{-1}\left( \frac{\partial ^{2}{\varvec{\Sigma} }}{\partial \varphi _{j}\partial \varphi _{k}}{\varvec{\Sigma} }^{-1}+\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}\left( -{\varvec{\Sigma} }^{-1} \frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{k}}{\varvec{\Sigma} }^{-1}\right) \right) \right) {\varvec{V}}, \quad j,\,k=1,\,2,\,3. \end{aligned}$$

Then, with ${\ddot{\ell} }_{{\varphi }_{1}{\varphi }_{2}}={\ddot{\ell} }_{{\varphi }_{2}{\varphi }_{1}},\,{\ddot{\ell} }_{{\varphi }_{1}{\varphi }_{3}}={\ddot{\ell} }_{{\varphi }_{3}{\varphi }_{1}}$ and ${\ddot{\ell} }_{{\varphi }_{2}{\varphi }_{3}}={\ddot{\ell} }_{{\varphi }_{3}{\varphi }_{2}},$ we have

$$\begin{aligned} \ddot{\ell }_{{\varphi }_1{\varphi }_1}&= \frac{1}{2}\text {tr}\left( ({\varvec{\Sigma }}^{-1})^2\right) -\frac{2}{\alpha ^2}{\varvec{V}}^\top \left( ({\varvec{\Sigma }}^{-1})^3+({\varvec{\Sigma}} ^{-1})^2\right) {\varvec{V}},\\ \ddot{\ell }_{{\varphi }_1{\varphi }_2}&= \frac{1}{2}{\text {tr}}\left( {\varvec{\Sigma}} ^{-1}{\varvec{R}}{\varvec{\Sigma }}^{-1}\right) -\frac{2}{\alpha ^2}{\varvec{V}}^\top \left( {\varvec{\Sigma}} ^{-1}{\varvec{R}}({\varvec{\Sigma}} ^{-1})^2+\left( {\varvec{\Sigma}} ^{-1}\right) ^2{\varvec{R}}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}},\\ \ddot{\ell }_{{\varphi }_1{\varphi }_3}&= \frac{\varphi _2}{2}{\text {tr}}\left( {\varvec{\Sigma}} ^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma}} ^{-1}\right) -\frac{2\varphi _2}{\alpha ^2}{\varvec{V}}^\top \left( {\varvec{\Sigma}} ^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}({\varvec{\Sigma}} ^{-1})^2+({\varvec{\Sigma }}^{-1})^2\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}},\\ \ddot{\ell }_{{\varphi }_2{\varphi }_2}&= \frac{1}{2}\text {tr}\left( {\varvec{\Sigma}} ^{-1}{\varvec{R}}{\varvec{\Sigma }}^{-1}{\varvec{R}}\right) -\frac{4}{\alpha ^2}{\varvec{V}}^\top \left( {\varvec{\Sigma}} ^{-1}{\varvec{R}}{\varvec{\Sigma }}^{-1}{\varvec{R}}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}},\\ \ddot{\ell }_{{\varphi }_2{\varphi }_3}&= -\frac{1}{2}\text {tr}\left( -\varphi _2{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma}} ^{-1}{\varvec{R}}+{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}\right) +\frac{2}{\alpha ^2}{\varvec{V}}^\top \left( -\varphi _2{\varvec{\Sigma}} ^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma}} ^{-1}{\varvec{R}}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}}\\&\quad+\frac{2}{\alpha ^2}{\varvec{V}}^\top {\varvec{\Sigma}} ^{-1}\left( \frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}+{\varvec{R}}\left( -\varphi _2{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}\right) \right) {\varvec{V}},\\ \ddot{\ell }_{{\varphi }_3{\varphi }_3}&= -\frac{\varphi _2}{2}{\text {tr}}\left( -\varphi _2{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma}} ^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}+{\varvec{\Sigma }}^{-1}\frac{\partial ^2{\varvec{R}}}{\partial \varphi _3^2}\right) -\frac{2\varphi _2^2}{\alpha ^2}{\varvec{V}}^\top \left( {\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma}} ^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}}\\&\quad+\frac{2\varphi _2}{\alpha ^2}{\varvec{V}}^\top {\varvec{\Sigma }}^{-1}\left( \frac{\partial ^2{\varvec{R}}}{\partial \varphi _3^2}{\varvec{\Sigma }}^{-1}-\varphi _2\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{R}}}{\partial \varphi _3}{\varvec{\Sigma }}^{-1}\right) {\varvec{V}}. \end{aligned}$$

