Squaring the Circle and Cubing the Sphere: Circular and Spherical Copulas

Abstract

Do there exist circular and spherical copulas in ${\mathbb{R}}^d$? That is, do there exist circularly symmetric distributions on the unit disk in ${\mathbb{R}}^2$ and spherically symmetric distributions on the unit ball in ${\mathbb{R}}^d$, $d\ge 3$, whose one-dimensional marginal distributions are uniform? The answer is yes for $d=2$ and 3, where the circular and spherical copulas are unique and can be determined explicitly, but no for $d\ge 4$. A one-parameter family of elliptical bivariate copulas is obtained from the unique circular copula in ${\mathbb{R}}^2$ by oblique coordinate transformations. Copulas obtained by a non-linear transformation of a uniform distribution on the unit ball in ${\mathbb{R}}^d$ are also described, and determined explicitly for $d=2$.

1. Introduction

Do there exist spherically symmetric distributions on the closed unit ball $B_d$ in ${\mathbb{R}}^d$ that have uniform one-dimensional marginal distributions on $[-1,1]$? A distribution on $B_d$ with this property may be said to “square the circle” when $d=2$ and to “cube the sphere” when $d\ge 3$.

The cumulative distribution function (cdf) of a multivariate distribution on the unit cube ${[0,1]}^d$ whose marginal distributions are uniform $[0,1]$ is commonly called a copula; see Nelsen [1] for an accessible introduction to this topic. However, although it is customary to confine attention to distributions on the unit cube, our interest is in spherically symmetric (= orthogonally invariant) distributions on $B_d$ with uniform marginal distributions. Therefore we take “copula” to mean a multivariate cdf on the centered cube $C_d:={[-1,1]}^d$ with uniform $[-1,1]$ marginals.

For $d=2$ (resp., $d\ge 3$), such a copula, if it exists, will be called a circular copula (resp., spherical copula) if it is the cdf of a circularly symmetric (resp., spherically symmetric) distribution on the unit disk $B_2$ (resp., unit ball $B_d$).

It will be noted in Section 2 and Section 3 that circular and spherical copulas are unique if they exist, but exist only for dimensions $d=2$ and $d=3$. The proof of non-existence for $d\ge 4$ is remarkably simple. Explicit expressions for these copulas are given in Section 3 and Section 4 respectively.

In Section 5, a new one-parameter family of bivariate copulas called elliptical copulas is obtained from the unique circular copula in ${\mathbb{R}}^2$ by oblique coordinate transformations. Finally, in Section 6, copulas obtained by a non-linear transformation of a uniform distribution on the unit ball in ${\mathbb{R}}^d$ are described, and determined explicitly for $d=2$.

2. Uniqueness and Existence of Circular and Spherical Copulas

Proposition 2.1.

Circular and spherical copulas are unique if they exist. (This result is well-known (e.g., Feller [2], pp. 31–33, who uses “random direction” to indicate the uniform distribution of $U\in \partial B_3$), and reappears frequently (e.g., Arellano-Valle [3], Theorem 3.1). The essence of the result goes back at least to Schoenberg [4].)

Proof. 

If a circular or spherical copula exists on $C_d$, it is the cdf of a random vector $Z\equiv (Z_1,\cdots ,Z_d)$ with a spherically symmetric distribution on $B_d$ and with each $Z_i\sim \mathrm{uniform}[-1,1]$. The latter implies that Z has no atom at the origin, i.e., $P[Z=0]=0$, so we may consider the “polar coordinates” representation $Z=R·U$, where $R=\parallel Z\parallel \le 1$ and $U=Z/\parallel Z\parallel$. It is well known (e.g., Cambanis et al. [5], Lemmas 1 and 2) that the random unit vector $U\equiv (U_1,\cdots ,U_d)$ is independent of R and is uniformly distributed on the unit sphere $\partial B_d$, which implies that each $U_i^2\sim \mathrm{Beta}(1/2,(d-1)/2)$. Since $Z_i=R{U}_i$, we have that

$$log\left(Z_i^2\right)=log\left(R^2\right)+log\left(U_i^2\right).$$

(1)

Because R and $U_i$ are independent, it follows that the characteristic function of $log\left(R^2\right)$ is the quotient of the characteristic functions of $log\left(Z_i^2\right)$ and $log\left(U_1^2\right)$. Thus the distribution of $log\left(R^2\right)$, and therefore that of R, is uniquely determined by the distributions of $Z_i^2$ and $U_i^2$, which are already specified above. Thus the the joint distribution of $(R,U)$ is uniquely determined, hence so is the distribution of Z, hence so its cdf = copula. □

The existence of spherical copulas is easy to determine in three or more dimensions:

Proposition 2.2.

Spherical copulas do not exist for $d\ge 4$. For $d=3$, the unique spherical copula is generated by the uniform distribution on the unit sphere $\partial B_3:=\{(x_1,x_2,x_3)\mid x_1^2+x_2^2+x_3^2=1\}$.

Proof. 

Let Z be as in the proof of Proposition 2.1. Then

$$\frac{1}{3}=E\left(Z_i^2\right)=E\left(R^2\right)E\left(U_i^2\right)\le \frac{1}{d}$$

(2)

since $Z_i\sim \mathrm{uniform}[-1,1]$, $0\le R\le 1$, and $U_i^2\sim \mathrm{Beta}(1/2,(d-1)/2)$. Thus $d\le 3$, so a spherical copula cannot exist when $d\ge 4$.

Furthermore, if a spherical copula is to exist for $d=3$, it follows from (2) that its generating random vector $Z\in B_3$ must satisfy $E\left(R^2\right)=1$, hence $R=1$ with probability one. This can occur only if Z is uniformly distributed on the unit sphere $\partial B_3$. But it is well known (this follows from the fact that the area of a spherical zone is proportion to its altitude—cf. Feller [2], Proposition (i), p. 30) that this distribution does indeed have uniform marginal distributions on $[-1,1]$, hence generates the unique spherical copula for $d=3$. □

3. The Bivariate Case: The Unique Circular Copula

The following three questions constitute an engaging classroom exercise.

Question 1. Let $(X,Y)$ be a random vector uniformly distributed on the unit disk (= ball) $B_2$ in ${\mathbb{R}}^2$. Find the marginal probability distributions of X and Y.

Answer: One can easily show that X has the “semi-circular” probability density function (pdf) given by

$$f\left(x\right)=\frac{2}{\pi }\sqrt{1-x^2},-1\le x\le 1.$$

(3)

(See Figure 1.) By symmetry, Y has the same pdf as X.

Question 2. Let $(X,Y)$ be a random vector uniformly distributed on the unit circle $\partial B_2$ in ${\mathbb{R}}^2$. Find the marginal probability distributions of X and Y.

Answer: We can represent $(X,Y)$ as $(cos\Theta ,sin\Theta )$ where $\Theta \sim \mathrm{uniform}[0,2\pi )$. It follows readily that X has pdf

$$f\left(x\right)=\frac{1}{\pi \sqrt{1-x^2}},-1<x<1.$$

(4)

(See Figure 1.) By symmetry, Y has the same pdf as X.

