A test for directional-linear independence, with applications to wildfire orientation and size

Abstract

A nonparametric test for assessing the independence between a directional random variable (circular or spherical, as particular cases) and a linear one is proposed in this paper. The statistic is based on the squared distance between nonparametric kernel density estimates and its calibration is done by a permutation approach. The size and power characteristics of various variants of the test are investigated and compared with those for classical correlation-based tests of independence in an extensive simulation study. Finally, the best-performing variant of the new test is applied in the analysis of the relation between the orientation and size of Portuguese wildfires.

Appendix

1.1 A Proof of Lemma 1

Proof The closed expression (just involving matrix computations) for T _n is obtained by splitting the calculation into three addends:

$$ \begin{aligned} T_n=&\int\limits_{\varOmega_{q}\times{\mathbb{R}}}{\left(\hat f_{({\mathbf{X}},Z);h,g}({\mathbf{x}},z)-\hat f_{{\mathbf{X}};h}({\mathbf{x}})\hat f_{Z;g}(z)\right)^2}{\omega_{q}{(d\mathbf{x})}}{dz} \\ = \, &\int\limits_{\varOmega_{q}\times{\mathbb{R}}}\left(\frac{c_{h,q}(L)}{ng}\sum_{i=1}^n L\left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right)K \left(\frac{z-Z_i}{g}\right)\right.\\ &\bigg.-\hat f_{{\mathbf{X}};h}({\mathbf{x}})\hat f_{Z;g}(z)\bigg)^2{\omega_{q}{(d\mathbf{x})}}{dz} \\ =&\sum_{i=1}^n\sum_{j=1}^n \int\limits_{\varOmega_{q}\times{\mathbb{R}}}\frac{c_{h,q}(L)^2}{n^2g^2} L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right) K \left(\frac{z-Z_i}{g}\right) \\ &\cdot L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_j}{h^2}\right) K \left(\frac{z-Z_j}{g}\right){\omega_{q}{(d\mathbf{x})}}{dz} \\ &-2\sum_{i=1}^n \int\limits_{\varOmega_{q}\times{\mathbb{R}}}\frac{c_{h,q}(L)}{ng} L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right) K \left(\frac{z-Z_i}{g}\right)\\ &\cdot\hat f_{{\mathbf{X}};h}({\mathbf{x}})\hat f_{Z;g}(z){\omega_{q}{(d\mathbf{x})}}{dz} \\ &+\int\limits_{\varOmega_{q}\times{\mathbb{R}}}{\hat f_{{\mathbf{X}};h}({\mathbf{x}})^2\hat f_{Z;g}(z)^2}{\omega_{q}{(d\mathbf{x})}}{dz} \\ = \, &(6)-(7)+(8). \end{aligned} $$

The first addend is:

$$ \begin{aligned} (6)= \, &\frac{c_{h,q}(L)^2}{n^2g^2}\sum_{i=1}^n\sum_{j=1}^n\int\limits_{\varOmega_{q}\times{\mathbb{R}}} L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right) K \left(\frac{z-Z_i}{g}\right) \\ &\cdot L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_j}{h^2}\right) K \left(\frac{z-Z_j}{g}\right){\omega_{q}{(d\mathbf{x})}}{dz} \\ =&\frac{c_{h,q}(L)^2}{n^2g^2}\sum_{i=1}^n\sum_{j=1}^n\int\limits_{\varOmega_{q}} {e^{-2/h^2}e^{{\mathbf{x}}'({\mathbf{X}}_i+{\mathbf{X}}_j)/h^2}}{\omega_{q}{(d\mathbf{x})}} \\ &\cdot \int\limits_{{\mathbb{R}}}{K \left(\frac{z-Z_i}{g}\right) K \left(\frac{z-Z_j}{g}\right)}{dz} \\ =&\frac{c_{h,q}(L)^2}{n^2}e^{-2/h^2}\sum_{i=1}^n\sum_{j=1}^n\frac{\phi_{\sqrt{2}g}(Z_i-Z_j)} {C_q \left(\|{\mathbf{X}}_i+{\mathbf{X}}_j\|/h^2\right)} \\ =&\frac{C_q \left(1/h^2\right)^2}{n^2}\sum_{i=1}^n\sum_{j=1}^n \frac{\phi_{\sqrt{2}g}(Z_i-Z_j)}{C_q \left(\|{\mathbf{X}}_i+{\mathbf{X}}_j\|/h^2\right)}. \end{aligned} $$