Appendix 3: Expected information matrix

The expected Fisher information matrix is ${\varvec{I}}({\varvec{\theta} }) = {\text {E}}(-{\varvec{\ddot{\ell} }}({\varvec{\theta} })).$ For the BS log-linear spatial model, this matrix is given by

$$\begin{aligned} {\varvec{I}}({\varvec{\theta} })=\left( \begin{array}{ccc} I_{\alpha \alpha } & I_{\alpha \mu } & {\varvec{I}}_{\alpha {\varvec{\varphi} }}\\ I_{\mu \alpha } & I_{\mu \mu } & {\varvec{I}}_{\mu {\varvec{\varphi} }}\\ {\varvec{I}}_{\varvec{\varphi }\alpha } &{\varvec{I}}_{\varvec{\varphi }\mu }&{\varvec{I}}_{\varvec{\varphi }{\varvec{\varphi} }} \end{array} \right) . \end{aligned}$$

Since the model error ${\varvec{\varepsilon} }\sim {\text {log}}{\text{-BS}}_{n}(\alpha {\mathbf{1}},\,{\mathbf{0}},\,{\varvec{\Sigma} }),$ we have that $({2}/{\alpha }){\varvec{V}}\sim {\text {N}}_{n}({\mathbf{0}},\,{\varvec{\Sigma} }),$ with ${\varvec{V}}$ given in (19). Then (see Muirhead 1982), $W = ({2}/{\alpha }){\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}({2}/{\alpha }){\varvec{V}}\sim \chi _{n}^{2},$ where $\chi _{n}^{2}$ denotes the chi-squared distribution with n degrees of freedom, and then, ${\text {E}}(W)=n.$ Thus,

$$\begin{aligned} I_{\alpha \alpha }={\text {E}}\left(-{\ddot{\ell} }_{\alpha \alpha }\right)={\text {E}}\left( - \frac{n}{\alpha ^{2}}+\frac{12}{\alpha ^{4}}{\varvec{V}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}\right) =\frac{2n}{\alpha ^{2}}. \end{aligned}$$

Moreover, by using ${\text {E}}\left( {\varvec{X}}^{\top} {\varvec{A}}{\varvec{X}}\right) =( {\text {E}}( {\varvec{X}})) ^{\top} {\varvec{A}}( {\text {E}}( {\varvec{X}})) +{\text {tr}}( {\varvec{A}}{\varvec{C}}),$ where ${\varvec{C}}$ is the covariance matrix of ${\varvec{X}}$ (see Kendrick 2002), ${\varvec{I}}_{\varvec{\varphi }{\varvec{\varphi} }} = (I_{\varphi _{j} \varphi _{k}})$ is a symmetric $3\times 3$ matrix with elements

$$\begin{aligned} I_{\varphi _{j} \varphi _{k}}&= {\text {E}}\left(-{\ddot{\ell} }_{\varphi _{j}\varphi _{k}}\right)=\frac{1}{2}{\text {tr}}\left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma }}}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1} \frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{k}}\right) , \quad j,\,k=1,\,2,\,3. \end{aligned}$$