In both cases, the joint distribution of $(X,Y)$ is circularly symmetric, that is, invariant under all orthogonal transformations of ${\mathbb{R}}^2$. A comparison of the shapes of the pdfs in Figure 1 suggest a third question:

Question 3. Does a circularly symmetric bivariate distribution with uniform $[-1,1]$ marginals exist on $B_2$? If so, it determines a circular copula on $C_2$, which is unique by Proposition 2.1. This also follows from uniqueness results for the Abel transform; see, e.g., Bracewell [6].

Answer: Optimistically, let’s seek an absolutely continuous solution. That is, we seek a bivariate pdf on $B_2$ of the form

$$f(x,y)=g(x^2+y^2)$$

such that the marginal pdf

$$f\left(x\right)\equiv {\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)dy,-1<x<1$$

is constant in x. Here g is a nonnegative function on $(0,1)$ that must satisfy

$$2\pi {\int }_0^{1}rg\left(r^2\right)dr=1$$

(5)

in order that $\int {\int }_{B_2}f(x,y)dxdy=1$ (transform to polar coordinates: $(x,y)\to (r,\theta )$).

To determine a suitable g, first set $h\left(t\right)=g(1-t)$, then let $u=y/\sqrt{1-x^2}$ to obtain

$$\begin{array}{ccc}f\left(x\right)\hfill & =\hfill & {\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}h(1-x^2-y^2)dy=\sqrt{1-x^2}{\int }_{-1}^{1}h\left((1-u^2)(1-x^2)\right)du\hfill \\ & =\hfill & 2\sqrt{1-x^2}{\int }_0^{1}h\left((1-u^2)(1-x^2)\right)du.\hfill \end{array}$$

If we take $h\left(t\right)=c{t}^{-1/2}$ then clearly $f\left(x\right)$ does not depend on x, and choosing $c=1/2\pi$ satisfies (5). Thus the bivariate pdf (see Figure 2)

$$f(x,y)=\frac{1}{2\pi \sqrt{1-x^2-y^2}},x^2+y^2<1$$

(6)

determines a circularly symmetric bivariate distribution on $B_2$ and yields the desired circular copula.

Question 4. Having determined the unique circularly symmetric distribution (6) on $B_2$ with uniform marginals, what is the corresponding cdf $F(x,y)$, that is, what is the corresponding circular copula?

Answer (see Theorem 3.1): The circular symmetry of $(X,Y)$ implies that its distribution is invariant under sign changes, i.e., $(X,Y)\stackrel{d}{=}(\pm X,\pm Y)$. By the following lemma, the cdf $F(x,y)\equiv P[X\le x,Y\le y]$ on $C_2\equiv {[-1,1]}^2$ can be expressed in terms of $F_0(x,y)$, its truncation to the first quadrant:

$$F_0(x,y)\equiv P[0\le X\le x,0\le Y\le y]$$

(7)

for $0\le x,y\le 1$, and also in terms of the complementary cdf $\overline{F}(x,y)\equiv P[X>x,Y>y]$ for $0\le x,y\le 1$. Because $(X,Y)\stackrel{d}{=}(\pm X,\pm Y)$ and has uniform $[-1,1]$ marginals,

$$\begin{array}{ccc}{F}_0(x,y)\hfill & =\hfill & P[0\le X\le 1,0\le Y\le 1]-P[X>x,0\le Y\le 1]\hfill \\ & & -P[0\le X\le 1,Y>y]+P[X>x,Y>y]\hfill \\ & =\hfill & \frac{1}{4}-\left(\frac{1-x}{4}\right)-\left(\frac{1-y}{4}\right)+\overline{F}(x,y)\hfill \\ & =\hfill & \frac{x+y-1}{4}+\overline{F}(x,y),0\le x,y\le 1.\hfill \end{array}$$

(8)

Lemma 3.1.

Let $(X,Y)$ be a bivariate random vector on $C_2$ with uniform $[-1,1]$ marginal distributions and sign-change invariance, i.e., $(X,Y)\stackrel{d}{=}(\pm X,\pm Y)$. Then for $(x,y)\in C_2$,

$$\begin{array}{ccc}F(x,y)\hfill & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)F_0\left(\right|x|,|y\left|\right)\hfill \end{array}$$

(9)

$$\begin{array}{cc}=\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)\left[\frac{\left|x\right|+\left|y\right|-1}{4}+\overline{F}\left(\right|x|,|y\left|\right)\right]\hfill \end{array}$$

(10)

where $\sigma \left(w\right)=\mathrm{sign}\left(w\right)$ if $w\ne 0$ and $\sigma \left(0\right)=0$.

Proof. 

To obtain (9), consider four cases:

Case 1: $0\le x,y\le 1$. Because $(X,Y)$ is sign-change invariant and has uniform $[-1,1]$ marginals,

$$\begin{array}{ccc}F(x,y)\hfill & =\hfill & P[0<X\le x,0<Y\le y]+P[0<X\le x,Y\le 0]\hfill \\ & & +P[X\le 0,0<Y\le y]+P[X\le 0,Y\le 0]\hfill \\ & =\hfill & F_0(x,y)+\frac{x}{4}+\frac{y}{4}+\frac{1}{4}\hfill \\ & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)F_0\left(\right|x|,|y\left|\right).\hfill \end{array}$$

Case 2: $-1\le x\le 0\le y\le 1$. Similarly,

$$\begin{array}{ccc}F(x,y)\hfill & =\hfill & P[X\le 0,0<Y\le y]-P[x<X\le 0,0\le Y\le y]\hfill \\ & & +P[X\le 0,Y\le 0]-P[x<X\le 0,Y\le 0]\hfill \\ & =\hfill & \frac{y}{4}-F_0(-x,y)+\frac{1}{4}-\frac{(-x)}{4}\hfill \\ & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)F_0\left(\right|x|,|y\left|\right).\hfill \end{array}$$

Case 3: $-1\le y\le 0\le x\le 1$. Similarly,

$$\begin{array}{ccc}F(x,y)\hfill & =\hfill & P[0<X\le x,Y\le 0]-P[0\le X\le x,y<Y\le 0]\hfill \\ & & +P[X\le 0,Y\le 0]-P[X\le 0,y<Y\le 0]\hfill \\ & =\hfill & \frac{x}{4}-F_0(x,-y)+\frac{1}{4}-\frac{(-y)}{4}\hfill \\ & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)F_0\left(\right|x|,|y\left|\right).\hfill \end{array}$$

Case 4: $-1\le x,y\le 0$. Similarly,

$$\begin{array}{ccc}F(x,y\hfill & )=\hfill & P[X\le 0,Y\le 0]-P[x<X\le 0,Y\le 0]\hfill \\ & & -P[X\le 0,y<Y\le 0]+P[x<X\le 0,y<Y\le 0]\hfill \\ & =\hfill & \frac{1}{4}-\frac{(-x)}{4}-\frac{(-y)}{4}+F_0(-x,-y)\hfill \\ & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)F_0\left(\right|x|,|y\left|\right).\hfill \end{array}$$

Finally, (10) follows from (9) by (8). □

Thus, to determine the circular copula $F(x,y)$ for the pdf (6), it suffices to determine the complementary cdf $\overline{F}(x,y)$ for $0\le x,y\le 1$ and apply (10). Because $\overline{F}(x,y)=0$ when $x^2+y^2\ge 1$, we need only consider the case where $x^2+y^2<1$.

