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Maximal asymmetry of bivariate copulas and consequences to measures of dependence

Maximal asymmetry of bivariate copulas and consequences to measures of dependence 1IntroductionTwo random variables XXand YYwith joint distribution function HHare called exchangeable if and only if the pairs (X,Y)\left(X,Y)and (Y,X)\left(Y,X)have the same distribution, or equivalently, if H(x,y)=H(y,x)H\left(x,y)=H(y,x)holds for all xxand yy. The study of exchangeable random variables has exhibited a lot of interest in statistics (see, for instance, [8] and references therein). In case XXand YYare identically distributed and have distribution function FF, then (X,Y)\left(X,Y)is exchangeable if and only if the underlying copula AAcoincides with its transpose At{A}^{t}(defined as At(x,y)=A(y,x){A}^{t}\left(x,y)=A(y,x)). Hence, in what follows we consider continuous and identically distributed random variables XXand YY. While the class of continuous exchangeable random variables XXand YYis uniquely characterized by the class of symmetric copulas, the exact opposite, i.e., maximal non-exchangeability of random variables, strongly depends on the choice of measure quantifying the degree of non-exchangeability. One natural measure of non-exchangeability was studied by Nelsen [21] as well as by Klement and Mesiar [15], who independently showed that d∞(A,At)≔supx,y∈[0,1]∣A(x,y)−A(y,x)∣≤13{d}_{\infty }\left(A,{A}^{t}):= \mathop{\sup }\limits_{x,y\in \left[0,1]}| A\left(x,y)-A(y,x)| \le \frac{1}{3}holds for every A∈CA\in {\mathcal{C}}and introduced the d∞{d}_{\infty }-based measure δ:C→[0,1]\delta :{\mathcal{C}}\to \left[0,1]via δ(A)≔3d∞(A,At)\delta \left(A):= 3{d}_{\infty }\left(A,{A}^{t}). Moreover, they characterized all copulas A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry and showed that these copulas always model slightly negatively correlated random variables XXand YYin the sense of Spearman’s ρ\rho . More precisely, δ(A)=1\delta \left(A)=1implies ρ(A)∈−59,−13\rho \left(A)\in \left[-\frac{5}{9},-\frac{1}{3}\right]. Similar results also hold for different measures of concordance (see [17]).Considering other metrics on the space of copulas yields alternative measures of non-exchangeability ([13,25]): In [13] the stronger conditioning-based metric D1{D}_{1}introduced in [27] was studied and the authors proved (among other things) that every copula A∈CA\in {\mathcal{C}}with maximal D1{D}_{1}-asymmetry (i.e., D1(A,At)=12{D}_{1}\left(A,{A}^{t})=\frac{1}{2}) is not maximal asymmetric with respect to d∞{d}_{\infty }and that no maximal d∞{d}_{\infty }-asymmetric copula is maximal asymmetric with respect to D1{D}_{1}.Building upon the results in [13] we here further investigate the family of copulas with maximal D1{D}_{1}-asymmetry, derive additional novel characterizations in terms of the Markov-product of copulas (see [3]), and study various topological properties; inter alia we prove that the family of mutually completely dependent copulas with maximal D1{D}_{1}-asymmetry is dense in the set of all copulas with maximal D1{D}_{1}-asymmetry. Furthermore, we extend the concept of maximal D1{D}_{1}-asymmetry to the general Dp{D}_{p}-metrics (p∈[1,∞)p\in {[}1,\infty )), defined by (1)Dp(A,B)≔∫[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣pdλ(x)dλ(y)1p,{D}_{p}\left(A,B):= {\left(\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y]){| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)}^{\tfrac{1}{p}},where KA(⋅,⋅),KB(⋅,⋅){K}_{A}\left(\cdot ,\cdot ),{K}_{B}\left(\cdot ,\cdot )denote the Markov kernels (regular conditional distributions) of A,B∈CA,B\in {\mathcal{C}}, respectively. Although all Dp{D}_{p}-metrics induce the same topology, we show the surprising result that maximal D1{D}_{1}-asymmetry is not equivalent to maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty ). In fact, copulas with maximal Dp{D}_{p}-asymmetry with p∈(1,∞)p\in \left(1,\infty )are always mutually completely dependent and maximal asymmetric w.r.t. D1{D}_{1}.Moreover, we tackle the question on the degree of dependence of copulas exhibiting maximal asymmetry with respect to d∞{d}_{\infty }or Dp{D}_{p}for every p∈[1,∞]p\in \left[1,\infty ]. Since measures of concordance are generally not suitable for quantifying dependence (see, for instance, [11]) we consider the dependence measures ζ1{\zeta }_{1}introduced in [27] and further studied in [10,11], as well as ξ\xi , defined in [4] and reinvestigated in [2]. Both measures have recently attracted a lot of interest (see, e.g., [1,10,11,14,24,26]) since, in contrast to standard methods like Spearman’s ρ\rho or Kendall’s τ\tau , these measures are 1 if and only if YYis a function of XXand 0 if and only if XXand YYare independent; moreover, they can be estimated consistently without underlying smoothness assumptions. We prove that when considering maximal d∞{d}_{\infty }-asymmetry ζ1∈56,1{\zeta }_{1}\in \left[\frac{5}{6},1\right]and ξ∈23,1\xi \in \left[\frac{2}{3},1\right]hold, and in the case of maximal D1{D}_{1}-asymmetry ζ1∈34,1{\zeta }_{1}\in \left[\frac{3}{4},1\right]and ξ∈12,1\xi \in \left(\frac{1}{2},1\right]follows. In other words, maximal non-exchangeable random variables (in the sense of d∞{d}_{\infty }or Dp{D}_{p}) always imply a high degree of dependence w.r.t. ζ1{\zeta }_{1}and ξ\xi .The rest of this article is organized as follows: Section 2 gathers preliminaries and notations that will be used throughout the article. In Section 3, we study possible values of ζ1{\zeta }_{1}and ξ\xi for maximal d∞{d}_{\infty }-asymmetric copulas and discuss an example illustrating differences of ζ1{\zeta }_{1}and ξ\xi in the context of ordinal sums. In Section 4, we revisit copulas with maximal D1{D}_{1}-asymmetry and derive several topological properties. Extensions on maximal Dp{D}_{p}-asymmetry for p∈[1,∞]p\in \left[1,\infty ]and some interrelations are established in Section 5. Consequences on the dependence measures ζ1{\zeta }_{1}and ξ\xi conclude the article (Section 6). Various examples and graphics illustrate both the obtained results and the ideas underlying the proofs.2Notation and preliminariesFor every metric space (Ω,d)\left(\Omega ,d)the Borel σ\sigma -field in Ω\Omega will be denoted by ℬ(Ω){\mathcal{ {\mathcal B} }}\left(\Omega ), λ\lambda will denote the Lebesgue measure on ℬ(R){\mathcal{ {\mathcal B} }}\left({\mathbb{R}}). T{\mathcal{T}}will denote the class of all measurable λ\lambda -preserving transformations on [0,1]\left[0,1], i.e., T={T:[0,1]→[0,1]measurable withλ(T−1(E))=λ(E)∀E∈ℬ([0,1])},{\mathcal{T}}=\left\{T:\left[0,1]\to \left[0,1]\hspace{0.33em}\hspace{0.1em}\text{measurable with}\hspace{0.1em}\hspace{0.33em}\lambda \left({T}^{-1}\left(E))=\lambda \left(E)\hspace{1.0em}\forall E\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])\right\},and Tb{{\mathcal{T}}}_{b}the subclass of all bijective T∈TT\in {\mathcal{T}}. Throughout the article C{\mathcal{C}}will denote the family of all two-dimensional copulas, P{\mathcal{P}}the family of all doubly-stochastic measures (for background on copulas and doubly stochastic measures we refer to [6,22] and references therein). Furthermore, MMdenotes the upper Fréchet Hoeffding bound, Π\Pi the product copula, and WWthe lower Fréchet Hoeffding bound. Additionally, the completely dependent copula induced by a measure-preserving transformation h∈Th\in {\mathcal{T}}will be denoted by Ch{C}_{h}(see [27], Definition 9). The family of all completely dependent copulas will be denoted by Ccd{{\mathcal{C}}}_{cd}and the family of all mutually completely dependent copulas by Cmcd≔{Ch∈Ccd:h∈Tb}{{\mathcal{C}}}_{mcd}:= \left\{{C}_{h}\in {{\mathcal{C}}}_{cd}:h\in {{\mathcal{T}}}_{b}\right\}. For every copula C∈CC\in {\mathcal{C}}the corresponding doubly stochastic measure will be denoted by μC{\mu }_{C}. As usual, d∞{d}_{\infty }denotes the uniform metric on C{\mathcal{C}}, i.e., d∞(A,B)≔max(x,y)∈[0,1]2∣A(x,y)−B(x,y)∣{d}_{\infty }\left(A,B):= \mathop{\max }\limits_{\left(x,y)\in {\left[0,1]}^{2}}| A\left(x,y)-B\left(x,y)| for every A,B∈CA,B\in {\mathcal{C}}. It is well-known that (C,d∞)\left({\mathcal{C}},{d}_{\infty })is a compact metric space (see [6]).In what follows, Markov kernels will play an important role. A mapping K:R×ℬ(R)→[0,1]K:{\mathbb{R}}\times {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})\to \left[0,1]is called a Markov kernel from (R,ℬ(R))\left({\mathbb{R}},{\mathcal{ {\mathcal B} }}\left({\mathbb{R}}))to (R,ℬ(R))\left({\mathbb{R}},{\mathcal{ {\mathcal B} }}\left({\mathbb{R}}))if the mapping x↦K(x,B)x\mapsto K\left(x,B)is measurable for every fixed B∈ℬ(R)B\in {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})and the mapping B↦K(x,B)B\mapsto K\left(x,B)is a probability measure for every fixed x∈Rx\in {\mathbb{R}}. A Markov kernel K:R×ℬ(R)→[0,1]K:{\mathbb{R}}\times {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})\to \left[0,1]is called regular conditional distribution of a (real-valued) random variable YYgiven (another random variable) XXif for every B∈ℬ(R)B\in {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})K(X(ω),B)=E(1B∘Y∣X)(ω)K\left(X\left(\omega ),B)={\mathbb{E}}\left({{\mathbb{1}}}_{B}\circ Y| X)\left(\omega )holds P{\mathbb{P}}-a.s. It is well-known that a regular conditional distribution of YYgiven XXexists and is unique PX{{\mathbb{P}}}^{X}-almost sure (where PX{{\mathbb{P}}}^{X}denotes the distribution of XX, i.e., the push-forward of P{\mathbb{P}}via XX). For every A∈CA\in {\mathcal{C}}(a version of) the corresponding regular conditional distribution (i.e., the regular conditional distribution of YYgiven XXin the case that (X,Y)∼A\left(X,Y)\hspace{0.33em} \sim \hspace{0.33em}A) will be denoted by KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot ). Note that for every A∈CA\in {\mathcal{C}}and Borel sets E,F∈ℬ([0,1])E,F\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])we have (2)∫EKA(x,F)dλ(x)=μA(E×F)and∫[0,1]KA(x,F)dλ(x)=λ(F).\mathop{\int }\limits_{E}{K}_{A}\left(x,F){\rm{d}}\lambda \left(x)={\mu }_{A}\left(E\times F)\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}\mathop{\int }\limits_{\left[0,1]}{K}_{A}\left(x,F){\rm{d}}\lambda \left(x)=\lambda \left(F).For more details and properties of conditional expectations and regular conditional distributions we refer to [12,16]. Expressing copulas in terms of their corresponding regular conditional distribution yields metrics stronger than d∞{d}_{\infty }(see [27]) and defined by (3)Dp(A,B)≔∫[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣pdλ(x)dλ(y)1p,{D}_{p}\left(A,B):= {\left(\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y]){| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)}^{\tfrac{1}{p}},(4)D∞(A,B)≔supy∈[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣dλ(x).{D}_{\infty }\left(A,B):= \mathop{\sup }\limits_{y\in \left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| \hspace{0.33em}{\rm{d}}\lambda \left(x).To simplify notation we will also write ΦA,B(y)≔∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣dλ(x){\Phi }_{A,B}(y):= {\int }_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| {\rm{d}}\lambda \left(x). We will also work with D∂{D}_{\partial }, defined by D∂(A,B)≔D1(A,B)+D1(At,Bt),{D}_{\partial }\left(A,B):= {D}_{1}\left(A,B)+{D}_{1}\left({A}^{t},{B}^{t}),whereby At{A}^{t}denotes the transpose of A∈CA\in {\mathcal{C}}. The metric D∂{D}_{\partial }can be seen as metrization of the so-called ∂\partial -convergence, introduced and studied in [18,19]. In [27], it is shown that (C,D1)\left({\mathcal{C}},{D}_{1})is a complete and separable metric space with diameter 1/2 and that the topology induced by D1{D}_{1}is strictly finer than the one induced by d∞{d}_{\infty }. For further background on D1{D}_{1}and D∂{D}_{\partial }as well as for possible extensions to the multivariate setting we refer to [6,7, 10,27] and references therein.The D1{D}_{1}-based dependence measure ζ1{\zeta }_{1}(introduced in [27] and further investigated in [10,11]) is defined as ζ1(X,Y)≔ζ1(A)≔3D1(A,Π),{\zeta }_{1}\left(X,Y):= {\zeta }_{1}\left(A):= 3{D}_{1}\left(A,\Pi ),whereby (X,Y)\left(X,Y)has copula A∈CA\in {\mathcal{C}}. In the sequel, we will also consider the dependence measure ξ\xi (first introduced in [4] and reinvestigated in [2]) defined as ξ(X,Y)≔∫Var(E(1{Y≥t}∣X))dμ(t)∫Var(1{Y≥t})dμ(t),\xi \left(X,Y):= \frac{\int {\rm{Var}}\left({\mathbb{E}}\left({{\mathbb{1}}}_{\left\{Y\ge t\right\}}| X)){\rm{d}}\mu \left(t)}{\int {\rm{Var}}\left({{\mathbb{1}}}_{\left\{Y\ge t\right\}}){\rm{d}}\mu \left(t)},where μ\mu is the law of YY. In the copula setting, it is straightforward to verify that ξ\xi can be expressed in terms of D2{D}_{2}and that ξ(X,Y)≔ξ(A)=6D22(A,Π)\xi \left(X,Y):= \xi \left(A)=6{D}_{2}^{2}\left(A,\Pi )holds. Both dependence measures attain values in [0,1]\left[0,1]and are 0 if and only if A=ΠA=\Pi , and 1 if and only if AAis completely dependent.Letting Sh(A){S}_{h}\left(A)denote the generalized shuffle of AAw.r.t. the first coordinate, implicitly defined via the corresponding doubly stochastic measure μA{\mu }_{A}by μSh(A)(E×F)≔μA(h−1(E)×F),{\mu }_{{{\mathcal{S}}}_{h}\left(A)}\left(E\times F):= {\mu }_{A}\left({h}^{-1}\left(E)\times F),for all E,F∈ℬ([0,1])E,F\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])(see, e.g., [5,9]), the following simple result holds:Lemma 2.1Let h∈Tbh\in {{\mathcal{T}}}_{b}be a λ\lambda -preserving bijection. Then ζ1(Sh(A))=ζ1(A){\zeta }_{1}\left({S}_{h}\left(A))={\zeta }_{1}\left(A)and ξ(Sh(A))=ξ(A)\xi \left({S}_{h}\left(A))=\xi \left(A)hold for every A∈CA\in {\mathcal{C}}.ProofAccording to Lemma 3.1 in [9] for h∈Tbh\in {{\mathcal{T}}}_{b}the Markov kernel of Sh(A){{\mathcal{S}}}_{h}\left(A)can be expressed as KSh(A)(x,[0,y])=KA(h−1(x),[0,y]){K}_{{{\mathcal{S}}}_{h}\left(A)}\left(x,\left[0,y])={K}_{A}\left({h}^{-1}\left(x),\left[0,y])and for p∈[1,∞)p\in {[}1,\infty )we obtain Dpp(Sh(A),Π)=∫[0,1]∫[0,1]∣KSh(A)(x,[0,y])−y∣pdλ(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(x),[0,y])−y∣pdλ(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(x),[0,y])−y∣pdλh(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(h(x)),[0,y])−y∣pdλ(x)dλ(y)=Dpp(A,Π),\begin{array}{rcl}{D}_{p}^{p}\left({{\mathcal{S}}}_{h}\left(A),\Pi )& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{{\mathcal{S}}}_{h}\left(A)}\left(x,\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(x),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(x),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}{\lambda }^{h}\left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(h\left(x)),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)={D}_{p}^{p}\left(A,\Pi ),\end{array}which proves the assertion.