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 ≥ 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 ≥ 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 ≥ 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 ≥ 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 B₂ (resp., unit ball B_d).

It will be noted in Sections 2 and 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 ≥ 4 is remarkably simple. Explicit expressions for these copulas are given in Sections 3 and 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 ∈ ∂B₃), 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 = (Z₁, . . . , Z_d) with a spherically symmetric distribution on B_d and with each Z_i ~ 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 = ||Z|| ≤ 1 and U = Z/||Z||. It is well known (e.g., Cambanis et al. [5], Lemmas 1 and 2) that the random unit vector U ≡ (U₁, . . . , U_d) is independent of R and is uniformly distributed on the unit sphere ∂B_d, which implies that each $U_i^2\sim \text{Beta}(1∕2,(d-1)∕2)$. Since Z_i = RU_i, we have that

$$\text{log}\left(Z_i^2\right)=\text{log}\left(R^2\right)+\text{log}\left(U_i^2\right).$$ (1)

Because R and U_i are independent, it follows that the characteristic function of log(R²) is the quotient of the characteristic functions of $\text{log}\left(Z_i^2\right)$ and $\text{log}\left(U_1^2\right)$. Thus the distribution of log(R²), 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 ≥ 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 ~ uniform[–1, 1], 0 ≤ R ≤ 1, and $U_i^2\sim \text{Beta}(1∕2,(d-1)∕2)$. Thus d ≤ 3, so a spherical copula cannot exist when d ≥ 4.

Furthermore, if a spherical copula is to exist for d = 3, it follows from (2) that its generating random vector Z ∈ B₃ must satisfy E(R²) = 1, hence R = 1 with probability one. This can occur only if Z is uniformly distributed on the unit sphere ∂B₃. 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) B₂ 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.

Figure 1.

The densities (3) (lower, blue) and (4) (upper, purple).

Question 2. Let (X, Y) be a random vector uniformly distributed on the unit circle ∂B₂ in ${\mathbb{R}}^2$. Find the marginal probability distributions of X and Y.

Answer: We can represent (X, Y) as (cos Θ, sin Θ) where Θ ~ uniform[0, 2π). 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 B₂? If so, it determines a circular copula on C₂, 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 B₂ 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 ${\iint }_{B_2}f(x,y)dxdy=1$ (transform to polar coordinates: (x, y) → (r, θ)).

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

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

If we take h(t) = ct^–1/2 then clearly f(x) does not depend on x, and choosing c = 1/2π 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 B₂ and yields the desired circular copula.

Figure 2.

Circularly symmetric bivariate density (6) on B₂.

Question 4. Having determined the unique circularly symmetric distribution (6) on B₂ 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) ≡ P[X ≤ x, Y ≤ y] on C₂ ≡ [–1, 1]² can be expressed in terms of F₀(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 ≤ x, y ≤ 1, and also in terms of the complementary cdf F̄(x, y) ≡ P[X > x, Y > y] for 0 ≤ x, y ≤ 1. Because $(X,Y)\stackrel{d}{=}(\pm X,\pm Y)$ and has uniform [–1, 1] marginals,

$$\begin{array}{cc}\hfill F_0(x,y)& =P[0\le X\le 1,0\le Y\le 1]-P[X>x,0\le Y\le 1]-P[0\le X\le 1,Y>y]+P[X>x,Y>y]\hfill \\ \hfill & =\frac{1}{4}-\left(\frac{1-x}{4}\right)-\left(\frac{1-y}{4}\right)+\stackrel{‒}{F}(x,y)\hfill \\ \hfill & =\frac{x+y-1}{4}+\stackrel{‒}{F}(x,y),0\le x,y\le 1.\hfill \end{array}$$ (8)

Lemma 3.1. Let (X, Y) be a bivariate random vector on C₂ with uniform [–1, 1] marginal distributions and sign-change invariance, i.e., $(X,Y)\stackrel{d}{=}(\pm X,\pm Y)$. Then for (x, y) ∈ C₂,

$$F(x,y)=\frac{x+y+1}{4}+\sigma \left(xy\right)F_0(\mid x\mid ,\mid y\mid )$$ (9)

$$=\frac{x+y+1}{4}+\sigma \left(xy\right)\left[\frac{\mid x\mid +\mid y\mid -1}{4}+\stackrel{‒}{F}(\mid x\mid ,\mid y\mid )\right]$$ (10)

where σ(w) = sign(w) if w ≠ 0 and σ(0) = 0.