First approach: When $0\le x,y\le 1$ and $x^2+y^2<1$, $\overline{F}(x,y)$ can be expressed as follows. By using Figure 3 we find that

$$\begin{array}{ccc}\overline{F}(x,y)\hfill & =\hfill & \frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_y^{\sqrt{1-s^2}}\frac{1}{\sqrt{1-s^2-t^2}}dt\right\}ds\hfill \\ & =\hfill & \frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_y^{\sqrt{1-s^2}}\frac{1}{\sqrt{(1-\frac{t^2}{1-s^2})}}\frac{dt}{\sqrt{1-s^2}}\right\}ds\hfill \\ & =\hfill & \frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_{\frac{y}{\sqrt{1-s^2}}}^1\frac{dv}{\sqrt{1-v^2}}\right\}ds\hfill \\ & =\hfill & \frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds.\hfill \end{array}$$

(11)

However, we were unable to evaluate this integral directly.

Second approach: Fortunately, we have found a solution in the molecular biology and optics literatures, where the problem of finding the area of the intersection of two spherical caps on the unit sphere $\partial B_3$ has been addressed. The following general result is due to Tovchigrechko and Vakser [7] and also appears in Oat and Sander [8].

Lemma 3.2.

Let $S_1$ and $S_2$ be spherical caps on $\partial B_3$. Let $r_1$ and $r_2$ denote their angular radii and let d denote the angular distance between their centers ($0<d\le \pi$). Assume that $0<r_1,r_2\le \pi /2$ and $d\le r_1+r_2$, so that the intersection $S_1\cap S_2\ne \varnothing$ and consists of a single “diangle”; (see Figure 4 and Figure 5.) Then Area $(S_1\cap S_2)$ is given by

$$\begin{array}{ccc}A(r_1,r_2;d)\hfill & =\hfill & 2\pi -2\pi cos\left(r_1\right)-2\pi cos\left(r_2\right)-2arccos\left(\frac{cos\left(d\right)-cos\left(r_1\right)cos\left(r_2\right)}{sin\left(r_1\right)sin\left(r_2\right)}\right)\hfill \\ & & +2cos\left(r_1\right)arccos\left(\frac{cos\left(d\right)cos\left(r_1\right)-cos\left(r_2\right)}{sin\left(d\right)sin\left(r_1\right)}\right)\hfill \\ & & +2cos\left(r_2\right)arccos\left(\frac{cos\left(d\right)cos\left(r_2\right)-cos\left(r_1\right)}{sin\left(d\right)sin\left(r_2\right)}\right).\hfill \end{array}$$

(12)

This result can be applied to obtain our desired circular copula as follows.

If $(X,Y,Z)$ is uniformly distributed on $\partial B_3$, then the event $\{X>x,Y>y\}$ corresponds to the intersection of the two spherical caps $\{X>x\}$ and $\{Y>y\}$, so $P[X>x,Y>y]$ is given by the area $A(x,y)$ of this intersection divided by the total area of $\partial B_3$, i.e., by $4\pi$. (See Figure 4 and Figure 5.) Also, the joint distribution of $(X,Y)$ is circularly symmetric on the unit disk $B_2$ and has uniform marginals, so must be the unique such bivariate distribution, namely the distribution with pdf (6).

Thus, for $0\le x,y\le 1$ and $x^2+y^2<1$, our desired complementary cdf is given by

$$\begin{array}{ccc}\overline{F}(x,y)\hfill & =\hfill & \frac{1}{4\pi }A(x,y)\hfill \end{array}$$

(13)

$$\begin{array}{cc}=\hfill & \frac{1}{4\pi }A(arccos\left(x\right),arccos\left(y\right);\pi /2)\hfill \\ =\hfill & \frac{1}{2}-\frac{x}{2}-\frac{y}{2}-\frac{1}{2\pi }arccos\left(-\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)\hfill \\ & +\frac{x}{2\pi }arccos\left(\frac{-y}{\sqrt{1-x^2}}\right)+\frac{y}{2\pi }arccos\left(\frac{-x}{\sqrt{1-y^2}}\right)\hfill \end{array}$$

(14)

$$\begin{array}{cc}\equiv \hfill & \frac{1-x-y}{4}+\alpha (x,y)\hfill \end{array}$$

(15)

where for $0\le x,y\le 1$ and $x^2+y^2<1$,

$$\begin{array}{ccc}\alpha (x,y)\hfill & =\hfill & \frac{1}{2\pi }\left[xarcsin\left(\frac{y}{\sqrt{1-x^2}}\right)+yarcsin\left(\frac{x}{\sqrt{1-y^2}}\right)\right.\hfill \\ & & \left.-arcsin\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)\right].\hfill \end{array}$$

(16)

Theorem 3.1.

The unique circular copula on $C_2$ is given by

$$F(x,y)=\frac{x+y+1}{4}+\alpha (x,y)$$

(17)

where $\alpha (x,y)$ is defined by (16) for $x^2+y^2<1$ and by

$$\alpha (x,y)=\sigma \left(xy\right)·\left(\frac{\left|x\right|+\left|y\right|-1}{4}\right)$$

(18)

for $x^2+y^2\ge 1$. Note that Equations (16) and (18) agree when $x^2+y^2=1$ and both are sign-change equivariant on $C_2$: for all $(x,y)\in C_2$ and all $ϵ,\delta =\pm 1$,

$$\alpha (ϵx,\delta y)=ϵ\delta ·\alpha (x,y).$$

(19)

Proof. 

From Equations (10) and (15), when $x^2+y^2<1$ we have

$$\begin{array}{ccc}F(x,y)\hfill & =\hfill & \frac{x+y+1}{4}+\sigma \left(xy\right)\alpha \left(\right|x|,|y\left|\right)\hfill \end{array}$$

(20)

$$\begin{array}{cc}=\hfill & \frac{x+y+1}{4}+\alpha (x,y)\hfill \end{array}$$

(21)

by (19). When $x^2+y^2\ge 1$, $\overline{F}\left(\right|x|,|y\left|\right)=0$ so (17) again holds by (10) and (18). □

See Figure 6 for a plot of the resulting copula (on ${[-1,1]}^2$).

4. The Trivariate Case: the Unique Spherical Copula

Question 5. Having determined the unique spherically symmetric distribution on $B_3$ with uniform marginals, namely, the uniform distribution on the unit sphere $\partial B_3$, what is the corresponding cdf $F(x,y,z)$ on $C_3$, i.e., the unique spherical copula?

Answer: As in Section 3, let $(X,Y,Z)$ be uniformly distributed on $\partial B_3$, so that $F(x,y,z)=P[X\le x,$ $Y\le y,Z\le z]$. Again we first determine the complementary cdf $\overline{F}(x,y,z)\equiv P[X>x,Y>y,$ $Z>z]$ for $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$, the intersection of the first octant of $C_3$ with the interior of $B_3$. Here the event $\{X>x,Y>y,Z>z\}$ corresponds to the intersection of the three spherical caps $\{X>x\}$, $\{Y>y\}$, and $\{Z>z\}$ on $\partial B_3$, so $\overline{F}(x,y,z)$ is the area $A(x,y,z)$ of this intersection divided by the total area $4\pi$ of $\partial B_3$.