□In the sequel, we will also work with rearrangements [23] (see [26] for an elegant application of rearrangements in the copula context). We call f∗:[0,1]→R{f}^{\ast }:\left[0,1]\to {\mathbb{R}}the decreasing rearrangement of a Borel measurable function f:[0,1]→Rf:\left[0,1]\to {\mathbb{R}}if it fulfills f∗(t)≔inf{x∈R:λ({z∈[0,1]:f(z)>x})≤t}{f}^{\ast }\left(t):= {\rm{\inf }}\left\{x\in {\mathbb{R}}:\lambda \left(\left\{z\in \left[0,1]:f\left(z)\gt x\right\})\le t\right\}. The stochastically increasing (SI)-rearrangement A↑{A}^{\uparrow }of AAis then defined as A↑(x,y)≔∫[0,x]KA(t,[0,y])∗dλ(t),{A}^{\uparrow }\left(x,y):= \mathop{\int }\limits_{\left[0,x]}{K}_{A}{\left(t,\left[0,y])}^{\ast }{\rm{d}}\lambda \left(t),whereby the rearrangement is applied on the first coordinate of KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot ), i.e., for every fixed y∈[0,1]y\in \left[0,1]the rearranged Markov kernel is defined via KA(t,[0,y])∗≔inf{x∈[0,1]:λ({z∈[0,1]:KA(z,[0,y])>x})≤t}{K}_{A}{\left(t,\left[0,y])}^{\ast }:= {\rm{\inf }}\left\{x\in \left[0,1]:\lambda \left(\left\{z\in \left[0,1]:{K}_{A}\left(z,\left[0,y])\gt x\right\})\le t\right\}. In [26], it was shown that A↑{A}^{\uparrow }is an SI copula and both dependence measures ζ1{\zeta }_{1}and ξ\xi are invariant w.r.t. to the rearrangement, i.e., they fulfill ζ1(A↑)=ζ1(A){\zeta }_{1}\left({A}^{\uparrow })={\zeta }_{1}\left(A)and ξ(A↑)=ξ(A)\xi \left({A}^{\uparrow })=\xi \left(A), respectively. Recall that a copula AAis called SI if there exists a Borel set Λ⊆[0,1]\Lambda \subseteq \left[0,1]with λ(Λ)=1\lambda \left(\Lambda )=1such that for any y∈[0,1]y\in \left[0,1]the mapping x↦KA(x,[0,y])x\mapsto {K}_{A}\left(x,\left[0,y])is non-increasing on Λ\Lambda . The family of all SI copulas will be denoted by C↑{{\mathcal{C}}}^{\uparrow }. For further information we refer to [22] and references therein.Given A,B∈CA,B\in {\mathcal{C}}a new copula denoted by A∗BA\ast Bcan be constructed via the so-called star/Markov product A∗BA\ast B(see [3]) by (5)(A∗B)(x,y)≔∫[0,1]∂2A(x,t)∂1B(t,y)dλ(t),\left(A\ast B)\left(x,y):= \mathop{\int }\limits_{\left[0,1]}{\partial }_{2}A\left(x,t){\partial }_{1}B\left(t,y){\rm{d}}\lambda \left(t),where ∂1A(x,y){\partial }_{1}A\left(x,y)denotes the partial derivative of AAwith respect to the first coordinate. The star product A∗BA\ast Bis always a copula, i.e., no smoothness assumptions on A,BA,Bare required. Translating to the Markov kernel setting the star product corresponds to the well-known composition of Markov kernels and the following lemma holds:Lemma 2.2[29] Suppose that A,B∈CA,B\in {\mathcal{C}}and let KA,KB{K}_{A},{K}_{B}denote the Markov kernels of A and B, respectively. Then the Markov kernel KA∘KB{K}_{A}\circ {K}_{B}, defined by(6)(KA∘KB)(x,F)≔∫[0,1]KB(y,F)KA(x,dy),\left({K}_{A}\circ {K}_{B})\left(x,F):= \mathop{\int }\limits_{\left[0,1]}{K}_{B}(y,F){K}_{A}\left(x,{\rm{d}}y),is a regular conditional distribution of A∗BA\ast B.3Maximal d∞{d}_{\infty }-asymmetric copulas and their extent of dependence with respect to ζ1{\zeta }_{1}and ξ\xi Since ordinal sums will play an important role in what follows, we briefly recall their definition. We follow [6] and let I⊆NI\subseteq {\mathbb{N}}be some finite index set, ((ai,bi))i∈I{(\left({a}_{i},{b}_{i}))}_{i\in I}be a family of non-overlapping intervals with 0≤ai<bi≤10\le {a}_{i}\lt {b}_{i}\le 1for each i∈Ii\in Isuch that ⋃i∈I[ai,bi]=[0,1]{\bigcup }_{i\in I}\left[{a}_{i},{b}_{i}]=\left[0,1]holds. Furthermore, (Ci)i∈I{\left({C}_{i})}_{i\in I}denotes a family of bivariate copulas. Then the copula CCdefined by C(x,y)=ai+(bi−ai)Cix−aibi−ai,y−aibi−ai,(x,y)∈(ai,bi)2min{x,y}elsewhereC\left(x,y)=\left\{\begin{array}{ll}{a}_{i}+\left({b}_{i}-{a}_{i}){C}_{i}\left(\frac{x-{a}_{i}}{{b}_{i}-{a}_{i}},\frac{y-{a}_{i}}{{b}_{i}-{a}_{i}}\right),\hspace{1.0em}& \left(x,y)\in {\left({a}_{i},{b}_{i})}^{2}\\ \min \left\{x,y\right\}\hspace{1.0em}& \hspace{0.1em}\text{elsewhere}\hspace{0.1em}\end{array}\right.is an ordinal sum, and we write C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}. The following lemma gathers some useful formulas for D1{D}_{1}and D22{D}_{2}^{2}, which will be used in the sequel.Lemma 3.1Let C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}be an ordinal sum with I≔{1,…,n}I:= \left\{1,\ldots ,n\right\}for some n∈Nn\in {\mathbb{N}}. ThenD22(C,Π)=∑i=1n((bi−ai)2D22(Ci,Π))+f(a1,…,an,b1,…,bn),D1(C,Π)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]∣KCi(x,[0,y])−(ai+(bi−ai)y)∣dλ(x)dλ(y)+g(a1,…,an,b1,…,bn),\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}({\left({b}_{i}-{a}_{i})}^{2}{D}_{2}^{2}\left({C}_{i},\Pi ))+f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\\ {D}_{1}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{C}_{i}}\left(x,\left[0,y])-\left({a}_{i}+\left({b}_{i}-{a}_{i})y)| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)+g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\end{array}whereby f and g are given by f(a1,…,an,b1,…,bn)≔∑i=1n(bi−ai)23+(bi−ai)(1−bi)−13f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}):= {\sum }_{i=1}^{n}\left(\frac{{\left({b}_{i}-{a}_{i})}^{2}}{3}+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right)-\frac{1}{3}and g(a1,…,an,b1,…,bn)≔∑i=1n(bi−ai)12−bi+bi22+ai22g({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}):= {\sum }_{i=1}^{n}\left({b}_{i}-{a}_{i})\left(\frac{1}{2}-{b}_{i}+\frac{{b}_{i}^{2}}{2}+\frac{{a}_{i}^{2}}{2}\right), respectively.ProofThe definition of D22{D}_{2}^{2}yields D22(C,Π)=∫[0,1]∫[0,1](KC(x,[0,y])−KΠ(x,[0,y]))2dλ(x)dλ(y)=∫[0,1]∫[0,1]KC(x,[0,y])2dλ(x)dλ(y)−2∫[0,1]y∫[0,1]KC(x,[0,y])dλ(x)dλ(y)+∫[0,1]∫[0,1]y2dλ(x)dλ(y)=∫[0,1]∫[0,1]KC(x,[0,y])2dλ(x)dλ(y)−13\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{({K}_{C}\left(x,\left[0,y])-{K}_{\Pi }\left(x,\left[0,y]))}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)-2\mathop{\displaystyle \int }\limits_{\left[0,1]}y\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}\left(x,\left[0,y]){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{y}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)-\frac{1}{3}\end{array}for every C∈CC\in {\mathcal{C}}. Using the fact that (without loss of generality) the Markov kernel KC(x,[0,y]){K}_{C}\left(x,\left[0,y])of CCis 0 below the squares (ai,bi)2{\left({a}_{i},{b}_{i})}^{2}and 1 above (ai,bi)2{\left({a}_{i},{b}_{i})}^{2}, and applying change of coordinates yields D22(C,Π)+13=∑i=1n∫(ai,bi)∫(ai,bi)KC(x,[0,y])2dλ(x)dλ(y)+∫(bi,1)∫(ai,bi)1dλ(x)dλ(y)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]KCi(x,[0,y])2dλ(x)dλ(y)+(bi−ai)(1−bi)=∑i=1n(bi−ai)2D22(Ci,Π)+13+(bi−ai)(1−bi).\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )+\frac{1}{3}& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left(\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \int }\limits_{\left({b}_{i},1)}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}1{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{C}_{i}}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\left({D}_{2}^{2}\left({C}_{i},\Pi )+\frac{1}{3}\right)+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right).\end{array}Analogously, we obtainD1(C,Π)=∑i=1n∫[0,1]∫(ai,bi)∣KC(x,[0,y])−y∣dλ(x)dλ(y)=∑i=1n∫(ai,bi)∫(ai,bi)KCix−aibi−ai,0,y−aibi−ai−ydλ(x)dλ(y)+∑i=1n∫(bi,1)∫(ai,bi)(1−y)dλ(x)dλ(y)+∑i=1n∫(0,ai)∫(ai,bi)ydλ(x)dλ(y)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]∣KCi(x,[0,y])−(ai+(bi−ai)y)∣dλ(x)dλ(y)+g(a1,…,an,b1,…,bn),\begin{array}{rcl}{D}_{1}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}| {K}_{C}(x,\left[0,y\left])-y| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\left|{K}_{{C}_{i}}\left(\frac{x-{a}_{i}}{{b}_{i}-{a}_{i}},\left[0,\frac{y-{a}_{i}}{{b}_{i}-{a}_{i}}\right]\right)-y\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & & +\mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left({b}_{i},1)}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\left(1-y)\hspace{0.33em}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left(0,{a}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}y\hspace{0.33em}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{C}_{i}}\left(x,\left[0,y])-\left({a}_{i}+\left({b}_{i}-{a}_{i})y)| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)+g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\end{array}with g(a1,…,an,b1,…,bn)g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})as in the theorem.□As a direct consequence, the dependence measure ξ\xi of ordinal sums can easily be expressed in terms of ξ(Ci)\xi \left({C}_{i}):Corollary 3.2Let C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}be an ordinal sum with I≔{1,…,n}I:= \left\{1,\ldots ,n\right\}for some n∈Nn\in {\mathbb{N}}. Thenξ(C)=∑i=1n(bi−ai)2ξ(Ci)+6f(a1,…,an,b1,…,bn)\xi \left(C)=\mathop{\sum }\limits_{i=1}^{n}{\left({b}_{i}-{a}_{i})}^{2}\xi \left({C}_{i})+6f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})holds, where f is defined according to Lemma 3.1 and only depends on the partition.The following example shows that ordinal sums can be used to construct copulas attaining every possible dependence value w.r.t. to ξ\xi and ζ1{\zeta }_{1}.Example 3.3Consider Cs=(⟨ai,bi,Ci⟩)i∈{1,2}{C}_{s}={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in \left\{1,2\right\}}, whereby a1≔0,a2≔s{a}_{1}:= 0,{a}_{2}:= s, b1≔s{b}_{1}:= s, and b2≔1{b}_{2}:= 1for s∈[0,1]s\in \left[0,1]and set C1=Π{C}_{1}=\Pi as well as C2=M{C}_{2}=M. Figure 1 depicts the support of μCs{\mu }_{{C}_{s}}for different choices of s∈[0,1]s\in \left[0,1]. Using Corollary 3.2 we have ξ(Cs)=(1−s)2+6s23+s(1−s)+(1−s)23−13=1−s2\xi \left({C}_{s})={\left(1-s)}^{2}+6\left(\frac{{s}^{2}}{3}+s\left(1-s)+\frac{{\left(1-s)}^{2}}{3}-\frac{1}{3}\right)=1-{s}^{2}. Therefore, the map φ:[0,1]→[0,1]\varphi :\left[0,1]\to \left[0,1]defined by s↦ξ(Cs)s\mapsto \xi \left({C}_{s})is continuous and onto. The same holds for ζ1(Cs)=1−s3{\zeta }_{1}\left({C}_{s})=1-{s}^{3}.Figure 1Mass distribution of the doubly stochastic measure μCs{\mu }_{{C}_{s}}for s=0.3s=0.3(left panel) and s=0.8s=0.8(right panel) considered in Example 3.3. For the dependence measures ξ\xi and ζ1{\zeta }_{1}we obtain ξ(C0.3)=0.91\xi \left({C}_{0.3})=0.91and ξ(C0.8)=0.36\xi \left({C}_{0.8})=0.36as well as ζ1(C0.3)=0.973{\zeta }_{1}\left({C}_{0.3})=0.973and ζ1(C0.8)=0.488{\zeta }_{1}\left({C}_{0.8})=0.488.Before deriving some first results concerning the range of the dependence measures ξ(A)\xi \left(A)and ζ1(A){\zeta }_{1}\left(A)for maximal d∞{d}_{\infty }-asymmetric copulas AA, we recall the characterizations of maximal d∞{d}_{\infty }-asymmetry derived in [21,15]: d∞(A,At){d}_{\infty }\left(A,{A}^{t})is maximal if and only if A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0and A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}or At23,13=0{A}^{t}\left(\frac{2}{3},\frac{1}{3}\right)=0and At13,23=13{A}^{t}\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}. Without loss of generality we may focus on the case A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}and A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0. Since AAis doubly stochastic in this case we obviously have μA0,13×13,23=μA13,23×23,1=μA23,1×0,13=13{\mu }_{A}\left(\left[0,\frac{1}{3}\right]\times \left[\frac{1}{3},\frac{2}{3}\right]\right)={\mu }_{A}\left(\left[\frac{1}{3},\frac{2}{3}\right]\times \left[\frac{2}{3},1\right]\right)={\mu }_{A}\left(\left[\frac{2}{3},1\right]\times \left[0,\frac{1}{3}\right]\right)=\frac{1}{3}. As a direct consequence, we can find copulas A1,A2,A3∈C{A}_{1},{A}_{2},{A}_{3}\in {\mathcal{C}}fulfilling (7)μA=13μA1f12+13μA2f23+13μA3f31,{\mu }_{A}=\frac{1}{3}{\mu }_{{A}_{1}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{A}_{2}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{A}_{3}}^{{f}_{31}},whereby the functions fij:[0,1]2→i−13,i3×j−13,j3{f}_{ij}:{\left[0,1]}^{2}\to \left[\frac{i-1}{3},\frac{i}{3}\right]\times \left[\frac{j-1}{3},\frac{j}{3}\right]are given by fij(x,y)=x+i−13,y+j−13{f}_{ij}\left(x,y)=\left(\frac{x+i-1}{3},\frac{y+j-1}{3}\right)for each (i,j)∈{1,2,3}2\left(i,j)\in {\left\{1,2,3\right\}}^{2}(and μAfij{\mu }_{A}^{{f}_{ij}}denotes the push-forward of μA{\mu }_{A}via fij{f}_{ij}).Theorem 3.4If A∈CA\in {\mathcal{C}}has maximal d∞{d}_{\infty }-asymmetry, i.e., if δ(A)=3d∞(A,At)=1\delta \left(A)=3{d}_{\infty }\left(A,{A}^{t})=1holds, then ξ\xi satisfies ξ(A)∈23,1\xi \left(A)\in \left[\frac{2}{3},1\right]. Moreover, for every s∈23,1s\in \left[\frac{2}{3},1\right]there exists a copula A with δ(A)=1\delta \left(A)=1fulfilling ξ(A)=s\xi \left(A)=s.ProofWe may assume that A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}and A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0. Then there exist copulas A1,A2,A3∈C{A}_{1},{A}_{2},{A}_{3}\in {\mathcal{C}}such that μA=13μA2f12+13μA3f23+13μA1f31{\mu }_{A}=\frac{1}{3}{\mu }_{{A}_{2}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{A}_{3}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{A}_{1}}^{{f}_{31}}holds. Defining h:[0,1]→[0,1]h:\left[0,1]\to \left[0,1]by h(x)=13+xifx∈0,23x−23ifx∈23,1,h\left(x)=\left\{\begin{array}{ll}\frac{1}{3}+x\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[0,\frac{2}{3}\right]\\ x-\frac{2}{3}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{2}{3},1\right],\end{array}\right.we have h∈Tbh\in {{\mathcal{T}}}_{b}and Sh(A)=i−13,i3,Aii∈{1,2,3}{{\mathcal{S}}}_{h}\left(A)={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Applying Lemmas 2.1 and 3.2 we therefore obtain ξ(A)=ξ(Sh(A))=6f(a1,…,an,b1,…,bn)+∑i=1319ξ(Ai)=23+∑i=1319ξ(Ai)≥23,\xi \left(A)=\xi \left({{\mathcal{S}}}_{h}\left(A))=6f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})+\mathop{\sum }\limits_{i=1}^{3}\frac{1}{9}\xi \left({A}_{i})=\frac{2}{3}+\mathop{\sum }\limits_{i=1}^{3}\frac{1}{9}\xi \left({A}_{i})\ge \frac{2}{3},with equality if and only if Ai=Π{A}_{i}=\Pi for every i=1,2,3i=1,2,3.Defining As{A}_{s}by μAs≔13μCsf12+13μCsf23+13μCsf31{\mu }_{{A}_{s}}:= \frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{31}}with Cs{C}_{s}as in Example 3.3 yields ξ(As)=ξ(Sh(As))=13ξ(Cs)+23.\xi \left({A}_{s})=\xi \left({{\mathcal{S}}}_{h}\left({A}_{s}))=\frac{1}{3}\xi \left({C}_{s})+\frac{2}{3}.Considering ξ(Cs)=1−s2∈[0,1]\xi \left({C}_{s})=1-{s}^{2}\in \left[0,1]for s∈[0,1]s\in \left[0,1]and using the same arguments as in Example 3.3 it follows that for every s0∈23,1{s}_{0}\in \left[\frac{2}{3},1\right]we find a copula A∈CA\in {\mathcal{C}}with ξ(A)=s0\xi \left(A)={s}_{0}and 3d∞(A,At)=δ(A)=13{d}_{\infty }\left(A,{A}^{t})=\delta \left(A)=1.□Since ζ1{\zeta }_{1}and ξ\xi are similar by construction, one might expect the analogous statements for ζ1{\zeta }_{1}. Note, however, that a different proof is needed since according to Lemma 3.1 the formulas for D1{D}_{1}are more involved.Theorem 3.5If A∈CA\in {\mathcal{C}}has maximal d∞{d}_{\infty }-asymmetry, then ζ1{\zeta }_{1}satisfies ζ1(A)∈56,1{\zeta }_{1}\left(A)\in \left[\frac{5}{6},1\right]. Furthermore, for every s∈56,1s\in \left[\frac{5}{6},1\right]there exists a copula AAwith δ(A)=1\delta \left(A)=1fulfilling ζ1(A)=s{\zeta }_{1}\left(A)=s.ProofProceeding as in the proof of Theorem 3.4 we obtain Sh(A)=i−13,i3,Aii∈{1,2,3}{{\mathcal{S}}}_{h}\left(A)={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Considering the (SI)-rearrangement Sh(A)↑{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }of Sh(A){{\mathcal{S}}}_{h}\left(A)it is clear that Sh(A)↑{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }is an ordinal sum again and can be expressed as Sh(A)↑=i−13,i3,Ai↑i∈{1,2,3}{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}^{\uparrow }\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Since every Ai↑{A}_{i}^{\uparrow }is SI and hence fulfills Ai↑(x,y)≥Π↑(x,y)=Π(x,y){A}_{i}^{\uparrow }\left(x,y)\ge {\Pi }^{\uparrow }\left(x,y)=\Pi \left(x,y)for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and every i∈{1,2,3}i\in \left\{1,2,3\right\}(see, e.g., [22][Section 5.2]), we obtain that Sh(A)↑≥CΠ≔i−13,i3,Πi∈{1,2,3}{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }\ge {C}_{\Pi }:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},\Pi \right\rangle \right)}_{i\in \left\{1,2,3\right\}}holds pointwise. Due to the fact that ζ1{\zeta }_{1}is monotone w.r.t. the pointwise order on C↑{{\mathcal{C}}}^{\uparrow }and ζ1{\zeta }_{1}is invariant w.r.t. to (SI)-rearrangements (see [26]), we obtain ζ1(A)=ζ1(Sh(A))=ζ1(Sh(A)↑)≥ζ1(CΠ)=56,{\zeta }_{1}\left(A)={\zeta }_{1}\left({{\mathcal{S}}}_{h}\left(A))={\zeta }_{1}\left({{\mathcal{S}}}_{h}{\left(A)}^{\uparrow })\ge {\zeta }_{1}\left({C}_{\Pi })=\frac{5}{6},where the last equality follows from Lemma 3.1 (the detailed calculations are deferred to Appendix A). To show the second assertion we can proceed analogously to the proof of Theorem 3.4 and use shrunk copies of the copula Cs{C}_{s}defined in Example 3.3 (see Appendix A).