Locally Anisotropic Nonstationary Covariance Functions on the Sphere

Abstract

Rapid developments in satellite remote-sensing technology have enabled the collection of geospatial data on a global scale, hence increasing the need for covariance functions that can capture spatial dependence on spherical domains. We propose a general method of constructing nonstationary, locally anisotropic covariance functions on the sphere based on covariance functions in $\mathbb {R}^3$. We also provide theorems that specify the conditions under which the resulting correlation function is isotropic or axially symmetric. For large datasets on the sphere commonly seen in modern applications, the Vecchia approximation is used to achieve higher scalability on statistical inference. The importance of flexible covariance structures is demonstrated numerically using simulated data and a precipitation dataset. Supplementary materials accompanying this paper appear online.

A Proofs

Proof of Theorem 1

If $\gamma _1 ({\textbf{s}})=\gamma _2 ({\textbf{s}}) \equiv \gamma$ is constant, then

$$\begin{aligned} \mathbf{D(\gamma )}&=\left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \gamma &{} 0 \\ 0 &{} 0 &{} \gamma \end{array} \right) ,\\ {\tilde{\mathbf{\Sigma }}(s)}&= \mathcal {R}_x (\kappa (s)) \mathbf{D(\gamma (s))} \mathcal {R}_x (\kappa (s))'\\&= \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \cos \kappa (s) &{} -\sin \kappa (s) \\ 0 &{} \sin \kappa (s) &{} \cos \kappa (s) \end{array} \right) \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \gamma &{} 0 \\ 0 &{} 0 &{} \gamma \end{array} \right) \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \cos \kappa (s) &{} \sin \kappa (s) \\ 0 &{} -\sin \kappa (s) &{} \cos \kappa (s) \end{array} \right) \\&= \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \gamma &{} 0 \\ 0 &{} 0 &{} \gamma \end{array} \right) . \end{aligned}$$

To compute $\mathbf{\Sigma (s)}=\mathcal {R}_z (l) \mathcal {R}_y (L) {\tilde{\mathbf{\Sigma }}(s)} \mathcal {R}_y (L)' \mathcal {R}_z (l)'$, we first compute

$$\begin{aligned} \textbf{A}:=\mathcal {R}_z (l) \mathcal {R}_y (L)&= \left( \begin{array}{ccc} \cos (l) &{} -\sin (l) &{} 0 \\ \sin (l) &{} \cos (l) &{} 0 \\ 0 &{} 0 &{} 1 \end{array} \right) \left( \begin{array}{ccc} \cos (L) &{} 0 &{} \sin (L) \\ 0 &{} 1 &{} 0 \\ -\sin (L) &{} 0 &{} \cos (L) \end{array} \right) \\&= \left( \begin{array}{ccc} \cos (l) \cos (L) &{} -\sin (l) &{} \cos (l) \sin (L) \\ \sin (l) \cos (L) &{} \cos (l) &{} \sin (l) \sin (L) \\ -\sin (L) &{} 0 &{} \cos (L) \end{array} \right) ,\\ \mathbf{\Sigma (s)}=\textbf{A} \tilde{\varvec{\Sigma }}(\textbf{s}) \mathbf{A'}&= \left( \begin{array}{ccc} (1-\gamma )x^2+\gamma &{} (1-\gamma )xy &{} (1-\gamma )xz \\ (1-\gamma )xy &{} (1-\gamma )y^2+\gamma &{} (1-\gamma )yz \\ (1-\gamma )xz &{} (1-\gamma )yz &{} (1-\gamma )z^2+\gamma \end{array} \right) \\&= (1-\gamma )\left( \begin{array}{c} x \\ y \\ z \end{array} \right) \left( \begin{array}{ccc} x&y&z\end{array} \right) + \gamma \textbf{I}_3\\&= (1-\gamma ){\tilde{\textbf{s}} \tilde{s}'} + \gamma \textbf{I}_3, \end{aligned}$$

