A class of copulas with piecewise linear horizontal sections

Journal of Statistical Planning and Inference 139 (2009) 3908 -- 3920

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A class of copulas with piecewise linear horizontal sections José Antonio Rodríguez-Lallena∗ Departamento de Estadística y Matemática Aplicada, Universidad de Almería, 04120 La Cañada de San Urbano (Almería), Spain

A R T I C L E

I N F O

A B S T R A C T

We introduce the class of bivariate copulas with piecewise linear horizontal sections whose graph is composed, at most, of two segments. It is a wide class of copulas which contains some known copulas. We study several properties of the copulas in the new class concerning absolute continuity, singular components, measures of association, concordance ordering, dependence concepts and symmetry. Finally, we provide several examples. © 2009 Elsevier B.V. All rights reserved.

Available online 22 May 2009 MSC: 62H05 62H20 Keywords: Absolute continuity Concordance ordering Copulas Dependence concepts Kendall's tau Singular components Spearman's rho Symmetry properties Tail dependence

1. Introduction First, we recall some concepts and results about (bivariate) copulas: see Nelsen (2006) for details. Let I denote the interval [0, 1]. A copula is a function C: I2 → I which satisfies the boundary conditions C(t, 0) = C(0, t) = 0

and

C(t, 1) = C(1, t) = t

for all t ∈ I

(1)

and the 2-increasing property, i.e., C(x , y ) − C(x , y) − C(x, y ) + C(x, y) ⱖ 0 for all x, x , y, y ∈ I such that x ⱕ x and y ⱕ y . Equivalently, a copula is the restriction to I2 of a continuous bivariate distribution function whose marginals are uniform on I. Recall also that W ⱕ C ⱕ M for every copula C, where W and M are the copulas defined by W(x, y) = max(x + y − 1, 0) and M(x, y) = min(x, y) for all (x, y) ∈ I2 . The importance of copulas has been increasing during their almost 50 years of existence because of their applications in several fields of research. Their relevance comes primarily from Sklar's theorem (Sklar, 1959), which states that the joint distribution function H of a random pair (X, Y), with marginal distributions F and G, is given by a copula C, uniquely determined on Ran F× Ran G, as follows: H(x, y) = C(F(x), G(y))

for all (x, y) ∈ [−∞, ∞]2 .

∗ Tel.: +34 950 015507; fax: +34 950 015167. E-mail address: [email protected]. 0378-3758/$ - see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.jspi.2009.05.032

(2)

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y0

M(x,y0)

Π(x,y0)

W(x,y0)

1 Fig. 1. The graphs of the horizontal sections of W,  and M.

For instance, if X and Y are independent we can take C = , where  is the copula defined by (x, y) = xy for all (x, y) ∈ I2 . Conversely, given F, G and C, the function H defined by (2) is a bivariate distribution function with marginals F and G. Hence copulas are a tool to build bivariate distributions with given marginals. Given a copula C, the transposed copula of C is the copula C t defined by C t (x, y) = C(y, x) for all (x, y) ∈ I2 . A copula C is symmetric if C t = C. Let X and Y be continuous random variables with joint distribution function H, marginals F and G and copula C (via Sklar's theorem). It is known that X and Y are exchangeable if, and only if, F = G and C is symmetric. In the literature we can find several general methods of constructing copulas which satisfy certain required properties. One of those methods is the construction of copulas with prescribed horizontal or vertical sections. The horizontal sections of a copula C are the functions fy0 (y0 ∈ I) defined on I by fy0 (x) = C(x, y0 ). The vertical sections of C are similarly defined. Since the vertical sections of a copula C coincide with the horizontal sections of C t , we can restrict our study to copulas with prescribed horizontal sections. For some y0 ∈ I, the graphs of the horizontal sections of the copulas W,  and M are depicted in Fig. 1. One of the statistical interpretations of the horizontal sections of a copula is the following: if X and Y are random variables uniformly distributed on I, C is the copula of (X, Y) and y0 ∈ I, then the horizontal section fy0 of C is proportional to the conditional distribution x → Pr[X ⱕ x|Y ⱕ y0 ], namely: fy0 (x) = y0 · Pr[X ⱕ x|Y ⱕ y0 ]

for all x ∈ I.

(3)

The purpose of this paper is to introduce and study a new class of copulas with prescribed horizontal sections which extends the class of copulas examined in Section 2.5 of Rodríguez-Lallena (1996). In the literature we can find different examples of copulas with piecewise linear horizontal sections (composed by one, two or more segments), such as—among others—the Fréchet and Mardia family of copulas (Fréchet, 1958; Mardia, 1970), the shuffles of Min (Mikusinski et al., 1992) and the class of asymmetric  semilinear copulas (De Baets et al., 2007). In this paper we study the copulas with piecewise linear horizontal sections whose graph is made up, at most, of two segments. The simplest examples of these copulas are , M and W. Other known examples are some shuffles of Min, some members of the Fréchet and Mardia family of copulas and every asymmetric semilinear copula. Moreover, it is common that the classes of copulas with prescribed horizontal (respectively, vertical) sections of a specific type, contain copulas with piecewise linear vertical (respectively, horizontal) sections whose graph is composed, at most, of two segments: for instance, the maximum and minimum copulas of those classes. This is the case of copulas with quadratic, cubic or two different types of sinusoidal horizontal or vertical sections: see Quesada-Molina and Rodríguez-Lallena (1995), Nelsen et al. (1997) and Rodríguez-Lallena (1996). Corollary 2.6 in Rodríguez-Lallena and Úbeda-Flores (2004) suggests that this fact can be extended to almost any class of copulas with prescribed horizontal or vertical sections. Finally, other examples of the new class of copulas can be found in Rodríguez-Lallena (1996) and Úbeda-Flores (1998). The new class contains a great variety of copulas, so we expect that such class may be useful in a number of applications. For instance, the new class provides very different ways to construct bivariate models satisfying a diversity of properties, since that class contains: absolutely continuous copulas, singular copulas and copulas with an absolutely continuous component and a singular component; families of copulas for which the population versions of the measures of association Kendall's tau and Spearman's rho take any value in [−1, 1]; copulas satisfying quite different properties of positive and negative dependence, etc. The new class of copulas is mainly composed of asymmetric copulas, but it also contains families of symmetric copulas (see Example 1 in Section 4). So that class of copulas may be useful to model both the exchangeability and the nonexchangeability of two random variables (but mainly the last one). In Section 2 we characterize the class of copulas with piecewise linear horizontal sections whose graph is made up, at most, of two segments. In Section 3, we characterize the absolute continuity and obtain expressions for the population versions of Kendall's tau and Spearman's rho associated with those copulas; we also study the concordance ordering, several dependence concepts and some symmetry properties for such copulas. Finally, in Section 4 we examine several families of copulas belonging to the new class.