□While the minimum value of ξ\xi for a copula A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry is attained if and only if Ai=Π{A}_{i}=\Pi for every i=1,2,3i=1,2,3in equation (7), ζ1{\zeta }_{1}exhibits a different behavior as demonstrated in the following example:Example 3.6Let A1∈C↑{A}_{1}\in {{\mathcal{C}}}^{\uparrow }be defined by A1(x,y)=xy+12x(1−x)y(1−y).{A}_{1}\left(x,y)=xy+\frac{1}{2}x\left(1-x)y\left(1-y).Then a version of the corresponding Markov kernel of A1{A}_{1}is given by KA1(x,[0,y])=y+12(2x−1)y(y−1){K}_{{A}_{1}}\left(x,\left[0,y])=y+\frac{1}{2}\left(2x-1)y(y-1). Furthermore, we set A3=A1{A}_{3}={A}_{1}and A2=Π{A}_{2}=\Pi and let AAdenote the ordinal sum given by A≔i−13,i3,Aii∈{1,2,3}A:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}and CΠ{C}_{\Pi }be the ordinal sum given by CΠ≔i−13,i3,Πi∈{1,2,3}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},\Pi \right\rangle \right)}_{i\in \left\{1,2,3\right\}}(Figure 2).By construction we have A≠CΠA\ne {C}_{\Pi }, however, considering ∫[0,1]∫[0,1]12(2x−1)y(y−1)dλ(x)dλ(y)=0\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\frac{1}{2}\left(2x-1)y(y-1){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=0yields ∫[0,1]∫[0,1]KΠ(x,[0,y])−y3dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3+12(2x−1)y(y−1)dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3+12(2x−1)y(y−1)dλ(x)dλ(y)=∫[0,1]∫[0,1]KA1(x,[0,y])−y3dλ(x)dλ(y)\begin{array}{rcl}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{\Pi }\left(x,\left[0,y])-\frac{y}{3}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\frac{2y}{3}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left(\frac{2y}{3}+\frac{1}{2}\left(2x-1)y(y-1)\right){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|\frac{2y}{3}+\frac{1}{2}\left(2x-1)y(y-1)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{{A}_{1}}\left(x,\left[0,y])-\frac{y}{3}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\end{array}and analogously we obtain ∫[0,1]∫[0,1]KΠ(x,[0,y])−23+y3dλ(x)dλ(y)=∫[0,1]∫[0,1]KA1(x,[0,y])−23+y3dλ(x)dλ(y).\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|{K}_{\Pi }\left(x,\left[0,y])-\left(\frac{2}{3}+\frac{y}{3}\right)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|{K}_{{A}_{1}}\left(x,\left[0,y])-\left(\frac{2}{3}+\frac{y}{3}\right)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y).Applying Lemma 3.1 we obtain ζ1(A)=ζ1(CΠ)=56{\zeta }_{1}\left(A)={\zeta }_{1}\left({C}_{\Pi })=\frac{5}{6}.Figure 2Density of the SI copulas CΠ{C}_{\Pi }(left panel) and AA(right panel) considered in Example 3.6. Although A(x,y)≥CΠ(x,y)A\left(x,y)\ge {C}_{\Pi }\left(x,y)holds for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and there exists some (x,y)\left(x,y)with A(x,y)>CΠ(x,y)A\left(x,y)\gt {C}_{\Pi }\left(x,y)we have ζ1(A)=ζ1(CΠ){\zeta }_{1}\left(A)={\zeta }_{1}\left({C}_{\Pi }). On the contrary, ξ\xi fulfills ξ(A)>ξ(CΠ)\xi \left(A)\gt \xi \left({C}_{\Pi }).Remark 3.7Considering the monotonicity of ζ1{\zeta }_{1}with respect to the pointwise order on C↑{{\mathcal{C}}}^{\uparrow }as proved in [26], Example 3.6 shows that there exist copulas A,B∈C↑A,B\in {{\mathcal{C}}}^{\uparrow }with A≤BA\le Bpointwise and A(x,y)<B(x,y)A\left(x,y)\lt B\left(x,y)for some (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}fulfilling ζ1(A)=ζ1(B){\zeta }_{1}\left(A)={\zeta }_{1}\left(B).4Maximal D1{D}_{1}-asymmetry of copulas revisitedIn this section, we complement characterizations of copulas with maximal D1{D}_{1}-asymmetry going back to [13] and derive some topological properties of subclasses. To be consistent with the notation in [13], the family of copulas with maximal D1{D}_{1}-asymmetry is denoted by Cκ=1≔{A∈C:κ(A)≔2D1(A,At)=1}⊆C,{{\mathcal{C}}}^{\kappa =1}:= \{A\in {\mathcal{C}}:\kappa \left(A):= 2{D}_{1}\left(A,{A}^{t})=1\}\subseteq {\mathcal{C}},the subclass of mutually completely dependent copulas is denoted by Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}. We start with the family of mutually completely dependent copulas and show closedness w.r.t. the metric D∂{D}_{\partial }.Proposition 4.1The set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is closed in (C,D∂)\left({\mathcal{C}},{D}_{\partial }).ProofLet (Ahn)n∈N{({A}_{{h}_{n}})}_{n\in {\mathbb{N}}}be a sequence of mutually completely dependent copulas with D∂{D}_{\partial }-limit AA. Since according to [27] the family of completely dependent copulas is closed w.r.t. D1{D}_{1}we obtain A∈CcdA\in {{\mathcal{C}}}_{cd}and At∈Ccd{A}^{t}\in {{\mathcal{C}}}_{cd}. Using [27, Lemma 10] there exist λ\lambda -preserving transformations g,g′∈Tg,g^{\prime} \in {\mathcal{T}}such that a version of the Markov kernel KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot )and KAt(⋅,⋅){K}_{{A}^{t}}\left(\cdot ,\cdot )is given by KA(x,E)=1E(g(x)){K}_{A}\left(x,E)={{\mathbb{1}}}_{E}\left(g\left(x))and KAt(x,E)=1E(g′(x)){K}_{{A}^{t}}\left(x,E)={{\mathbb{1}}}_{E}\left(g^{\prime} \left(x)), respectively. Furthermore, since a copula AAis completely dependent if and only if it is left-invertible w.r.t. the ∗\ast -product (see [27]) we have M=At∗AM={A}^{t}\ast A. Applying Lemma 2.2 therefore yields that g∘g′(x)=id(x)g\circ g^{\prime} \left(x)=id\left(x)for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]. Using the fact that ggis surjective λ\lambda -almost everywhere, there exists a λ\lambda -preserving and bijective transformation h∈Tbh\in {{\mathcal{T}}}_{b}such that h=gh=gholds λ\lambda -a.e., implying A=Ah∈CmcdA={A}_{h}\in {{\mathcal{C}}}_{mcd}. It remains to show that D1(Ah,Aht)=12{D}_{1}\left({A}_{h},{A}_{h}^{t})=\frac{1}{2}, which can be done as follows. Using [13][Theorem 3.5] and the triangle inequality we obtain 12=∫[0,1]∣hn−hn−1∣dλ(x)≤∫[0,1]∣hn−h∣dλ(x)+∫[0,1]∣h−h−1∣dλ(x)+∫[0,1]∣h−1−hn−1∣dλ(x)\frac{1}{2}=\mathop{\int }\limits_{\left[0,1]}| {h}_{n}-{h}_{n}^{-1}| {\rm{d}}\lambda \left(x)\le \mathop{\int }\limits_{\left[0,1]}| {h}_{n}-h| {\rm{d}}\lambda \left(x)+\mathop{\int }\limits_{\left[0,1]}| h-{h}^{-1}| {\rm{d}}\lambda \left(x)+\mathop{\int }\limits_{\left[0,1]}| {h}^{-1}-{h}_{n}^{-1}| {\rm{d}}\lambda \left(x)for every n∈Nn\in {\mathbb{N}}. Applying [27][Proposition 15 (ii)] yields D1(Ah,Aht)=∫[0,1]∣h(x)−h−1(x)∣dλ(x)≥12−[D1(Ahn,Ah)+D1(Ahnt,Aht)]=12−D∂(Ahn,Ah).{D}_{1}\left({A}_{h},{A}_{h}^{t})=\mathop{\int }\limits_{\left[0,1]}| h\left(x)-{h}^{-1}\left(x)| {\rm{d}}\lambda \left(x)\ge \frac{1}{2}-{[}{D}_{1}\left({A}_{{h}_{n}},{A}_{h})+{D}_{1}\left({A}_{{h}_{n}}^{t},{A}_{h}^{t})]=\frac{1}{2}-{D}_{\partial }\left({A}_{{h}_{n}},{A}_{h}).Together with the fact that the maximal distance cannot exceed 12\frac{1}{2}it follows that D1(Ah,Aht)=12{D}_{1}\left({A}_{h},{A}_{h}^{t})=\frac{1}{2}, which completes the proof.□The following example shows that the set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is not closed w.r.t. the metric D1{D}_{1}.Example 4.2Let Ahn∈Cmcd{A}_{{h}_{n}}\in {{\mathcal{C}}}_{mcd}be the mutually completely dependent copula induced by the bijective measure-preserving transformation hn:[0,1]→[0,1]{h}_{n}:\left[0,1]\to \left[0,1], given by hn(x)=x+j−1nifx∈j−1n,jnandj∈1,…,n2x−1+jnifx∈j−1n,jnandj∈n2+1,…,n1ifx=1{h}_{n}\left(x)=\left\{\begin{array}{ll}x+\frac{j-1}{n}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[\frac{j-1}{n},\frac{j}{n}\right)\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}j\in \left\{1,\ldots ,\frac{n}{2}\right\}\\ x-1+\frac{j}{n}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[\frac{j-1}{n},\frac{j}{n}\right)\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}j\in \left\{\frac{n}{2}+1,\ldots ,n\right\}\\ 1\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x=1\end{array}\right.for all n∈2Nn\in 2{\mathbb{N}}and let Ah∈Ccd{A}_{h}\in {{\mathcal{C}}}_{cd}be the completely dependent copula induced by the λ\lambda -preserving transformation h:[0,1]→[0,1]h:\left[0,1]\to \left[0,1]given by h(x)≔2x(mod1)h\left(x):= 2x\left(mod1)(see Figure 1 in [9]). Setting Cn≔i−14,i4,Ah2ni∈{1,2,3,4}{C}_{n}:= {\left(\left\langle \frac{i-1}{4},\frac{i}{4},{A}_{{h}_{2n}}\right\rangle \right)}_{i\in \left\{1,2,3,4\right\}}and C≔i−14,i4,Ahi∈{1,2,3,4}C:= {\left(\left\langle \frac{i-1}{4},\frac{i}{4},{A}_{h}\right\rangle \right)}_{i\in \left\{1,2,3,4\right\}}we have Cn∈Cmcd{C}_{n}\in {{\mathcal{C}}}_{mcd}and C∈CcdC\in {{\mathcal{C}}}_{cd}and according to [9, Example 3.3] it is straightforward to verify that limn→∞D1(Cn,C)=0{\mathrm{lim}}_{n\to \infty }{D}_{1}\left({C}_{n},C)=0. As a next step, we reorder the shrunk copulas to obtain maximal D1{D}_{1}-asymmetry. Let ffdenote the λ\lambda -preserving interval exchange transformation f:[0,1]→[0,1]f:\left[0,1]\to \left[0,1]defined by f(x)≔(x−14)114,1(x)+(x+34)10,14(x)f\left(x):= \left(x-\frac{1}{4}){{\mathbb{1}}}_{\left(\tfrac{1}{4},1\right]}\left(x)+\left(x+\frac{3}{4}){{\mathbb{1}}}_{\left[0,\tfrac{1}{4}\right]}\left(x)and, furthermore, let Sf(Cn)∈Cmcd{{\mathcal{S}}}_{f}\left({C}_{n})\in {{\mathcal{C}}}_{mcd}and Sf(C)∈Ccd{{\mathcal{S}}}_{f}\left(C)\in {{\mathcal{C}}}_{cd}denote the respective shuffles (Figure 3). Due to the fact that the metric D1{D}_{1}is shuffle-invariant w.r.t. bijective transformations (using the same arguments as in the proof of Lemma 2.1) yields limn→∞D1(Sf(Cn),Sf(C))=limn→∞D1(Cn,C)=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({{\mathcal{S}}}_{f}\left({C}_{n}),{{\mathcal{S}}}_{f}\left(C))=\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({C}_{n},C)=0.Setting U≔0,14∪34,1U:= \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]and considering property (3) of Theorem 4.1 in [13] (see also Theorem 4.4(iii) in the sequel) we directly obtain that Sf(Cn){{\mathcal{S}}}_{f}\left({C}_{n})and Sf(C){{\mathcal{S}}}_{f}\left(C)are maximal asymmetric w.r.t. D1{D}_{1}, which shows that Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is not closed w.r.t. the metric D1{D}_{1}.Figure 3The support of the copulas μSf(Cn){\mu }_{{{\mathcal{S}}}_{f}\left({C}_{n})}(black) for n=4n=4(left panel) and n=8n=8(right panel) as well as the copula μSf(C){\mu }_{{{\mathcal{S}}}_{f}\left(C)}(magenta) as considered in Example 4.2.Leaving the subclass of mutually completely dependent copulas we will now derive novel and handy characterizations of copulas with maximal D1{D}_{1}-asymmetry and then show some topological properties. The following lemma, showing that the ∗\ast -product cannot increase the Dp{D}_{p}-distance, will be useful in the sequel. The result has already been stated for D1{D}_{1}in a slightly different context in [28].Lemma 4.3For every A,B,C∈CA,B,C\in {\mathcal{C}}, the following inequality holds for every p∈[1,∞)p\in {[}1,\infty ): Dpp(A∗B,A∗C)≤Dpp(B,C).{D}_{p}^{p}\left(A\ast B,A\ast C)\le {D}_{p}^{p}\left(B,C).ProofApplying Lemma 2.2, Jensen’s inequality, disintegration and using the fact that μA{\mu }_{A}is doubly stochastic we obtain Dpp(A∗B,A∗C)=∫[0,1]∫[0,1]∫[0,1]KB(t,[0,y])KA(x,dt)−∫[0,1]KC(t,[0,y])KA(x,dt)pdλ(x)dλ(y)≤∫[0,1]∫[0,1]∫[0,1]∣KB(t,[0,y])−KC(t,[0,y])∣pKA(x,dt)dλ(x)dλ(y)=∫[0,1]∫[0,1]2∣KB(t,[0,y])−KC(t,[0,y])∣pdμA(x,t)dλ(y)=∫[0,1]∫[0,1]∣KB(t,[0,y])−KC(t,[0,y])∣pdλ(t)dλ(y)=Dpp(B,C),\begin{array}{rcl}{D}_{p}^{p}\left(A\ast B,A\ast C)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{\left|\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{B}\left(t,\left[0,y]){K}_{A}\left(x,{\rm{d}}t)-\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}\left(t,\left[0,y]){K}_{A}\left(x,{\rm{d}}t)\right|}^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{K}_{A}\left(x,{\rm{d}}t){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{{\left[0,1]}^{2}}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{\rm{d}}{\mu }_{A}\left(x,t){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{\rm{d}}\lambda \left(t){\rm{d}}\lambda (y)={D}_{p}^{p}\left(B,C),\end{array}which completes the proof.□The next theorem gathers several equivalent characterizations of copulas having maximal D1{D}_{1}-asymmetry (see [13]), and the novel ones established here are (v) and (vi).Theorem 4.4For every A∈CA\in {\mathcal{C}}the following statements are equivalent: (i)κ(A)=1\kappa \left(A)=1,(ii)ΦA,At(12)=1{\Phi }_{A,{A}^{t}}\left(\frac{1}{2})=1(or equivalently, AAhas maximal D∞{D}_{\infty }-asymmetry),(iii)there exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with the following properties: λU∩0,12=λU∩12,1=14,μAU×0,12=12,μA0,12×U=0,\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right)=\lambda \left(U\cap \left[\frac{1}{2},1\right]\right)=\frac{1}{4},\hspace{1em}{\mu }_{A}\left(U\times \left[0,\frac{1}{2}\right]\right)=\frac{1}{2},\hspace{1em}{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=0,(iv)there exist sets U1,U2∈ℬ([0,1]){U}_{1},{U}_{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with U1⊆0,12{U}_{1}\subseteq \left[0,\frac{1}{2}\right], U2⊆12,1{U}_{2}\subseteq \left(\frac{1}{2},1\right], λ(U1)=λ(U2)=14\lambda \left({U}_{1})=\lambda \left({U}_{2})=\frac{1}{4}and V1≔0,12⧹U1{V}_{1}:= \left[0,\frac{1}{2}\right]\setminus {U}_{1}and V2≔12,1⧹U2{V}_{2}:= \left(\frac{1}{2},1\right]\setminus {U}_{2}, and copulas C1,C2,C3,C4∈C{C}_{1},{C}_{2},{C}_{3},{C}_{4}\in {\mathcal{C}}such that the following identityA(x,y)=14[C1(F1(x),G1(y))+C2(G1(x),G2(y))+C3(F2(x),F1(y))+C4(G2(x),F2(y))]A\left(x,y)=\frac{1}{4}{[}{C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))]holds, whereby Fi(x)≔4λ(Ui∩[0,x]){F}_{i}\left(x):= 4\lambda \left({U}_{i}\cap \left[0,x])and Gi(x)≔4λ(Vi∩[0,x]){G}_{i}\left(x):= 4\lambda \left({V}_{i}\cap \left[0,x])for i=1,2i=1,2.(v)(A∗A)12,12=0\left(A\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=0,(vi)D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}.ProofThe equivalences of (i), (ii), (iii), and (iv) have already been proved in [13]. Note that the equivalence in property (ii) directly follows from the facts that ΦA,At{\Phi }_{A,{A}^{t}}is Lipschitz continuous with Lipschitz constant 2 and the function ΦA,At:[0,1]→[0,1]{\Phi }_{A,{A}^{t}}:\left[0,1]\to \left[0,1]fulfills ΦA,At(y)≤min{2y,2(1−y)}{\Phi }_{A,{A}^{t}}(y)\le \min \left\{2y,2\left(1-y)\right\}for every y∈[0,1]y\in \left[0,1](see Lemma 5 in [27]). To show that (i) and (v) are equivalent we may proceed as follows: Suppose that κ(A)=1\kappa \left(A)=1. Then according to property (iii) there exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(U)=12\lambda \left(U)=\frac{1}{2}such that KAx,0,12=1{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=1for every x∈Ux\in U. Applying equation (2) and disintegration yields another Borel set V⊆UcV\subseteq {U}^{c}with λ(V)=12\lambda \left(V)=\frac{1}{2}and KAx,0,12=0{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0for every x∈Vx\in V. Setting V˜≔Uc⧹V\tilde{V}:= {U}^{c}\setminus V, then obviously λ(V˜)=0\lambda \left(\tilde{V})=0holds, and applying Lemma 2.2 we obtain μA∗A0,12×0,12=∫[0,12]∫[0,1]KA(s,0,12)KA(x,ds)dλ(x)=∫[0,12]∫U1KA(x,ds)+∫V0KA(x,ds)+∫V˜KA(s,0,12)KA(x,ds)dλ(x)≤∫[0,12]KA(x,U)dλ(x)+∫[0,12]KA(x,V˜)dλ(x)≤μA0,12×U+μA([0,1]×V˜)=0+λ(V˜)=0.\begin{array}{rcl}{\mu }_{A\ast A}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\left(\mathop{\displaystyle \int }\limits_{U}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{V}0\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{\tilde{V}}{K}_{A}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s)\right){\rm{d}}\lambda \left(x)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,\tilde{V}){\rm{d}}\lambda \left(x)\\ & \le & {\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)+{\mu }_{A}(\left[0,1]\times \tilde{V})=0+\lambda \left(\tilde{V})=0.\end{array}Suppose now that (A∗A)12,12=0\left(A\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=0holds. Then according to equation (5) we have ∫[0,1]KAtx,0,12KAx,0,12dλ(x)=0,\mathop{\int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)=0,so there exists a set Λ∈ℬ([0,1])\Lambda \in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(Λ)=1\lambda \left(\Lambda )=1such that KAtx,0,12KAx,0,12=0{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0holds for all x∈Λx\in \Lambda . Considering min{a,b}=12(a+b−∣a−b∣)\min \left\{a,b\right\}=\frac{1}{2}(a+b-| a-b| )therefore yields ΦA,At12=∫[0,1]∣KAx,0,12−KAtx,0,12∣dλ(x)=∫[0,1]KAx,0,12dλ(x)+∫[0,1]KAtx,0,12dλ(x)−2∫[0,1]min{KAx,0,12,KAtx,0,12}dλ(x)=12+12−2∫Λmin{KAx,0,12,KAtx,0,12}dλ(x)=12+12−2⋅0=1.