where $x=\cos (L)\cos (l)$, $y=\cos (L)\sin (l)$, $z=\sin (L)$ are the (x, y, z)-coordinates of a three-dimensional Cartesian coordinate system. Then

$$\begin{aligned} |\mathbf{\Sigma (s)}|&= det\{(1-\gamma )\ {\tilde{\textbf{s}} \tilde{s}'} + \gamma \textbf{I}_3\}\\ {}&= \gamma ^3 \cdot det\left\{ \textbf{I}_3 +\frac{1-\gamma }{\gamma }{\tilde{\textbf{s}} \tilde{s}'}\right\} \\ {}&= \gamma ^3 \cdot det\left\{ 1 +\frac{1-\gamma }{\gamma }{\tilde{\textbf{s}}' \tilde{s}}\right\} \\ {}&= \gamma ^3 \cdot det\left\{ 1 +\frac{1-\gamma }{\gamma } \cdot 1\right\} \\ {}&= \gamma ^2 \end{aligned}$$

does not depend on $\textbf{s}$. And for $i \ne j$,

$$\begin{aligned} (\mathbf{\Sigma (s_i)+\Sigma (s_j)})^{-1} = \left( (1-\gamma ){\tilde{\textbf{s}}_i \tilde{s}_i'} + (1-\gamma ){\tilde{\textbf{s}}_j \tilde{s}_j'} + 2\gamma \textbf{I}_3 \right) ^{-1}. \end{aligned}$$

WLOG, we ignore the constant coefficients inside the inverse, and then

$$\begin{aligned} (\mathbf{\Sigma (s_i)+\Sigma (s_j)})^{-1}&= \left( {\tilde{\textbf{s}}_i \tilde{s}_i'} + {\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3 \right) ^{-1}\\&= ({\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3)^{-1} - ({\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3)^{-1} {\tilde{\textbf{s}}_i}\\ {}&\quad [1+{\tilde{\textbf{s}}_i}'({\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3)^{-1} {\tilde{\textbf{s}}_i}]^{-1} {\tilde{\textbf{s}}_i}' ({\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3)^{-1}. \end{aligned}$$

Let $\textbf{B}:= ({\tilde{\textbf{s}}}_j \tilde{s}_j' + \textbf{I}_3)^{-1}$, and so

$$\begin{aligned} (\mathbf{\Sigma (s_i)+\Sigma (s_j)})^{-1}&= \textbf{B} - \textbf{B} {\tilde{\textbf{s}}}_i (1+{\tilde{\textbf{s}}}_i' B \tilde{s}_i)^{-1}{\tilde{\textbf{s}}}_i' B=\textbf{B} - \frac{\textbf{B} {\tilde{\textbf{s}}}_\textbf{i} {\tilde{\textbf{s}}}_\mathbf{i'} \textbf{B}}{1+{\tilde{\textbf{s}}}_i' B \tilde{s}_i},\\ q^2({\textbf{s}_i,s_j})&\propto ({\tilde{\textbf{s}}}_i - \tilde{s}_j)' (\mathbf{\Sigma (s_i) + \Sigma (s_j)})^{-1} ({\tilde{\textbf{s}}}_i - \tilde{s}_j)\\&\propto ({\tilde{\textbf{s}}}_i - \tilde{s}_j)' \textbf{B} ({\tilde{\textbf{s}}}_i - \tilde{s}_j)-\frac{1}{1+{\tilde{\textbf{s}}_i' B \tilde{s}_i}} ({\tilde{\textbf{s}}}_i - \tilde{s}_j)' \textbf{B} {\tilde{\textbf{s}}}_\textbf{i} {\tilde{\textbf{s}}}_\mathbf{i'} \textbf{B} ({\tilde{\textbf{s}}}_i - \tilde{s}_j). \end{aligned}$$