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y

M

(a(y),b(y))

C

W

1 Fig. 2. Graph of a horizontal section composed of two segments.

2. Copulas with piecewise linear horizontal sections composed, at most, of two segments In this section, we characterize the class of copulas with piecewise linear horizontal sections whose graph is composed, at most, of two segments. Every horizontal section of this type is given by a unique point (a(y), b(y)) (y ∈ I); so the graph of this section is the polygonal joining the points (0, 0), (a(y), b(y)) and (1, 1) (see Fig. 2). If the graph of the horizontal section is a unique segment, then, at first, any point of the segment could be taken as (a(y), b(y)). The points (a(y), b(y)), with y ∈ I, yield two functions a and b defined on I, which we assume continuous. Since the points (a(y), b(y)) are in the parallelogram bounded by the horizontal sections of W and M, the functions a and b must satisfy the following two conditions: a(y) ∈ I for every y ∈ I

(4)

max(a(y) + y − 1, 0) ⱕ b(y) ⱕ min(a(y), y) for every y ∈ I.

(5)

and

Let C: I2 → I be the function whose horizontal sections are the piecewise linear functions determined by the points (a(y), b(y)), with y ∈ I. It is easy to check that C is given by ⎧ b(y) ⎪ , 0 ⱕ x < a(y), x ⎪ ⎪ ⎨ a(y) C(x, y) = b(y), x = a(y), ⎪ ⎪ ⎪ ⎩ y − (1 − x) y − b(y) , a(y) < x ⱕ 1. 1 − a(y)

(6)

Then, the functions a and b will be called the generating functions of C. Remark that we could replace the first two lines in (6) by “xb(y)/a(y), 0 ⱕ x ⱕ a(y)” with the convention that xb(y)/a(y) = 0 whenever x = a(y) = 0. An analogous argument can be applied to the last two lines in (6). The set of copulas of the form (6)—where the generating functions are supposed to be continuous—will be denoted by P. In the sequel, when we consider a copula C in P, we will assume that their generating functions will be denoted by a and b, as in (6). A simple probabilistic interpretation of the copulas in P is the following one: Let (X, Y) be a random pair with copula C in P and marginals uniform on I. Let y0 ∈ I, such that a(y0 ) ∈ (0, 1). Then, from equality (3), the conditional distribution function x → Pr[X ⱕ x|Y ⱕ y0 ] spreads uniformly a mass of b(y0 )/(y0 a(y0 )) on the interval [0, a(y0 )] and a mass of (y0 − b(y0 ))/(y0 (1 − a(y0 ))) on the interval [a(y0 ), 1]. Otherwise, if a(y0 ) ∈ {0, 1}, such conditional distribution is uniform on I. The main purpose of this section is to characterize the copulas of the set P. First, we provide a preliminary lemma which provides several immediate consequences of conditions (4) and (5). Lemma 1. Let a and b be two real continuous functions defined on I and satisfying both conditions (4) and (5). Then, the following properties hold: 1. 2. 3. 4. 5.

y − b(y) ⱕ 1 − a(y) for every y ∈ I; b(0) = 0; b(1) = a(1); if a(y) = 0 for some y ∈ I, then b(y) = 0 and if a(y) = 1 for some y ∈ I, then b(y) = y.

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The following result shows that every function of the form (6), under the hypotheses of Lemma 1, satisfies certain conditions needed to be a copula. Lemma 2. Let a and b be two real continuous functions defined on I satisfying both conditions (4) and (5). If C is the function defined by (6), then: 1. C is a continuous function from I2 onto I and 2. C satisfies the boundary conditions (1). Proof. Since a and b are continuous functions, we only need to prove the continuity of C at the points of the form (a(y), y), with y ∈ I, which is a simple analytical exercise (note that C(a(y), y) = b(y) for every y ∈ I). And the proof of Lemma 2 follows immediately from parts 2, 3, 4 and 5 of Lemma 1.  Therefore, a function C satisfying the hypotheses of Lemma 2 is a copula if, and only if, C is 2-increasing. Now we present the main result of this paper. Theorem 3. Let a, b and C be as in Lemma 2. Let r, s be the functions defined by r(y) =

b(y) a(y)

s(y) =

y − b(y) 1 − a(y)

for every y ∈ I such that a(y)  0,

(7)

for every y ∈ I such that a(y)  1.

(8)

Then, C is a copula if, and only if, the following three conditions hold: 1. both r and s are nondecreasing functions on their respective domains; 2. s(y) ⱕ r(y ) for every y, y ∈ I such that y < y and a(y) < a(y ) and 3. r(y) ⱕ s(y ) for every y, y ∈ I such that y < y and a(y) > a(y ). Proof. The 2-increasingness of C is equivalent to the nondecreasingness of every function Gyy defined on I by Gyy (x) = C(x, y ) − C(x, y) (0 ⱕ y < y ⱕ 1). Observe that all functions Gyy are piecewise linear. Let y, y ∈ I be such that y < y . We consider two cases. (1) If a(y) ⱕ a(y ), after straightforward computations we have that ⎧ ⎨ x(r(y ) − r(y)), Gyy (x) = x(r(y ) − s(y)) − y + s(y), ⎩  y − y − (1 − x)(s(y ) − s(y)),

0 ⱕ x < a(y), a(y) ⱕ x ⱕ a(y ), a(y ) < x ⱕ 1.

(2) If a(y) > a(y ), we obtain ⎧ 0 ⱕ x < a(y ), ⎨ x(r(y ) − r(y)),    Gyy (x) = x(s(y ) − r(y)) + y − s(y ), a(y ) ⱕ x ⱕ a(y), ⎩  y − y − (1 − x)(s(y ) − s(y)), a(y) < x ⱕ 1. Hence the result follows.  The next result provides sufficient conditions for the functions a and b in order that the function C defined by (6) is a copula. Such conditions are rather general, and they often provide an easier way to check that C is a copula (see Examples 1, 3 and 4 in Section 4). Theorem 4. Let a, b and C be as in Lemma 2. Suppose that the functions a and b satisfy the following conditions: 1. There exists a partition 0 = y0 < y1 < y2 < · · · < yn = 1 of I such that, for each i = 1, 2, . . . , n, we have that a and b are differentiable on (yi−1 , yi ) and either a (y) < 0, a (y) = 0, or a (y) > 0 for every y ∈ (yi−1 , yi ). 2. a (y)(b(y) − ya(y)) ⱖ 0 for every y ∈ I such that a (y) exists. 3. a (y)b(y)/a(y) ⱕ b (y) whenever a(y)  0 and both a (y) and b (y) exist. 4. b (y) ⱕ 1 + a (y)(y − b(y))/(1 − a(y)) whenever a(y)  1 and both a (y) and b (y) exist. Then, the function C is a copula.