In addition, ${\varvec{I}}_{\varvec{\varphi }\alpha } = {\varvec{I}}_{\alpha {\varvec{\varphi} }}^{\top} = (I_{\varphi _{1} \alpha },\, I_{\varphi _{2} \alpha },\, I_{\varphi _{3} \alpha })^{\top}$ is a $3\times 1$ vector with elements given by

$$\begin{aligned} I_{\varphi _{j}\alpha }={\text {E}}\left(-{\ddot{\ell} }_{\varphi _{j}\alpha }\right)=\frac{1}{\alpha }{\text {E}}\left( \left( \frac{2}{\alpha }{\varvec{V}}\right) ^{\top} \left( {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\right) \left( \frac{2}{\alpha }{\varvec{V}}\right) \right) =\frac{1}{\alpha }{\text {tr}}\left({ \varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _j}\right) , \quad j=1,\,2,\,3, \end{aligned}$$

where ${\text {tr}}({\varvec{A}})$ denotes the trace of the matrix ${\varvec{A}}.$ To obtain the elements $I_{\alpha \mu },\, I_{\mu \mu }$ and $I_{\mu {\varvec{\varphi} }},$ for ${(y_{i}-\mu )}/{2}$ expected to be small enough with $\cosh (\cdot )\approx 1$, using expansion in Taylor series for $\cosh (\cdot )$ and $\alpha$ to be small enough, we have

$$\begin{aligned} I_{\alpha \mu }&= I_{\mu \alpha }={\text {E}}\left( -{\ddot{\ell} }_{\alpha \mu }\right) ={\text {E}}\left( \frac{4}{\alpha ^{3}}{\varvec{U}}^{\top} {\varvec{\Sigma} }^{-1}{\varvec{V}}\right) \approx {\text {E}}\left( \frac{4}{\alpha ^{3}}{\mathbf{1}}^{\top} {\varvec{\Sigma }}^{-1}{\varvec{V}}\right) =0,\\ I_{\mu \mu }&= {\text {E}}\left(-{\ddot{\ell} }_{\mu \mu }\right)\approx \frac{1}{4}{\text {E}}\left( \left( \frac{2}{\alpha }{\varvec{V}}\right) ^{\top} {\varvec{\Sigma} }^{-1}\left( \frac{2}{\alpha }{\varvec{V}}\right) \right) +{\text {E}}\left( \frac{1}{\alpha ^{2}}{\mathbf{1}}^{\top} {\varvec{\Sigma} }^{-1}{\mathbf{1}}-\frac{n}{4}\right) =\frac{1}{\alpha ^{2}}{\mathbf{1}}^{\top} {\varvec{\Sigma }}^{-1}{\mathbf{1}}. \end{aligned}$$

Furthermore, ${\varvec{I}}_{\varvec{\varphi }\mu } = {\varvec{I}}_{\mu {\varvec{\varphi} }}^{\top} = (I_{\varphi _{1}\mu },\, I_{\varphi _{2}\mu },\, I_{\varphi _{3}\mu })^{\top}$ is a $3\times 1$ vector with elements given by

$$\begin{aligned} I_{\mu \varphi _{j}}={\text {E}}\left(-{\ddot{\ell} }_{\mu \varphi _{j}}\right)\approx \frac{1}{\alpha }{\text {E}}\left( \left( {\mathbf{1}}^{\top} {\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\right) \left( \frac{2}{\alpha }{\varvec{V}}\right) \right) =0, \quad j=1,\,2,\,3. \end{aligned}$$

Appendix 4: Score vector ${\varvec{U}}({\varvec{\omega} })$ and matrix ${\varvec{G}}({\varvec{\omega} })$

