Bivariate Cardioid Distributions

Abstract

Bivariate generalizations of the cardioid distribution have been constructed by the mixture method. Two types of mixtures have been considered. These mixture models allow the joint probability density functions to be represented as Fourier series easily, and this facilitates parameter estimation. A test of uniformity or isotropy of a given set of directions for one bivariate model is developed. For another bivariate model, a test of independence is derived. A data set on orientation of nests of 50 scrub birds and creek directions has also been analysed.

8 Appendix

Derivation of bivariate cardioid distribution Model II (b):

The bivariate beta distribution of [20, 21] has pdf g(x, y) given by

$$\begin{aligned} g(x,y) = \frac{\Gamma (a+b+c)}{\Gamma (a)\Gamma (b)\Gamma (c)} \frac{x^{a-1}y^{b-1}(1-x)^{b+c-1}(1-y)^{a+c-1}}{(1-xy)^{a+b+c}}, 0\le x,y<1 \end{aligned}$$

with a, b and $c>0$. The X and Y marginals are beta distributions B(x; a, c) and B(y; b, c), respectively. Note that this bivariate beta distribution is defined over $0\le x,y<1$ as opposed to bivariate beta distribution in (10) where $0\le x+y<1$.

This bivariate beta distribution has infinite series representation ([21])

$$\begin{aligned} g(x,y)= \sum _{i=0}^{\infty }de_i\frac{x^{a+i-1}(1-x)^{b+c-1}y^{b+i-1}(1-y)^{a+c-1}}{B(a+i,b+c)B(b+i,a+c)} \end{aligned}$$

(29)

where $B(a,b,c)= \frac{\Gamma (a)\Gamma (b)\Gamma (c)}{\Gamma (a+b+c)}$, $d = \frac{\Gamma (a+c)\Gamma (b+c)}{\Gamma (c)\Gamma (a+b+c)}$ and $e_i = \frac{(a)_i(b)_i\Gamma (c)}{(a+b+c)_i i!}$

By using (29),

$$\begin{aligned} {\begin{matrix} &{} f(\theta _1,\theta _2)= \sum _{i=0}^{\infty }de_i \int _{0}^{1}\{1+2x\delta _1\cos (\theta _1-\mu )\}\frac{x^{a+i-1}(1-x)^{b+c-1}}{B(a+i, b+c)}dx \\ &{} . \int _{0}^{1}\{1+2y\delta _2\cos (\theta _2-\mu )\}\frac{y^{b+i-1}(1-y)^{a+c-1}}{B(b+i,a+c)}dy \end{matrix}} \end{aligned}$$

(30)

Since

$$\begin{aligned} \int _{0}^{1}\{1+2x\delta _1\cos (\theta _1-\mu )\} \frac{x^{a+i-1}(1-x)^{b+c-1}}{B(a+i, b+c)}dx = 1+2 \frac{a+i}{a+b+c+i}\delta _1\cos (\theta _1-\mu ), \end{aligned}$$

$$\begin{aligned} \int _{0}^{1}\{1+2y\delta _2\cos (\theta _2-\mu )\} \frac{y^{b+i-1}(1-y)^{a+c-1}}{B(b+i, a+c)}dy = 1+2 \frac{b+i}{a+b+c+i}\delta _2\cos (\theta _2-\mu ) \end{aligned}$$

pdf (30) becomes

$$\begin{aligned} {\begin{matrix} &{}f(\theta _1,\theta _2)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &{}= \sum _{i=0}^{\infty }de_i \Big \{1+2\frac{a+i}{a+b+c+i}\delta _1\cos (\theta _1-\mu )\Big \}\Big \{1+2\frac{b+i}{a+b+c+i}\delta _2\cos (\theta _2-\mu )\Big \} \\ &{}= \sum _{i=0}^{\infty }de_i \Big \{1+2\frac{a+i}{a+b+c+i}\delta _1\cos (\theta _1-\mu )+2\frac{b+i}{a+b+c+i}\delta _2\cos (\theta _2-\mu ) \\ &{}+ 4\frac{a+i}{a+b+c+i}\frac{b+i}{a+b+c+i} \delta _1\delta _2\cos (\theta _1-\mu )\cos (\theta _2-\mu )\Big \}. \end{matrix}} \end{aligned}$$