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Proof. From condition 1, it is clear that a(y) might be zero only on intervals [yi , yk ], where 0 ⱕ i ⱕ k ⱕ n (if i = k such interval degenerates to a single point). In such case, we have that a (y) < 0 for every y ∈ (yi−1 , yi ) (if i > 0) and a (y) > 0 for every y ∈ (yk , yk+1 ) (if k < n). Similarly, a(y) might be equal to 1 only on intervals [yi , yk ], where 0 ⱕ i ⱕ k ⱕ n. In this case, a (y) > 0 for every y ∈ (yi−1 , yi ) (if i > 0) and a (y) < 0 for every y ∈ (yk , yk+1 ) (if k < n). Moreover, for every interval (yi−1 , yi ) where a (y) be positive (or negative), it is clear that a(y) ∈ (0, 1) for every y ∈ (yi−1 , yi ). Now, in order to prove that C is a copula, we show that the three conditions of Theorem 3 hold. First, we prove that the function r defined by (7) is nondecreasing on its domain D = {y ∈ I : a(y)  0}. If i is in {1, 2, . . . , n} and (yi−1 , yi ) ⊂ D, then, from condition 3, we have that r (y) = (b (y)a(y) − b(y)a (y))/a2 (y) ⱖ 0 for all y ∈ (yi−1 , yi ). Thus, the function r is nondecreasing on (yi−1 , yi ). In fact, since r is continuous on D, we can conclude that r is nondecreasing on every interval included in D. If D is an interval, then the nondecreasingness of r is already proved. Otherwise, suppose that (, ) and (, ), with  ⱕ , are two consecutive intervals in D (i.e., a(y) = 0 for every y ∈ [, ]). It is clear that there exists  > 0 such that a (y) < 0 for every y ∈ [ − , ] and a (y ) > 0 for every y ∈ [,  + ]. Hence, by applying condition 2 twice, we have r(y) = b(y)/a(y) ⱕ y < y ⱕ b(y )/a(y ) = r(y ). Now we can conclude that r is nondecreasing on its domain. We also have to prove that the function s defined by (8) is nondecreasing on its domain E = {y ∈ I : a(y)  1}. If i is in {1, 2, . . . , n} and (yi−1 , yi ) ⊂ E, then

s (y) =

(1 − b (y))(1 − a(y)) + a (y)(y − b(y)) 1 − a2 (y)

for all y ∈ (yi−1 , yi ). From condition 4, we have that s (y) ⱖ 0 on this interval, whence s is nondecreasing on (yi−1 , yi ). Since s is continuous on E, we have that s is nondecreasing on every interval included in E. If E is an interval, then there is nothing to prove. Otherwise, suppose that (, ) and (, ), with  ⱕ , are two consecutive intervals in E (i.e., a(y) = 1 for every y ∈ [, ]). Thus, there exists  > 0 such that a (y) > 0 for every y ∈ [ − , ] and a (y ) < 0 for every y ∈ [,  + ]. Hence, by applying condition 2 twice, we have s(y) = (y − b(y))/(1 − a(y)) ⱕ y < y = (y − y a(y ))/(1 − a(y )) ⱕ (y − b(y ))/(1 − a(y )) = s(y ), as desired. Finally, we prove condition 2 of Theorem 3 (the condition 3 of Theorem 3 can be proved similarly). Let y, y ∈ I such that y < y and a(y) < a(y ). Then, there exist y1 , y2 ∈ I such that y ⱕ y1 < y2 ⱕ y and a (v) > 0 for all v ∈ [y1 , y2 ]. By applying condition 2 twice and the nondecreasingness of r and s, we have that s(y) ⱕ s(y1 ) ⱕ y1 < y2 ⱕ r(y2 ) ⱕ r(y ), as desired.  3. Properties of the new class of copulas We now study several properties of the copulas of the set P introduced in Section 2. They concern absolute continuity, singular components, measures of association, concordance ordering, dependence concepts and symmetry. The first result of this section may be seen as an immediate consequence of the geometric method of constructing copulas introduced in Section 2. Theorem 5. Let C be a copula in P, and let y ∈ (0, 1). Then: 1. C(x, y) = (x, y) for every x ∈ I if, and only if, b(y) = ya(y); 2. C(x, y) = M(x, y) for every x ∈ I if, and only if, b(y) = a(y) = y and 3. C(x, y) = W(x, y) for every x ∈ I if, and only if, b(y) = 0 and a(y) = 1 − y. As a consequence, C =  if, and only if, b(y) = ya(y) for every y ∈ I; C = M if, and only if, b(y) = a(y) = y for every y ∈ I and C = W if, and only if, a(y) = 1 − y and b(y) = 0 for every y ∈ I. Given a real continuous function a defined on I and satisfying condition (4), we consider the set Ia = {y ∈ I : a(y) ∈ (0, 1)}, which will be useful along the paper. The following theorem shows that most copulas of the set P have a singular component. However, a variety of families of absolutely continuous copulas can also be found in P. In the next section we provide examples of all types. Theorem 6. Let C be a copula in P such that its generating functions a and b are absolutely continuous. Then, C is absolutely continuous if, and only if, the following equality holds: 

a (y)(b(y) − ya(y)) dy = 0. a(y)(1 − a(y)) Ia

(9)

Otherwise, C has a singular component whose mass is concentrated on the set {(a(y), y) : y ∈ Ia }; and that mass is given by the integral in the first member of (9).

J.A. Rodríguez-Lallena / Journal of Statistical Planning and Inference 139 (2009) 3908 -- 3920

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Proof. Since a and b are absolutely continuous, we have that ⎧  b (y)a(y) − b(y)a (y) ⎪ ⎪ 2 , 0 ⱕ x < a(y), ⎨ * C a2 (y) (x, y) =  (y))(1 − a(y)) + a (y)(y − b(y)) (1 − b ⎪ *x*y ⎪ , a(y) < x ⱕ 1, ⎩ (1 − a(y))2 whenever a (y) and b (y) exists. Now, we need to compute the following integral:  I=

1



0

1

0

2

* C (x, y) dx dy. *x*y

If Ia = I, then  1 1 b (y)a(y) − b(y)a (y) (1 − b (y))(1 − a(y)) + a (y)(y − b(y)) dx dy dxdy + 2 a (y) 0 0 0 a(y) (1 − a(y))2  1  1  1   1 b(y)a (y) a (y)(y − b(y)) = b (y) dy − (1 − b (y)) dy + dy + dy a(y) 1 − a(y) 0 0 0 0   1  1 a (y)(b(y) − ya(y)) a (y)(y − b(y)) b(y)a (y) =1− − dy = 1 − dy. a(y) 1 − a(y) a(y)(1 − a(y)) 0 0 

1

I=



a(y)

If Ia I, then we write   I=

2

* C (x, y) dxdy + *x*y

1

Ia 0



 I\Ia 0

1

2

* C (x, y) dx dy. *x*y

Similarly to the previous computation, we have  

1

Ia 0

2

* C (x, y) dx dy = (Ia ) − *x*y



a (y)(b(y) − ya(y)) dy, a(y)(1 − a(y)) Ia

where (Ia ) is the Lebesgue measure of Ia . To compute the second integral, suppose that y ∈ I\Ia . Then, from parts 4 and 5 of Lemma 1, we have b(y) = ya(y); and hence, from part 1 of Theorem 5, C(x, y) = xy for every x ∈ I. Thus, 



2

* C (x, y) dx dy = (I\Ia ) = 1 − (Ia ), *x*y

1

I\Ia 0

whence the result follows.