Proof. To obtain (9), consider four cases:

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

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

Case 2: –1 ≤ x ≤ 0 ≤ y ≤ 1. Similarly,

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

Case 3: –1 ≤ y ≤ 0 ≤ x ≤ 1. Similarly,

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

Case 4: –1 ≤ x, y ≤ 0. Similarly,

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

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 F̄(x, y) for 0 ≤ x, y ≤ 1 and apply (10). Because F̄(x, y) = 0 when x² + y² ≥ 1, we need only consider the case where x² + y² < 1.

First approach: When 0 ≤ x, y ≤ 1 and x² + y² < 1, F̄(x, y) can be expressed as follows. By using Figure 3 we find that

$$\begin{array}{cc}\hfill \stackrel{‒}{F}(x,y)& =\frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_y^{\sqrt{1-s^2}}\frac{1}{\sqrt{1-s^2-t^2}}dt\right\}ds\hfill \\ \hfill & =\frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_y^{\sqrt{1-s^2}}\frac{1}{\sqrt{(1-\frac{t^2}{1-s^2})}}\frac{dt}{\sqrt{1-s^2}}\right\}ds\hfill \\ \hfill & =\frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left\{{\int }_{\frac{y}{\sqrt{1-s^2}}}^1\frac{dv}{\sqrt{1-v^2}}\right\}ds\hfill \\ \hfill & =\frac{1}{2\pi }{\int }_x^{\sqrt{1-y^2}}\left[\frac{\pi }{2}-\text{arcsin}\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds.\hfill \end{array}$$ (11)

However, we were unable to evaluate this integral directly.

Figure 3.

Region of integration, 2-dimensional case.

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 ∂B₃ 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 S₁ and S₂ be spherical caps on ∂B₃. Let r₁ and r₂ denote their angular radii and let d denote the angular distance between their centers (0 < d ≤ π). Assume that 0 < r₁, r₂ ≤ π/2 and d ≤ r₁+r₂, so that the intersection $S_1\cap S_2\ne \varnothing$ and consists of a single “diangle”; (see Figures 4 and 5.) Then Area($S_1\cap S_2$) is given by

$$A(r_1,r_2;d)=2\pi -2\pi \text{cos}\left(r_1\right)-2\pi \text{cos}\left(r_2\right)-2\text{arccos}\left(\frac{\text{cos}\left(d\right)-\text{cos}\left(r_1\right)\text{cos}\left(r_2\right)}{\text{sin}\left(r_1\right)\text{sin}\left(r_2\right)}\right)+2\text{cos}\left(r_1\right)\text{arccos}\left(\frac{\text{cos}\left(d\right)\text{cos}\left(r_1\right)-\text{cos}\left(r_2\right)}{\text{sin}\left(d\right)\text{sin}\left(r_1\right)}\right)+2\text{cos}\left(r_2\right)\text{arccos}\left(\frac{\text{cos}\left(d\right)\text{cos}\left(r_2\right)-\text{cos}\left(r_1\right)}{\text{sin}\left(d\right)\text{sin}\left(r_2\right)}\right).$$ (12)

Figure 4.

Intersection of two spherical caps.

Figure 5.

Intersection of two spherical caps, circular representation (modified from Tovchigrechko and Vakser [7]).