So, computation of $q({\textbf{s}_i,s_j})$ only involves terms ${\tilde{\textbf{s}}_i' B \tilde{s}_i}$, ${\tilde{\textbf{s}}_j' B \tilde{s}_j}$ and ${\tilde{\textbf{s}}_i' B \tilde{s}_j}$. Because

$$\begin{aligned} \textbf{B}=({\tilde{\textbf{s}}_j \tilde{s}_j'} + \textbf{I}_3)^{-1}=\textbf{I}_3-{\tilde{\textbf{s}}_j} (1+{\tilde{\textbf{s}}_j' \tilde{s}_j})^{-1} {\tilde{\textbf{s}}_j'}=\textbf{I}_3-\frac{1}{2}{\tilde{\textbf{s}}_j \tilde{s}_j'}, \end{aligned}$$

we have

$$\begin{aligned} {\tilde{\textbf{s}}}_i' B \tilde{s}_i&= {\tilde{\textbf{s}}}_i' \tilde{s}_i - \frac{1}{2}({\tilde{\textbf{s}}}_i' \tilde{s}_j)^2 = 1-\frac{1}{2}({\tilde{\textbf{s}}}_i' \tilde{s}_j)^2\\ {\tilde{\textbf{s}}_j' B \tilde{s}_j}&= {\tilde{\textbf{s}}_j' \tilde{s}_j} - \frac{1}{2}({\tilde{\textbf{s}}_j' \tilde{s}_j})^2 = 1-\frac{1}{2}=\frac{1}{2}\\ {\tilde{\textbf{s}}_i' B \tilde{s}_j}&= {\tilde{\textbf{s}}_i' \tilde{s}_j} - \frac{1}{2}({\tilde{\textbf{s}}_i' \tilde{s}_j}) ({\tilde{\textbf{s}}_j' \tilde{s}_j}) = {\tilde{\textbf{s}}_i' \tilde{s}_j}-\frac{1}{2}{\tilde{\textbf{s}}_i' \tilde{s}_j}=\frac{1}{2}{\tilde{\textbf{s}}_i' \tilde{s}_j}. \end{aligned}$$

Further,

$$\begin{aligned} {\tilde{\textbf{s}}_i' \tilde{s}_j} = \left[ ({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})-{\tilde{\textbf{s}}_i' \tilde{s}_i} - {\tilde{\textbf{s}}_j' \tilde{s}_j}\right] /2 = \left[ ({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})-2\right] /2. \end{aligned}$$

So, $q({\textbf{s}_i,\textbf{s}_j})$ just depends on the distance $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$. For the normalization term $c({\textbf{s}_i, \textbf{s}_j})$, since we have proved that $|\mathbf{\Sigma (s_i)}|=|\mathbf{\Sigma (s_j)}| \equiv \gamma ^2$,

$$\begin{aligned} \begin{aligned} c({\textbf{s}_i, \textbf{s}_j})&= |\mathbf{\Sigma (s_i)}|^{1/4} |\mathbf{\Sigma (s_j)}|^{1/4} |(\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})/2|^{-1/2}\\&\propto |(\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})^{-1}|^{1/2}\\&\propto \left( |(\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})^{-1}| \cdot |({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})|/[({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})] \right) ^{1/2}\\&\propto \left( |(\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})^{-1}| \cdot det\{({\tilde{\textbf{s}}_i - \tilde{s}_j})({\tilde{\textbf{s}}_i - \tilde{s}_j})'\}/[({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})] \right) ^{1/2}\\&\propto \left( det\left\{ (\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})^{-1} ({\tilde{\textbf{s}}_i - \tilde{s}_j})({\tilde{\textbf{s}}_i - \tilde{s}_j})'\right\} /[({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})] \right) ^{1/2}\\&\propto \left( det\left\{ ({\tilde{\textbf{s}}_i - \tilde{s}_j})' (\mathbf{\Sigma (s_i)}+\mathbf{\Sigma (s_j)})^{-1} ({\tilde{\textbf{s}}_i - \tilde{s}_j})\right\} /[({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})] \right) ^{1/2}\\&\propto \left\{ \frac{q^2({\textbf{s}_i,\textbf{s}_j})}{({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})} \right\} ^{1/2}. \end{aligned} \end{aligned}$$

(10)

We have proved that $q({\textbf{s}_i,\textbf{s}_j})$ just depends on $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$, so $c({\textbf{s}_i, \textbf{s}_j})$ also only depends on the distance $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$.