Observe that hypothesis 2 in Theorem 4 is coherent with the fact that the integral in Eq. (9) must be nonnegative: under that hypothesis the integrand function in (9) is nonnegative. Kendall's tau and Spearman's rho are the two most commonly used nonparametric measures of association for the components of a continuous random pair (X, Y). The population versions of such measures depend only on the copula C of the random pair (X, Y), and are given by

C = 1 − 4



1 0



1 0

*C *C (x, y) (x, y) dx dy and C = 12 *x *y

 0

1

 0

1

C(x, y) dx dy − 3,

respectively (Nelsen, 2006). Copulas with piecewise linear horizontal sections such as some shuffles of Min have been proposed as examples of copulas C for which ( C , C ) is in the boundary of the ( , ) region (Nelsen, 2006). In general, the class of copulas introduced in Section 2 may be useful to find very different relationships between and : see several examples in Section 4; moreover, some results in Fredricks and Nelsen (2007) can be applied to copulas of the new class. The following theorem provides expressions of Kendall's tau and Spearman's rho for such copulas. Theorem 7. Let C be a copula in P. Then C is given by

C = 6

 0

1

(b(y) − ya(y)) dy

and, if the generating functions a and b are absolutely continuous, then C is given by



 b2 (y) (y − b(y))2  C = 2 + a (y) dy. b(y) − yb (y) + a(y) 1 − a(y) Ia

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Proof. The proof is straightforward for C . For C , we need to decompose the integral as in the proof of Theorem 6, namely:  1 1  1   1  *C *C *C *C *C *C (x, y) (x, y) dx dy = (x, y) (x, y) dx dy + (x, y) (x, y) dx dy. *y *y *y 0 0 *x Ia 0 *x I\Ia 0 *x Then we obtain the result after some simple computations.



Given two copulas C1 and C2 , C2 is said to be more concordant than C1 , and we write C1 ≺ C2 , if C1 (x, y) ⱕ C2 (x, y) for every (x, y) ∈ I2 (Joe, 1997). The next result, whose proof is straightforward, characterizes the fact that two copulas in P are ordered by ≺. Theorem 8. Let C1 and C2 be two copulas in P. For i = 1, 2, let ai and bi be the generating functions of Ci , and let ri and si be the functions defined by (7) and (8). Then, C1 ≺ C2 if, and only if, r1 (y) ⱕ r2 (y) whenever a1 (y)a2 (y)  0, and s1 (y) ⱖ s2 (y) whenever (1 − a1 (y))(1 − a2 (y))  0. The next results characterize those continuous random pairs with copula C in P which satisfy certain positive dependence properties (similar results can be obtained for the corresponding negative dependence concepts). For the definitions of those dependence properties we refer to Joe (1997) and Nelsen (2006). The proofs of our results rely on the characterizations of such dependence properties in terms of the copula associated with the random pair: see Nelsen (2006). As for Theorem 5, the following result may be seen as an immediate consequence of the geometric method of constructing copulas shown in Section 2. Theorem 9. Let (X, Y) be a continuous random pair whose associated copula C is in P. Then PQD(X, Y) (X and Y are positively quadrant dependent), i.e.,  ≺ C, if, and only if, b(y) ⱖ ya(y) for all y ∈ I. In the next theorems we study other positive dependence concepts which are generally stronger than positive quadrant dependence. However, the first result shows that, for copulas in the set P, some of those concepts are equivalent to positive quadrant dependence. Theorem 10. Let (X, Y) be a continuous random pair whose copula C is in P. Then, the following statements are equivalent: 1. 2. 3. 4.

PQD(X, Y); LTD(Y|X) (Y is left tail decreasing in X); RTI(Y|X) (Y is right tail increasing in X) and SI(Y|X) (Y is stochastically increasing in X).

Proof. It is known that SI(Y|X) implies both LTD(Y|X) and RTI(Y|X), and that any of these concepts implies PQD(X, Y). It is also known that SI(Y|X) if, and only if, the function fy (x) (recall that fy (x) = C(x, y)) is nonincreasing on its domain for every y ∈ I. And this last statement is clearly equivalent to the fact that r(y) ⱖ s(y) for every y ∈ Ia (r and s are again the functions defined by (7) and (8)). For every y ∈ Ia , we have that r(y) ⱖ s(y) if, and only if, b(y) ⱖ ya(y); and b(y) = ya(y) for every y ∈ I\Ia . So, from Theorem 9, we have that SI(Y|X) if, and only if, PQD(X, Y), whence the proof follows.  The following result characterizes two other concepts of positive dependence for copulas in P. Theorem 11. Let (X, Y) be a continuous random pair whose copula C is in P. Let r and s be the functions defined by (7) and (8). Then: 1. LTD(X|Y) if, and only if, both functions −r(y)/y and s(y)/y are nondecreasing on their respective domains. 2. RTI(X|Y) if, and only if, both functions (r(y) − 1)/(1 − y) and (1 − s(y))/(1 − y) are nondecreasing on their respective domains. Proof. We prove the second part of this theorem: the first part can be proved in a similar way. It is known that RTI(X|Y) if, and only if, for each x ∈ I, the function y → (x − C(x, y))/(1 − y) is nonincreasing on [0, 1). Observe that ⎧ x(1 − r(y)) ⎪ , 0 ⱕ x < a(y), ⎪ ⎪ ⎪ ⎪ 1−y x − C(x, y) ⎨ x − b(y) = , x = a(y), ⎪ 1−y 1−y ⎪ ⎪ ⎪ ⎪ ⎩ x(1 − s(y)) − y + s(y) , a(y) < x ⱕ 1. 1−y Let y, y ∈ [0, 1) such that y < y . We should prove that x − C(x, y ) x − C(x, y) ⱖ 1−y 1 − y

(10)

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for every x ∈ (0, 1) (Eq. (10) holds immediately for x = 0 and x = 1). Suppose that a(y) ⱕ a(y ) (the case a(y) > a(y ) can be proved similarly). Inequality (10) holds for every x ∈ (0, a(y)] (here we suppose a(y) > 0; the case a(y) = 0 is considered below) if, and only if, x(1 − r(y))/(1 − y) ⱖ x(1 − r(y ))/(1 − y ) for every x ∈ (0, a(y)], i.e., if, and only if, r(y) − 1 r(y ) − 1 ⱖ . 1−y 1 − y

(11)

Inequality (10) holds for every x ∈ (a(y), a(y )) (in the case that a(y) < a(y )) if, and only if, x(1 − s(y)) − y + s(y) x(1 − r(y )) ⱖ 1−y 1 − y

(12)

for every x ∈ (a(y), a(y )). Since the two members in inequality (12) are linear functions in x, that inequality holds for every x ∈ (a(y), a(y )) if, and only if, it holds for the extreme points a(y) and a(y ). For x = a(y) inequality (12) reads (a(y) − b(y))/(1 − y) ⱖ a(y)(1 − r(y ))/(1 − y ). From part 4 of Lemma 1, this inequality is trivially satisfied when a(y) = 0; and, if a(y) > 0, then it is equivalent to (1 − r(y))/(1 − y) ⱖ (1 − r(y ))/(1 − y ), which is (11). For x = a(y ) inequality (12) reads a(y )(1 − s(y)) − y + s(y) a(y )(1 − r(y )) ⱖ . 1−y 1 − y

(13)