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

If (X, Y, Z) is uniformly distributed on ∂B₃, 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 ∂B₃, i.e., by 4π. (See Figures 4 and 5.) Also, the joint distribution of (X, Y) is circularly symmetric on the unit disk B₂ and has uniform marginals, so must be the unique such bivariate distribution, namely the distribution with pdf (6).

Thus, for 0 ≤ x, y ≤ 1 and x² + y² < 1, our desired complementary cdf is given by

$$\stackrel{‒}{F}(x,y)=\frac{1}{4\pi }A(x,y)$$ (13)

$$\begin{array}{cc}\hfill & =\frac{1}{4\pi }A(\text{arccos}\left(x\right),\text{arccos}\left(y\right);\pi ∕2)\hfill \\ \hfill & =\frac{1}{2}-\frac{x}{2}-\frac{y}{2}-\frac{1}{2\pi }\text{arccos}\left(-\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)+\frac{x}{2\pi }\text{arccos}\left(\frac{-y}{\sqrt{1-x^2}}\right)+\frac{y}{2\pi }\text{arccos}\left(\frac{-x}{\sqrt{1-y^2}}\right)\hfill \end{array}$$ (14)

$$\equiv \frac{1-x-y}{4}+\alpha (x,y)$$ (15)

where for 0 ≤ x, y ≤ 1 and x² + y² < 1,

$$\alpha (x,y)=\frac{1}{2\pi }\left[x\text{arcsin}\left(\frac{y}{\sqrt{1-x^2}}\right)+y\text{arcsin}\left(\frac{x}{\sqrt{1-y^2}}\right)-\text{arcsin}\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)\right].$$ (16)

Theorem 3.1. The unique circular copula on C₂ is given by

$$F(x,y)=\frac{x+y+1}{4}+\alpha (x,y)$$ (17)

where α(x, y) is defined by (16) for x² + y² < 1 and by

$$\alpha (x,y)=\sigma \left(xy\right)\cdot \left(\frac{\mid x\mid +\mid y\mid -1}{4}\right)$$ (18)

for x² + y² ≥ 1. Note that Equations (16) and (18) agree when x² + y² = 1 and both are sign-change equivariant on C₂: for all (x, y) ∈ C₂ and all ε, δ = ±1,

$$\alpha (∊x,\delta y)=∊\delta \cdot \alpha (x,y).$$ (19)

Proof. From Equations (10) and (15), when x² + y² < 1 we have

$$F(x,y)=\frac{x+y+1}{4}+\sigma \left(xy\right)\alpha (\mid x\mid ,\mid y\mid )$$ (20)

$$=\frac{x+y+1}{4}+\alpha (x,y)$$ (21)

by (19). When x² + y² ≥ 1, F̄(|x|, |y|) = 0 so (17) again holds by (10) and (18).

See Figure 6 for a plot of the resulting copula (on [–1, 1]²).

Figure 6.

The copula (17) in Theorem 3.1.

4. The Trivariate Case: the Unique Spherical Copula

Question 5. Having determined the unique spherically symmetric distribution on B₃ with uniform marginals, namely, the uniform distribution on the unit sphere ∂B₃, what is the corresponding cdf F(x, y, z) on C₃, i.e., the unique spherical copula?

Answer: As in Section 3, let (X, Y, Z) be uniformly distributed on ∂B₃, so that F(x, y, z) = P[X ≤ x, Y ≤ y, Z ≤ z]. Again we first determine the complementary cdf F̄(x, y, z) ≡ P[X > x, Y > y, Z > z] for 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1, the intersection of the first octant of C₃ with the interior of B₃. 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 ∂B₃, so F̄(x, y, z) is the area A(x, y, z) of this intersection divided by the total area 4π of ∂B₃.

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 $\stackrel{‒}{F}(x,y,z)\equiv \frac{1}{4\pi }A(x,y,z)$ directly for the trivariate case, as now described.