Overall, we can show

$$\begin{aligned} \rho _{NS}({\textbf{s}_i,\textbf{s}_j})=c({\textbf{s}_i,\textbf{s}_j})\rho (q({\textbf{s}_i,\textbf{s}_j})) \end{aligned}$$

only depends on the distance $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$, where $\rho (q)$ is a valid isotropic correlation function. So, $\rho _{NS}({\textbf{s}_i,\textbf{s}_j})$ is isotropic. $\square$

Proof of Theorem 2

If $\kappa ({\textbf{s}}) \equiv 0$ and $\gamma _1(\cdot )$, $\gamma _2(\cdot )$ depend on L only, then $\mathcal {R}_x (\kappa (s)) \equiv \mathcal {R}_x (0) = \textbf{I}_3$. Then

$$\begin{aligned} {\tilde{\mathbf{\Sigma }}(s)} = \mathbf{D(\gamma (s))}=\left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \gamma _1(L) &{} 0 \\ 0 &{} 0 &{} \gamma _2(L) \end{array} \right) =\left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} \gamma _1(L) &{} 0 \\ 0 &{} 0 &{} \gamma _1(L) \end{array} \right) + \left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \gamma _2(L)-\gamma _1(L) \end{array} \right) . \end{aligned}$$

Due to the results in Theorem 1, we have

$$\begin{aligned}{} & {} {\varvec{\Sigma }}({\textbf{s}})=\mathcal {R}_z (l) \mathcal {R}_y (L) {\tilde{\mathbf{\Sigma }}(s)} \mathcal {R}_y (L)' \mathcal {R}_z (l)'\\ {}{} & {} \quad =(1-\gamma _1(L)){\tilde{\textbf{s}} \tilde{s}'} + \gamma _1(L) \textbf{I}_3 + (\gamma _2(L)-\gamma _1(L)) {\tilde{\textbf{s}}^*} ({\tilde{\textbf{s}}^*})', \end{aligned}$$

where

$$\begin{aligned} {\tilde{\textbf{s}}^*}=\left( \begin{array}{c} \cos (l) \sin (L) \\ \sin (l) \sin (L) \\ \cos (L) \end{array} \right) \ \, \ \ ({\tilde{\textbf{s}}^{*}})' ({\tilde{\textbf{s}}^*})=1. \end{aligned}$$