From part 5 of Lemma 1, inequality (13) is trivially satisfied when a(y ) = 1. If a(y ) < 1, then we have (a(y )(1 − s(y)) − y + s(y))/(1 − y) = a(y ) + (a(y ) − 1)(y − s(y))/(1 − y), whence inequality (13) is equivalent to the following one: (y − s(y))/(1 − y) ⱕ a(y )(y − r(y ))/((1 − y )(a(y ) − 1)). It is easy to check that a(y )(y − r(y ))/((1 − y )(a(y ) − 1)) = (y − s(y ))/(1 − y ), whence inequality (13) is equivalent to 1 + (y − s(y))/(1 − y) ⱕ 1 + (y − s(y ))/(1 − y ), i.e., to the inequality 1 − s(y) 1 − s(y ) ⱕ . 1−y 1 − y

(14)

Inequality (10) holds for every x ∈ [a(y ), 1) (the case a(y ) = 1 was considered above) if, and only if, (x(1 − s(y)) − y + s(y))/(1 − y) ⱖ (x(1 − s(y )) − y + s(y ))/(1 − y ) for every x ∈ [a(y ), 1). It is easy to check that the last inequality is equivalent to (1 − x)(1 − s(y))/(1 − y) ⱕ (1 − x)(1 − s(y ))/(1 − y ). Thus, inequality (10) holds for every x ∈ [a(y ), 1) if, and only if, inequality (14) holds, which completes the proof of part 2.  In the following lemmas and theorems an additional hypothesis for the generating function a is sometimes required, which does not suppose a real restriction for our purposes. Namely, a should be a piecewise monotone function, i.e., there exists a partition 0 = y0 < y1 < y2 < · · · < yn = 1 of I such that, for each i = 1, 2, . . . , n, a is either strictly monotone or constant on [yi−1 , yi ]. Observe that this hypothesis is similar to condition 1 of Theorem 4. Next we prove two lemmas which will be useful for further theorems of this paper. Lemma 12. Let (X, Y) be a continuous random pair whose copula C is in P. Suppose that C  . Then the following statements hold: 1. If LTD(X|Y), then there exists y0 ∈ (0, 1] such that b(y) > ya(y) for all y ∈ (0, y0 ) and b(y) = ya(y) for all y ∈ [y0 , 1]. 2. If RTI(X|Y), then there exists y0 ∈ [0, 1) such that b(y) = ya(y) for all y ∈ [0, y0 ] and b(y) > ya(y) for all y ∈ (y0 , 1). 3. If SI(X|Y), then b(y) > ya(y) for all y ∈ (0, 1); moreover, if a, b and C are absolutely continuous, then the generating function a must be constant. 4. If a is piecewise monotone and b(y) > ya(y) on some open interval J, then a is nondecreasing on J. Proof. First we prove part 2. Since RTI(X|Y) implies PQD(X, Y), we have b(y) ⱖ ya(y) for every y ∈ [0, 1] (recall Theorem 9). Since C  , and from Lemma 1, there exists y ∈ (0, 1) ∩ Ia such that b(y) > ya(y). It is easy to check that b(y) > ya(y) is equivalent to (r(y) − 1)/(1 − y) > − 1; and, since (1 − s(y))/(1 − y) = 1 + (b(y) − ya(y))/((1 − y)(1 − a(y))), it is also equivalent to (1 − s(y))/(1 − y) > 1. From part 2 of Theorem 11, (r(y ) − 1)/(1 − y ) > − 1 for every y ∈ [y, 1) such that a(y )  0, and (1 − s(y ))/(1 − y ) > 1 for every y ∈ [y, 1) such that a(y )  1; whence b(y ) > y a(y ) for every y ∈ [y, 1) (a(y ) ∈ (0, 1) for every y ∈ [y, 1)). As a consequence, there exist y0 ∈ [0, 1) such that b(y) = ya(y) for every y ∈ [0, y0 ] and b(y) > ya(y) for all y ∈ [y0 , 1). The proof of part 1 is similar. If SI(X|Y), then we have that b(y) > ya(y) for all y ∈ (0, 1) as an immediate consequence of parts 1 and 2 (SI(X|Y) implies both LTD(X|Y) and RTI(X|Y)). The last assertion of part 3 follows from Eq. (9). In order to prove property 4, suppose that a is not nondecreasing on J. Then, there exist y1 , y2 ∈ J such that y1 < y2 and a(y1 ) > a(y2 ). Since a is piecewise monotone, there exists an interval (c, d) ⊂ (y1 , y2 ) such that a is decreasing on (c, d). Then, from part 3 in Theorem 3, we have that r(c ) < s(d) for every c ∈ (c, d). Since the inequality b(y) > ya(y) is equivalent to s(y) < y < r(y), then we obtain c < r(c ) < s(d) < d for every c ∈ (c, d). By taking limits in the last inequality when c → d we obtain a contradiction. So a is nondecreasing on J, as claimed.  Lemma 13. Let (X, Y) be a continuous random pair whose copula C is in P. Let f and g be strictly increasing functions on Ran X and Ran Y, respectively. It is known that the copulas of the pairs (f (X), −g(Y)), (−f (X), g(Y)) and (−f (X), −g(Y)) are, respectively, given by

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˙ y) = x − C(x, 1 − y), C(x, ¨ y) = y − C(1 − x, y) and C(x, ˆ y) = x + y − 1 + C(1 − x, 1 − y) (Nelsen, 2006). Then, C, ˙ C¨ and Cˆ are in P; and C(x, the generating functions of these copulas are, respectively, given by ˙ a(y) = a(1 − y), ¨ a(y) = 1 − a(y),

˙ b(y) = a(1 − y) − b(1 − y), ¨ b(y) = y − b(y),

ˆ a(y) = 1 − a(1 − y),

ˆ b(y) = y + b(1 − y) − a(1 − y)