Score vector used in the local influence method is given by ${\varvec{U}}({\varvec{\omega} })=\left( {\partial {\varvec{V}}_{\varvec{\omega }}}/{\partial {\varvec{\omega} }^{\top} }\right) ^{\top} \left( 2{\varvec{\Sigma} }^{-1}{\varvec{V}_{\omega }}\right) +({1}/{2}){\varvec{A}}{\varvec{T}_{\omega }},$where ${\varvec{T}_{\omega }}=({\varvec{T}}_{\omega _{1}},\ldots ,{\varvec{T}}_{\omega _{n}}),$ with ${\varvec{T}}_{\omega _{i}}=\tanh({(y_{i}+{\varvec{A}}_{i}{\varvec{\omega} }-\mu )}/{2}),$ and ${\varvec{A}}_{i}$ is the ith row of the matrix ${\varvec{A}}.$ For $\cosh ({(y_{i}+{\varvec{a}}_{i}{\varvec{\omega} }-\mu )}/{2})\approx 1,$ with $i=1,\ldots ,n,$ we get

$$\begin{aligned} \frac{\partial {\varvec{V}}_{\varvec{\omega }}}{\partial {\varvec{\omega} }^{\top} } =\frac{1}{2} \left( \begin{array}{ccc} \cosh \left( \frac{y_{1}+{\varvec{A}_{1}}{\varvec{\omega} }-\mu }{2}\right) \sigma _{11} &\cdots & \cosh \left( \frac{y_{1}+{\varvec{A}_{1}}{\varvec{\omega} }-\mu }{2}\right) \sigma _{1n}\\ \vdots & \ddots & \vdots \\ \cosh \left( \frac{y_{n}+{\varvec{A}_{n}}{\varvec{\omega} }-\mu }{2}\right) \sigma _{n1} & \cdots & \cosh \left( \frac{y_{n}+{\varvec{A}_{n}}{\varvec{\omega} }-\mu }{2}\right) \sigma _{nn}\\ \end{array} \right) =\frac{1}{2}{\varvec{A}}, \end{aligned}$$

where $\sigma _{ij}$ is the (i, j) element of the matrix ${\varvec{A}},$ for $i,\,j=1,\ldots ,n.$ Furthermore, for $\cosh ({(y_{i}+{\varvec{a}}_{i}{\varvec{\omega} }-\mu )}/{2})\approx 1,$ with $i=1,\ldots ,n,$ ${\varvec{T}_{\omega }}$ can be approximated by ${\varvec{V}_{\omega }},$ from which it follows that ${\varvec{U}}({\varvec{\omega} })=({1}/{2}){\varvec{A}}{\varvec{V}_{\omega }}-({2}/{\alpha ^{2}}){\varvec{A}}{\varvec{\Sigma} }^{-1}{\varvec{V}_{\omega }}.$ Thus,

$$\begin{aligned} {\varvec{U}}({\varvec{\omega }}){\varvec{U}}^{\top} ({\varvec{\omega} })&= \left( \frac{1}{2}{\varvec{A}}{\varvec{V}_{\omega }}-\frac{2}{\alpha ^{2}}{\varvec{A}}{\varvec{\Sigma} }^{-1}{\varvec{V}_{\omega }}\right) \left( \frac{1}{2}{\varvec{A}}{\varvec{V}_{\omega }}-\frac{2}{\alpha ^{2}}{\varvec{A}}{\varvec{\Sigma} }^{-1}{\varvec{V}_{\omega }}\right) ^{\top} \\&= \frac{\alpha ^{2}}{16}{\varvec{A}}\left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) \left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) ^{\top} {\varvec{A}}-\frac{1}{4}{\varvec{A}}\left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) \left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) ^{\top} {\varvec{\Sigma }}^{-1}{\varvec{A}}\\&\quad-\frac{1}{4}{\varvec{A}}{\varvec{\Sigma} }^{-1}\left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) \left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) ^{\top} {\varvec{A}}+\frac{1}{\alpha ^{2}}{\varvec{A}}{\varvec{\Sigma} }^{-1}\left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) \left( \frac{2}{\alpha }{\varvec{V}_{\omega }}\right) ^{\top} {\varvec{\Sigma} }^{-1}{\varvec{A}}. \end{aligned}$$