Thus,

$$\begin{aligned} |\mathbf{\Sigma (s)}|&= det\{(1-\gamma _1(L))\ {\tilde{\textbf{s}} \tilde{s}'} + \gamma _1(L) \textbf{I}_3+ (\gamma _2(L)-\gamma _1(L)) {\tilde{\textbf{s}}^*} ({\tilde{\textbf{s}}^*})'\}\\&= \gamma _1(L)^3 \cdot det \left\{ \frac{1-\gamma _1(L)}{\gamma _1(L)} {\tilde{\textbf{s}} \tilde{s}'} + \frac{\gamma _2(L)-\gamma _1(L)}{\gamma _1(L)}{\tilde{\textbf{s}}^*} ({\tilde{\textbf{s}}^*})'+\textbf{I}_3 \right\} \\&= \gamma _1(L)^3 \cdot det \left\{ \left( \begin{array}{cc} \frac{1-\gamma _1(L)}{\gamma _1(L)} {\tilde{\textbf{s}}}&\frac{\gamma _2(L)-\gamma _1(L)}{\gamma _1(L)}{\tilde{\textbf{s}}^*} \end{array} \right) \left( \begin{array}{c} {\tilde{\textbf{s}}'} \\ ({\tilde{\textbf{s}}^*})' \end{array} \right) +\textbf{I}_3 \right\} \\&= \gamma _1(L)^3 \cdot det \left\{ \left( \begin{array}{c} {\tilde{\textbf{s}}'} \\ ({\tilde{\textbf{s}}^*})' \end{array} \right) \left( \begin{array}{cc} \frac{1-\gamma _1(L)}{\gamma _1(L)} {\tilde{\textbf{s}}}&\frac{\gamma _2(L)-\gamma _1(L)}{\gamma _1(L)}{\tilde{\textbf{s}}^*} \end{array} \right) +\textbf{I}_2 \right\} \\&= \gamma _1(L)^3 \cdot \left| \begin{array}{cc} \frac{1-\gamma _1(L)}{\gamma _1(L)}+1 &{} 2\sin (L)\cos (L)\\ 2\sin (L)\cos (L) &{} \frac{\gamma _2(L)-\gamma _1(L)}{\gamma _1(L)}+1 \end{array} \right| \\&= \gamma _1(L)\gamma _2(L) - 4\gamma _1(L)^3 \sin ^2(L) \cos ^2(L) \end{aligned}$$

only depend on L. WLOG, ignore $\gamma _1(L)$, $\gamma _2(L)$ again (they only depend on L),

$$\begin{aligned}&(\mathbf{\Sigma (s_i)+\Sigma (s_j)})^{-1} \\ {}&\quad = \left[ {\tilde{\textbf{s}}_i \tilde{s}_i'} + {\tilde{\textbf{s}}_i^*} ({\tilde{\textbf{s}}_i^*})' + {\tilde{\textbf{s}}_j \tilde{s}_j'} + {\tilde{\textbf{s}}_j^*} ({\tilde{\textbf{s}}_j^*})' + \textbf{I}_3 \right] ^{-1}\\ {}&= \left[ \left( \begin{array}{cc} {\tilde{\textbf{s}}_i}&{\tilde{\textbf{s}}_i^*} \end{array} \right) \left( \begin{array}{c} {\tilde{\textbf{s}}_i'} \\ ({\tilde{\textbf{s}}_i^*})' \end{array} \right) + \left( \begin{array}{cc} {\tilde{\textbf{s}}_j}&{\tilde{\textbf{s}}_j^*} \end{array} \right) \left( \begin{array}{c} {\tilde{\textbf{s}}_j'} \\ ({\tilde{\textbf{s}}_j^*})' \end{array} \right) + \textbf{I}_3 \right] ^{-1}\\&= \textbf{V}^{-1} - \textbf{V}^{-1} \left( \begin{array}{cc} {\tilde{\textbf{s}}_i}&{\tilde{\textbf{s}}_i^*} \end{array} \right) \left[ \textbf{I}_2 + \left( \begin{array}{c} {\tilde{\textbf{s}}_i'} \\ ({\tilde{\textbf{s}}_i^*})' \end{array} \right) \textbf{V}^{-1} \left( \begin{array}{cc} {\tilde{\textbf{s}}_i}&{\tilde{\textbf{s}}_i^*} \end{array} \right) \right] ^{-1} \left( \begin{array}{c} {\tilde{\textbf{s}}_i'} \\ ({\tilde{\textbf{s}}_i^*})' \end{array} \right) \textbf{V}^{-1}\\&= \textbf{V}^{-1} - \textbf{V}^{-1} \left( \begin{array}{cc} {\tilde{\textbf{s}}_i}&{\tilde{\textbf{s}}_i^*} \end{array} \right) \left[ \textbf{I}_2 + \left( \begin{array}{cc} {\tilde{\textbf{s}}_i'} \textbf{V}^{-1} {\tilde{\textbf{s}}_i} &{} {\tilde{\textbf{s}}_i'} \textbf{V}^{-1} {\tilde{\textbf{s}}_i^*} \\ ({\tilde{\textbf{s}}_i^*})' \textbf{V}^{-1} {\tilde{\textbf{s}}_i} &{} ({\tilde{\textbf{s}}_i^*})' \textbf{V}^{-1} {\tilde{\textbf{s}}_i^*} \end{array} \right) \right] ^{-1} \left( \begin{array}{c} {\tilde{\textbf{s}}_i'} \\ ({\tilde{\textbf{s}}_i^*})' \end{array} \right) \textbf{V}^{-1}, \end{aligned}$$