for every y ∈ I. The proof of this lemma is straightforward. The following theorem characterizes three more concepts of positive dependence. Theorem 14. Let (X, Y) be a continuous random pair whose copula C is in P. Suppose that C   and the generating function a is piecewise monotone. Then: 1. SI(X|Y) if, and only if, (a) b(y) > ya(y) for all y ∈ (0, 1), (b) the function r given by (7) is concave and (c) the function s given by (8) is convex. 2. LCSD(X, Y) (X and Y are left corner set decreasing) if, and only if, LTD(X|Y). 3. RCSI(X, Y) (X and Y are right corner set increasing) if, and only if, RTI(X|Y). Proof. It is known that SI(X|Y) if, and only if, for any x ∈ I, the vertical section of C, given by gx (y) = C(x, y), is concave (i.e., gx (y) is nonincreasing on its domain). Suppose that SI(X|Y). Then, part (a) in 1 follows from part 3 of Lemma 12; and, from part 4 of such lemma, we have that the generating function a is nondecreasing on I. If x ∈ I, we define a(−1) (x) = sup{y ∈ I : a(y) < x} (if the set {y ∈ I : a(y) < x} is empty, we define a(−1) (x) = 0) and a[−1] (x) = sup{y ∈ I : a(y) ⱕ x} (but a[−1] (x) = 0 if the set {y ∈ I : a(y) ⱕ x} is empty). It is easy to check that: y < a(−1) (x) if, and only if, x > a(y); y > a[−1] (x) if, and only if, x < a(y) and a(−1) (x) ⱕ y ⱕ a[−1] (x) if, and only if, x = a(y). Thus, ⎧ ⎨ y − (1 − x)s(y), y < a(−1) (x), gx (y) = C(x, y) = b(y), (15) a(−1) (x) ⱕ y ⱕ a[−1] (x), ⎩ xr(y), y > a[−1] (x). From this equation, since gx is concave, it is easy to derive that s is convex on [0, a(−1) (1)) and r is concave on (a[−1] (0), 1]. Since b(y) > ya(y) for all y ∈ (0, 1), we have Ia = (0, 1), whence a(−1) (1) = 1 and a[−1] (0) = 0. Therefore parts (b) and (c) in 1 hold. Conversely, suppose that (a), (b) and (c) hold. Then, we have again that a is nondecreasing on I; and we have to prove that gx is nonincreasing on its domain for every x ∈ I. Observe that ⎧ ⎨ 1 − (1 − x)s (y), y < a(−1) (x),  gx (y) = b (y), (16) a(−1) (x) < y < a[−1] (x), ⎩  y > a[−1] (x) xr (y), almost everywhere. From (b) and (c), we have that gx is nonincreasing on each of the intervals [0, a(−1) (x)) and (a[−1] (x), 1] (almost everywhere). In general, we can compute the limits: lim

gx (y) = b (a(−1) (x)) − (a(−1) (x) − b(a(−1) (x))a (a(−1) (x))/(1 − x),

lim

gx (y) = b (a[−1] (x)) − b(a[−1] (x))a (a[−1] (x))/x.

y→a(−1) (x)−

y→a[−1] (x)+

If a[−1] (x) = a(−1) (x), then the first limit is equal to the second one when a (a[−1] (x)) = 0; otherwise (a (a[−1] (x)) > 0), the first limit is greater than the second one if, and only if, (a(−1) (x) − b(a(−1) (x))/(1 − x) ⱕ b(a(−1) (x))/x, i.e., b(a(−1) (x)) ⱖ a(−1) (x)x = a(−1) (x)a(a(−1) (x)), which occurs from hypothesis (a). Thus, in this case gx is nonincreasing on its domain. Otherwise, if a(−1) (x) < a[−1] (x), we have that a is constant on the interval [a(−1) (x), a[−1] (x)]; since (b/a) is nonincreasing on this interval, it is easy to see that b is also nonincreasing on [a(−1) (x), a[−1] (x)]. Thus, lim

y→a(−1) (x)−

gx (y) = b (a(−1) (x)) ⱖ b (a[−1] (x)) =

lim

y→a[−1] (x)+

gx (y),

i.e., gx is nonincreasing on its domain, which completes the proof of part 1. In order to prove parts 2 and 3, recall that LCSD(X, Y) if, and only if, C(x, y)C(x , y ) ⱖ C(x, y )C(x , y) for every x, x , y, y ∈ I such that x < x and y < y ; and RCSI(X, Y) if, and only if, ˆ  , y) ˆ y )C(x ˆ y)C(x ˆ  , y ) ⱖ C(x, C(x,

(17)

for every ∈ I such that and where Cˆ is the copula defined in Lemma 13. We prove part 3 (part 2 can be proved in a similar way). We only need to prove that RTI(X|Y) implies RCSI(X, Y) under our hypotheses. From parts 2 and 4 in x, x , y, y

x < x

y < y ,

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Lemma 12, there exists y0 ∈ [0, 1) such that b(y) = ya(y) for all y ∈ [0, y0 ], b(y) > ya(y) for all y ∈ (y0 , 1) and a is nondecreasing on ˆ y) = xy whenever y ∈ [0, y0 ] and C(x, ˆ y) > xy whenever y ∈ (y0 , 1). Let a, ˜ b˜ be the functions defined by [y0 , 1]. Moreover, C(x,  ˜ a(y) =

a(y0 ), a(y),

0 ⱕ y < y0 y0 ⱕ y ⱕ 1

and

 a(y0 )y, ˜ b(y) = b(y),

0 ⱕ y < y0 , y0 ⱕ y ⱕ 1.

Then, it is clear that a˜ and b˜ are generating functions of C, and a˜ is nondecreasing on I. We now prove that RCSI(X, Y) holds by using the new pair of generating functions. Observe that ⎧ ˜ − y), ˜ 0 ⱕ x < 1 − a(1 ⎨ xu(y), ˜ − y) − a˜ (1 − y), x = 1 − a(1 ˆ y) = y + b(1 ˜ − y), C(x, ⎩ ˜ ˜ − y) < x ⱕ 1, y − (1 − x)v(y), 1 − a(1

(18)

where ˜ u(y) =

˜ − y) − a˜ (1 − y) y + b(1 1 − a˜ (1 − y)

and

˜ v(y) =

˜ − y) ˜ − y) − b(1 a(1 . ˜ − y) a(1

In order to prove (17) for every x, x , y, y ∈ I such that x < x and y < y , we consider the following cases: (1) x ⱕ 1 − a˜ (1 − y); ˜ − y) < x ⱕ 1 − a(1 ˜ − y ); (3) x ⱕ 1 − a(1 ˜ − y) ⱕ 1 − a(1 ˜ − y ) < x ; (4) 1 − a˜ (1 − y) < x < x ⱕ 1 − a(1 ˜ − y ); (5) (2) x ⱕ 1 − a(1 ˜ − y ) < x and (6) 1 − a(1 ˜ − y) ⱕ 1 − a(1 ˜ − y ) < x < x .For the case (1) inequality (17) holds immediately. For ˜ − y) < x ⱕ 1 − a(1 1 − a(1 ˜ ˜ ˜ ˜ ⱖ y−(1−x )v(y), which is equivalent to (1− a˜ (1−y))u(y) ⱖ y−(1−x )v(y); and the case (2) inequality (17) holds if, and only if, x u(y)  − (1 − x )v(y ˜  )) ⱖ u(y ˜ ˜  )(y − (1 − x )v(y)), ˜ this last inequality holds trivially. For the case (3) inequality (17) holds if, and only if, u(y)(y which holds for all x if, and only if, it holds for x = 1 and x = 1 − a˜ (1 − y ); then, after some computation, we can conclude that ˜ inequality (17) holds if, and only if, the function u(y)/y is nonincreasing on its domain. Similar reasonings to those in the first ˜ three cases prove that: inequality (17) holds for the case (4); it holds for the case (5) if, and only if, u(y)/y is nonincreasing and ˜ ˜ v(y)/y is nondecreasing on their respective domains; and it holds for the case (6) if, and only if, v(y)/y is nondecreasing on its ˜ ˜ domain. Finally, it is easy to check that u(y)/y is nonincreasing if, and only if, (1 − s(y))/(1 − y) is nondecreasing; and v(y)/y is nondecreasing if, and only if, (r(y) − 1)/(1 − y) is nondecreasing. Hence, from part 2 of Theorem 11, the result follows.  The following result characterizes the strongest concept of positive dependence which we deal with in this paper. Theorem 15. Let (X, Y) be a random pair whose copula C is in P. Suppose that C and their generating functions a and b are absolutely continuous, and that a is piecewise monotone. Then PLR(X, Y) (X and Y are positively likelihood ratio dependent) if, and only if, a is constant and b is concave. Proof. Suppose that PLR(X, Y) holds. Then we also have SI(X|Y). From part 3 of Lemma 12 and from Theorem 6, there exists a constant K ∈ I such that a(y) = K for all y ∈ I. Then, the density of C is given by ⎧

b (y)