We begin by extending (11) to obtain an integral expression for F̄(x, y, z) when 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1. We require the fact that

$$0\le a,b\le 1\text{and}{a}^2+b^2=1\text{implies arcsin}\left(a\right)+\text{arcsin}\left(b\right)=\pi ∕2.$$ (22)

Lemma 4.1. If 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1, then

$$\stackrel{‒}{F}(x,y,z)=\frac{1}{4\pi }{\int }_x^{\sqrt{1-y^2-z^2}}\left[\frac{\pi }{2}-\text{arcsin}\left(\frac{y}{\sqrt{1-s^2}}\right)-\text{arcsin}\left(\frac{z}{\sqrt{1-s^2}}\right)\right]ds.$$ (23)

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

Proof. Since X² + Y² + Z² = 1 and $(X,Y,Z)\stackrel{d}{=}(X,Y,-Z)$, it follows from (6) by using Figure 7 that

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

Now apply (22) to obtain (23).

Figure 7.

Region of integration, 3-dimensional case, Lemma 4.1.

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 ≤ x, y ≤ 1 and x² + y² < 1,

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

where α(x, y) is given by (16). Because $z\le \sqrt{1-x^2-y^2}\le \sqrt{1-y^2}$ when 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1, it follows that

$$\frac{1}{2\pi }{\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-\text{arcsin}\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds=\frac{\sqrt{1-x^2-y^2}-z}{4}+\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},y).$$ (24)

Therefore from (23) and (24), if 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1 then

$$4\pi \stackrel{‒}{F}(x,y,z)={\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-\text{arcsin}\left(\frac{x}{\sqrt{1-s^2}}\right)\right]ds+{\int }_z^{\sqrt{1-x^2-y^2}}\left[\frac{\pi }{2}-\text{arcsin}\left(\frac{y}{\sqrt{1-s^2}}\right)\right]ds-\frac{\pi }{2}\left[\sqrt{1-x^2-y^2}-z\right]=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+2\pi \left[\alpha (z,x)-\alpha (\sqrt{1-x^2-y^2},x)\right]+\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+2\pi \left[\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},y)\right]-\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+2\pi \left[\alpha (z,x)+\alpha (z,y)-\alpha (\sqrt{1-x^2-y^2},x)-\alpha (\sqrt{1-x^2-y^2},y)\right]=\frac{\pi }{2}(\sqrt{1-x^2-y^2}-z)+x\text{arcsin}\left(\frac{z}{\sqrt{1-x^2}}\right)+z\text{arcsin}\left(\frac{x}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)+y\text{arcsin}\left(\frac{z}{\sqrt{1-y^2}}\right)+z\text{arcsin}\left(\frac{y}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{yz}{\sqrt{(1-y^2)(1-z^2)}}\right)-x\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-\sqrt{1-x^2-y^2}\text{arcsin}\left(\frac{x}{\sqrt{x^2+y^2}}\right)+\text{arcsin}\left(\frac{x\sqrt{1-x^2-y^2}}{\sqrt{(1-x^2)(x^2+y^2)}}\right)-y\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)-\sqrt{1-x^2-y^2}\text{arcsin}\left(\frac{y}{\sqrt{x^2+y^2}}\right)+\text{arcsin}\left(\frac{y\sqrt{1-x^2-y^2}}{\sqrt{(1-y^2)(x^2+y^2)}}\right).$$