where

$$\begin{aligned} \textbf{V} = \left( \begin{array}{cc} {\tilde{\textbf{s}}_j}&{\tilde{\textbf{s}}_j^*} \end{array} \right) \left( \begin{array}{c} {\tilde{\textbf{s}}_j'} \\ ({\tilde{\textbf{s}}_j^*})' \end{array} \right) + \textbf{I}_3. \end{aligned}$$

Then

$$\begin{aligned} q^2({\textbf{s}_i,\textbf{s}_j}) \propto&({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j})' (\mathbf{\Sigma (s_i)+\Sigma (s_j)})^{-1} ({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j})\\ \propto&({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j})' \textbf{V}^{-1} ({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j}) - ({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j})' \textbf{V}^{-1} \left( \begin{array}{cc} {\tilde{\textbf{s}}_i}&{\tilde{\textbf{s}}_i^*} \end{array} \right) \\ {}&\left[ \textbf{I}_2 + \left( \begin{array}{cc} {\tilde{\textbf{s}}_i'} \textbf{V}^{-1} {\tilde{\textbf{s}}_i} &{} {\tilde{\textbf{s}}_i'} \textbf{V}^{-1} {\tilde{\textbf{s}}_i^*} \\ ({\tilde{\textbf{s}}_i^*})' \textbf{V}^{-1} {\tilde{\textbf{s}}_i} &{} ({\tilde{\textbf{s}}_i^*})' \textbf{V}^{-1} {\tilde{\textbf{s}}_i^*} \end{array} \right) \right] ^{-1}\cdot \\&\left( \begin{array}{c} {\tilde{\textbf{s}}_i'} \\ ({\tilde{\textbf{s}}_i^*})' \end{array} \right) \textbf{V}^{-1}({\tilde{\textbf{s}}_i}-{\tilde{\textbf{s}}_j}). \end{aligned}$$

Because

$$\begin{aligned} \textbf{V}^{-1}&= \textbf{I}_3 - \left( \begin{array}{cc} {\tilde{\textbf{s}}_j}&{\tilde{\textbf{s}}_j^*} \end{array} \right) \left[ \textbf{I}_2 + \left( \begin{array}{cc} 1 &{} {\tilde{\textbf{s}}_j}' {\tilde{\textbf{s}}_j^*}\\ ({\tilde{\textbf{s}}_j})' {\tilde{\textbf{s}}_j} &{} 1 \end{array} \right) \right] ^{-1} \left( \begin{array}{c} {\tilde{\textbf{s}}_j'} \\ ({\tilde{\textbf{s}}_j^*})' \end{array} \right) \\&= \textbf{I}_3 - \frac{1}{4-({\tilde{\textbf{s}}_j}' {\tilde{\textbf{s}}_j^*})^2} \left( \begin{array}{cc} {\tilde{\textbf{s}}_j}&{\tilde{\textbf{s}}_j^*} \end{array} \right) \left[ \left( \begin{array}{cc} 2 &{} -{\tilde{\textbf{s}}_j}' {\tilde{\textbf{s}}_j^*}\\ -({\tilde{\textbf{s}}_j})' {\tilde{\textbf{s}}_j} &{} 2 \end{array} \right) \right] ^{-1} \left( \begin{array}{c} {\tilde{\textbf{s}}_j'} \\ ({\tilde{\textbf{s}}_j^*})' \end{array} \right) , \end{aligned}$$