⎪ 2 ⎨ , 0 ⱕ x < K, * C K g(x, y) = (x, y) =  ⎪ *x*y ⎩ 1 − b (y) , K < x ⱕ 1. 1−K

It is known that PLR(X, Y) holds if, and only if, we have g(x, y)g(x , y ) ⱖ g(x, y )g(x , y) for every x, x , y, y ∈ I such that x < x and y < y . This inequality holds immediately when x < K and x > K; otherwise (x ⱕ K ⱕ x ), as it is easy to check, g(x, y)g(x , y ) ⱖ g(x, y )g(x , y) if, and only if, b (y) ⱖ b (y ), whence the conclusion follows.  Let (X, Y) be a continuous random pair with copula C. Recall that the lower and upper tail dependence parameters, when   they exist, can be, respectively, computed as follows: L = C (0+) and U = 2 − C (1−), where C is the diagonal section of C, i.e., C (t) = C(t, t) for all t ∈ I. Observe that, if C ∈ P, then ⎧ b(t) ⎪ 0 ⱕ t < a(t), ⎪ ⎪ t a(t) , ⎨ C (t) = b(t), t = a(t), ⎪ ⎪ ⎪ ⎩ t − (1 − t) t − b(t) , a(t) < t ⱕ 1. 1 − a(t) It is not possible to provide general expressions for the parameters L and U associated with copulas in P, because the limits in their definitions should be computed, if there exist, in a different way depending on the generating function a (we will see some examples in Section 4). Finally, next theorem studies the symmetry in a continuous random pair (X, Y) whose copula is in the set P. See Nelsen (2006) for a complete review of some symmetry properties, namely: marginal symmetry, radial symmetry and joint symmetry.

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Theorem 16. Let (x0 , y0 ) be a point in R2 and let (X, Y) be a continuous random pair whose copula C is in P. Suppose that (X, Y) is marginally symmetric about (x0 , y0 ). Then: 1. (X, Y) is radially symmetric about (x0 , y0 ) if, and only if, either C =  or the following two equalities hold for every y ∈ I: a(y) + a(1 − y) = 1 and b(1 − y) − b(y) = 1 − y − a(y). 2. (X, Y) is jointly symmetric about (x0 , y0 ) if and only if C =  (X and Y are independent). ˆ and (X, Y) is jointly symmetric about (x0 , y0 ) Proof. It is known that (X, Y) is radially symmetric about (x0 , y0 ) if, and only if, C = C; ¨ Now, the proofs of both parts are straightforward consequences of Lemma 13.  if, and only if, C = C˙ = C. 4. Examples In this section we provide several families of copulas in the class P which exhibit quite different properties and structures. These families constitute a small sample among the great variety of copulas that we can find in P. All the examples in this section are introduced by means of (6), given the generating functions a and b. First, we introduce a family of symmetric absolutely continuous copulas. Example 1. Let a and b be the functions defined by a(y) =  and

b(y) = (1 − )y + min(y, )

for all y ∈ I,

where  ∈ (0, 1) and max(/( − 1), ( − 1)/ ) ⱕ ⱕ 1. Then, from Theorem 4, we obtain the following biparametric family of copulas: ⎧ (1 − ) ⎪ ⎪ xy, max(x, y) ⱕ , xy + ⎪ ⎪  ⎪ ⎨ xy + x(1 − y), x ⱕ  < y, C, (x, y) = ⎪ y ⱕ  < x, xy + y(1 − x), ⎪ ⎪ ⎪ ⎪ ⎩ xy +  (1 − x)(1 − y),  < min(x, y). 1− By applying Theorem 6 we have that all these copulas are absolutely continuous. And they are clearly symmetric. In the following paragraphs we comment some other properties of this biparametric family of copulas. Observe that the range for contains the interval [0, 1]. When  = 12 we have the biggest range for , which is the interval [−1, 1]; if  increases or decreases from 12 , then the first endpoint of the range for increases (this range tends to be reduced to the interval [0, 1] when  → 0+ or  → 1− ). Theorem 7 produces the following results: C = 3(1 − ) and C, = 23 C . Thus, the maximum and minimum values for ,

,

and in this family are C1/2,1 = 34 , C1/2,1 = 12 and C1/2,−1 = − 34 , C1/2,−1 = − 12 , respectively. For every 0 ∈ (0, 1), the parametric family of copulas C0 , is positively ordered, i.e., C0 , 1 ≺ C0 , 2 whenever 1 ⱕ 2 . Observe that, for every  ∈ (0, 1), C, =  if, and only if, = 0. Therefore, if (X, Y) is a continuous random pair with copula C, , then PQD(X, Y) whenever ∈ [0, 1] (and X and Y are negatively quadrant dependent whenever ∈ [max(/( − 1), ( − 1)/ ), 0]). From Theorem 15 we have PLR(X, Y) if, and only if, ∈ [0, 1], whence all the other concepts of positive dependence studied in this paper hold when ∈ [0, 1]. If < 0, we can obtain

analogous conclusions for the respective concepts of negative dependence. Moreover, the random pair (X, Y) does not present tail dependence in the sense that L = U = 0. Finally, from Theorem 16, when the pair (X, Y) is marginally symmetric about a point in R2 , we have that (X, Y) is radially symmetric if, and only if, ( − 1/2) = 0. The next example provides a family of singular copulas: Example 2. Let  ∈ (0, 1). If we take as generating functions     −y y− y− a(y) = max , and b(y) = max 0,  1− 1−

for all y ∈ I,

then we obtain the following family of copulas: C (x, y) = min(x, max(0, y − (1 − x))),

(x, y) ∈ I2 .

From Theorem 3, it is easy to prove that the functions C are indeed copulas; and, from Theorem 6, these copulas are singular: a (y)(b(y) − ya(y)) =1 a(y)(1 − a(y))

for every y ∈ (0, 1)\{},

J.A. Rodríguez-Lallena / Journal of Statistical Planning and Inference 139 (2009) 3908 -- 3920

whence  Ia

3919

a (y)(b(y) − ya(y)) = 1. a(y)(1 − a(y))

This family of copulas C is negatively ordered, since C2 ≺ C1 whenever 1 ⱕ 2 . As a limit, we can define C0 = M and C1 = W. Theorem 7 produces the following results: C = C = 1 − 2; thus, the whole range for and is attained by this family of copulas. Since b(y) − ya(y) < 0 when y ∈ (0, ) and b(y) − ya(y) > 0 when y ∈ (, 1), the copulas C cannot be associated with quadrant dependence random pairs (except for the extreme cases C0 and C1 ). Moreover, the tail dependence parameters for a continuous random pair with copula C are L = 0 and U = 1 − . Finally, since a(0) + a(1) = 2  1, these copulas cannot be associated with radially symmetric random pairs. Unlike copulas in Examples 1 and 2, copulas of the family introduced in the following example have an absolutely continuous component and also a singular component. Example 3. Let  ∈ I. If the generating functions are given by a(y) = y2 and b(y) = y2 + (1 − )y3 for every y ∈ I, then we obtain the family of functions C (x, y) = xy +





1+y

min(x, y2 ) − xy2 ,

(x, y) ∈ I2 .