By (22), however,

$$\sqrt{1-x^2-y^2}\text{arcsin}\left(\frac{x}{\sqrt{x^2+y^2}}\right)+\sqrt{1-x^2-y^2}\text{arcsin}\left(\frac{y}{\sqrt{x^2+y^2}}\right)=\sqrt{1-x^2-y^2}\left(\frac{\pi }{2}\right),$$

so if we define h(x, y) by

$$h(x,y)=\text{arcsin}\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)+\text{arcsin}\left(\frac{x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\right)+\text{arcsin}\left(\frac{y\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}\sqrt{x^2+y^2}}\right)$$ (25)

for 0 ≤ x, y ≤ 1 and x² + y² < 1, then

$$4\pi \stackrel{‒}{F}(x,y,z)=-\frac{\pi }{2}z+h(x,y)-\text{arcsin}\left(\frac{xy}{\sqrt{(1-x^2)(1-y^2)}}\right)+x\text{arcsin}\left(\frac{z}{\sqrt{1-x^2}}\right)+z\text{arcsin}\left(\frac{x}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)+y\text{arcsin}\left(\frac{z}{\sqrt{1-y^2}}\right)+z\text{arcsin}\left(\frac{y}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{yz}{\sqrt{(1-y^2)}(1-z^2)}\right)-x\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-y\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)=-\frac{\pi }{2}z+h(x,y)+\alpha (x,y)-x\text{arcsin}\left(\frac{y}{\sqrt{1-x^2}}\right)-y\text{arcsin}\left(\frac{x}{\sqrt{1-y^2}}\right)+x\text{arcsin}\left(\frac{z}{\sqrt{1-x^2}}\right)+z\text{arcsin}\left(\frac{x}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{xz}{\sqrt{(1-x^2)(1-z^2)}}\right)+y\text{arcsin}\left(\frac{z}{\sqrt{1-y^2}}\right)+z\text{arcsin}\left(\frac{y}{\sqrt{1-z^2}}\right)-\text{arcsin}\left(\frac{yz}{\sqrt{(1-y^2)(1-z^2)}}\right)-x\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}}\right)-y\text{arcsin}\left(\frac{\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}}\right)$$

where α(x, y) is given by (16). Now (22) gives

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

so the above simplifies to

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

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

and

$$\begin{array}{cc}\hfill \text{sin}(\alpha +\beta )& =\text{sin}\alpha \text{cos}\beta +\text{cos}\alpha \text{sin}\beta \hfill \\ \hfill & =\frac{x\sqrt{1-x^2-y^2}}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\frac{x}{\sqrt{1-y^2}\sqrt{x^2+y^2}}+\frac{y}{\sqrt{1-x^2}\sqrt{x^2+y^2}}\frac{y\sqrt{1-x^2-y^2}}{\sqrt{1-y^2}\sqrt{x^2+y^2}}\hfill \\ \hfill & =\frac{\sqrt{1-x^2-y^2}}{\sqrt{(1-x^2)(1-y^2)}}\hfill \\ \hfill & =\text{sin}\gamma \hfill \end{array}$$

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

$$\stackrel{‒}{F}(x,y,z)=\frac{1-x-y-z}{8}+\frac{\Delta (x,y,z)}{2}$$ (28)

for 0 ≤ x, y, z ≤ 1 and x² + y² + z² < 1.

We now apply (28) to obtain the cdf F(x, y, z) for all (x, y, z) ∈ C₃. For this, extend the definition of Δ in (27) to all (x, y, z) ∈ C₃ by means of (16) and (18).

Theorem 4.1. The unique spherical copula F(x, y, z) on C₃ is given as follows: for x² + y² + z² < 1,

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

Proof. See [9].

5. A One-Parameter Family of Elliptical Copulas

Let (X, Y) ~ f(x, y) in (6), the unique circularly symmetric distribution on the unit disk B₂ with uniform [–1, 1] marginals. For any angle γ ε (–π/2, π/2), consider the transformed variables

$$U=X,V_{\gamma }=X\text{sin}\gamma +Y\text{cos}\gamma .$$ (29)

By the circular symmetry of $(X,Y),V_{\gamma }\stackrel{d}{=}Y\sim \text{uniform}[-1,1]$, so the random vector (U, V_γ) again generates a copula on the centered square C₂. Denote the pdf and cdf of (U, V_γ) by f_γ(u, v) and F_γ(u, v) respectively. Then {F_γ | γ ε (–π/2, π/2)} is a one-parameter family of elliptical copulas, so-called because the support of (U, V_γ) is the ellipse

$$E_{\gamma }≔\{(u,v)\mid u^2+v^2-2uv\text{sin}\gamma \le {\text{cos}}^2\gamma \}.$$ (30)

(Note that E₀ = B₂.) From (29), the correlation coefficient of U and V_γ is given simply by

$$\rho (U,V_{\gamma })=\text{sin}\gamma$$ (31)

so γ indicates the degree of linear dependence between U and V_γ.