Recall that two approaches were proposed in Section 3 to obtain the area $A(x,y)$ of the intersection of two circular caps $\{X>x\}$ and $\{Y>y\}$. The first approach led to the integral (11) that we were unable to evaluate explicitly, so we adopted a second approach based on the geometric Lemma 3.2 of Tovchigrechko and Vakser [7]. Andrey Tovchigrechko has kindly suggested a method for extending Lemma 3.2 to the case of three spherical caps in general position, which if carried out would yield an explicit expression for $A(x,y,z)$. However, we have found that because the axes of our three caps are mutually orthogonal, the two approaches just mentioned for the bivariate case can be combined to obtain $\overline{F}(x,y,z)\equiv \frac{1}{4\pi }A(x,y,z)$ directly for the trivariate case, as now described.

We begin by extending (11) to obtain an integral expression for $\overline{F}(x,y,z)$ when $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$. We require the fact that

$$0\le a,b\le 1\mathrm{and}{a}^2+b^2=1\mathrm{implies}arcsin\left(a\right)+arcsin\left(b\right)=\pi /2.$$

(22)

Lemma 4.1.

If $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$, then

$$\overline{F}(x,y,z)=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)-arcsin\left(\frac{z}{\sqrt{1-s^2}}\right)\right]ds.$$

(23)

Because $(X,Y,Z)$ is exchangeable, (23) remains valid under any permutation of $x,y,z$ on the right-hand side.

Proof. 

Since $X^2+Y^2+Z^2=1$ and $(X,Y,Z)\stackrel{d}{=}(X,Y,-Z)$, it follows from (6) by using Figure 7 that

$$\begin{array}{c}{\displaystyle P[X>x,Y>y,Z>z]}\hfill \\ \hspace{1em}\hspace{1em}=\frac{1}{2}P[X>x,Y>y,X^2+Y^2<1-z^2]\hfill \\ \hspace{1em}\hspace{1em}=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left\{{\int }_y^{\sqrt{1-s^2-z^2}}\frac{1}{\sqrt{1-s^2-t^2}}dt\right\}ds\hfill \\ \hspace{1em}\hspace{1em}=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left\{{\int }_y^{\sqrt{1-s^2-z^2}}\frac{1}{\sqrt{(1-\frac{t^2}{1-s^2})}}\frac{dt}{\sqrt{1-s^2}}\right\}ds\hfill \\ \hspace{1em}\hspace{1em}=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left\{{\int }_{\frac{y}{\sqrt{1-s^2}}}^{\sqrt{\frac{1-s^2-z^2}{1-s^2}}}\frac{dv}{\sqrt{1-v^2}}\right\}ds\hfill \\ \hspace{1em}\hspace{1em}=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left[arcsin\left(\sqrt{\frac{1-s^2-z^2}{1-s^2}}\right)-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds.\hfill \end{array}$$

Now apply (22) to obtain (23). □

As noted above, the integral in (23) appears difficult to evaluate explicitly but the following indirect argument succeeds. Recall from (11) and (15) that when $0\le x,y\le 1$ and $x^2+y^2<1$,

$$\begin{array}{ccc}\overline{F}(x,y)\hfill & =\hfill & \frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds\hfill \\ & =\hfill & \frac{1-x-y}{4}+\alpha (x,y)\hfill \end{array}$$

where $\alpha (x,y)$ is given by (16). Because $z\le \sqrt{1-x^2-y^2}\le \sqrt{1-y^2}$ when $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$, it follows that

$$\begin{array}{c}{\displaystyle \frac{1}{2\pi }{\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds}\hfill \\ \hspace{1em}\hspace{1em}=\frac{\sqrt{1-x^2-y^2}-z}{4}+\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},y).\hfill \end{array}$$

(24)

Therefore from (23) and (24), if $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$ then

$$\begin{array}{c}{\displaystyle 4\pi \overline{F}(x,y,z)}\hfill \\ \hspace{1em}\hspace{1em}={\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{x}{\sqrt{1-s^2}}\right)\right]ds\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+{\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-arcsin\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds-\frac{\pi }{2}\left[\sqrt{1-x^2-y^2}-z\right]\hfill \\ \hspace{1em}\hspace{1em}=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+2\pi \left[\alpha (z,x)-\alpha (\sqrt{1-x^2-y^2},x)\right]\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+2\pi \left[\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},y)\right]\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)\hfill \\ \hspace{1em}\hspace{1em}=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+2\pi \left[\alpha (z,x)+\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},x)-\alpha (\sqrt{1-x^2-y^2},y)\right]\hfill \\ \hspace{1em}\hspace{1em}=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+xarcsin\left(\frac{z}{\sqrt{1-x^2}}\right)+zarcsin\left(\frac{x}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+yarcsin\left(\frac{z}{\sqrt{1-y^2}}\right)+zarcsin\left(\frac{y}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{yz}{\sqrt{(1-y^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-xarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-\sqrt{1-x^2-y^2}arcsin\left(\frac{x}{\sqrt{x^2+y^2}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+arcsin\left(\frac{x\sqrt{1-x^2-y^2}}{\sqrt{(1-x^2)(x^2+y^2)}}\right)-yarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-\sqrt{1-x^2-y^2}arcsin\left(\frac{y}{\sqrt{x^2+y^2}}\right)+arcsin\left(\frac{y\sqrt{1-x^2-y^2}}{\sqrt{(1-y^2)(x^2+y^2)}}\right).\hfill \end{array}$$

By (22), however,

$$\begin{array}{ccc}\sqrt{1-x^2-y^2}arcsin\left(\frac{x}{\sqrt{x^2+y^2}}\right)\hfill & +\hfill & \sqrt{1-x^2-y^2}arcsin\left(\frac{y}{\sqrt{x^2+y^2}}\right)\hfill \\ & =\hfill & \sqrt{1-x^2-y^2}\left(\frac{\pi }{2}\right),\hfill \end{array}$$

so if we define $h(x,y)$ by

$$\begin{array}{ccc}h(x,y)\hfill & =\hfill & arcsin\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)+arcsin\left(\frac{x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\right)\hfill \\ & & +arcsin\left(\frac{y\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}\sqrt{x^2+y^2}}\right)\hfill \end{array}$$