Therefore, ${\varvec{G}}({\varvec{\omega} })={\text {E}}\left( {\varvec{U}}({\varvec{\omega} }){\varvec{U}}^{\top} ({\varvec{\omega} })\right) = {\varvec{A}}\left( \frac{\alpha }{4}{\varvec{\Sigma} }^{{\frac{1}{2}}}-{\frac{1}{\alpha}}{\varvec{\Sigma} }^{{-\frac{1}{2}}}\right) ^{2}{\varvec{A}}.$ To find the appropriate perturbation, according to the methodology proposed by Zhu et al. (2007), it is necessary to find ${\varvec{A}},$ such that ${\varvec{G}}({\varvec{\omega} })=c{\varvec{I}}_{n},$ for $c>0.$ Considering $c=1,$ then ${\varvec{A}}$ must satisfy $(({\alpha }/{4}){\varvec{\Sigma} }^{({1}/{2})}-({1}/{\alpha }){\varvec{\Sigma} }^{-\frac{1}{2}})^2=({\varvec{A}}^{-1})^2.$ A solution of the equation above is given by ${\varvec{A}}=(({\alpha }/{4}){\varvec{\Sigma} }^{\frac{1}{2}}-({1}/{\alpha }){\varvec{\Sigma} }^{-\frac{1}{2}})^{-1}.$ Then, $\tilde{\varvec{\omega }}=(({\alpha }/{4}){\varvec{\Sigma} }^{\frac{1}{2}}-({1}/{\alpha }){\varvec{\Sigma} }^{-\frac{1}{2}})^{-1}{\varvec{\omega }}$ is an appropriate perturbation for the BS log-linear spatial model.

Appendix 5: The perturbation matrix

The perturbation matrix for the BS log-linear spatial model obtained from (20) is given by

$$\begin{aligned} \Delta _{\mu }=\frac{\partial \ell ({\varvec{\theta} }|{\varvec{\omega} })}{\partial \mu \partial {\varvec{\omega} }^{\top} }=\frac{\partial {\varvec{V}}_{\varvec{\omega} }^{\top} }{\partial \mu }\left( \frac{-2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}=- \frac{1}{2}{\varvec{U}}_{\varvec{\omega} }^{\top} \left( \frac{-2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}, \end{aligned}$$

where ${\varvec{U}_{\omega }}=({\varvec{U}}_{\omega _{1}},\ldots ,{\varvec{U}}_{\varvec \omega _{n}}),$ with ${\varvec{U}}_{\omega _{i}}=\cosh (({y_{i}+{\varvec{a}_{i}{\varvec{\omega}} }-\mu )}/{2}),$ for $i=1,\ldots ,n.$ Thus, the results presented in (32), (33) and (34) are obtained, for $\cosh ({(y_{i}+{\varvec{a}_{i}\varvec\omega }-\mu )}/{2})\approx 1,$ as

$$\begin{aligned} \Delta _{\mu }&= \frac{\partial \ell ({\varvec{\theta} }|{\varvec{\omega} })}{\partial \mu \partial {\varvec{\omega} }^{\top} }\approx {\mathbf{1}}^{\top} \left( \frac{1}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}-\frac{1}{4}{\varvec{I}_{n}}\right) {\varvec{A}},\\ \Delta _{\alpha }&= \frac{\partial \ell ({\varvec{\theta} }|{\varvec{\omega} })}{\partial \alpha \partial {\varvec{\omega} }^{\top} }=\frac{\partial {\varvec{V}}_{\varvec{\omega} }^{\top} }{\partial \alpha }\left( - \frac{2}{\alpha ^{2}}{\varvec{\Sigma }}^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}\\&\quad+{\varvec{V}}_{\varvec{\omega} }^{\top} \frac{\partial }{\partial \alpha }\left( \left( \frac{-2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}\right) \\&={\varvec{D}}^{\top} \left( \frac{-1}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{4}{\varvec{I}_{n}}\right) {\varvec{A}}\\&\quad+\left( \frac{4}{\alpha ^{3}}{\varvec{\Sigma} }^{-1}{\varvec{A}}+\left( -\frac{2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}\left( \frac{1}{\alpha ^{2}}{\varvec{\Sigma} }^{-\frac{1}{2}}+\frac{1}{4}{\varvec{\Sigma} }^{\frac{1}{2}}\right) {\varvec{A}}\right) , \end{aligned}$$