From Theorem 4 (after some computations), it can be shown that C is a copula for every  ∈ I. Observe that C0 = . If 0 <  ⱕ 1, then it is easy to check (by using Theorem 6) that C has a singular component which is concentrated along the curve √ {(x, x) : x ∈ I}, and whose C -measure is given by 2(1 − ln 2). Theorem 7 yields the following results:

C =

 2

and C =

2

(17 − 24 ln 2) +  . 3

It is clear that the family C is positively ordered: C1 ≺ C2 if, and only if, 1 ⱕ 2 . Therefore,  ≺ C for all  ∈ I, whence we have PQD(X, Y) for every continuous random pair (X, Y) with copula C . From Theorem 11, we also have RTI(X|Y), but not LTD(X|Y) (if 0 <  ⱕ 1). Now, the tail dependence parameters are L = 0 and U = /2. Finally, it is immediate to check that the copulas C , with 0 <  ⱕ 1, cannot be associated with radially symmetric random pairs. In the three previous examples the lower tail dependence parameter is L = 0. However, this equality does not hold for every copula in the set P, as the following example shows. Example 4. Let  ∈ I. If the generating functions are given by a(y) = b(y) = y, for all y ∈ I, then we obtain the family of functions  x, 0 ⱕ x ⱕ y, y − y C (x, y) = , y < x ⱕ 1. y − (1 − x) 1 − y From Theorem 4, it is easy to prove that C is a copula for every  ∈ I. Now, the tail dependence parameters for a continuous random pair with copula C are L =  and U = 0. For the last example we need to recall that a function : I → I is the diagonal section of some copula if, and only if,  is a diagonal, i.e., (a) (1) = 1, (b) 0 ⱕ (t ) − (t) ⱕ 2(t − t) for all t, t ∈ I with t < t and (c) (t) ⱕ t for all t ∈ I: see Fredricks and Nelsen (1997) or Nelsen (2006). Example 5. Let a be the function defined by a(y) = y for all y ∈ I, and let C be the function given by (6), for some function b satisfying condition (5). Observe that, if C is a copula, then b = C . Then, let b =  be a diagonal. Now, from Theorem 3, we can obtain that a function C with such generating functions is a copula if, and only if, (i) r(y) = (y)/y is nondecreasing on (0, 1], (ii) s(y) = (y − (y))/(1 − y) is nondecreasing on [0, 1) and (iii) (y − (y))/(1 − y) ⱕ (y )/y for all y, y ∈ I with y < y . It is easy to see that, under conditions (i) and (ii), condition (iii) can be replaced by the following: (iv) (y) ⱖ y2 for all y ∈ I. Observe that C has the following form:

C(x, y) =

⎧ (y) ⎪ ⎪ ⎪x y , ⎨

0 ⱕ x < y,

(y), x = y, ⎪ ⎪ ⎪ ⎩ y − y − (y) (1 − x), y < x ⱕ 1. 1−y

(19)

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J.A. Rodríguez-Lallena / Journal of Statistical Planning and Inference 139 (2009) 3908 -- 3920

So C is the asymmetric semilinear copula A introduced and studied in De Baets et al. (2007). From Theorem 6, it is easy to prove that A is absolutely continuous if, and only if, (y) = y2 for all y ∈ I, i.e.,  is the unique absolutely continuous semilinear copula A . By using Theorem 7 we have the expressions for the measures and associated with the copulas A obtained in De Baets 1 et al. (2007) (but there is a small mistake in this paper: the expression for should be = 6 0 (x) dx − 2).  Observe that the asymmetric semilinear copulas A are positively ordered in the following sense: A ≺ A if, and only if,  ⱕ  . Hence, if (X, Y) is a continuous random pair with copula A , then we have PQD(X, Y). Finally, from Theorem 16, when the pair (X, Y) is marginally symmetric about a point in R2 , we have that (X, Y) is radially symmetric if, and only if, (y) − (1 − y) = 2y − 1 for all y ∈ I. For instance, this property is satisfied for the following family of diagonals:  (y) = y + (1 − )y2 (y ∈ I), with ∈ I (observe that  also satisfies conditions (i), (ii) and (iv) for all ∈ I). Acknowledgements The author acknowledges the support of the Junta de Andalucía and the Ministerio de Educación y Ciencia (Spain), under Research Project MTM2006-12218 (co-funded by FEDER, European Union). References De Baets, B., De Meyer, H., Mesiar, R., 2007. Asymmetric semilinear copulas. Kybernetika 43, 221–233. Fréchet, M., 1958. Remarques au sujet de la note précédente. C. R. Acad. Sci. Paris, Sér. I Math. 246, 2719–2720. Fredricks, G.A., Nelsen, R.B., 1997. Copulas constructed from diagonal sections. In: Beneš, V., Štepán, J. (Eds.), Distributions with Given Marginals and Moment Problems. Kluwer, Dordrecht, pp. 129–136. Fredricks, G.A., Nelsen, R.B., 2007. On the relationship between Spearman's rho and Kendall's tau for pairs of continuous random variables. J. Statist. Plann. Inference 137, 2143–2150. Joe, H., 1997. Multivariate Models and Dependence Concepts. Chapman & Hall, New York. Mardia, K.V., 1970. Families of Bivariate Distributions. Charles Griffin and Co., London. Mikusinski, P., Sherwood, H., Taylor, M.D., 1992. Shuffles of Min. Stochastica 13, 61–74.  Nelsen, R.B., 2006. An Introduction to Copulas. second ed. Springer, New York. Nelsen, R.B., Quesada-Molina, J.J., Rodríguez-Lallena, J.A., 1997. Bivariate copulas with cubic sections. J. Nonparametr. Statist. 7, 205–220. Quesada-Molina, J.J., Rodríguez-Lallena, J.A., 1995. Bivariate copulas with quadratic sections. J. Nonparametr. Statist. 5, 323–337. Rodríguez-Lallena, J.A., 1996. Estudio de la compatibilidad y diseño de nuevas familias en la teoría de cópulas. Aplicaciones. Servicio de Publicaciones de la Universidad de Almería, Spain. Rodríguez-Lallena, J.A., Úbeda-Flores, M., 2004. A new class of bivariate copulas. Statist. Probab. Lett. 66, 315–325. Sklar, A., 1959. Fonctions de répartition à n dimensions et leurs marges. Publ. Inst. Statist. Univ. Paris 8, 229–231. Úbeda-Flores, M., 1998. Introducción a la teoría de cópulas. Aplicaciones. Predoctoral Research Dissertation, Universidad de Almería, Spain.