\begin{array}{rcl}{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)-{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)| {\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)-2\mathop{\displaystyle \int }\limits_{\left[0,1]}\min \left\{{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right),{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)\right\}{\rm{d}}\lambda \left(x)\\ & =& \frac{1}{2}+\frac{1}{2}-2\mathop{\displaystyle \int }\limits_{\Lambda }\min \left\{{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right),{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)\right\}{\rm{d}}\lambda \left(x)=\frac{1}{2}+\frac{1}{2}-2\cdot 0=1.\end{array}To show the equivalence of (i) and (vi) first assume that D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}. Then applying Lemma 4.3 directly yields D1(A,At)≥12{D}_{1}\left(A,{A}^{t})\ge \frac{1}{2}, hence κ(A)=1\kappa \left(A)=1. On the other hand, if κ(A)=1\kappa \left(A)=1holds we may proceed as follows: There exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(U)=12\lambda \left(U)=\frac{1}{2}and KAx,0,12=1{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=1as well as KAtx,0,12=0{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)=0for every x∈Ux\in U. Using disintegration and equation (2) there exists a Borel set V⊆UcV\subseteq {U}^{c}with λ(V)=12\lambda \left(V)=\frac{1}{2}and KAx,0,12=0{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0and KAtx,0,12=1{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)=1for every x∈Vx\in V. As before, set V˜≔Uc⧹V\tilde{V}:= {U}^{c}\setminus V. Applying Lemma 2.2 yields A∗At12,12=μA∗At0,12×0,12=∫[0,12]∫[0,1]KAts,0,12KA(x,ds)dλ(x)≤∫[0,12]∫Uc1KA(x,ds)dλ(x)=∫[0,12]KA(x,Uc)dλ(x)=μA0,12×Uc=12−μA0,12×U=12\begin{array}{rcl}A\ast {A}^{t}\left(\frac{1}{2},\frac{1}{2}\right)& =& {\mu }_{A\ast {A}^{t}}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)=\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{{U}^{c}}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,{U}^{c}){\rm{d}}\lambda \left(x)\\ & =& {\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times {U}^{c}\right)=\frac{1}{2}-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\frac{1}{2}\end{array}as well as μA∗At0,12×0,12=∫[0,12]∫U0KA(x,ds)+∫V1KA(x,ds)+∫V˜KAts,0,12KA(x,ds)dλ(x)≥∫[0,12]KA(x,V)dλ(x)=μA0,12×V=12−μA0,12×Vc=12−μA0,12×U+μA0,12×V˜≥12.\begin{array}{rcl}{\mu }_{A\ast {A}^{t}}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\left(\mathop{\displaystyle \int }\limits_{U}0\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{V}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{\tilde{V}}{K}_{{A}^{t}}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s)\right){\rm{d}}\lambda \left(x)\\ & \ge & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,V){\rm{d}}\lambda \left(x)={\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times V\right)=\frac{1}{2}-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times {V}^{c}\right)\\ & =& \frac{1}{2}-\left({\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)+{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times \tilde{V}\right)\right)\ge \frac{1}{2}.\end{array}Together with property (v) there exist Borel sets Δ1⊆0,12{\Delta }_{1}\subseteq \left[0,\frac{1}{2}\right]and Δ2⊆12,1{\Delta }_{2}\subseteq \left(\frac{1}{2},1\right]with λ(Δ1)=λ(Δ2)=12\lambda \left({\Delta }_{1})=\lambda \left({\Delta }_{2})=\frac{1}{2}such that KA∗Ax1,0,12=0,KA∗Ax2,0,12=1andKA∗Atx1,0,12=1,KA∗Atx2,0,12=0{K}_{A\ast A}\left({x}_{1},\left[0,\frac{1}{2}\right]\right)=0,\hspace{1em}{K}_{A\ast A}\left({x}_{2},\left[0,\frac{1}{2}\right]\right)=1\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}{K}_{A\ast {A}^{t}}\left({x}_{1},\left[0,\frac{1}{2}\right]\right)=1,\hspace{1em}{K}_{A\ast {A}^{t}}\left({x}_{2},\left[0,\frac{1}{2}\right]\right)=0for every x1∈Δ1{x}_{1}\in {\Delta }_{1}and x2∈Δ2{x}_{2}\in {\Delta }_{2}, which gives ΦA∗A,A∗At12=∫[0,1]∣KA∗Ax,0,12−KA∗Atx,0,12∣dλ(x)=λ(Δ1)+λ(Δ2)=1.{\Phi }_{A\ast A,A\ast {A}^{t}}\left(\frac{1}{2}\right)=\mathop{\int }\limits_{\left[0,1]}| {K}_{A\ast A}\left(x,\left[0,\frac{1}{2}\right]\right)-{K}_{A\ast {A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)| {\rm{d}}\lambda \left(x)=\lambda \left({\Delta }_{1})+\lambda \left({\Delta }_{2})=1.Since y↦ΦA,B(y)y\mapsto {\Phi }_{A,B}(y)is Lipschitz continuous with Lipschitz constant 2 (see [27] [Lemma 5]), the property that D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}follows immediately and the proof is complete.□Remark 4.5For mutually completely dependent copulas Ah∈Cmcd{A}_{h}\in {{\mathcal{C}}}_{mcd}, property (vi) of Theorem 4.4 simplifies to κ(Ah)=1if and only ifD1(Ah∗Ah,M)=D1(Ah2,M)=‖h2−id‖1=12,\kappa \left({A}_{h})=1\hspace{1em}\hspace{0.1em}\text{if and only if}\hspace{0.1em}\hspace{0.33em}{D}_{1}\left({A}_{h}\ast {A}_{h},M)={D}_{1}\left({A}_{{h}^{2}},M)=\Vert {h}^{2}-id{\Vert }_{1}=\frac{1}{2},where the second equality of the right-hand side directly follows from [27] [Proposition 15 (ii)]. Furthermore, considering property (v) of Theorem 4.4 one might conjecture that (A∗At)12,12=12\left(A\ast {A}^{t})\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}also implies κ(A)=1\kappa \left(A)=1. For the symmetric copula MM, however, it is clear that M12,12=12M\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}as well as M∗Mt=M∗M=MM\ast {M}^{t}=M\ast M=Mholds.Not surprisingly, the following result holds.Proposition 4.6The set Cκ=1{{\mathcal{C}}}^{\kappa =1}is closed in (C,D∂)\left({\mathcal{C}},{D}_{\partial }).ProofLet (An)n∈N{({A}_{n})}_{n\in {\mathbb{N}}}be a sequence of maximal D1{D}_{1}-asymmetric copulas fulfilling limn→∞D∂(An,A)=0{\mathrm{lim}}_{n\to \infty }{D}_{\partial }\left({A}_{n},A)=0for some A∈CA\in {\mathcal{C}}. Then applying Theorem 4.4, the triangle inequality and the fact that D∂{D}_{\partial }-convergence implies both limn→∞ΦAn,A12=0{\mathrm{lim}}_{n\to \infty }{\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)=0and limn→∞ΦAnt,At12=0{\mathrm{lim}}_{n\to \infty }{\Phi }_{{A}_{n}^{t},{A}^{t}}\left(\frac{1}{2}\right)=0we obtain 1=ΦAn,Ant12≤ΦAn,A12+ΦA,At12+ΦAt,Ant121={\Phi }_{{A}_{n},{A}_{n}^{t}}\left(\frac{1}{2}\right)\le {\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)+{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)+{\Phi }_{{A}^{t},{A}_{n}^{t}}\left(\frac{1}{2}\right)and hence ΦA,At12≥1−limn→∞ΦAn,A12−limn→∞ΦAt,Ant12=1.□\hspace{11em}{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)\ge 1-\mathop{\mathrm{lim}}\limits_{n\to \infty }{\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)-\mathop{\mathrm{lim}}\limits_{n\to \infty }{\Phi }_{{A}^{t},{A}_{n}^{t}}\left(\frac{1}{2}\right)=1.\hspace{12em}\square Remark 4.7Proposition 4.6 certainly is not surprising, however, the following result is. Key for proving the statement is property (v) of Theorem 4.4.Theorem 4.8The set Cκ=1{{\mathcal{C}}}^{\kappa =1}is closed in (C,D1)\left({\mathcal{C}},{D}_{1}).ProofSuppose that A,A1,A2,…A,{A}_{1},{A}_{2},\ldots are copulas, that κ(An)=1\kappa \left({A}_{n})=1for every n∈Nn\in {\mathbb{N}}, and that limn→∞D1(An,A)=0{\mathrm{lim}}_{n\to \infty }{D}_{1}\left({A}_{n},A)=0. Since the ∗\ast -product is jointly continuous w.r.t. D1{D}_{1}(see [29]) we have limn→∞D1(An∗An,A∗A)=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({A}_{n}\ast {A}_{n},A\ast A)=0.Considering that D1{D}_{1}-convergence implies d∞{d}_{\infty }-convergence, limn→∞(An∗An)12,12=A∗A12,12{\mathrm{lim}}_{n\to \infty }\left({A}_{n}\ast {A}_{n})\left(\frac{1}{2},\frac{1}{2}\right)=A\ast A\left(\frac{1}{2},\frac{1}{2}\right)follows, and applying Theorem 4.4 the proof is complete.□Analogous to the fact that shuffles are dense in (C,d∞)\left({\mathcal{C}},{d}_{\infty })the set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is dense in (Cκ=1,d∞)\left({{\mathcal{C}}}^{\kappa =1},{d}_{\infty }).Theorem 4.9The set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is dense in (Cκ=1,d∞)\left({{\mathcal{C}}}^{\kappa =1},{d}_{\infty }).ProofFix ε>0\varepsilon \gt 0and let A∈Cκ=1A\in {{\mathcal{C}}}^{\kappa =1}be a copula with maximal D1{D}_{1}-asymmetry. According to property (iv) in Theorem 4.4 there exist sets U1,U2,V1,V2∈ℬ([0,1]){U}_{1},{U}_{2},{V}_{1},{V}_{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])and copulas C1,C2,C3,C4∈C{C}_{1},{C}_{2},{C}_{3},{C}_{4}\in {\mathcal{C}}such that A(x,y)=14[C1(F1(x),G1(y))+C2(G1(x),G2(y))+C3(F2(x),F1(y))+C4(G2(x),F2(y))],A\left(x,y)=\frac{1}{4}{[}{C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))],whereby Fi(x)≔4λ(Ui∩[0,x])=4∫[0,x]1Ui(s)λ(s){F}_{i}\left(x):= 4\lambda \left({U}_{i}\cap \left[0,x])=4{\int }_{\left[0,x]}{{\mathbb{1}}}_{{U}_{i}}\left(s)\lambda \left(s)and Gi(x)≔4λ(Vi∩[0,x])=4∫[0,x]1Vi(s)dλ(s){G}_{i}\left(x):= 4\lambda \left({V}_{i}\cap \left[0,x])=4{\int }_{\left[0,x]}{{\mathbb{1}}}_{{V}_{i}}\left(s){\rm{d}}\lambda \left(s)for i∈{1,2}i\in \left\{1,2\right\}. It is well-known that Cmcd{{\mathcal{C}}}_{mcd}(in fact even the family of straight shuffles) is dense in (C,d∞)\left({\mathcal{C}},{d}_{\infty })(see, e.g., [6][Corollary 4.1.16]), hence, we can find mutually completely dependent copulas Ch1,Ch2,Ch3,Ch4∈Cmcd{C}_{{h}_{1}},{C}_{{h}_{2}},{C}_{{h}_{3}},{C}_{{h}_{4}}\in {{\mathcal{C}}}_{mcd}with d∞(Ci,Chi)<ε{d}_{\infty }\left({C}_{i},{C}_{{h}_{i}})\lt \varepsilon for every i∈{1,2,3,4}i\in \left\{1,2,3,4\right\}. Defining A˜\tilde{A}by A˜(x,y)≔14[Ch1(F1(x),G1(y))+Ch2(G1(x),G2(y))+Ch3(F2(x),F1(y))+Ch4(G2(x),F2(y))],\tilde{A}\left(x,y):= \frac{1}{4}{[}{C}_{{h}_{1}}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{{h}_{2}}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{{h}_{3}}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{{h}_{4}}\left({G}_{2}\left(x),{F}_{2}(y))],and applying Theorem 4.4 we conclude that A˜\tilde{A}has maximal D1{D}_{1}-asymmetry too. Furthermore, using the triangle inequality we obtain supx,y∈[0,1]∣A(x,y)−A˜(x,y)∣≤14supx,y∈[0,1]∣C1(F1(x),G1(y))−Ch1(F1(x),G1(y))∣+14supx,y∈[0,1]∣C2(G1(x),G2(y))−Ch2(G1(x),G2(y))∣+14supx,y∈[0,1]∣C3(F2(x),F1(y))−Ch3(F2(x),F1(y))∣+14supx,y∈[0,1]∣C4(G2(x),F2(y))−Ch4(G2(x),F2(y))∣≤14∑i=14d∞(Ci,Chi)<ε4+ε4+ε4+ε4=ε.\begin{array}{rcl}\mathop{\sup }\limits_{x,y\in \left[0,1]}| A\left(x,y)-\tilde{A}\left(x,y)| & \le & \frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))-{C}_{{h}_{1}}\left({F}_{1}\left(x),{G}_{1}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))-{C}_{{h}_{2}}\left({G}_{1}\left(x),{G}_{2}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))-{C}_{{h}_{3}}\left({F}_{2}\left(x),{F}_{1}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))-{C}_{{h}_{4}}\left({G}_{2}\left(x),{F}_{2}(y))| \\ & \le & \frac{1}{4}\mathop{\displaystyle \sum }\limits_{i=1}^{4}{d}_{\infty }\left({C}_{i},{C}_{{h}_{i}})\lt \frac{\varepsilon }{4}+\frac{\varepsilon }{4}+\frac{\varepsilon }{4}+\frac{\varepsilon }{4}=\varepsilon .\end{array}As final step we have to show A˜∈Cmcd\tilde{A}\in {{\mathcal{C}}}_{mcd}, which can be done as follows: Fix y∈[0,1]y\in \left[0,1], then by applying Lemma 1 in [20] using the fact that KChi(x,[0,y]){K}_{{C}_{{h}_{i}}}\left(x,\left[0,y])is given by KChi(x,[0,y])=1[0,y](hi(x)){K}_{{C}_{{h}_{i}}}\left(x,\left[0,y])={{\mathbb{1}}}_{\left[0,y]}\left({h}_{i}\left(x))for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]and every i∈{1,…,4}i\in \left\{1,\ldots ,4\right\}, the Markov kernel KA˜(x,[0,y]){K}_{\tilde{A}}\left(x,\left[0,y])of A˜\tilde{A}can be expressed by KA˜(x,[0,y])=1[0,G1(y)](h1∘F1(x))forx∈U11[0,G2(y)](h2∘G1(x))forx∈V11[0,F1(y)](h3∘F2(x))forx∈U21[0,F2(y)](h4∘G2(x))forx∈V2{K}_{\tilde{A}}\left(x,\left[0,y])=\left\{\begin{array}{ll}{{\mathbb{1}}}_{\left[0,{G}_{1}(y)]}\hspace{0.33em}({h}_{1}\circ {F}_{1}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {U}_{1}\\ {{\mathbb{1}}}_{\left[0,{G}_{2}(y)]}\hspace{0.33em}({h}_{2}\circ {G}_{1}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {V}_{1}\\ {{\mathbb{1}}}_{\left[0,{F}_{1}(y)]}\hspace{0.33em}({h}_{3}\circ {F}_{2}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {U}_{2}\\ {{\mathbb{1}}}_{\left[0,{F}_{2}(y)]}\hspace{0.33em}({h}_{4}\circ {G}_{2}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {V}_{2}\end{array}\right.for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]. Since {U1,U2,V1,V2}\left\{{U}_{1},{U}_{2},{V}_{1},{V}_{2}\right\}form a partition of [0,1]\left[0,1], for each y∈[0,1]y\in \left[0,1]we have that KA˜(x,[0,y])∈{0,1}{K}_{\tilde{A}}\left(x,\left[0,y])\in \left\{0,1\right\}for λ\lambda -a.e. x∈[0,1]x\in \left[0,1], which is equivalent to A˜\tilde{A}being completely dependent (see [3]). Using the same arguments we also obtain for every y∈[0,1]y\in \left[0,1]that KA˜t(x,[0,y])=(∂1A˜t)(x,y)=(∂2A˜)(y,x)∈{0,1}{K}_{{\tilde{A}}^{t}}\left(x,\left[0,y])=\left({\partial }_{1}{\tilde{A}}^{t})\left(x,y)=\left({\partial }_{2}\tilde{A})(y,x)\in \left\{0,1\right\}for λ\lambda -a.e. x∈[0,1]x\in \left[0,1], i.e., A˜t{\tilde{A}}^{t}is completely dependent too. Altogether, we have shown that A˜∈Cmcd\tilde{A}\in {{\mathcal{C}}}_{mcd}, which completes the proof.□5Maximal Dp{D}_{p}-asymmetrySince the metrics Dp{D}_{p}, p∈[1,∞]p\in \left[1,\infty ]induce the same topology on C{\mathcal{C}}one could conjecture that maximal Dp{D}_{p}-asymmetry might be the same as maximal D1{D}_{1}-asymmetry. We will falsify this idea and start with three simple lemmata.Lemma 5.1[27] Suppose that h1,h2{h}_{1},{h}_{2}are λ\lambda -preserving transformations on [0,1]\left[0,1]and let Ah1{A}_{{h}_{1}}, Ah2{A}_{{h}_{2}}denote the corresponding completely dependent copulas. ThenDpp(Ah1,Ah2)=D1(Ah1,Ah2)=‖h1−h2‖1{D}_{p}^{p}\left({A}_{{h}_{1}},{A}_{{h}_{2}})={D}_{1}\left({A}_{{h}_{1}},{A}_{{h}_{2}})=\Vert {h}_{1}-{h}_{2}{\Vert }_{1}holds for every p∈(1,∞)p\in \left(1,\infty ).Lemma 5.2The metric space (C,Dp)\left({\mathcal{C}},{D}_{p})has the following diameter: (1)diamDp(C)=2−1p{{\rm{diam}}}_{{D}_{p}}\left({\mathcal{C}})={2}^{-\tfrac{1}{p}}for p∈[1,∞)p\in {[}1,\infty ).(2)diamD∞(C)=1{{\rm{diam}}}_{{D}_{\infty }}\left({\mathcal{C}})=1.ProofAccording to Lemma 5 in [27] we have diamD1(C)=∫[0,1]min{2y,2(1−y)}dλ(y)=12.{{\rm{diam}}}_{{D}_{1}}\left({\mathcal{C}})=\mathop{\int }\limits_{\left[0,1]}\min \left\{2y,2\left(1-y)\right\}{\rm{d}}\lambda (y)=\frac{1}{2}.Since ∣KA(x,[0,y])−KB(x,[0,y])∣∈[0,1]| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| \in \left[0,1]it is straightforward to verify that (8)Dpp(A,B)≤D1(A,B)≤Dp(A,B){D}_{p}^{p}\left(A,B)\le {D}_{1}\left(A,B)\le {D}_{p}\left(A,B)holds for every A,B∈CA,B\in {\mathcal{C}}and p∈[1,∞)p\in {[}1,\infty ). As a direct consequence, we obtain Dp(A,B)≤D1(A,B)1p≤2−1p{D}_{p}\left(A,B)\le {D}_{1}{\left(A,B)}^{\tfrac{1}{p}}\le {2}^{-\tfrac{1}{p}}. On the other hand, there exist copulas A,B∈CA,B\in {\mathcal{C}}with Dp(A,B)=2−1p{D}_{p}\left(A,B)={2}^{-\tfrac{1}{p}}. Considering A=MA=Mand B=WB=Wand applying Lemma 5.1 yield Dpp(M,W)=D1(M,W)=12.{D}_{p}^{p}\left(M,W)={D}_{1}\left(M,W)=\frac{1}{2}.The assertion for p=∞p=\infty is a direct consequence of Lemma 5 in [27].