Proposition 5.1. The pdf of (U, V_γ) is given by

$$f_{\gamma }(u,v)=\frac{1}{2\pi \sqrt{{\text{cos}}^2\gamma -u^2-v^2+2uv\text{sin}\gamma }}{1}_{E_{\gamma }}(u,v).$$ (32)

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

$$x(u,v)=u,y(u,v)=\frac{v-u\text{sin}\gamma }{\text{cos}\left(\gamma \right)}$$ (33)

so

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

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

$$\begin{array}{cc}\hfill f_{\gamma }(u,v)& =f(x(u,v),y(u,v))\cdot J\hfill \\ \hfill & =\frac{1}{2\pi \sqrt{1-u^2-{(v-u\text{sin}\gamma )}^2∕{\text{cos}}^2\gamma }}{1}_{B_2}(x(u,v),y(u,v))\cdot \frac{1}{\text{cos}\gamma }\hfill \\ \hfill & =\frac{\text{cos}\gamma }{2\pi \sqrt{(1-u^2){\text{cos}}^2\gamma -{(v-u\text{sin}\gamma )}^2}}\frac{1}{\text{cso}\gamma }\cdot 1_{B_2}(x(u,v),y(u,v))\hfill \\ \hfill & =\frac{1}{2\pi \sqrt{{\text{cos}}^2\gamma -u^2-v^2+2uv\text{sin}\gamma }}{1}_{E_{\gamma }}(u,v).\hfill \end{array}$$

Figure 8 shows the density f_γ(u, v) with γ = π/4.

Figure 8.

The density f_γ(u, v) in (32) (with γ = π/4).

To describe the family of elliptical copulas F_γ, we extend the definitions (16) and (18) as follows. First, for (u, v) ∈ E_γ define

$${\alpha }_{\gamma }(u,v)=\frac{1}{2\pi }\left[u\text{arcsin}\left(\frac{v-u\text{sin}\gamma }{\text{cos}\gamma \sqrt{1-u^2}}\right)+v\text{arcsin}\left(\frac{u-v\text{sin}\gamma }{\text{cos}\gamma \sqrt{1-v^2}}\right)-\text{arcsin}\left(\frac{uv-\text{sin}\gamma }{\sqrt{(1-u^2)(1-v^2)}}\right)\right].$$ (34)

Note that α_γ reduces to α in (16) when γ = 0, i.e., when V_γ = Y. From (12),

$$A(\text{arccos}u,\text{arccos}v;\frac{\pi }{2}-\gamma )=(1-u-v)\pi +4\pi {\alpha }_{\gamma }(u,v).$$ (35)

Next, extend the definition of α_γ(u, v) to C₂ \ E_γ as follows (see Figure 9):

$${\alpha }_{\gamma }(u,v)=\{\begin{array}{cc}\frac{u+v-1}{4}\hfill & \text{if}(u,v)\in R_5\left(\gamma \right)≔(C_2\backslash E_{\gamma })\cap \{(u,v)\mid u+v>1+\text{sin}\gamma \},\hfill \\ \frac{u-v+1}{4}\hfill & \text{if}(u,v)\in R_6\left(\gamma \right)≔(C_2\backslash E_{\gamma })\cap \{(u,v)\mid v-u>1-\text{sin}\gamma \},\hfill \\ \frac{-u+v+1}{4}\hfill & \text{if}(u,v)\in R_7\left(\gamma \right)≔(C_2\backslash E_{\gamma })\cap \{(u,v)\mid v-u<\text{sin}\gamma -1\},\hfill \\ \frac{-u-v-1}{4}\hfill & \text{if}(u,v)\in R_8\left(\gamma \right)≔(C_2\backslash E_{\gamma })\cap \{(u,v)\mid u+v<-\text{sin}\gamma -1\}.\hfill \end{array}\phantom{\}}$$ (36)

Figure 9.