we can figure out that the computation of $q^2({\textbf{s}_i, \textbf{s}_j})$ only involves the following types of terms

$$\begin{aligned} \left\{ \begin{aligned} {\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_i} =&\, 1\\ {\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_i^*} =&\, {\tilde{\textbf{s}}_i}' \cdot \left[ (\tan (s_{i2})){\tilde{\textbf{s}}_i} + \left( 0,0,\cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})}\right) ' \right] \\ {}&= \tan (s_{i2}) + \sin (s_{i2})\left[ \cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})}\right] \\ ({\tilde{\textbf{s}}_i^*})' {\tilde{\textbf{s}}_i}^* =&1\\ ({\tilde{\textbf{s}}_i^*})' {\tilde{\textbf{s}}_j^*} \\ {}&= \left[ (\tan (s_{i2})){\tilde{\textbf{s}}_i} + \left( 0,0,\cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})} \right) ' \right] '\\ {}&\quad \cdot \left[ (\tan (s_{j2})){\tilde{\textbf{s}}_j} + \left( 0,0,\cos (s_{j2})-\frac{\sin ^2(s_{j2})}{\cos (s_{j2})} \right) ' \right] \\ =&\, \tan (s_{i2}) \tan (s_{j2}) ({\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j}) + \tan (s_{i2}) \sin (s_{i2})\left[ \cos (s_{j2})-\frac{\sin ^2(s_{j2})}{\cos (s_{j2})}\right] \\ {}&+ \tan (s_{j2}) \sin (s_{j2}) \left[ \cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})}\right] \\ {}&+ \left[ \cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})}\right] \left[ \cos (s_{j2})-\frac{\sin ^2(s_{j2})}{\cos (s_{j2})}\right] \\ {\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j^*} =&\tan (s_{j2})({\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j}) + \sin (s_{i2})\left[ \cos (s_{j2})-\frac{\sin ^2(s_{j2})}{\cos (s_{j2})}\right] \\ {\tilde{\textbf{s}}_j}' {\tilde{\textbf{s}}_i^*} =&\tan (s_{i2})({\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j}) + \sin (s_{j2})\left[ \cos (s_{i2})-\frac{\sin ^2(s_{i2})}{\cos (s_{i2})}\right] \\ {\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j} =&\left[ ({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})-2\right] /2. \end{aligned} \right. \end{aligned}$$

We can change the index i to j for the first three terms, and they are still valid. Thus, these values only depend on $s_{i2}$, $s_{j2}$ and ${\tilde{\textbf{s}}_i}' {\tilde{\textbf{s}}_j}$, and ${\tilde{\textbf{s}}_i}' {\tilde{ \textbf{s}}_j}$ can be expressed in terms of the distance $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$. The computation of $q^2({\textbf{s}_i,\textbf{s}_j})$ only depends on the distance $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$ and the longitudes $s_{i2}$, $s_{j2}$. Similar to (10) in the proof of Theorem 1, we can also show that $c({\textbf{s}_i,\textbf{s}_j})$ is a function of $({\tilde{\textbf{s}}_i - \tilde{s}_j})'({\tilde{\textbf{s}}_i - \tilde{s}_j})$, $s_{i2}$ and $s_{j2}$. Then $\rho _{NS}({\textbf{s}_i,\textbf{s}_j})=c({\textbf{s}_i,\textbf{s}_j})\rho (q({\textbf{s}_i,\textbf{s}_j})):=\rho _A ({\tilde{\textbf{s}}_i - \tilde{s}_j}, s_{i2}, s_{j2})$, so it is axially symmetric. $\square$