(25)

for $0\le x,y\le 1$ and $x^2+y^2<1$, then

$$\begin{array}{c}{\displaystyle 4\pi \overline{F}(x,y,z)}\hfill \\ \hspace{1em}\hspace{1em}=-\frac{\pi }{2}z+h(x,y)-arcsin\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+xarcsin\left(\frac{z}{\sqrt{1-x^2}}\right)+zarcsin\left(\frac{x}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+yarcsin\left(\frac{z}{\sqrt{1-y^2}}\right)+zarcsin\left(\frac{y}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{yz}{\sqrt{(1-y^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-xarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-yarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)\hfill \\ \hspace{1em}\hspace{1em}=-\frac{\pi }{2}z+h(x,y)+\alpha (x,y)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-xarcsin\left(\frac{y}{\sqrt{1-x^2}}\right)-yarcsin\left(\frac{x}{\sqrt{1-y^2}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+xarcsin\left(\frac{z}{\sqrt{1-x^2}}\right)+zarcsin\left(\frac{x}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+yarcsin\left(\frac{z}{\sqrt{1-y^2}}\right)+zarcsin\left(\frac{y}{\sqrt{1-z^2}}\right)-arcsin\left(\frac{yz}{\sqrt{(1-y^2)(1-z^2)}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-xarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-yarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)\hfill \end{array}$$

where $\alpha (x,y)$ is given by (16). Now (22) gives

$$\begin{array}{c}xarcsin\left(\frac{y}{\sqrt{1-x^2}}\right)+xarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)=x\left(\frac{\pi }{2}\right)\hfill \\ yarcsin\left(\frac{x}{\sqrt{1-y^2}}\right)+yarcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)=y\left(\frac{\pi }{2}\right)\hfill \end{array}$$

so the above simplifies to

$$4\pi \overline{F}(x,y,z)=-\frac{\pi }{2}(x+y+z)+h(x,y)+2\pi \Delta (x,y,z)$$

(26)

where

$$\Delta (x,y,z)=\alpha (x,y)+\alpha (x,z)+\alpha (y,z)$$

(27)

a symmetric function of $(x,y,z)$. By (22), however,

$$\begin{array}{ccc}h(x,y)\hfill & =\hfill & \frac{\pi }{2}-arcsin\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{(1-x^2)(1-y^2)}}\right)+arcsin\left(\frac{x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\right)\hfill \\ & & +arcsin\left(\frac{y\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}\sqrt{x^2+y^2}}\right)\hfill \\ & \equiv \hfill & \frac{\pi }{2}-\gamma +\alpha +\beta \hfill \end{array}$$

and

$$\begin{array}{c}{\displaystyle sin(\alpha +\beta )}\hfill \\ \hspace{1em}\hspace{1em}=sin\alpha cos\beta +cos\alpha sin\beta \hfill \\ \hspace{1em}\hspace{1em}=\frac{x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\frac{x}{\sqrt{1-y^2}\sqrt{x^2+y^2}}+\frac{y}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\frac{y\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}\sqrt{x^2+y^2}}\hfill \\ \hspace{1em}\hspace{1em}=\frac{\sqrt{1-x^2-y^2}}{\sqrt{(1-x^2)(1-y^2)}}\hfill \\ \hspace{1em}\hspace{1em}=sin\gamma \hfill \end{array}$$

so $\alpha +\beta =\gamma$, hence $h(x,y)\equiv \frac{\pi }{2}$ identically in $(x,y)$. Therefore we conclude that

$$\overline{F}(x,y,z)=\frac{1-x-y-z}{8}+\frac{\Delta (x,y,z)}{2}$$

(28)

for $0\le x,y,z\le 1$ and $x^2+y^2+z^2<1$.

We now apply (28) to obtain the cdf $F(x,y,z)$ for all $(x,y,z)\in C_3$. For this, extend the definition of $\Delta$ in (27) to all $(x,y,z)\in C_3$ by means of (16) and (18).

Theorem 4.1.

The unique spherical copula $F(x,y,z)$ on $C_3$ is given as follows:

for $x^2+y^2+z^2<1$,

$$F(x,y,z)=\left\{\begin{array}{cc}\frac{{\displaystyle 1+x+y+z}}{{\displaystyle 8}}+\frac{{\displaystyle \Delta (x,y,z)}}{{\displaystyle 2}},\hfill & \mathrm{if}{x}^2+y^2+z^2<1\hfill \\ \frac{{\displaystyle 1+x+y+z}}{{\displaystyle 8}}+\frac{{\displaystyle \Delta (x,y,z)}}{{\displaystyle 2}}\hfill \\ +\sigma \left(xyz\right)\left[\frac{{\displaystyle 1-\left|x\right|-\left|y\right|-\left|z\right|}}{{\displaystyle 8}}+\frac{{\displaystyle \Delta \left(\right|x|,|y|,|z\left|\right)}}{{\displaystyle 2}}\right],\hfill & \mathrm{if}{x}^2+y^2+z^2\ge 1.\hfill \end{array}\right.$$

Proof. 

See [9]. □

5. A One-Parameter Family of Elliptical Copulas

Let $(X,Y)\sim f(x,y)$ in (6), the unique circularly symmetric distribution on the unit disk $B_2$ with uniform $[-1,1]$ marginals. For any angle $\gamma \in (-\pi /2,\pi /2)$, consider the transformed variables

$$U=X,V_{\gamma }=Xsin\gamma +Ycos\gamma .$$

(29)

By the circular symmetry of $(X,Y)$, $V_{\gamma }\stackrel{d}{=}Y\sim \mathrm{uniform}[-1,1]$, so the random vector $(U,V_{\gamma })$ again generates a copula on the centered square $C_2$. Denote the pdf and cdf of $(U,V_{\gamma })$ by $f_{\gamma }(u,v)$ and $F_{\gamma }(u,v)$ respectively. Then $\{F_{\gamma }\mid \gamma \in (-\pi /2,\pi /2)\}$ is a one-parameter family of elliptical copulas, so-called because the support of $(U,V_{\gamma })$ is the ellipse

$$E_{\gamma }:=\{(u,v)\mid u^2+v^2-2uvsin\gamma \le {cos}^2\gamma \}.$$

(30)

(Note that $E_0=B_2$.) From (29), the correlation coefficient of U and $V_{\gamma }$ is given simply by

$$\rho (U,V_{\gamma })=sin\gamma$$

(31)

so $\gamma$ indicates the degree of linear dependence between U and $V_{\gamma }$.

Proposition 5.1.

The pdf of $(U,V_{\gamma })$ is given by

$$f_{\gamma }(u,v)=\frac{1}{2\pi \sqrt{{cos}^2\gamma -u^2-v^2+2uvsin\gamma }}{1}_{E_{\gamma }}(u,v).$$

(32)

Proof. 

The pdf can be obtained by a standard Jacobian computation. From (29),

$$x(u,v)=u,y(u,v)=\frac{v-usin\gamma }{cos\left(\gamma \right)}$$

(33)

so

$$\begin{array}{c}\begin{array}{cc}\frac{{\displaystyle \partial x(u,v)}}{{\displaystyle \partial u}}=1,\hfill & \frac{{\displaystyle \partial x(u,v)}}{{\displaystyle \partial v}}=0,\hfill \\ \frac{{\displaystyle \partial y(u,v)}}{{\displaystyle \partial u}}=-tan\gamma ,\hfill & \frac{{\displaystyle \partial y(u,v)}}{{\displaystyle \partial v}}=\frac{{\displaystyle 1}}{{\displaystyle cos\gamma }}.\hfill \end{array}\hfill \end{array}$$

Thus the Jacobian of the transformation is $J=1/cos\gamma$, so from (6) we obtain

$$\begin{array}{ccc}{f}_{\gamma }(u,v)\hfill & =\hfill & f\left(x\right(u,v),y(u,v\left)\right)·J\hfill \\ & =\hfill & \frac{1}{2\pi \sqrt{1-u^2-{(v-usin\gamma )}^2/{cos}^2\gamma }}{1}_{B_2}(x(u,v),y(u,v))·\frac{1}{cos\gamma }\hfill \\ & =\hfill & \frac{cos\gamma }{2\pi \sqrt{(1-u^2){cos}^2\gamma -{(v-usin\gamma )}^2}}\frac{1}{cos\gamma }·1_{B_2}(x(u,v),y(u,v))\hfill \\ & =\hfill & \frac{1}{2\pi \sqrt{{cos}^2\gamma -u^2-v^2+2uvsin\gamma }}{1}_{E_{\gamma }}(u,v).\hfill \end{array}$$

□

Figure 8 shows the density $f_{\gamma }(u,v)$ with $\gamma =\pi /4$.