Eight regions R₁(γ) – R₈(γ) for an elliptical copula (with = π/8).

Note that (34) and (36) agree on ∂E_γ, i.e., when u² + v² – 2uv sin γ = cos² γ. Also note that (36) reduces to α in (18) when γ = 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 C₂ 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,

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

Theorem 5.1. The cdf ≡ copula of (U, V_γ) 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)

Figure 12.

The copula F_γ(u, v) in (38) (with γ = π/4).

Proof. To find F_γ(u, v) we again use the formula (12) for the area of the intersection of two spherical caps on ∂B₃. 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 R₅(γ) – R₈(γ) are defined in (36) and (see Figure 9)

$$\begin{array}{cc}\hfill R_1\left(\gamma \right)& =E_{\gamma }\cap \{(u,v)\mid 0\le u,v\le 1\},\hfill \\ \hfill R_2\left(\gamma \right)& =E_{\gamma }\cap \{(u,v)\mid -1\le u\le 0\le v\le 1\},\hfill \\ \hfill R_3\left(\gamma \right)& =E_{\gamma }\cap \{(u,v)\mid -1\le v\le 0\le u\le 1\},\hfill \\ \hfill R_4\left(\gamma \right)& =E_{\gamma }\cap \{(u,v)\mid -1\le u,v\le 0\}.\hfill \end{array}$$

Case 1: (u, v) ∈ R₁(γ). By using Figure 10,

$$\begin{array}{cc}\hfill F_{\gamma }(u,v)& =P[U\le u,V_{\gamma }\le v]\hfill \\ \hfill & =1-P[U>u]-P[V_{\gamma }>v]+P[U>u,V_{\gamma }>v]\hfill \\ \hfill & =1-\left(\frac{1-u}{2}\right)-\left(\frac{1-v}{2}\right)+P[X>u,X\text{sin}\gamma +Y\text{cos}\gamma >v]\hfill \\ \hfill & =\frac{u+v}{2}+\frac{1}{4\pi }A(\text{arccos}u,\text{arccos}v;\pi ∕2-\gamma )\hfill \\ \hfill & =\frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\text{by}\left(35\right)\right].\hfill \end{array}$$

Figure 10.

The region [X > u, X sin(γ) + Y cos(γ) > v] for Case 1 (with γ = π/8).

Case 2: (u, v)) ∈ R₂(γ). Because $(X,Y)\stackrel{d}{=}(-X,Y)$ and using Figure 11

$$\begin{array}{cc}\hfill F_{\gamma }(u,v)& =P[U\le u]-P[U\le u,V_{\gamma }>v]\hfill \\ \hfill & =\frac{u+1}{2}-P[X\le u,X\text{sin}\gamma +Y\text{cos}\gamma >v]\hfill \\ \hfill & =\frac{u+1}{2}-P[-X\le u,-X\text{sin}\gamma +Y\text{cos}\gamma >v]\hfill \\ \hfill & =\frac{u+1}{2}-P[X\ge -u,X\text{sin}(-\gamma )+Y\text{cos}(-\gamma )>v]\hfill \\ \hfill & =\frac{u+1}{2}-\frac{1}{4\pi }A(\text{arccos}(-u),\text{arccos}v;\pi ∕2+\gamma )\hfill \\ \hfill & =\frac{u+v+1}{4}-{\alpha }_{-\gamma }(-u,v)\left[\text{by}\left(35\right)\right]\hfill \\ \hfill & =\frac{u+v+1}{4}+{\alpha }_{\gamma }(u,v)\left[\text{by}\left(34\right)\right].\hfill \end{array}$$

Figure 11.