□Slightly adapting the notation of the previous section we will now focus on the family Cκp=1{{\mathcal{C}}}^{{\kappa }_{p}=1}of all bivariate copulas with maximal Dp{D}_{p}-asymmetry, i.e., Cκp=1≔{A∈C:κp(A)≔21pDp(A,At)=1}{{\mathcal{C}}}^{{\kappa }_{p}=1}:= \left\{A\in {\mathcal{C}}:{\kappa }_{p}\left(A):= {2}^{\tfrac{1}{p}}{D}_{p}\left(A,{A}^{t})=1\right\}. Building upon Lemma 5.1 and Theorem 3.5 in [13] there are mutually completely dependent copulas A∈CmcdA\in {{\mathcal{C}}}_{mcd}such that κp(A)=1{\kappa }_{p}\left(A)=1is attained for every p∈[1,∞]p\in \left[1,\infty ]. In fact, the copula Ah{A}_{h}defined in Example 3.4 in [13] has maximal Dp{D}_{p}-asymmetry for every p∈[1,∞]p\in \left[1,\infty ]. The following lemma shows that a copula with maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty )also has maximal D1{D}_{1}-asymmetry.Lemma 5.3If A∈CA\in {\mathcal{C}}satisfies κp(A)=1{\kappa }_{p}\left(A)=1for some p∈(1,∞)p\in \left(1,\infty ), then κ1(A)=1{\kappa }_{1}\left(A)=1holds.ProofUsing the inequality Dpp(A,B)≤D1(A,B){D}_{p}^{p}\left(A,B)\le {D}_{1}\left(A,B)as well as the fact that D1(A,B)≤12{D}_{1}\left(A,B)\le \frac{1}{2}holds for all A,B∈CA,B\in {\mathcal{C}}we obtain 1=21pDp(A,At)≤(2D1(A,At))1p≤1,1={2}^{\tfrac{1}{p}}{D}_{p}\left(A,{A}^{t})\le {(2{D}_{1}\left(A,{A}^{t}))}^{\tfrac{1}{p}}\le 1,which yields D1(A,At)=12{D}_{1}\left(A,{A}^{t})=\frac{1}{2}.□The following example, however, shows that the reverse implication does not hold in general.Example 5.4Suppose that A∈CA\in {\mathcal{C}}corresponds to the uniform distribution on the union of the four squares (Figure 4) 0,14×14,24,14,24×24,34,24,34×34,1,34,1×0,14.\left(0,\frac{1}{4}\right)\times \left(\frac{1}{4},\frac{2}{4}\right),\left(\frac{1}{4},\frac{2}{4}\right)\times \left(\frac{2}{4},\frac{3}{4}\right),\left(\frac{2}{4},\frac{3}{4}\right)\times \left(\frac{3}{4},1\right),\left(\frac{3}{4},1\right)\times \left(0,\frac{1}{4}\right).Since AA(and At{A}^{t}) is a checkerboard copula (see, for instance, [11,18]) a version of the Markov kernel of AAis piecewise linear in yyfor fixed x∈[0,1]x\in \left[0,1]and does not depend on the choice of the point x∈i−14,i4x\in \left(\frac{i-1}{4},\frac{i}{4}\right)for every i∈{1,…,4}i\in \left\{1,\ldots ,4\right\}, (a version of) the corresponding Markov kernel is given by KA(x,[0,y])=(4y−1)114,24(y)+124,1(y)forx∈0,14(4y−2)124,34(y)+134,1(y)forx∈14,24(4y−3)134,1(y)forx∈24,34(4y)10,14(y)+114,1(y)forx∈34,1.{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}\left(4y-1){{\mathbb{1}}}_{\left(\tfrac{1}{4},\tfrac{2}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{2}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(0,\frac{1}{4}\right)\\ \left(4y-2){{\mathbb{1}}}_{\left(\tfrac{2}{4},\tfrac{3}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{3}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{1}{4},\frac{2}{4}\right)\\ \left(4y-3){{\mathbb{1}}}_{\left(\tfrac{3}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{2}{4},\frac{3}{4}\right)\\ \left(4y){{\mathbb{1}}}_{\left[0,\tfrac{1}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{1}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{3}{4},1\right).\end{array}\right.It is straightforward to verify κ1(A)=1{\kappa }_{1}\left(A)=1(e.g., by using property (iv) or property (v) in Theorem 4.4). On the other hand, simple calculations (see Appendix A) yield Dpp(A,At)=14+2∫[0,14](4x)pdλ(x)=14+24p+4{D}_{p}^{p}\left(A,{A}^{t})=\frac{1}{4}+2\mathop{\int }\limits_{\left[0,\frac{1}{4}]}{\left(4x)}^{p}{\rm{d}}\lambda \left(x)=\frac{1}{4}+\frac{2}{4p+4}for every p∈[1,∞)p\in {[}1,\infty ). As a direct consequence we obtain Dpp(A,At)<2−1{D}_{p}^{p}\left(A,{A}^{t})\lt {2}^{-1}for every p∈(1,∞)p\in \left(1,\infty ), i.e., although AAhas maximal D1{D}_{1}-asymmetry, it fails to have maximal Dp{D}_{p}-asymmetry.Figure 4Density of the copula AA(left panel) and the copula At{A}^{t}(right panel) considered in Example 5.4. The copula AAhas maximal D1{D}_{1}-asymmetry, i.e., κ1(A)=1{\kappa }_{1}\left(A)=1, nevertheless κp(A)<1{\kappa }_{p}\left(A)\lt 1holds for p∈(1,∞)p\in \left(1,\infty ).Contrary to D1{D}_{1}, the class Cκp=1{{\mathcal{C}}}^{{\kappa }_{p}=1}, p∈(1,∞)p\in \left(1,\infty )only contains mutually completely dependent copulas.Theorem 5.5If A∈CA\in {\mathcal{C}}has maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty ), then A is a mutually completely dependent copula.ProofIf κp(A)=1{\kappa }_{p}\left(A)=1we have κ1(A)=1{\kappa }_{1}\left(A)=1and Dp(A,At)=2−1p{D}_{p}\left(A,{A}^{t})={2}^{-\tfrac{1}{p}}, which implies 12=∫[0,1]∫[0,1]∣KA(x,[0,y])−KAt(x,[0,y])∣pdλ(x)dλ(y)≤∫[0,1]∫[0,1]∣KA(x,[0,y])−KAt(x,[0,y])∣dλ(x)dλ(y)=12.\begin{array}{rcl}\frac{1}{2}& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=\frac{1}{2}.\end{array}Therefore, we obtain ∣KA(x,[0,y])−KAt(x,[0,y])∣p=∣KA(x,[0,y])−KAt(x,[0,y])∣,{| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| }^{p}=| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| ,or equivalently, that (9)∣KA(x,[0,y])−KAt(x,[0,y])∣∈{0,1}| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| \in \left\{0,1\right\}holds for λ2{\lambda }^{2}-a.e. (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}. According to Lemma 5.3 and property (iii) in Theorem 4.4 there exist sets U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])and V∈ℬ([0,1])V\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])such that U∩V=∅U\cap V=\varnothing , λ(U)=λ(V)=12\lambda \left(U)=\lambda \left(V)=\frac{1}{2}, and KA(x,[0,y])=≤1for everyy∈0,121for everyy∈12,1KAt(x,[0,y])=0for everyy∈0,12≤1for everyy∈12,1,{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}\le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.\hspace{1.0em}{K}_{{A}^{t}}\left(x,\left[0,y])=\left\{\begin{array}{ll}0\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ \le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.,for every x∈Ux\in Uas well as KA(x,[0,y])=0for everyy∈0,12≤1for everyy∈12,1KAt(x,[0,y])=≤1for everyy∈0,121for everyy∈12,1,{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}0\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ \le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.\hspace{1.0em}{K}_{{A}^{t}}\left(x,\left[0,y])=\left\{\begin{array}{ll}\le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.,for every x∈Vx\in V. Fix x∈Ux\in Usuch that equation (9) holds and suppose that KA(x,[0,y])=y0∈(0,1){K}_{A}\left(x,\left[0,y])={y}_{0}\in \left(0,1)for some y∈0,12y\in \left[0,\frac{1}{2}\right]. Then due to equation (9) the Markov kernel of At{A}^{t}must satisfy KAt(x,[0,y])=y0{K}_{{A}^{t}}\left(x,\left[0,y])={y}_{0}, which is a contradiction to the fact that KAt(x,[0,y])=0{K}_{{A}^{t}}\left(x,\left[0,y])=0for every y∈0,12y\in \left[0,\frac{1}{2}\right]. Hence, we obtain that KA(x,[0,y])∈{0,1}{K}_{A}\left(x,\left[0,y])\in \left\{0,1\right\}for every y∈[0,1]y\in \left[0,1]. In an analogous way, we obtain that KAt(x,[0,y])∈{0,1}{K}_{{A}^{t}}\left(x,\left[0,y])\in \left\{0,1\right\}holds for every y∈[0,1]y\in \left[0,1]. Proceeding in the exactly same manner for x∈Vx\in Vwe obtain that for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]and every y∈[0,1]y\in \left[0,1]the Markov kernels of AAand At{A}^{t}satisfy KA(x,[0,y])∈{0,1}{K}_{A}\left(x,\left[0,y])\in \left\{0,1\right\}and KAt(x,[0,y])∈{0,1}{K}_{{A}^{t}}\left(x,\left[0,y])\in \left\{0,1\right\}. By Theorem 7.1 in [3] and Lemma 3.4 in [10], it follows that AAand At{A}^{t}are completely dependent, implying that AAis mutually completely dependent.□Altogether we have shown the following results:Corollary 5.6The following properties hold: (1)Cκ1=1=Cκ∞=1{{\mathcal{C}}}^{{\kappa }_{1}=1}={{\mathcal{C}}}^{{\kappa }_{\infty }=1}.(2)Cκ1=1⊋Cκp=1{{\mathcal{C}}}^{{\kappa }_{1}=1}\hspace{0.33em} \supsetneq \hspace{0.33em}{{\mathcal{C}}}^{{\kappa }_{p}=1}for every p∈(1,∞)p\in \left(1,\infty ).(3)Cmcdκ1=1=Cκp=1{{\mathcal{C}}}_{mcd}^{{\kappa }_{1}=1}={{\mathcal{C}}}^{{\kappa }_{p}=1}for every p∈(1,∞)p\in \left(1,\infty ).6Maximal Dp{D}_{p}-asymmetric copulas and their values for ζ1{\zeta }_{1}and ξ\xi In Section 3, we have shown that copulas A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry have very high dependence scores with respect to ζ1{\zeta }_{1}and ξ\xi . Here we now focus on the range of these dependence measures to maximal Dp{D}_{p}-asymmetric copulas.Theorem 6.1If A∈CA\in {\mathcal{C}}satisfies κp(A)=1{\kappa }_{p}\left(A)=1for some p∈(1,∞)p\in \left(1,\infty ), then ζ1(A)=ξ(A)=1{\zeta }_{1}\left(A)=\xi \left(A)=1.ProofSince ζ1(A){\zeta }_{1}\left(A)and ξ(A)\xi \left(A)are 1 if and only if AAis completely dependent, the assertion directly follows from Theorem 5.5.□For the case p=1p=1different values for ξ\xi and ζ1{\zeta }_{1}are possible.Theorem 6.2If A∈CA\in {\mathcal{C}}satisfies κ1(A)=1{\kappa }_{1}\left(A)=1, then ξ(A)∈12,1\xi \left(A)\in \left(\frac{1}{2},1\right]holds.ProofProceeding analogously to the proof of Theorem 4.4 we obtain (At∗A)12,12=12\left({A}^{t}\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}if κ1(A)=1{\kappa }_{1}\left(A)=1(see Appendix A). Since At∗A{A}^{t}\ast Ais a copula, we find copulas A1,A2∈C{A}_{1},{A}_{2}\in {\mathcal{C}}with (At∗A)=i−12,i2,Aii∈{1,2}≕A˜\left({A}^{t}\ast A)={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2\right\}}\hspace{0.33em}=: \hspace{0.33em}\tilde{A}. Setting CΠ≔i−12,i2,Πi∈{1,2}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{2},\frac{i}{2},\Pi \right\rangle \right)}_{i\in \left\{1,2\right\}}, μAt∗A≠μCΠ{\mu }_{{A}^{t}\ast A}\ne {\mu }_{{C}_{\Pi }}follows. In fact, according to Theorem 4.4 there exists a set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])such that λU∩0,12=λU∩12,1=14\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right)=\lambda \left(U\cap \left[\frac{1}{2},1\right]\right)=\frac{1}{4}and 0=μA0,12×U=∫0,12KA(x,U)dλ(x)0={\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\mathop{\int }\limits_{\left[0,\frac{1}{2}\right]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x). Hence, we can find Borel sets Λ1⊆0,12{\Lambda }_{1}\subseteq \left[0,\frac{1}{2}\right], Λ2⊆12,1{\Lambda }_{2}\subseteq \left(\frac{1}{2},1\right]such that λ(Λ1)=λ(Λ2)=12\lambda \left({\Lambda }_{1})=\lambda \left({\Lambda }_{2})=\frac{1}{2}and KA(x,U)=0{K}_{A}\left(x,U)=0for all x∈Λ1x\in {\Lambda }_{1}and KA(x,U)=1{K}_{A}\left(x,U)=1for all x∈Λ2x\in {\Lambda }_{2}. The set Λ3{\Lambda }_{3}defined by [0,1]⧹(Λ1∪Λ2)\left[0,1]\setminus \left({\Lambda }_{1}\cup {\Lambda }_{2})obviously fulfills λ(Λ3)=0\lambda \left({\Lambda }_{3})=0. Hence, we have μAt∗A(U×U)=∫U∫[0,1]KA(s,U)KAt(x,ds)dλ(x)=∫UKAt(x,Λ2)dλ(x)+∫U∫Λ3KA(s,U)KAt(x,ds)dλ(x)≥∫UKAt(x,Λ2)dλ(x)=μAt(U×Λ2)=μA(Λ2×U)=∫Λ2KA(x,U)dλ(x)=∫12,1KA(x,U)dλ(x)=μA12,1×U=λ(U)−μA0,12×U=λ(U)=12.\begin{array}{rcl}{\mu }_{{A}^{t}\ast A}\left(U\times U)& =& \mathop{\displaystyle \int }\limits_{U}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(s,U){K}_{{A}^{t}}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{U}{K}_{{A}^{t}}\left(x,{\Lambda }_{2}){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U}\mathop{\displaystyle \int }\limits_{{\Lambda }_{3}}{K}_{A}\left(s,U){K}_{{A}^{t}}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & \ge & \mathop{\displaystyle \int }\limits_{U}{K}_{{A}^{t}}\left(x,{\Lambda }_{2}){\rm{d}}\lambda \left(x)={\mu }_{{A}^{t}}\left(U\times {\Lambda }_{2})={\mu }_{A}\left({\Lambda }_{2}\times U)=\mathop{\displaystyle \int }\limits_{{\Lambda }_{2}}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left(\frac{1}{2},1\right]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)={\mu }_{A}\left(\left(\frac{1}{2},1\right]\times U\right)=\lambda \left(U)-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\lambda \left(U)=\frac{1}{2}.\end{array}On the other hand, μCΠ(U×U)=∫UKCΠ(x,U)dλ(x)=∫U∩[0,12]KCΠ(x,U)dλ(x)+∫U∩(12,1]KCΠ(x,U)dλ(x)=∫U∩[0,12]2λU∩0,12dλ(x)+∫U∩(12,1]2λ(U∩12,1)dλ(x)=14\begin{array}{rcl}{\mu }_{{C}_{\Pi }}\left(U\times U)& =& \mathop{\displaystyle \int }\limits_{U}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{U\cap \left[0,\frac{1}{2}]}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U\cap \left(\frac{1}{2},1]}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{U\cap \left[0,\frac{1}{2}]}2\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U\cap \left(\frac{1}{2},1]}2\lambda \left(U\cap \left(\frac{1}{2},1\right]){\rm{d}}\lambda \left(x)=\frac{1}{4}\end{array}holds, implying Ai≠Π{A}_{i}\ne \Pi for i=1,2i=1,2, hence, considering that according to Lemma 4.3 we have D22(A,Π)≥D22(At∗A,At∗Π)=D22(At∗A,Π){D}_{2}^{2}\left(A,\Pi )\ge {D}_{2}^{2}\left({A}^{t}\ast A,{A}^{t}\ast \Pi )={D}_{2}^{2}\left({A}^{t}\ast A,\Pi )and applying Corollary 3.2 finally yields 1≥ξ(A)≥ξ(A˜)=14ξ(A1)+14ξ(A2)+12>12.□\hspace{10.25em}1\ge \xi \left(A)\ge \xi \left(\tilde{A})=\frac{1}{4}\xi \left({A}_{1})+\frac{1}{4}\xi \left({A}_{2})+\frac{1}{2}\gt \frac{1}{2}.\hspace{15em}\square Theorem 6.3If A∈CA\in {\mathcal{C}}satisfies κ1(A)=1{\kappa }_{1}\left(A)=1, then ζ1(A)∈34,1{\zeta }_{1}\left(A)\in \left[\frac{3}{4},1\right]holds.ProofUsing the same arguments as in the proof of Theorem 6.2 we find copulas A1,A2∈C{A}_{1},{A}_{2}\in {\mathcal{C}}such that (At∗A)=i−12,i2,Aii∈{1,2}≕A˜\left({A}^{t}\ast A)={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2\right\}}\hspace{0.33em}=: \hspace{0.33em}\tilde{A}holds. Since A˜\tilde{A}is an ordinal sum it is clear that the (SI)-rearrangement A˜↑{\tilde{A}}^{\uparrow }of A˜\tilde{A}satisfies A˜↑=i−12,i2,Ai↑i∈{1,2}{\tilde{A}}^{\uparrow }={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}^{\uparrow }\right\rangle \right)}_{i\in \left\{1,2\right\}}. As SI copula, Ai↑{A}_{i}^{\uparrow }fulfills Ai↑(x,y)≥Π(x,y){A}_{i}^{\uparrow }\left(x,y)\ge \Pi \left(x,y)for all (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and i∈{1,2}i\in \left\{1,2\right\}, implying A˜↑(x,y)≥CΠ(x,y){\tilde{A}}^{\uparrow }\left(x,y)\ge {C}_{\Pi }\left(x,y)for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}, whereby CΠ{C}_{\Pi }is defined as CΠ≔i−12,i2,Πi∈{1,2}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{2},\frac{i}{2},\Pi \right\rangle \right)}_{i\in \left\{1,2\right\}}. Hence, using Lemma 4.3 we obtain 13≥D1(A,Π)≥D1(At∗A,At∗Π)=D1(At∗A,Π)=D1(A˜↑,Π)≥D1(CΠ,Π),\frac{1}{3}\ge {D}_{1}\left(A,\Pi )\ge {D}_{1}\left({A}^{t}\ast A,{A}^{t}\ast \Pi )={D}_{1}\left({A}^{t}\ast A,\Pi )={D}_{1}\left({\tilde{A}}^{\uparrow },\Pi )\ge {D}_{1}\left({C}_{\Pi },\Pi ),whereby we used the fact that D1(A,Π){D}_{1}\left(A,\Pi )is monotone with respect to the pointwise order in C↑{{\mathcal{C}}}^{\uparrow }(see [26]). Using Lemma 3.1 we obtain ζ1(CΠ)=3D1(CΠ,Π)=34∫[0,1]∫[0,1]y−y2dλ(x)dλ(y)+34∫[0,1]∫[0,1]y−12−y2dλ(x)dλ(y)+38=316+38−316+38=68=34,{\zeta }_{1}\left({C}_{\Pi })=3{D}_{1}\left({C}_{\Pi },\Pi )=\frac{3}{4}\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|y-\frac{y}{2}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\frac{3}{4}\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|y-\frac{1}{2}-\frac{y}{2}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\frac{3}{8}=\frac{3}{16}+\frac{3}{8}-\frac{3}{16}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4},which completes the proof.□The following example demonstrates that it is possible to find copulas A∈Cκ1=1A\in {{\mathcal{C}}}^{{\kappa }_{1}=1}such that ζ1(A){\zeta }_{1}\left(A)(or ξ(A)\xi \left(A), respectively) is arbitrarily close to the lower bound derived in Theorems 6.2 and 6.3.Example 6.4Let n∈Nn\in {\mathbb{N}}be a natural number with n≥3n\ge 3, set N≔2nN:= {2}^{n}and define the sets UN1,UN2,VN1,VN2∈ℬ([0,1]){U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])by UN1=⋃j=1N42j−2N,2j−1N,VN1=⋃j=1N42j−1N,2jN,UN2=⋃j=1N412+2j−2N,12+2j−1N,VN2=⋃j=1N412+2j−2N,12+2j−1N.\begin{array}{rcl}{U}_{N}^{1}& =& \mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{2j-2}{N},\frac{2j-1}{N}\right),\hspace{1em}{V}_{N}^{1}=\mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{2j-1}{N},\frac{2j}{N}\right),\\ {U}_{N}^{2}& =& \mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{1}{2}+\frac{2j-2}{N},\frac{1}{2}+\frac{2j-1}{N}\right),{V}_{N}^{2}=\mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{1}{2}+\frac{2j-2}{N},\frac{1}{2}+\frac{2j-1}{N}\right).