For the second addend,

$$ \begin{aligned} (7)= \, &2\sum_{i=1}^n \int\limits_{\varOmega_{q}\times{\mathbb{R}}}\frac{c_{h,q}(L)}{ng} L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right)K \left(\frac{z-Z_i}{g}\right) \\ &\cdot\hat f_{{\mathbf{X}};h}({\mathbf{x}})\hat f_{Z;g}(z){\omega_{q}{(d\mathbf{x})}}{dz} \\ = \, &2\frac{c_{h,q}(L)}{ng}\sum_{i=1}^n \int\limits_{\varOmega_{q}}{L\left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right)\hat f_{{\mathbf{X}};h}({\mathbf{x}})}{\omega_{q}{(d\mathbf{x})}} \\ &\cdot \int\limits_{{\mathbb{R}}}{K \left(\frac{z-Z_i}{g}\right)\hat f_{Z;g}(z)}{dz} \\ = \, &2\frac{c_{h,q}(L)}{ng}\sum_{i=1}^n \left\{\left[ \int\limits_{\varOmega_{q}}L\left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right)\right.\right. \\ &\left.\cdot\frac{c_{h,q}(L)}{n}\sum_{j=1}^n L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_j}{h^2}\right){\omega_{q}{(d\mathbf{x})}} \right] \\ & \left.\cdot \left[\int\limits_{{\mathbb{R}}}{ K \left(\frac{z-Z_i}{g}\right)\frac{1}{ng}\sum_{k=1}^n K\left(\frac{z-Z_k}{g}\right)}{dz}\right]\right\} \\ = &\frac{2}{n^3}\sum_{i=1}^n\left\{\Bigg[\int\limits_{\varOmega_{q}}{c_{h,q}(L)^2e^{-2/h^2} e^{{\mathbf{x}}'\left({\mathbf{X}}_i+{\mathbf{X}}_j\right)/h^2}}{\omega_{q}{(d\mathbf{x})}}\Bigg]\right. \\ &\left.\cdot \left(\sum_{k=1}^n\phi_{\sqrt{2}g}(Z_i-Z_k)\right)\right\} \\ = &\frac{2}{n^3}\sum_{i=1}^n\left\{\Bigg[\sum_{j=1}^n \frac{C_q \left(1/h^2\right)^2}{C_q \left(\|{\mathbf{X}}_i+{\mathbf{X}}_j\|/h^2\right)}\Bigg]\right. \\ &\left.\cdot\left(\sum_{k=1}^n\phi_{\sqrt{2}g}(Z_i-Z_k)\right)\right\}. \\ \end{aligned} $$

Finally, the third addend is obtained as

$$ \begin{aligned} (8)=& \int\limits_{\varOmega_{q}\times{\mathbb{R}}}{\hat f_{{\mathbf{X}};h}({\mathbf{x}})^2\hat f_{Z;g}(z)^2}{\omega_{q}{(d\mathbf{x})}}{dz} \\ =& \int\limits_{\varOmega_{q}}{\hat f_{{\mathbf{X}};h}({\mathbf{x}})^2}{{\mathbf{x}}} \int\limits_{{\mathbb{R}}}{\hat f_{Z;g}(z)^2}{dz} \\ =&\left[\sum_{i=1}^n\sum_{j=1}^n\frac{c_{h,q}(L)^2}{n^2} \int\limits_{\varOmega_{q}}L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_i}{h^2}\right)\right. \\ &\left.\cdot L \left(\frac{1-{\mathbf{x}}'{\mathbf{X}}_j}{h^2}\right){\omega_{q}{(d\mathbf{x})}}\right] \\ &\cdot\Bigg[\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2g^2} \int\limits_{{\mathbb{R}}}{ K\left(\frac{z-Z_i}{g}\right)K \left(\frac{z-Z_j}{g}\right)}{dz}\Bigg] \\ =&\Bigg[\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\frac{C_q \left(1/h^2\right)^2}{C_q \left(\|{\mathbf{X}}_i+{\mathbf{X}}_j\|/h^2\right)}\Bigg] \\ &\cdot \Bigg[\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\phi_{\sqrt{2}g}(Z_i-Z_j)\Bigg]. \\ \end{aligned} $$

From the previous results and after applying some matrix algebra, it turns out that

$$ \begin{aligned} T_n \, = \, &{\mathbf{1}}_n\bigg(\frac{1}{n^2}{\boldsymbol{\Uppsi}}(h)\circ{\boldsymbol{\Omega}}(g)-\frac{2}{n^3}{\boldsymbol{\Uppsi}}(h){\boldsymbol{\Omega}}(g) \\ &+\frac{1}{n^4}{\boldsymbol{\Uppsi}}(h){\mathbf{1}}_n{\mathbf{1}}_n'{\boldsymbol{\Omega}}(g)\bigg){\mathbf{1}}_n', \\ \end{aligned} $$

where

$$ \begin{aligned} {\boldsymbol{\Uppsi}}(h)=&\left(\frac{C_q\left(1/h^2\right)^2}{C_q \left(\|{\mathbf{X}}_i+{\mathbf{X}}_j\|/h^2\right)}\right)_{ij}, \\ {\boldsymbol{\Omega}}(g)=& \left(\phi_{\sqrt{2}g} \left(Z_i-Z_j\right)\right)_{ij}. \\ \end{aligned} $$