The region [X < u, X sin(γ) + Y cos(γ) > v] for Case 2 (with γ = π/8).

Case 3: (u, v) ∈ R₃(γ). Then (–u, –v) ∈ R₂(γ), so by Lemma 5.1 and Case 2,

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

Case 4: (u, v) ∈ R₄(γ). Then (–u, –v) ∈ R₁(γ), so by Lemma 5.1 and Case 1, the argument for Case 3 applies verbatim.

Case 5: (u, v) ∈ R₅(γ).

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

Case 6: (u, v) ∈ R₆(γ).

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

Case 7: (u, v) ∈ R₇(γ). Then (–u, –v) ∈ R₆(γ), so by Lemma 5.1 and Case 6, the argument for Case 3 applies verbatim.

Case 8: (u, v) ∈ R₈(γ). Then (–u, –v) ∈ R₅(γ), 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

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

it follows that the random variables

$$U≔\frac{X}{\sqrt{1-Y^2}},V≔\frac{Y}{\sqrt{1-X^2}}$$ (39)

satisfy

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

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

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

so the joint distribution of (U, V) generates a copula F(u, v) on the centered cube C₂ ≡ [–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(U) = E(V) = 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)

$$f(u,v)=\frac{1}{\pi }\frac{\sqrt{(1-u^2)(1-v^2)}}{{(1-u^2{v}^2)}^2}{1}_{C_2}(u,v).$$ (40)

Figure 13.

Joint density f(u, v) of (U, V) in (40).

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

$$u^2(1-y^2)=x^2,\text{and}{v}^2(1-x^2)=y^2.$$

Substitution of the second expression for y² into the left side of the first relation and vice versa yields

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

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

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

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

By symmetry it follows that the Jacobian is given by

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

and hence the determinant of J is given by

$$\mid J\mid =\frac{\sqrt{(1-u^2)(1-v^2)}}{{(1-u^2{v}^2)}^2}.$$

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

For 0 ≤ u, v ≤ 1, (u, v) ≠ (1, 1), let E₁(u) and E₂(v) be the ellipses

$$E_1\left(u\right)=\left\{(x,y)\mid \frac{x^2}{u^2}+y^2\le 1\phantom{\mid }\right\}$$ (42)

$$E_2\left(v\right)=\left\{(x,y)\mid x^2+\frac{y^2}{v^2}\le 1\phantom{\mid }\right\}.$$ (43)

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

Lemma 6.1.

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

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

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

Then

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

from which the result follows.

Figure 14.

Integration regions for Lemma 6.2.

Theorem 6.1. The copula (= cdf) corresponding to the pdf (40) is given by (see Figure 15)

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

Figure 15.

Nonlinear transformation copula F(u, v) in Theorem 6.1.

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) ∈ C₂,

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

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 B₃ 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 C₃. To find this copula one would need to determine $\text{Volume}(E_1\left(u\right)\cap E_2\left(v\right)\cap E_3\left(w\right))$, where now, for 0 ≤ u, v, w ≤ 1, E₁(u), E₂(v), and E₃(w) are the ellipsoids

$$\begin{array}{cc}\hfill E_1\left(u\right)& =\left\{(x,y,z)\mid \frac{x^2}{u^2}+y^2+z^2\le 1\phantom{\mid }\right\}\hfill \\ \hfill E_2\left(v\right)& =\left\{(x,y,z)\mid x^2+\frac{y^2}{v^2}+z^2\le 1\phantom{\mid }\right\}\hfill \\ \hfill E_3\left(w\right)& =\left\{(x,y,z)\mid x^2+y^2+\frac{y^2}{w^2}\le 1\phantom{\mid }\right\}.\hfill \end{array}$$