To describe the family of elliptical copulas $F_{\gamma }$, we extend the definitions (16) and (18) as follows. First, for $(u,v)\in E_{\gamma }$ define

$$\begin{array}{ccc}{\alpha }_{\gamma }(u,v)\hfill & =\hfill & \frac{1}{2\pi }\left[uarcsin\left(\frac{v-usin\gamma }{cos\gamma \sqrt{1-u^2}}\right)+varcsin\left(\frac{u-vsin\gamma }{cos\gamma \sqrt{1-v^2}}\right)\right.\hfill \\ & & \left.-arcsin\left(\frac{uv-sin\gamma }{\sqrt{(1-u^2)(1-v^2)}}\right)\right].\hfill \end{array}$$

(34)

Note that ${\alpha }_{\gamma }$ reduces to $\alpha$ in (16) when $\gamma =0$, i.e., when $V_{\gamma }=Y$. From (12),

$$A(arccosu,arccosv;\frac{\pi }{2}-\gamma )=(1-u-v)\pi +4\pi {\alpha }_{\gamma }(u,v).$$

(35)

Next, extend the definition of ${\alpha }_{\gamma }(u,v)$ to $C_2\backslash E_{\gamma }$ as follows (see Figure 9):

$${\alpha }_{\gamma }(u,v)=\left\{\begin{array}{cc}\frac{{\displaystyle u+v-1}}{{\displaystyle 4}}\hfill & \mathrm{if}(u,v)\in R_5\left(\gamma \right):=(C_2\backslash E_{\gamma })\cap \{(u,v)\mid u+v>1+sin\gamma \},\hfill \\ \frac{{\displaystyle u-v+1}}{{\displaystyle 4}}\hfill & \mathrm{if}(u,v)\in R_6\left(\gamma \right):=(C_2\backslash E_{\gamma })\cap \{(u,v)\mid v-u>1-sin\gamma \},\hfill \\ \frac{{\displaystyle -u+v+1}}{{\displaystyle 4}}\hfill & \mathrm{if}(u,v)\in R_7\left(\gamma \right):=(C_2\backslash E_{\gamma })\cap \{(u,v)\mid v-u<sin\gamma -1\},\hfill \\ \frac{{\displaystyle -u-v-1}}{{\displaystyle 4}}\hfill & \mathrm{if}(u,v)\in R_8\left(\gamma \right):=(C_2\backslash E_{\gamma })\cap \{(u,v)\mid u+v<-sin\gamma -1\}.\hfill \end{array}\right.$$

(36)

Note that (34) and (36) agree on $\partial E_{\gamma }$, i.e., when $u^2+v^2-2uvsin\gamma ={cos}^2\gamma$. Also note that (36) reduces to $\alpha$ in (18) when $\gamma =0$. The following lemma will be useful for the proof of Theorem 5.1.

Lemma 5.1.

Let $(U,V)$ be a bivariate random vector in $C_2$ with uniform $[-1,1]$ marginals that satisfies $(U,V)\stackrel{d}{=}(-U,-V)$. Then the cdf $F(u,v)$ satisfies

$$F(u,v)=\frac{u+v}{2}+F(-u,-v),(u,v)\in C_2.$$

(37)

Proof. 

By the symmetry condition,

$$\begin{array}{ccc}F(u,v)\hfill & =\hfill & P[-U\le u,-V\le v]=P[U\ge -u,V\ge -v]\hfill \\ & =\hfill & 1-P[U<-u]-P[V<-v]+P[U<-u,V<-v]\hfill \\ & =\hfill & 1-\left(\frac{-u+1}{2}\right)+\left(\frac{-v+1}{2}\right)+P[U\le -u,V\le -v]\hfill \\ & =\hfill & \frac{u+v}{2}+F(-u,-v).\hfill \end{array}$$

□

Theorem 5.1.

The cdf ≡ copula of $(U,V_{\gamma })$ is given by (see Figure 12)

$$F_{\gamma }(u,v)=\frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v),(u,v)\in C_2.$$

(38)

Proof. 

To find $F_{\gamma }(u,v)$ we again use the formula (12) for the area of the intersection of two spherical caps on $\partial B_3$. Here, unlike (14), the axes of the two caps are not necessarily perpendicular. The single formula (38) is obtained by considering the partition $C_2={\cup }_{i=1}^8{R}_i\left(\gamma \right)$, where $R_5\left(\gamma \right)-R_8\left(\gamma \right)$ are defined in (36) and (see Figure 9)

$$\begin{array}{ccc}{R}_1\left(\gamma \right)\hfill & =\hfill & E_{\gamma }\cap \{(u,v)\mid 0\le u,v\le 1\},\hfill \\ R_2\left(\gamma \right)\hfill & =\hfill & E_{\gamma }\cap \{(u,v)\mid -1\le u\le 0\le v\le 1\},\hfill \\ R_3\left(\gamma \right)\hfill & =\hfill & E_{\gamma }\cap \{(u,v)\mid -1\le v\le 0\le u\le 1\},\hfill \\ R_4\left(\gamma \right)\hfill & =\hfill & E_{\gamma }\cap \{(u,v)\mid -1\le u,v\le 0\}.\hfill \end{array}$$

Case 1: $(u,v)\in R_1\left(\gamma \right)$. By using Figure 10,

$$\begin{array}{ccc}{F}_{\gamma }(u,v)\hfill & =\hfill & P[U\le u,V_{\gamma }\le v]\hfill \\ & =\hfill & 1-P[U>u]-P[V_{\gamma }>v]+P[U>u,V_{\gamma }>v]\hfill \\ & =\hfill & 1-\left(\frac{1-u}{2}\right)-\left(\frac{1-v}{2}\right)+P[X>u,Xsin\gamma +Ycos\gamma >v]\hfill \\ & =\hfill & \frac{u+v}{2}+\frac{1}{4\pi }A(arccosu,arccosv;\pi /2-\gamma )\hfill \\ & =\hfill & \frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\mathrm{by}\left(35\right)\right].\hfill \end{array}$$

Case 2: $(u,v))\in R_2\left(\gamma \right)$. Because $(X,Y)\stackrel{d}{=}(-X,Y)$ and using Figure 11

$$\begin{array}{ccc}{F}_{\gamma }(u,v)\hfill & =\hfill & P[U\le u]-P[U\le u,V_{\gamma }>v]\hfill \\ & =\hfill & \frac{u+1}{2}-P[X\le u,Xsin\gamma +Ycos\gamma >v]\hfill \\ & =\hfill & \frac{u+1}{2}-P[-X\le u,-Xsin\gamma +Ycos\gamma >v]\hfill \\ & =\hfill & \frac{u+1}{2}-P[X\ge -u,Xsin(-\gamma )+Ycos(-\gamma )>v]\hfill \\ & =\hfill & \frac{u+1}{2}-\frac{1}{4\pi }A(arccos(-u),arccosv;\pi /2+\gamma )\hfill \\ & =\hfill & \frac{u+v+1}{4}-{\alpha }_{-\gamma }(-u,v)\left[\mathrm{by}\left(35\right)\right]\hfill \\ & =\hfill & \frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\mathrm{by}\left(34\right)\right].\hfill \end{array}$$