where ${\varvec{D}}=( D_{1},\ldots ,D_{n}) ^{\top},$ with $D_{i}={\varvec{l}}_{i}\varvec\omega$ and ${\varvec{l}}_{i}$ being the ith row of the matrix

$$\begin{aligned} {\varvec{L}} = {\varvec{A}}\left( \frac{1}{\alpha ^{2}}{\varvec{\Sigma} }^{-\frac{1}{2}}+\frac{1}{4}{\varvec{\Sigma} }^{\frac{1}{2}}\right) {\varvec{A}}. \end{aligned}$$

In addition,

$$\begin{aligned} \Delta _{\varphi _{j}}&=\frac{\partial \ell ({\varvec{\theta} }|{\varvec{\omega} })}{\partial {{\varphi} _{j}}\partial {\varvec{\omega} }^{\top} }=\frac{\partial {\varvec{V}}_{\varvec{\omega} }^{\top} }{\partial \varphi _{j}}\left( - \frac{2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) {\varvec{A}}\\&\quad+{\varvec{V}}_{\varvec{\omega} }^{\top} \left( \frac{2}{\alpha ^{2}}{\varvec{\Sigma }}^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}{\varvec{A}}+\left( -\frac{2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{2}{\varvec{I}_{n}}\right) \frac{\partial {\varvec{A}}}{\partial \varphi _{j}}\right) \\&= {\varvec{M}}^{\top} \left( \frac{-1}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}+\frac{1}{4}{\varvec{I}_{n}}\right) {\varvec{A}}+{\varvec{V}}_{\varvec{\omega} }^{\top} \left( \frac{2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}\frac{\partial {\varvec{\Sigma} }}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-1}\right) {\varvec{A}}\\&\quad+\left(\frac{1}{2}{\varvec{I}_{n}}- \frac{2}{\alpha ^{2}}{\varvec{\Sigma} }^{-1}\right) {\varvec{A}}\left( \frac{1}{\alpha }{\varvec{\Sigma }}^{-\frac{1}{2}}\frac{\partial {\varvec{\Sigma }}^{\frac{1}{2}}}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-\frac{1}{2}}+\frac{\alpha }{4}\frac{\partial {\varvec{\Sigma} }^{\frac{1}{2}}}{\partial \varphi _{j}}\right) {\varvec{A}},\quad j = 1, 2, 3, \end{aligned}$$

where ${\varvec{M}}=( m_{1},\ldots ,m_{n}) ^{\top},$ with $m_{i}={\varvec{l}}_{i}{\varvec{\omega} }$ and ${\varvec{l}}_{i}$ being the ith row of matrix

$$\begin{aligned} {\varvec{L}} = {\varvec{A}}\left({\frac{\alpha }{4}}\frac{\partial {\varvec{\Sigma} }^{\frac{1}{2}}}{\partial \varphi _{j}}-{\frac{1}{\alpha}}{\varvec{\Sigma} }^{-\frac{1}{2}}\frac{\partial {\varvec{\Sigma} }^{\frac{1}{2}}}{\partial \varphi _{j}}{\varvec{\Sigma} }^{-\frac{1}{2}}\right) {\varvec{A}}, \quad j=1,\,2,\,3. \end{aligned}$$

Details about ${\partial {\varvec{\Sigma} }^{\frac{1}{2}}}/{\partial \varphi _{j}}$ can be found in De Bastiani et al. (2015).