\end{array}Obviously, we have λ(UN1)=λ(UN2)=λ(VN1)=λ(VN2)=14\lambda \left({U}_{N}^{1})=\lambda \left({U}_{N}^{2})=\lambda \left({V}_{N}^{1})=\lambda \left({V}_{N}^{2})=\frac{1}{4}and λ(UN1∪UN2∪VN1∪VN2)=1\lambda \left({U}_{N}^{1}\cup {U}_{N}^{2}\cup {V}_{N}^{1}\cup {V}_{N}^{2})=1. Letting AN{A}_{N}denote the copula corresponding to the uniform distribution on the union of the four sets UN1×VN1{U}_{N}^{1}\times {V}_{N}^{1}, UN2×UN1{U}_{N}^{2}\times {U}_{N}^{1}, VN1×VN2{V}_{N}^{1}\times {V}_{N}^{2}and VN2×UN2{V}_{N}^{2}\times {U}_{N}^{2}(Figure 5), then by Theorem 4.4 AN{A}_{N}has maximal D1{D}_{1}-asymmetry. As next step we calculate the dependence measure ζ1(AN){\zeta }_{1}\left({A}_{N}).Applying Lemma 6.3 in [10] and using the fact that for every y∈[0,1]y\in \left[0,1]the identity of KAN(x1,[0,y])=KAN(x2,[0,y]){K}_{{A}_{N}}\left({x}_{1},\left[0,y])={K}_{{A}_{N}}\left({x}_{2},\left[0,y])holds for λ\lambda -a.e. x1,x2∈X{x}_{1},{x}_{2}\in X, whereby X∈{UN1,UN2,VN1,VN2}X\in \left\{{U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\right\}, we obtain ζ1(AN)3=D1(AN,Π)≤2N+1N∑j=1N∫[0,1]KANx,0,jN−jNdλ(x)=2N+∑X∈{UN1,UN2,VN1,VN2}1N∑j=1N∫XKANx,0,jN−jNdλ(x)︸≕m(X).\begin{array}{rcl}\frac{{\zeta }_{1}\left({A}_{N})}{3}& =& {D}_{1}\left({A}_{N},\Pi )\le \frac{2}{N}+\frac{1}{N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|{\rm{d}}\lambda \left(x)\\ & =& \frac{2}{N}+\displaystyle \sum _{X\in \left\{{U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\right\}}\mathop{\underbrace{\frac{1}{N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\mathop{\displaystyle \int }\limits_{X}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|{\rm{d}}\lambda \left(x)}}\limits_{=: m\left(X)}.\end{array}Considering X=UN1X={U}_{N}^{1}and x∈UN1x\in {U}_{N}^{1}, a version of the Markov kernel KAnx,0,jN{K}_{{A}_{n}}\left(x,\left[0,\frac{j}{N}\right]\right)is given by KANx,0,jN=2jNforj∈2,4,…,N22(j−1)Nforj∈1,3,…,N2−11forj>N2,{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)=\left\{\begin{array}{ll}\frac{2j}{N}\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\in \left\{2,4,\ldots ,\frac{N}{2}\right\}\\ \frac{2\left(j-1)}{N}\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\in \left\{1,3,\ldots ,\frac{N}{2}-1\right\}\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\gt \frac{N}{2},\end{array}\right.which yields m(UN1)=14N∑j=1NKANx,0,jN−jN=14N∑j∈{2,4,…,N∕2}2jN−jN+∑j∈{1,3,…,N∕2−1}2(j−1)N−jN+∑j=N2+1N1−jN=14N∑j=1N44jN−2jN+∑j=1N44(j−1)N−2j−1N+∑j=1N212−jN=14NN4+2N−12≤116+12N2.\begin{array}{rcl}m\left({U}_{N}^{1})& =& \frac{1}{4N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|\\ & =& \frac{1}{4N}\left(\displaystyle \sum _{j\in \left\{2,4,\ldots ,N/2\right\}}\left|\frac{2j}{N}-\frac{j}{N}\right|+\displaystyle \sum _{j\in \left\{1,3,\ldots ,N/2-1\right\}}\left|\frac{2\left(j-1)}{N}-\frac{j}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=\frac{N}{2}+1}^{N}\left(1-\frac{j}{N}\right)\right)\\ & =& \frac{1}{4N}\left(\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{4}}\left|\frac{4j}{N}-\frac{2j}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{4}}\left|\frac{4\left(j-1)}{N}-\frac{2j-1}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{2}}\left(\frac{1}{2}-\frac{j}{N}\right)\right)\\ & =& \frac{1}{4N}\left(\frac{N}{4}+\frac{2}{N}-\frac{1}{2}\right)\le \frac{1}{16}+\frac{1}{2{N}^{2}}.\end{array}In a similar manner, we obtain m(UN2)=116+18Nm\left({U}_{N}^{2})=\frac{1}{16}+\frac{1}{8N}, m(VN1)=116+18Nm\left({V}_{N}^{1})=\frac{1}{16}+\frac{1}{8N}, and m(VN2)≤116+12N2m\left({V}_{N}^{2})\le \frac{1}{16}+\frac{1}{2{N}^{2}}. Together with Theorem 6.3 it follows that 34≤ζ1(AN)≤34+274N+3N2,\frac{3}{4}\le {\zeta }_{1}\left({A}_{N})\le \frac{3}{4}+\frac{27}{4N}+\frac{3}{{N}^{2}},which shows that for sufficiently large N∈NN\in {\mathbb{N}}the dependence value ζ1(AN){\zeta }_{1}\left({A}_{N})is arbitrarily close to 34\frac{3}{4}.Using similar calculations (see Appendix A) yields 12<ξ(AN)≤12+aN,\frac{1}{2}\lt \xi \left({A}_{N})\le \frac{1}{2}+{a}_{N},whereby limN→∞aN=0{\mathrm{lim}}_{N\to \infty }{a}_{N}=0.Figure 5Density of the copula AN{A}_{N}(gray) as considered in Example 6.4 and support of the mutually completely dependent copula BN{B}_{N}(magenta) according to Remark 6.5 for N=8N=8(left panel) and N=32N=32(right panel).Remark 6.5Slightly modifying the construction from Example 6.4 (which corresponds to copying shrunk versions of the product copula Π\Pi in the small squares) we now construct the copula BN{B}_{N}by copying shrunk versions of MMin every square of the “diagonal” of each of the four sets UN1×VN1{U}_{N}^{1}\times {V}_{N}^{1}, UN2×UN1{U}_{N}^{2}\times {U}_{N}^{1}, VN1×VN2{V}_{N}^{1}\times {V}_{N}^{2}, and VN2×UN2{V}_{N}^{2}\times {U}_{N}^{2}as depicted in Figure 5 (magenta lines). The shuffle BN{B}_{N}is obviously maximal D1{D}_{1}-asymmetric and, being completely dependent, fulfills ζ1(BN)=1=ξ(BN){\zeta }_{1}\left({B}_{N})=1=\xi \left({B}_{N}). Hence, setting CNα≔αAN+(1−α)BN{C}_{N}^{\alpha }:= \alpha {A}_{N}+\left(1-\alpha ){B}_{N}for every α∈[0,1]\alpha \in \left[0,1](with AN{A}_{N}according to Example 6.4) obviously yields a maximal D1{D}_{1}-asymmetric copula CNα{C}_{N}^{\alpha }. Due to the fact that ζ1(CNα){\zeta }_{1}\left({C}_{N}^{\alpha })and ξ(CNα)\xi \left({C}_{N}^{\alpha })are continuous in α\alpha the intermediate value theorem implies that for every s∈[ζ1(AN),1]s\in \left[{\zeta }_{1}\left({A}_{N}),1]we can find a copula CNα{C}_{N}^{\alpha }with ζ1(CNα)=s{\zeta }_{1}\left({C}_{N}^{\alpha })=sand the same result holds for ζ1{\zeta }_{1}replaced by ξ\xi . In other words, each point in the intervals mentioned in Theorems 6.2 and 6.3 is attained.Remark 6.6We have been able neither to find a copula A∈Cκ1=1A\in {{\mathcal{C}}}^{{\kappa }_{1}=1}fulfilling ζ1(A)=34{\zeta }_{1}\left(A)=\frac{3}{4}, nor to prove that such a copula does not exist. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Dependence Modeling de Gruyter

Maximal asymmetry of bivariate copulas and consequences to measures of dependence

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de Gruyter
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© 2022 Florian Griessenberger and Wolfgang Trutschnig, published by De Gruyter
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2300-2298
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2300-2298
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10.1515/demo-2022-0115
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Abstract

1IntroductionTwo random variables XXand YYwith joint distribution function HHare called exchangeable if and only if the pairs (X,Y)\left(X,Y)and (Y,X)\left(Y,X)have the same distribution, or equivalently, if H(x,y)=H(y,x)H\left(x,y)=H(y,x)holds for all xxand yy. The study of exchangeable random variables has exhibited a lot of interest in statistics (see, for instance, [8] and references therein). In case XXand YYare identically distributed and have distribution function FF, then (X,Y)\left(X,Y)is exchangeable if and only if the underlying copula AAcoincides with its transpose At{A}^{t}(defined as At(x,y)=A(y,x){A}^{t}\left(x,y)=A(y,x)). Hence, in what follows we consider continuous and identically distributed random variables XXand YY. While the class of continuous exchangeable random variables XXand YYis uniquely characterized by the class of symmetric copulas, the exact opposite, i.e., maximal non-exchangeability of random variables, strongly depends on the choice of measure quantifying the degree of non-exchangeability. One natural measure of non-exchangeability was studied by Nelsen [21] as well as by Klement and Mesiar [15], who independently showed that d∞(A,At)≔supx,y∈[0,1]∣A(x,y)−A(y,x)∣≤13{d}_{\infty }\left(A,{A}^{t}):= \mathop{\sup }\limits_{x,y\in \left[0,1]}| A\left(x,y)-A(y,x)| \le \frac{1}{3}holds for every A∈CA\in {\mathcal{C}}and introduced the d∞{d}_{\infty }-based measure δ:C→[0,1]\delta :{\mathcal{C}}\to \left[0,1]via δ(A)≔3d∞(A,At)\delta \left(A):= 3{d}_{\infty }\left(A,{A}^{t}). Moreover, they characterized all copulas A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry and showed that these copulas always model slightly negatively correlated random variables XXand YYin the sense of Spearman’s ρ\rho . More precisely, δ(A)=1\delta \left(A)=1implies ρ(A)∈−59,−13\rho \left(A)\in \left[-\frac{5}{9},-\frac{1}{3}\right]. Similar results also hold for different measures of concordance (see [17]).Considering other metrics on the space of copulas yields alternative measures of non-exchangeability ([13,25]): In [13] the stronger conditioning-based metric D1{D}_{1}introduced in [27] was studied and the authors proved (among other things) that every copula A∈CA\in {\mathcal{C}}with maximal D1{D}_{1}-asymmetry (i.e., D1(A,At)=12{D}_{1}\left(A,{A}^{t})=\frac{1}{2}) is not maximal asymmetric with respect to d∞{d}_{\infty }and that no maximal d∞{d}_{\infty }-asymmetric copula is maximal asymmetric with respect to D1{D}_{1}.Building upon the results in [13] we here further investigate the family of copulas with maximal D1{D}_{1}-asymmetry, derive additional novel characterizations in terms of the Markov-product of copulas (see [3]), and study various topological properties; inter alia we prove that the family of mutually completely dependent copulas with maximal D1{D}_{1}-asymmetry is dense in the set of all copulas with maximal D1{D}_{1}-asymmetry. Furthermore, we extend the concept of maximal D1{D}_{1}-asymmetry to the general Dp{D}_{p}-metrics (p∈[1,∞)p\in {[}1,\infty )), defined by (1)Dp(A,B)≔∫[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣pdλ(x)dλ(y)1p,{D}_{p}\left(A,B):= {\left(\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y]){| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)}^{\tfrac{1}{p}},where KA(⋅,⋅),KB(⋅,⋅){K}_{A}\left(\cdot ,\cdot ),{K}_{B}\left(\cdot ,\cdot )denote the Markov kernels (regular conditional distributions) of A,B∈CA,B\in {\mathcal{C}}, respectively. Although all Dp{D}_{p}-metrics induce the same topology, we show the surprising result that maximal D1{D}_{1}-asymmetry is not equivalent to maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty ). In fact, copulas with maximal Dp{D}_{p}-asymmetry with p∈(1,∞)p\in \left(1,\infty )are always mutually completely dependent and maximal asymmetric w.r.t. D1{D}_{1}.Moreover, we tackle the question on the degree of dependence of copulas exhibiting maximal asymmetry with respect to d∞{d}_{\infty }or Dp{D}_{p}for every p∈[1,∞]p\in \left[1,\infty ]. Since measures of concordance are generally not suitable for quantifying dependence (see, for instance, [11]) we consider the dependence measures ζ1{\zeta }_{1}introduced in [27] and further studied in [10,11], as well as ξ\xi , defined in [4] and reinvestigated in [2]. Both measures have recently attracted a lot of interest (see, e.g., [1,10,11,14,24,26]) since, in contrast to standard methods like Spearman’s ρ\rho or Kendall’s τ\tau , these measures are 1 if and only if YYis a function of XXand 0 if and only if XXand YYare independent; moreover, they can be estimated consistently without underlying smoothness assumptions. We prove that when considering maximal d∞{d}_{\infty }-asymmetry ζ1∈56,1{\zeta }_{1}\in \left[\frac{5}{6},1\right]and ξ∈23,1\xi \in \left[\frac{2}{3},1\right]hold, and in the case of maximal D1{D}_{1}-asymmetry ζ1∈34,1{\zeta }_{1}\in \left[\frac{3}{4},1\right]and ξ∈12,1\xi \in \left(\frac{1}{2},1\right]follows. In other words, maximal non-exchangeable random variables (in the sense of d∞{d}_{\infty }or Dp{D}_{p}) always imply a high degree of dependence w.r.t. ζ1{\zeta }_{1}and ξ\xi .The rest of this article is organized as follows: Section 2 gathers preliminaries and notations that will be used throughout the article. In Section 3, we study possible values of ζ1{\zeta }_{1}and ξ\xi for maximal d∞{d}_{\infty }-asymmetric copulas and discuss an example illustrating differences of ζ1{\zeta }_{1}and ξ\xi in the context of ordinal sums. In Section 4, we revisit copulas with maximal D1{D}_{1}-asymmetry and derive several topological properties. Extensions on maximal Dp{D}_{p}-asymmetry for p∈[1,∞]p\in \left[1,\infty ]and some interrelations are established in Section 5. Consequences on the dependence measures ζ1{\zeta }_{1}and ξ\xi conclude the article (Section 6). Various examples and graphics illustrate both the obtained results and the ideas underlying the proofs.2Notation and preliminariesFor every metric space (Ω,d)\left(\Omega ,d)the Borel σ\sigma -field in Ω\Omega will be denoted by ℬ(Ω){\mathcal{ {\mathcal B} }}\left(\Omega ), λ\lambda will denote the Lebesgue measure on ℬ(R){\mathcal{ {\mathcal B} }}\left({\mathbb{R}}). T{\mathcal{T}}will denote the class of all measurable λ\lambda -preserving transformations on [0,1]\left[0,1], i.e., T={T:[0,1]→[0,1]measurable withλ(T−1(E))=λ(E)∀E∈ℬ([0,1])},{\mathcal{T}}=\left\{T:\left[0,1]\to \left[0,1]\hspace{0.33em}\hspace{0.1em}\text{measurable with}\hspace{0.1em}\hspace{0.33em}\lambda \left({T}^{-1}\left(E))=\lambda \left(E)\hspace{1.0em}\forall E\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])\right\},and Tb{{\mathcal{T}}}_{b}the subclass of all bijective T∈TT\in {\mathcal{T}}. Throughout the article C{\mathcal{C}}will denote the family of all two-dimensional copulas, P{\mathcal{P}}the family of all doubly-stochastic measures (for background on copulas and doubly stochastic measures we refer to [6,22] and references therein). Furthermore, MMdenotes the upper Fréchet Hoeffding bound, Π\Pi the product copula, and WWthe lower Fréchet Hoeffding bound. Additionally, the completely dependent copula induced by a measure-preserving transformation h∈Th\in {\mathcal{T}}will be denoted by Ch{C}_{h}(see [27], Definition 9). The family of all completely dependent copulas will be denoted by Ccd{{\mathcal{C}}}_{cd}and the family of all mutually completely dependent copulas by Cmcd≔{Ch∈Ccd:h∈Tb}{{\mathcal{C}}}_{mcd}:= \left\{{C}_{h}\in {{\mathcal{C}}}_{cd}:h\in {{\mathcal{T}}}_{b}\right\}. For every copula C∈CC\in {\mathcal{C}}the corresponding doubly stochastic measure will be denoted by μC{\mu }_{C}. As usual, d∞{d}_{\infty }denotes the uniform metric on C{\mathcal{C}}, i.e., d∞(A,B)≔max(x,y)∈[0,1]2∣A(x,y)−B(x,y)∣{d}_{\infty }\left(A,B):= \mathop{\max }\limits_{\left(x,y)\in {\left[0,1]}^{2}}| A\left(x,y)-B\left(x,y)| for every A,B∈CA,B\in {\mathcal{C}}. It is well-known that (C,d∞)\left({\mathcal{C}},{d}_{\infty })is a compact metric space (see [6]).In what follows, Markov kernels will play an important role. A mapping K:R×ℬ(R)→[0,1]K:{\mathbb{R}}\times {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})\to \left[0,1]is called a Markov kernel from (R,ℬ(R))\left({\mathbb{R}},{\mathcal{ {\mathcal B} }}\left({\mathbb{R}}))to (R,ℬ(R))\left({\mathbb{R}},{\mathcal{ {\mathcal B} }}\left({\mathbb{R}}))if the mapping x↦K(x,B)x\mapsto K\left(x,B)is measurable for every fixed B∈ℬ(R)B\in {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})and the mapping B↦K(x,B)B\mapsto K\left(x,B)is a probability measure for every fixed x∈Rx\in {\mathbb{R}}. A Markov kernel K:R×ℬ(R)→[0,1]K:{\mathbb{R}}\times {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})\to \left[0,1]is called regular conditional distribution of a (real-valued) random variable YYgiven (another random variable) XXif for every B∈ℬ(R)B\in {\mathcal{ {\mathcal B} }}\left({\mathbb{R}})K(X(ω),B)=E(1B∘Y∣X)(ω)K\left(X\left(\omega ),B)={\mathbb{E}}\left({{\mathbb{1}}}_{B}\circ Y| X)\left(\omega )holds P{\mathbb{P}}-a.s. It is well-known that a regular conditional distribution of YYgiven XXexists and is unique PX{{\mathbb{P}}}^{X}-almost sure (where PX{{\mathbb{P}}}^{X}denotes the distribution of XX, i.e., the push-forward of P{\mathbb{P}}via XX). For every A∈CA\in {\mathcal{C}}(a version of) the corresponding regular conditional distribution (i.e., the regular conditional distribution of YYgiven XXin the case that (X,Y)∼A\left(X,Y)\hspace{0.33em} \sim \hspace{0.33em}A) will be denoted by KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot ). Note that for every A∈CA\in {\mathcal{C}}and Borel sets E,F∈ℬ([0,1])E,F\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])we have (2)∫EKA(x,F)dλ(x)=μA(E×F)and∫[0,1]KA(x,F)dλ(x)=λ(F).