Case 3: $(u,v)\in R_3\left(\gamma \right)$. Then $(-u,-v)\in R_2\left(\gamma \right)$, so by Lemma 5.1 and Case 2,

$$\begin{array}{ccc}{F}_{\gamma }(u,v)\hfill & =\hfill & \frac{u+v}{2}+F_{\gamma }(-u,-v)\hfill \\ & =\hfill & \frac{u+v}{2}+\frac{-u-v+1}{4}+{\alpha }_{\gamma }(-u,-v)\hfill \\ & =\hfill & \frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\mathrm{by}\left(34\right)\right].\hfill \end{array}$$

Case 4: $(u,v)\in R_4\left(\gamma \right)$. Then $(-u,-v)\in R_1\left(\gamma \right)$, so by Lemma 5.1 and Case 1, the argument for Case 3 applies verbatim.

Case 5: $(u,v)\in R_5\left(\gamma \right)$.

$$\begin{array}{ccc}{F}_{\gamma }(u,v)\hfill & =\hfill & 1-P[U>u]-P[V_{\gamma }>v]\hfill \\ & =\hfill & 1-\left(\frac{1-u}{2}\right)-\left(\frac{1-v}{2}\right)\hfill \\ & =\hfill & \frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\mathrm{by}\left(36\right)\right].\hfill \end{array}$$

Case 6: $(u,v)\in R_6\left(\gamma \right)$.

$$\begin{array}{ccc}{F}_{\gamma }(u,v)\hfill & =\hfill & P[U\le u]\hfill \\ & =\hfill & \frac{u+1}{2}\hfill \\ & =\hfill & \frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\mathrm{by}\left(36\right)\right].\hfill \end{array}$$

Case 7: $(u,v)\in R_7\left(\gamma \right)$. Then $(-u,-v)\in R_6\left(\gamma \right)$, so by Lemma 5.1 and Case 6, the argument for Case 3 applies verbatim.

Case 8: $(u,v)\in R_8\left(\gamma \right)$. Then $(-u,-v)\in R_5\left(\gamma \right)$, so by Lemma 5.1 and Case 5, the argument for Case 3 applies verbatim. □

6. Copulas Derived from the Uniform Distribution on the Unit Ball

Up to now we have addressed the question of whether copulas can be generated by means of linear functions of a circularly symmetric or spherically symmetric random vector. Now we ask whether non-linear functions of such random vectors can generate copulas. We shall restrict attention to random vectors uniformly distributed over the unit ball $B_d$ and produce relatively simple non-linear functions that generate copulas on $C_d$.

We begin with the bivariate case. Suppose that $(X,Y)$ is distributed uniformly on the unit disk $B_2=\{(x,y)\in {\mathbb{R}}^2\mid x^2+y^2\le 1\}$. Because

$$\begin{array}{cccc}& X\mid Y\hfill & \sim \hfill & \mathrm{uniform}[-\sqrt{1-Y^2},\sqrt{1-Y^2}]\hfill \\ \mathrm{and}\hfill & Y\mid X\hfill & \sim \hfill & \mathrm{uniform}[-\sqrt{1-X^2},\sqrt{1-X^2}]\hfill \end{array}$$

it follows that the random variables

$$U:=\frac{X}{\sqrt{1-Y^2}},V:=\frac{Y}{\sqrt{1-X^2}}$$

(39)

satisfy

$$\begin{array}{ccc}U\mid Y\hfill & \sim \hfill & \mathrm{uniform}[-1,1]\hfill \\ V\mid X\hfill & \sim \hfill & \mathrm{uniform}[-1,1].\hfill \end{array}$$

Thus, U and Y are independent, V and X are independent, and unconditionally,

$$\begin{array}{ccc}U\hfill & \sim \hfill & \mathrm{uniform}[-1,1]\hfill \\ V\hfill & \sim \hfill & \mathrm{uniform}[-1,1]\hfill \end{array}$$

so the joint distribution of $(U,V)$ generates a copula $F_{(}u,v)$ on the centered cube $C_2\equiv {[-1,1]}^2$. Note that U and V are not linear functions of $(X,Y)$.

Question 6: Are U and V independent, and if not, what is the nature of their dependence?

Answer: Clearly U and V are uncorrelated, since $E\left(U\right)=E\left(V\right)=0$ and

$$E\left(UV\right)=E\left(\frac{XY}{\sqrt{(1-X^2)(1-Y^2)}}\right)=0$$

all by the circular symmetry of $(X,Y)$. However, the joint pdf and cdf of $(U,V)$ derived below show that they are not independent.

Proposition 6.1.

The joint density of $(U,V)$ is given by (see Figure 13)

$$\begin{array}{c}f(u,v)=\frac{1}{\pi }\frac{\sqrt{(1-u^2)(1-v^2)}}{{(1-u^2{v}^2)}^2}{1}_{C_2}(u,v).\hfill \end{array}$$

(40)

Proof. 

This pdf is again obtained via the Jacobian method. It follows from (39) that

$$\begin{array}{c}{u}^2(1-y^2)=x^2,and{v}^2(1-x^2)=y^2.\hfill \end{array}$$

Substitution of the second expression for $y^2$ into the left side of the first relation and vice versa yields

$$\begin{array}{c}{x}^2=\frac{u^2(1-v^2)}{1-u^2{v}^2},y^2=\frac{v^2(1-u^2)}{1-u^2{v}^2}\hfill \end{array}$$

so, since x and u (y and v) have the same signs by (39), we obtain

$$x\equiv x(u,v)=\frac{u\sqrt{1-v^2}}{\sqrt{1-u^2{v}^2}},y\equiv y(u,v)=\frac{v\sqrt{1-u^2}}{\sqrt{1-u^2{v}^2}}.$$

(41)

Thus

$$\begin{array}{ccc}\frac{\partial x}{\partial u}\hfill & =\hfill & \sqrt{1-v^2}\left[\frac{1}{\sqrt{1-u^2{v}^2}}+u{(1-u^2{v}^2)}^{-3/2}\left(u{v}^2\right)\right]\hfill \\ & =\hfill & \frac{\sqrt{1-v^2}}{\sqrt{1-u^2{v}^2}}\left\{1+\frac{u^2{v}^2}{1-u^2{v}^2}\right\}\hfill \\ & =\hfill & \frac{\sqrt{1-v^2}}{{(1-u^2{v}^2)}^{3/2}},\hfill \end{array}$$

$$\begin{array}{ccc}\frac{\partial x}{\partial v}\hfill & =\hfill & u\left[\frac{{(1-v^2)}^{-1/2}(-v)}{\sqrt{1-u^2{v}^2}}+\sqrt{1-v^2}{(1-u^2{v}^2)}^{-3/2}\left(u^{2}v\right)\right]\hfill \\ & =\hfill & \frac{uv}{\sqrt{1-v^2}\sqrt{1-u^2{v}^2}}\left\{-1+\frac{u^2(1-v^2)}{1-u^2{v}^2}\right\}\hfill \\ & =\hfill & -\frac{uv(1-u^2)}{\sqrt{1-v^2}{(1-u^2{v}^2)}^{3/2}}.\hfill \end{array}$$

By symmetry it follows that the Jacobian is given by

$$\begin{array}{c}J=\left(\begin{array}{cc}\frac{{\displaystyle \sqrt{1-v^2}}}{{\displaystyle {(1-u^2{v}^2)}^{3/2}}}& -\frac{{\displaystyle uv(1-u^2)}}{{\displaystyle \sqrt{1-v^2}{(1-u^2{v}^2)}^{3/2}}}\\ -\frac{{\displaystyle uv(1-v^2)}}{{\displaystyle \sqrt{1-u^2}{(1-u^2{v}^2)}^{3/2}}}& \frac{{\displaystyle \sqrt{1-u^2}}}{{\displaystyle {(1-u^2{v}^2)}^{3/2}}}\end{array}\right)\end{array}$$

and hence the determinant of J is given by

$$\begin{array}{c}\left|J\right|=\frac{\sqrt{(1-u^2)(1-v^2)}}{{(1-u^2{v}^2)}^2}.\hfill \end{array}$$

Because the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi }{1}_{B_2}(x,y)$, the result (40) follows. □

For $0\le u,v\le 1$, $(u,v)\ne (1,1)$, let $E_1\left(u\right)$ and $E_2\left(v\right)$ be the ellipses

$$\begin{array}{ccc}{E}_1\left(u\right)\hfill & =\hfill & \left\{(x,y)|\frac{x^2}{u^2}+y^2\le 1\right\}\hfill \end{array}$$

(42)

$$\begin{array}{ccc}{E}_2\left(v\right)\hfill & =\hfill & \left\{(x,y)|x^2+\frac{y^2}{v^2}\le 1\right\}.\hfill \end{array}$$

(43)

The next lemma leads to the cdf $F(u,v)$ corresponding to the pdf (40).

Lemma 6.1.

$$\mathrm{Area}(E_1\left(u\right)\cap E_2\left(v\right))=2uarcsin\left(\frac{v\sqrt{1-u^2}}{\sqrt{1-u^2{v}^2}}\right)+2varcsin\left(\frac{u\sqrt{1-v^2}}{\sqrt{1-u^2{v}^2}}\right).$$

Proof. 

Define the points $o,a,b,c,d,d,f,g$ as follows: see Figure 14,

$$\begin{array}{ccc}o\hfill & =\hfill & (0,0)\hfill \\ a\hfill & =\hfill & (x(u,v),y(u,v))=\left(\frac{u\sqrt{1-v^2}}{\sqrt{1-u^2{v}^2}},\frac{v\sqrt{1-u^2}}{\sqrt{1-u^2{v}^2}}\right)\hfill \\ b\hfill & =\hfill & (u,0)\hfill \\ c\hfill & =\hfill & (0,v)\hfill \\ d\hfill & =\hfill & (\sqrt{1-y^2(u,v)},y(u,v))\hfill \\ e\hfill & =\hfill & (x(u,v),\sqrt{1-x^2(u,v)})\hfill \\ f\hfill & =\hfill & (1,0)\hfill \\ g\hfill & =\hfill & (0,1).\hfill \end{array}$$

Then

$$\begin{array}{ccc}\frac{1}{4}\mathrm{Area}(E_1\left(u\right)\cap E_2\left(v\right))\hfill & =\hfill & \mathrm{Area}\left(oab\right)+\mathrm{Area}\left(oac\right)\hfill \\ & =\hfill & u\mathrm{Area}\left(odf\right)+v\mathrm{Area}\left(oeg\right)\hfill \\ & =\hfill & \frac{u}{2}arcsin\left(y(u,v)\right)+\frac{v}{2}arcsin\left(x(u,v)\right)\hfill \end{array}$$

from which the result follows. □

Theorem 6.1.

The copula (= cdf) corresponding to the pdf (40) is given by (see Figure 15)

$$F(u,v)=\frac{u+v+1}{4}+\frac{u}{2\pi }arcsin\left(\frac{v\sqrt{1-u^2}}{\sqrt{1-u^2{v}^2}}\right)+\frac{v}{2\pi }arcsin\left(\frac{u\sqrt{1-v^2}}{\sqrt{1-u^2{v}^2}}\right),(u,v)\in C_2.$$

Proof. 

Because $(U,V)$ is sign-change invariant and has uniform $[-1,1]$ marginals, it follows from (7) and (9) in Lemma 3.1 and from (39) that for $(u,v)\in C_2$,

$$\begin{array}{ccc}F(u,v)\hfill & =\hfill & \frac{u+v+1}{4}+\sigma \left(uv\right)P[0\le U\le |u|,0\le V\le |v\left|\right]\hfill \\ & =\hfill & \frac{u+v+1}{4}+\frac{\sigma \left(uv\right)}{4}P[U^2\le u^2,V^2\le v^2]\hfill \\ & =\hfill & \frac{u+v+1}{4}+\frac{\sigma \left(uv\right)}{4}P[(X,Y)\in E_1\left(u\right)\cap E_2\left(v\right)]\hfill \\ & =\hfill & \frac{u+v+1}{4}+\frac{\sigma \left(uv\right)}{4\pi }\mathrm{Area}(E_1\left(\right|u\left|\right)\cap E_2\left(\right|v\left|\right)).\hfill \end{array}$$

The result now follows from Lemma 6.1. □

Remark: 

The construction (39) extends readily to generate a copula on $C_d$. For $d=3$, for example, let $(X,Y,Z)$ be uniformly distributed on the unit ball $B_3$ and define

$$U:=\frac{X}{\sqrt{1-Y^2-Z^2}},V:=\frac{Y}{\sqrt{1-X^2-Z^2}},W:=\frac{Z}{\sqrt{1-X^2-Y^2}}.$$

Then the marginal distributions of U, V, and W are each uniform $[-1,1]$ so the cdf $G(u,v,w)$ is a copula on $C_3$. To find this copula one would need to determine $\mathrm{Volume}(E_1\left(u\right)\cap E_2\left(v\right)\cap E_3\left(w\right))$, where now, for $0\le u,v,w\le 1$, $E_1\left(u\right)$, $E_2\left(v\right)$, and $E_3\left(w\right)$ are the ellipsoids

$$\begin{array}{ccc}{E}_1\left(u\right)\hfill & =\hfill & \left\{(x,y,z)|\frac{x^2}{u^2}+y^2+z^2\le 1\right\}\hfill \\ E_2\left(v\right)\hfill & =\hfill & \left\{(x,y,z)|x^2+\frac{y^2}{v^2}+z^2\le 1\right\}\hfill \\ E_3\left(w\right)\hfill & =\hfill & \left\{(x,y,z)|x^2+y^2+\frac{y^2}{w^2}\le 1\right\}.\hfill \end{array}$$