\mathop{\int }\limits_{E}{K}_{A}\left(x,F){\rm{d}}\lambda \left(x)={\mu }_{A}\left(E\times F)\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}\mathop{\int }\limits_{\left[0,1]}{K}_{A}\left(x,F){\rm{d}}\lambda \left(x)=\lambda \left(F).For more details and properties of conditional expectations and regular conditional distributions we refer to [12,16]. Expressing copulas in terms of their corresponding regular conditional distribution yields metrics stronger than d∞{d}_{\infty }(see [27]) and defined by (3)Dp(A,B)≔∫[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣pdλ(x)dλ(y)1p,{D}_{p}\left(A,B):= {\left(\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y]){| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)}^{\tfrac{1}{p}},(4)D∞(A,B)≔supy∈[0,1]∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣dλ(x).{D}_{\infty }\left(A,B):= \mathop{\sup }\limits_{y\in \left[0,1]}\mathop{\int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| \hspace{0.33em}{\rm{d}}\lambda \left(x).To simplify notation we will also write ΦA,B(y)≔∫[0,1]∣KA(x,[0,y])−KB(x,[0,y])∣dλ(x){\Phi }_{A,B}(y):= {\int }_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| {\rm{d}}\lambda \left(x). We will also work with D∂{D}_{\partial }, defined by D∂(A,B)≔D1(A,B)+D1(At,Bt),{D}_{\partial }\left(A,B):= {D}_{1}\left(A,B)+{D}_{1}\left({A}^{t},{B}^{t}),whereby At{A}^{t}denotes the transpose of A∈CA\in {\mathcal{C}}. The metric D∂{D}_{\partial }can be seen as metrization of the so-called ∂\partial -convergence, introduced and studied in [18,19]. In [27], it is shown that (C,D1)\left({\mathcal{C}},{D}_{1})is a complete and separable metric space with diameter 1/2 and that the topology induced by D1{D}_{1}is strictly finer than the one induced by d∞{d}_{\infty }. For further background on D1{D}_{1}and D∂{D}_{\partial }as well as for possible extensions to the multivariate setting we refer to [6,7, 10,27] and references therein.The D1{D}_{1}-based dependence measure ζ1{\zeta }_{1}(introduced in [27] and further investigated in [10,11]) is defined as ζ1(X,Y)≔ζ1(A)≔3D1(A,Π),{\zeta }_{1}\left(X,Y):= {\zeta }_{1}\left(A):= 3{D}_{1}\left(A,\Pi ),whereby (X,Y)\left(X,Y)has copula A∈CA\in {\mathcal{C}}. In the sequel, we will also consider the dependence measure ξ\xi (first introduced in [4] and reinvestigated in [2]) defined as ξ(X,Y)≔∫Var(E(1{Y≥t}∣X))dμ(t)∫Var(1{Y≥t})dμ(t),\xi \left(X,Y):= \frac{\int {\rm{Var}}\left({\mathbb{E}}\left({{\mathbb{1}}}_{\left\{Y\ge t\right\}}| X)){\rm{d}}\mu \left(t)}{\int {\rm{Var}}\left({{\mathbb{1}}}_{\left\{Y\ge t\right\}}){\rm{d}}\mu \left(t)},where μ\mu is the law of YY. In the copula setting, it is straightforward to verify that ξ\xi can be expressed in terms of D2{D}_{2}and that ξ(X,Y)≔ξ(A)=6D22(A,Π)\xi \left(X,Y):= \xi \left(A)=6{D}_{2}^{2}\left(A,\Pi )holds. Both dependence measures attain values in [0,1]\left[0,1]and are 0 if and only if A=ΠA=\Pi , and 1 if and only if AAis completely dependent.Letting Sh(A){S}_{h}\left(A)denote the generalized shuffle of AAw.r.t. the first coordinate, implicitly defined via the corresponding doubly stochastic measure μA{\mu }_{A}by μSh(A)(E×F)≔μA(h−1(E)×F),{\mu }_{{{\mathcal{S}}}_{h}\left(A)}\left(E\times F):= {\mu }_{A}\left({h}^{-1}\left(E)\times F),for all E,F∈ℬ([0,1])E,F\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])(see, e.g., [5,9]), the following simple result holds:Lemma 2.1Let h∈Tbh\in {{\mathcal{T}}}_{b}be a λ\lambda -preserving bijection. Then ζ1(Sh(A))=ζ1(A){\zeta }_{1}\left({S}_{h}\left(A))={\zeta }_{1}\left(A)and ξ(Sh(A))=ξ(A)\xi \left({S}_{h}\left(A))=\xi \left(A)hold for every A∈CA\in {\mathcal{C}}.ProofAccording to Lemma 3.1 in [9] for h∈Tbh\in {{\mathcal{T}}}_{b}the Markov kernel of Sh(A){{\mathcal{S}}}_{h}\left(A)can be expressed as KSh(A)(x,[0,y])=KA(h−1(x),[0,y]){K}_{{{\mathcal{S}}}_{h}\left(A)}\left(x,\left[0,y])={K}_{A}\left({h}^{-1}\left(x),\left[0,y])and for p∈[1,∞)p\in {[}1,\infty )we obtain Dpp(Sh(A),Π)=∫[0,1]∫[0,1]∣KSh(A)(x,[0,y])−y∣pdλ(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(x),[0,y])−y∣pdλ(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(x),[0,y])−y∣pdλh(x)dλ(y)=∫[0,1]∫[0,1]∣KA(h−1(h(x)),[0,y])−y∣pdλ(x)dλ(y)=Dpp(A,Π),\begin{array}{rcl}{D}_{p}^{p}\left({{\mathcal{S}}}_{h}\left(A),\Pi )& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{{\mathcal{S}}}_{h}\left(A)}\left(x,\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(x),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(x),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}{\lambda }^{h}\left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left({h}^{-1}\left(h\left(x)),\left[0,y])-y\hspace{-0.25em}{| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)={D}_{p}^{p}\left(A,\Pi ),\end{array}which proves the assertion.□In the sequel, we will also work with rearrangements [23] (see [26] for an elegant application of rearrangements in the copula context). We call f∗:[0,1]→R{f}^{\ast }:\left[0,1]\to {\mathbb{R}}the decreasing rearrangement of a Borel measurable function f:[0,1]→Rf:\left[0,1]\to {\mathbb{R}}if it fulfills f∗(t)≔inf{x∈R:λ({z∈[0,1]:f(z)>x})≤t}{f}^{\ast }\left(t):= {\rm{\inf }}\left\{x\in {\mathbb{R}}:\lambda \left(\left\{z\in \left[0,1]:f\left(z)\gt x\right\})\le t\right\}. The stochastically increasing (SI)-rearrangement A↑{A}^{\uparrow }of AAis then defined as A↑(x,y)≔∫[0,x]KA(t,[0,y])∗dλ(t),{A}^{\uparrow }\left(x,y):= \mathop{\int }\limits_{\left[0,x]}{K}_{A}{\left(t,\left[0,y])}^{\ast }{\rm{d}}\lambda \left(t),whereby the rearrangement is applied on the first coordinate of KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot ), i.e., for every fixed y∈[0,1]y\in \left[0,1]the rearranged Markov kernel is defined via KA(t,[0,y])∗≔inf{x∈[0,1]:λ({z∈[0,1]:KA(z,[0,y])>x})≤t}{K}_{A}{\left(t,\left[0,y])}^{\ast }:= {\rm{\inf }}\left\{x\in \left[0,1]:\lambda \left(\left\{z\in \left[0,1]:{K}_{A}\left(z,\left[0,y])\gt x\right\})\le t\right\}. In [26], it was shown that A↑{A}^{\uparrow }is an SI copula and both dependence measures ζ1{\zeta }_{1}and ξ\xi are invariant w.r.t. to the rearrangement, i.e., they fulfill ζ1(A↑)=ζ1(A){\zeta }_{1}\left({A}^{\uparrow })={\zeta }_{1}\left(A)and ξ(A↑)=ξ(A)\xi \left({A}^{\uparrow })=\xi \left(A), respectively. Recall that a copula AAis called SI if there exists a Borel set Λ⊆[0,1]\Lambda \subseteq \left[0,1]with λ(Λ)=1\lambda \left(\Lambda )=1such that for any y∈[0,1]y\in \left[0,1]the mapping x↦KA(x,[0,y])x\mapsto {K}_{A}\left(x,\left[0,y])is non-increasing on Λ\Lambda . The family of all SI copulas will be denoted by C↑{{\mathcal{C}}}^{\uparrow }. For further information we refer to [22] and references therein.Given A,B∈CA,B\in {\mathcal{C}}a new copula denoted by A∗BA\ast Bcan be constructed via the so-called star/Markov product A∗BA\ast B(see [3]) by (5)(A∗B)(x,y)≔∫[0,1]∂2A(x,t)∂1B(t,y)dλ(t),\left(A\ast B)\left(x,y):= \mathop{\int }\limits_{\left[0,1]}{\partial }_{2}A\left(x,t){\partial }_{1}B\left(t,y){\rm{d}}\lambda \left(t),where ∂1A(x,y){\partial }_{1}A\left(x,y)denotes the partial derivative of AAwith respect to the first coordinate. The star product A∗BA\ast Bis always a copula, i.e., no smoothness assumptions on A,BA,Bare required. Translating to the Markov kernel setting the star product corresponds to the well-known composition of Markov kernels and the following lemma holds:Lemma 2.2[29] Suppose that A,B∈CA,B\in {\mathcal{C}}and let KA,KB{K}_{A},{K}_{B}denote the Markov kernels of A and B, respectively. Then the Markov kernel KA∘KB{K}_{A}\circ {K}_{B}, defined by(6)(KA∘KB)(x,F)≔∫[0,1]KB(y,F)KA(x,dy),\left({K}_{A}\circ {K}_{B})\left(x,F):= \mathop{\int }\limits_{\left[0,1]}{K}_{B}(y,F){K}_{A}\left(x,{\rm{d}}y),is a regular conditional distribution of A∗BA\ast B.3Maximal d∞{d}_{\infty }-asymmetric copulas and their extent of dependence with respect to ζ1{\zeta }_{1}and ξ\xi Since ordinal sums will play an important role in what follows, we briefly recall their definition. We follow [6] and let I⊆NI\subseteq {\mathbb{N}}be some finite index set, ((ai,bi))i∈I{(\left({a}_{i},{b}_{i}))}_{i\in I}be a family of non-overlapping intervals with 0≤ai<bi≤10\le {a}_{i}\lt {b}_{i}\le 1for each i∈Ii\in Isuch that ⋃i∈I[ai,bi]=[0,1]{\bigcup }_{i\in I}\left[{a}_{i},{b}_{i}]=\left[0,1]holds. Furthermore, (Ci)i∈I{\left({C}_{i})}_{i\in I}denotes a family of bivariate copulas. Then the copula CCdefined by C(x,y)=ai+(bi−ai)Cix−aibi−ai,y−aibi−ai,(x,y)∈(ai,bi)2min{x,y}elsewhereC\left(x,y)=\left\{\begin{array}{ll}{a}_{i}+\left({b}_{i}-{a}_{i}){C}_{i}\left(\frac{x-{a}_{i}}{{b}_{i}-{a}_{i}},\frac{y-{a}_{i}}{{b}_{i}-{a}_{i}}\right),\hspace{1.0em}& \left(x,y)\in {\left({a}_{i},{b}_{i})}^{2}\\ \min \left\{x,y\right\}\hspace{1.0em}& \hspace{0.1em}\text{elsewhere}\hspace{0.1em}\end{array}\right.is an ordinal sum, and we write C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}. The following lemma gathers some useful formulas for D1{D}_{1}and D22{D}_{2}^{2}, which will be used in the sequel.Lemma 3.1Let C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}be an ordinal sum with I≔{1,…,n}I:= \left\{1,\ldots ,n\right\}for some n∈Nn\in {\mathbb{N}}. ThenD22(C,Π)=∑i=1n((bi−ai)2D22(Ci,Π))+f(a1,…,an,b1,…,bn),D1(C,Π)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]∣KCi(x,[0,y])−(ai+(bi−ai)y)∣dλ(x)dλ(y)+g(a1,…,an,b1,…,bn),\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}({\left({b}_{i}-{a}_{i})}^{2}{D}_{2}^{2}\left({C}_{i},\Pi ))+f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\\ {D}_{1}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{C}_{i}}\left(x,\left[0,y])-\left({a}_{i}+\left({b}_{i}-{a}_{i})y)| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)+g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\end{array}whereby f and g are given by f(a1,…,an,b1,…,bn)≔∑i=1n(bi−ai)23+(bi−ai)(1−bi)−13f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}):= {\sum }_{i=1}^{n}\left(\frac{{\left({b}_{i}-{a}_{i})}^{2}}{3}+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right)-\frac{1}{3}and g(a1,…,an,b1,…,bn)≔∑i=1n(bi−ai)12−bi+bi22+ai22g({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}):= {\sum }_{i=1}^{n}\left({b}_{i}-{a}_{i})\left(\frac{1}{2}-{b}_{i}+\frac{{b}_{i}^{2}}{2}+\frac{{a}_{i}^{2}}{2}\right), respectively.ProofThe definition of D22{D}_{2}^{2}yields D22(C,Π)=∫[0,1]∫[0,1](KC(x,[0,y])−KΠ(x,[0,y]))2dλ(x)dλ(y)=∫[0,1]∫[0,1]KC(x,[0,y])2dλ(x)dλ(y)−2∫[0,1]y∫[0,1]KC(x,[0,y])dλ(x)dλ(y)+∫[0,1]∫[0,1]y2dλ(x)dλ(y)=∫[0,1]∫[0,1]KC(x,[0,y])2dλ(x)dλ(y)−13\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{({K}_{C}\left(x,\left[0,y])-{K}_{\Pi }\left(x,\left[0,y]))}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)-2\mathop{\displaystyle \int }\limits_{\left[0,1]}y\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}\left(x,\left[0,y]){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{y}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)-\frac{1}{3}\end{array}for every C∈CC\in {\mathcal{C}}. Using the fact that (without loss of generality) the Markov kernel KC(x,[0,y]){K}_{C}\left(x,\left[0,y])of CCis 0 below the squares (ai,bi)2{\left({a}_{i},{b}_{i})}^{2}and 1 above (ai,bi)2{\left({a}_{i},{b}_{i})}^{2}, and applying change of coordinates yields D22(C,Π)+13=∑i=1n∫(ai,bi)∫(ai,bi)KC(x,[0,y])2dλ(x)dλ(y)+∫(bi,1)∫(ai,bi)1dλ(x)dλ(y)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]KCi(x,[0,y])2dλ(x)dλ(y)+(bi−ai)(1−bi)=∑i=1n(bi−ai)2D22(Ci,Π)+13+(bi−ai)(1−bi).\begin{array}{rcl}{D}_{2}^{2}\left(C,\Pi )+\frac{1}{3}& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left(\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}{K}_{C}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \int }\limits_{\left({b}_{i},1)}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}1{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{C}_{i}}{\left(x,\left[0,y])}^{2}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\left({D}_{2}^{2}\left({C}_{i},\Pi )+\frac{1}{3}\right)+\left({b}_{i}-{a}_{i})\left(1-{b}_{i})\right).\end{array}Analogously, we obtainD1(C,Π)=∑i=1n∫[0,1]∫(ai,bi)∣KC(x,[0,y])−y∣dλ(x)dλ(y)=∑i=1n∫(ai,bi)∫(ai,bi)KCix−aibi−ai,0,y−aibi−ai−ydλ(x)dλ(y)+∑i=1n∫(bi,1)∫(ai,bi)(1−y)dλ(x)dλ(y)+∑i=1n∫(0,ai)∫(ai,bi)ydλ(x)dλ(y)=∑i=1n(bi−ai)2∫[0,1]∫[0,1]∣KCi(x,[0,y])−(ai+(bi−ai)y)∣dλ(x)dλ(y)+g(a1,…,an,b1,…,bn),\begin{array}{rcl}{D}_{1}\left(C,\Pi )& =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}| {K}_{C}(x,\left[0,y\left])-y| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\left|{K}_{{C}_{i}}\left(\frac{x-{a}_{i}}{{b}_{i}-{a}_{i}},\left[0,\frac{y-{a}_{i}}{{b}_{i}-{a}_{i}}\right]\right)-y\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & & +\mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left({b}_{i},1)}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}\left(1-y)\hspace{0.33em}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\mathop{\displaystyle \sum }\limits_{i=1}^{n}\mathop{\displaystyle \int }\limits_{\left(0,{a}_{i})}\mathop{\displaystyle \int }\limits_{\left({a}_{i},{b}_{i})}y\hspace{0.33em}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \sum }\limits_{i=1}^{n}\left({\left({b}_{i}-{a}_{i})}^{2}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{{C}_{i}}\left(x,\left[0,y])-\left({a}_{i}+\left({b}_{i}-{a}_{i})y)| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\right)+g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n}),\end{array}with g(a1,…,an,b1,…,bn)g\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})as in the theorem.□As a direct consequence, the dependence measure ξ\xi of ordinal sums can easily be expressed in terms of ξ(Ci)\xi \left({C}_{i}):Corollary 3.2Let C=(⟨ai,bi,Ci⟩)i∈IC={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in I}be an ordinal sum with I≔{1,…,n}I:= \left\{1,\ldots ,n\right\}for some n∈Nn\in {\mathbb{N}}. Thenξ(C)=∑i=1n(bi−ai)2ξ(Ci)+6f(a1,…,an,b1,…,bn)\xi \left(C)=\mathop{\sum }\limits_{i=1}^{n}{\left({b}_{i}-{a}_{i})}^{2}\xi \left({C}_{i})+6f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})holds, where f is defined according to Lemma 3.1 and only depends on the partition.The following example shows that ordinal sums can be used to construct copulas attaining every possible dependence value w.r.t. to ξ\xi and ζ1{\zeta }_{1}.Example 3.3Consider Cs=(⟨ai,bi,Ci⟩)i∈{1,2}{C}_{s}={(\langle {a}_{i},{b}_{i},{C}_{i}\rangle )}_{i\in \left\{1,2\right\}}, whereby a1≔0,a2≔s{a}_{1}:= 0,{a}_{2}:= s, b1≔s{b}_{1}:= s, and b2≔1{b}_{2}:= 1for s∈[0,1]s\in \left[0,1]and set C1=Π{C}_{1}=\Pi as well as C2=M{C}_{2}=M. Figure 1 depicts the support of μCs{\mu }_{{C}_{s}}for different choices of s∈[0,1]s\in \left[0,1]. Using Corollary 3.2 we have ξ(Cs)=(1−s)2+6s23+s(1−s)+(1−s)23−13=1−s2\xi \left({C}_{s})={\left(1-s)}^{2}+6\left(\frac{{s}^{2}}{3}+s\left(1-s)+\frac{{\left(1-s)}^{2}}{3}-\frac{1}{3}\right)=1-{s}^{2}. Therefore, the map φ:[0,1]→[0,1]\varphi :\left[0,1]\to \left[0,1]defined by s↦ξ(Cs)s\mapsto \xi \left({C}_{s})is continuous and onto. The same holds for ζ1(Cs)=1−s3{\zeta }_{1}\left({C}_{s})=1-{s}^{3}.Figure 1Mass distribution of the doubly stochastic measure μCs{\mu }_{{C}_{s}}for s=0.3s=0.3(left panel) and s=0.8s=0.8(right panel) considered in Example 3.3. For the dependence measures ξ\xi and ζ1{\zeta }_{1}we obtain ξ(C0.3)=0.91\xi \left({C}_{0.3})=0.91and ξ(C0.8)=0.36\xi \left({C}_{0.8})=0.36as well as ζ1(C0.3)=0.973{\zeta }_{1}\left({C}_{0.3})=0.973and ζ1(C0.8)=0.488{\zeta }_{1}\left({C}_{0.8})=0.488.Before deriving some first results concerning the range of the dependence measures ξ(A)\xi \left(A)and ζ1(A){\zeta }_{1}\left(A)for maximal d∞{d}_{\infty }-asymmetric copulas AA, we recall the characterizations of maximal d∞{d}_{\infty }-asymmetry derived in [21,15]: d∞(A,At){d}_{\infty }\left(A,{A}^{t})is maximal if and only if A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0and A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}or At23,13=0{A}^{t}\left(\frac{2}{3},\frac{1}{3}\right)=0and At13,23=13{A}^{t}\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}. Without loss of generality we may focus on the case A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}and A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0. Since AAis doubly stochastic in this case we obviously have μA0,13×13,23=μA13,23×23,1=μA23,1×0,13=13{\mu }_{A}\left(\left[0,\frac{1}{3}\right]\times \left[\frac{1}{3},\frac{2}{3}\right]\right)={\mu }_{A}\left(\left[\frac{1}{3},\frac{2}{3}\right]\times \left[\frac{2}{3},1\right]\right)={\mu }_{A}\left(\left[\frac{2}{3},1\right]\times \left[0,\frac{1}{3}\right]\right)=\frac{1}{3}. As a direct consequence, we can find copulas A1,A2,A3∈C{A}_{1},{A}_{2},{A}_{3}\in {\mathcal{C}}fulfilling (7)μA=13μA1f12+13μA2f23+13μA3f31,{\mu }_{A}=\frac{1}{3}{\mu }_{{A}_{1}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{A}_{2}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{A}_{3}}^{{f}_{31}},whereby the functions fij:[0,1]2→i−13,i3×j−13,j3{f}_{ij}:{\left[0,1]}^{2}\to \left[\frac{i-1}{3},\frac{i}{3}\right]\times \left[\frac{j-1}{3},\frac{j}{3}\right]are given by fij(x,y)=x+i−13,y+j−13{f}_{ij}\left(x,y)=\left(\frac{x+i-1}{3},\frac{y+j-1}{3}\right)for each (i,j)∈{1,2,3}2\left(i,j)\in {\left\{1,2,3\right\}}^{2}(and μAfij{\mu }_{A}^{{f}_{ij}}denotes the push-forward of μA{\mu }_{A}via fij{f}_{ij}).Theorem 3.4If A∈CA\in {\mathcal{C}}has maximal d∞{d}_{\infty }-asymmetry, i.e., if δ(A)=3d∞(A,At)=1\delta \left(A)=3{d}_{\infty }\left(A,{A}^{t})=1holds, then ξ\xi satisfies ξ(A)∈23,1\xi \left(A)\in \left[\frac{2}{3},1\right]. Moreover, for every s∈23,1s\in \left[\frac{2}{3},1\right]there exists a copula A with δ(A)=1\delta \left(A)=1fulfilling ξ(A)=s\xi \left(A)=s.ProofWe may assume that A13,23=13A\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}and A23,13=0A\left(\frac{2}{3},\frac{1}{3}\right)=0. Then there exist copulas A1,A2,A3∈C{A}_{1},{A}_{2},{A}_{3}\in {\mathcal{C}}such that μA=13μA2f12+13μA3f23+13μA1f31{\mu }_{A}=\frac{1}{3}{\mu }_{{A}_{2}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{A}_{3}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{A}_{1}}^{{f}_{31}}holds. Defining h:[0,1]→[0,1]h:\left[0,1]\to \left[0,1]by h(x)=13+xifx∈0,23x−23ifx∈23,1,h\left(x)=\left\{\begin{array}{ll}\frac{1}{3}+x\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[0,\frac{2}{3}\right]\\ x-\frac{2}{3}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{2}{3},1\right],\end{array}\right.we have h∈Tbh\in {{\mathcal{T}}}_{b}and Sh(A)=i−13,i3,Aii∈{1,2,3}{{\mathcal{S}}}_{h}\left(A)={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Applying Lemmas 2.1 and 3.2 we therefore obtain ξ(A)=ξ(Sh(A))=6f(a1,…,an,b1,…,bn)+∑i=1319ξ(Ai)=23+∑i=1319ξ(Ai)≥23,\xi \left(A)=\xi \left({{\mathcal{S}}}_{h}\left(A))=6f\left({a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n})+\mathop{\sum }\limits_{i=1}^{3}\frac{1}{9}\xi \left({A}_{i})=\frac{2}{3}+\mathop{\sum }\limits_{i=1}^{3}\frac{1}{9}\xi \left({A}_{i})\ge \frac{2}{3},with equality if and only if Ai=Π{A}_{i}=\Pi for every i=1,2,3i=1,2,3.Defining As{A}_{s}by μAs≔13μCsf12+13μCsf23+13μCsf31{\mu }_{{A}_{s}}:= \frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{12}}+\frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{23}}+\frac{1}{3}{\mu }_{{C}_{s}}^{{f}_{31}}with Cs{C}_{s}as in Example 3.3 yields ξ(As)=ξ(Sh(As))=13ξ(Cs)+23.\xi \left({A}_{s})=\xi \left({{\mathcal{S}}}_{h}\left({A}_{s}))=\frac{1}{3}\xi \left({C}_{s})+\frac{2}{3}.Considering ξ(Cs)=1−s2∈[0,1]\xi \left({C}_{s})=1-{s}^{2}\in \left[0,1]for s∈[0,1]s\in \left[0,1]and using the same arguments as in Example 3.3 it follows that for every s0∈23,1{s}_{0}\in \left[\frac{2}{3},1\right]we find a copula A∈CA\in {\mathcal{C}}with ξ(A)=s0\xi \left(A)={s}_{0}and 3d∞(A,At)=δ(A)=13{d}_{\infty }\left(A,{A}^{t})=\delta \left(A)=1.□Since ζ1{\zeta }_{1}and ξ\xi are similar by construction, one might expect the analogous statements for ζ1{\zeta }_{1}. Note, however, that a different proof is needed since according to Lemma 3.1 the formulas for D1{D}_{1}are more involved.Theorem 3.5If A∈CA\in {\mathcal{C}}has maximal d∞{d}_{\infty }-asymmetry, then ζ1{\zeta }_{1}satisfies ζ1(A)∈56,1{\zeta }_{1}\left(A)\in \left[\frac{5}{6},1\right]. Furthermore, for every s∈56,1s\in \left[\frac{5}{6},1\right]there exists a copula AAwith δ(A)=1\delta \left(A)=1fulfilling ζ1(A)=s{\zeta }_{1}\left(A)=s.ProofProceeding as in the proof of Theorem 3.4 we obtain Sh(A)=i−13,i3,Aii∈{1,2,3}{{\mathcal{S}}}_{h}\left(A)={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Considering the (SI)-rearrangement Sh(A)↑{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }of Sh(A){{\mathcal{S}}}_{h}\left(A)it is clear that Sh(A)↑{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }is an ordinal sum again and can be expressed as Sh(A)↑=i−13,i3,Ai↑i∈{1,2,3}{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }={\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}^{\uparrow }\right\rangle \right)}_{i\in \left\{1,2,3\right\}}. Since every Ai↑{A}_{i}^{\uparrow }is SI and hence fulfills Ai↑(x,y)≥Π↑(x,y)=Π(x,y){A}_{i}^{\uparrow }\left(x,y)\ge {\Pi }^{\uparrow }\left(x,y)=\Pi \left(x,y)for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and every i∈{1,2,3}i\in \left\{1,2,3\right\}(see, e.g., [22][Section 5.2]), we obtain that Sh(A)↑≥CΠ≔i−13,i3,Πi∈{1,2,3}{{\mathcal{S}}}_{h}{\left(A)}^{\uparrow }\ge {C}_{\Pi }:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},\Pi \right\rangle \right)}_{i\in \left\{1,2,3\right\}}holds pointwise. Due to the fact that ζ1{\zeta }_{1}is monotone w.r.t. the pointwise order on C↑{{\mathcal{C}}}^{\uparrow }and ζ1{\zeta }_{1}is invariant w.r.t. to (SI)-rearrangements (see [26]), we obtain ζ1(A)=ζ1(Sh(A))=ζ1(Sh(A)↑)≥ζ1(CΠ)=56,{\zeta }_{1}\left(A)={\zeta }_{1}\left({{\mathcal{S}}}_{h}\left(A))={\zeta }_{1}\left({{\mathcal{S}}}_{h}{\left(A)}^{\uparrow })\ge {\zeta }_{1}\left({C}_{\Pi })=\frac{5}{6},where the last equality follows from Lemma 3.1 (the detailed calculations are deferred to Appendix A). To show the second assertion we can proceed analogously to the proof of Theorem 3.4 and use shrunk copies of the copula Cs{C}_{s}defined in Example 3.3 (see Appendix A).□While the minimum value of ξ\xi for a copula A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry is attained if and only if Ai=Π{A}_{i}=\Pi for every i=1,2,3i=1,2,3in equation (7), ζ1{\zeta }_{1}exhibits a different behavior as demonstrated in the following example:Example 3.6Let A1∈C↑{A}_{1}\in {{\mathcal{C}}}^{\uparrow }be defined by A1(x,y)=xy+12x(1−x)y(1−y).{A}_{1}\left(x,y)=xy+\frac{1}{2}x\left(1-x)y\left(1-y).Then a version of the corresponding Markov kernel of A1{A}_{1}is given by KA1(x,[0,y])=y+12(2x−1)y(y−1){K}_{{A}_{1}}\left(x,\left[0,y])=y+\frac{1}{2}\left(2x-1)y(y-1). Furthermore, we set A3=A1{A}_{3}={A}_{1}and A2=Π{A}_{2}=\Pi and let AAdenote the ordinal sum given by A≔i−13,i3,Aii∈{1,2,3}A:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2,3\right\}}and CΠ{C}_{\Pi }be the ordinal sum given by CΠ≔i−13,i3,Πi∈{1,2,3}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{3},\frac{i}{3},\Pi \right\rangle \right)}_{i\in \left\{1,2,3\right\}}(Figure 2).By construction we have A≠CΠA\ne {C}_{\Pi }, however, considering ∫[0,1]∫[0,1]12(2x−1)y(y−1)dλ(x)dλ(y)=0\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\frac{1}{2}\left(2x-1)y(y-1){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=0yields ∫[0,1]∫[0,1]KΠ(x,[0,y])−y3dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3+12(2x−1)y(y−1)dλ(x)dλ(y)=∫[0,1]∫[0,1]2y3+12(2x−1)y(y−1)dλ(x)dλ(y)=∫[0,1]∫[0,1]KA1(x,[0,y])−y3dλ(x)dλ(y)\begin{array}{rcl}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{\Pi }\left(x,\left[0,y])-\frac{y}{3}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\frac{2y}{3}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left(\frac{2y}{3}+\frac{1}{2}\left(2x-1)y(y-1)\right){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|\frac{2y}{3}+\frac{1}{2}\left(2x-1)y(y-1)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{{A}_{1}}\left(x,\left[0,y])-\frac{y}{3}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\end{array}and analogously we obtain ∫[0,1]∫[0,1]KΠ(x,[0,y])−23+y3dλ(x)dλ(y)=∫[0,1]∫[0,1]KA1(x,[0,y])−23+y3dλ(x)dλ(y).\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|{K}_{\Pi }\left(x,\left[0,y])-\left(\frac{2}{3}+\frac{y}{3}\right)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|{K}_{{A}_{1}}\left(x,\left[0,y])-\left(\frac{2}{3}+\frac{y}{3}\right)\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y).Applying Lemma 3.1 we obtain ζ1(A)=ζ1(CΠ)=56{\zeta }_{1}\left(A)={\zeta }_{1}\left({C}_{\Pi })=\frac{5}{6}.Figure 2Density of the SI copulas CΠ{C}_{\Pi }(left panel) and AA(right panel) considered in Example 3.6. Although A(x,y)≥CΠ(x,y)A\left(x,y)\ge {C}_{\Pi }\left(x,y)holds for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and there exists some (x,y)\left(x,y)with A(x,y)>CΠ(x,y)A\left(x,y)\gt {C}_{\Pi }\left(x,y)we have ζ1(A)=ζ1(CΠ){\zeta }_{1}\left(A)={\zeta }_{1}\left({C}_{\Pi }). On the contrary, ξ\xi fulfills ξ(A)>ξ(CΠ)\xi \left(A)\gt \xi \left({C}_{\Pi }).Remark 3.7Considering the monotonicity of ζ1{\zeta }_{1}with respect to the pointwise order on C↑{{\mathcal{C}}}^{\uparrow }as proved in [26], Example 3.6 shows that there exist copulas A,B∈C↑A,B\in {{\mathcal{C}}}^{\uparrow }with A≤BA\le Bpointwise and A(x,y)<B(x,y)A\left(x,y)\lt B\left(x,y)for some (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}fulfilling ζ1(A)=ζ1(B){\zeta }_{1}\left(A)={\zeta }_{1}\left(B).4Maximal D1{D}_{1}-asymmetry of copulas revisitedIn this section, we complement characterizations of copulas with maximal D1{D}_{1}-asymmetry going back to [13] and derive some topological properties of subclasses. To be consistent with the notation in [13], the family of copulas with maximal D1{D}_{1}-asymmetry is denoted by Cκ=1≔{A∈C:κ(A)≔2D1(A,At)=1}⊆C,{{\mathcal{C}}}^{\kappa =1}:= \{A\in {\mathcal{C}}:\kappa \left(A):= 2{D}_{1}\left(A,{A}^{t})=1\}\subseteq {\mathcal{C}},the subclass of mutually completely dependent copulas is denoted by Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}. We start with the family of mutually completely dependent copulas and show closedness w.r.t. the metric D∂{D}_{\partial }.Proposition 4.1The set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is closed in (C,D∂)\left({\mathcal{C}},{D}_{\partial }).ProofLet (Ahn)n∈N{({A}_{{h}_{n}})}_{n\in {\mathbb{N}}}be a sequence of mutually completely dependent copulas with D∂{D}_{\partial }-limit AA. Since according to [27] the family of completely dependent copulas is closed w.r.t. D1{D}_{1}we obtain A∈CcdA\in {{\mathcal{C}}}_{cd}and At∈Ccd{A}^{t}\in {{\mathcal{C}}}_{cd}. Using [27, Lemma 10] there exist λ\lambda -preserving transformations g,g′∈Tg,g^{\prime} \in {\mathcal{T}}such that a version of the Markov kernel KA(⋅,⋅){K}_{A}\left(\cdot ,\cdot )and KAt(⋅,⋅){K}_{{A}^{t}}\left(\cdot ,\cdot )is given by KA(x,E)=1E(g(x)){K}_{A}\left(x,E)={{\mathbb{1}}}_{E}\left(g\left(x))and KAt(x,E)=1E(g′(x)){K}_{{A}^{t}}\left(x,E)={{\mathbb{1}}}_{E}\left(g^{\prime} \left(x)), respectively. Furthermore, since a copula AAis completely dependent if and only if it is left-invertible w.r.t. the ∗\ast -product (see [27]) we have M=At∗AM={A}^{t}\ast A. Applying Lemma 2.2 therefore yields that g∘g′(x)=id(x)g\circ g^{\prime} \left(x)=id\left(x)for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]. Using the fact that ggis surjective λ\lambda -almost everywhere, there exists a λ\lambda -preserving and bijective transformation h∈Tbh\in {{\mathcal{T}}}_{b}such that h=gh=gholds λ\lambda -a.e., implying A=Ah∈CmcdA={A}_{h}\in {{\mathcal{C}}}_{mcd}. It remains to show that D1(Ah,Aht)=12{D}_{1}\left({A}_{h},{A}_{h}^{t})=\frac{1}{2}, which can be done as follows. Using [13][Theorem 3.5] and the triangle inequality we obtain 12=∫[0,1]∣hn−hn−1∣dλ(x)≤∫[0,1]∣hn−h∣dλ(x)+∫[0,1]∣h−h−1∣dλ(x)+∫[0,1]∣h−1−hn−1∣dλ(x)\frac{1}{2}=\mathop{\int }\limits_{\left[0,1]}| {h}_{n}-{h}_{n}^{-1}| {\rm{d}}\lambda \left(x)\le \mathop{\int }\limits_{\left[0,1]}| {h}_{n}-h| {\rm{d}}\lambda \left(x)+\mathop{\int }\limits_{\left[0,1]}| h-{h}^{-1}| {\rm{d}}\lambda \left(x)+\mathop{\int }\limits_{\left[0,1]}| {h}^{-1}-{h}_{n}^{-1}| {\rm{d}}\lambda \left(x)for every n∈Nn\in {\mathbb{N}}. Applying [27][Proposition 15 (ii)] yields D1(Ah,Aht)=∫[0,1]∣h(x)−h−1(x)∣dλ(x)≥12−[D1(Ahn,Ah)+D1(Ahnt,Aht)]=12−D∂(Ahn,Ah).{D}_{1}\left({A}_{h},{A}_{h}^{t})=\mathop{\int }\limits_{\left[0,1]}| h\left(x)-{h}^{-1}\left(x)| {\rm{d}}\lambda \left(x)\ge \frac{1}{2}-{[}{D}_{1}\left({A}_{{h}_{n}},{A}_{h})+{D}_{1}\left({A}_{{h}_{n}}^{t},{A}_{h}^{t})]=\frac{1}{2}-{D}_{\partial }\left({A}_{{h}_{n}},{A}_{h}).Together with the fact that the maximal distance cannot exceed 12\frac{1}{2}it follows that D1(Ah,Aht)=12{D}_{1}\left({A}_{h},{A}_{h}^{t})=\frac{1}{2}, which completes the proof.□The following example shows that the set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is not closed w.r.t. the metric D1{D}_{1}.Example 4.2Let Ahn∈Cmcd{A}_{{h}_{n}}\in {{\mathcal{C}}}_{mcd}be the mutually completely dependent copula induced by the bijective measure-preserving transformation hn:[0,1]→[0,1]{h}_{n}:\left[0,1]\to \left[0,1], given by hn(x)=x+j−1nifx∈j−1n,jnandj∈1,…,n2x−1+jnifx∈j−1n,jnandj∈n2+1,…,n1ifx=1{h}_{n}\left(x)=\left\{\begin{array}{ll}x+\frac{j-1}{n}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[\frac{j-1}{n},\frac{j}{n}\right)\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}j\in \left\{1,\ldots ,\frac{n}{2}\right\}\\ x-1+\frac{j}{n}\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x\in \left[\frac{j-1}{n},\frac{j}{n}\right)\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}j\in \left\{\frac{n}{2}+1,\ldots ,n\right\}\\ 1\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}x=1\end{array}\right.for all n∈2Nn\in 2{\mathbb{N}}and let Ah∈Ccd{A}_{h}\in {{\mathcal{C}}}_{cd}be the completely dependent copula induced by the λ\lambda -preserving transformation h:[0,1]→[0,1]h:\left[0,1]\to \left[0,1]given by h(x)≔2x(mod1)h\left(x):= 2x\left(mod1)(see Figure 1 in [9]). Setting Cn≔i−14,i4,Ah2ni∈{1,2,3,4}{C}_{n}:= {\left(\left\langle \frac{i-1}{4},\frac{i}{4},{A}_{{h}_{2n}}\right\rangle \right)}_{i\in \left\{1,2,3,4\right\}}and C≔i−14,i4,Ahi∈{1,2,3,4}C:= {\left(\left\langle \frac{i-1}{4},\frac{i}{4},{A}_{h}\right\rangle \right)}_{i\in \left\{1,2,3,4\right\}}we have Cn∈Cmcd{C}_{n}\in {{\mathcal{C}}}_{mcd}and C∈CcdC\in {{\mathcal{C}}}_{cd}and according to [9, Example 3.3] it is straightforward to verify that limn→∞D1(Cn,C)=0{\mathrm{lim}}_{n\to \infty }{D}_{1}\left({C}_{n},C)=0. As a next step, we reorder the shrunk copulas to obtain maximal D1{D}_{1}-asymmetry. Let ffdenote the λ\lambda -preserving interval exchange transformation f:[0,1]→[0,1]f:\left[0,1]\to \left[0,1]defined by f(x)≔(x−14)114,1(x)+(x+34)10,14(x)f\left(x):= \left(x-\frac{1}{4}){{\mathbb{1}}}_{\left(\tfrac{1}{4},1\right]}\left(x)+\left(x+\frac{3}{4}){{\mathbb{1}}}_{\left[0,\tfrac{1}{4}\right]}\left(x)and, furthermore, let Sf(Cn)∈Cmcd{{\mathcal{S}}}_{f}\left({C}_{n})\in {{\mathcal{C}}}_{mcd}and Sf(C)∈Ccd{{\mathcal{S}}}_{f}\left(C)\in {{\mathcal{C}}}_{cd}denote the respective shuffles (Figure 3). Due to the fact that the metric D1{D}_{1}is shuffle-invariant w.r.t. bijective transformations (using the same arguments as in the proof of Lemma 2.1) yields limn→∞D1(Sf(Cn),Sf(C))=limn→∞D1(Cn,C)=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({{\mathcal{S}}}_{f}\left({C}_{n}),{{\mathcal{S}}}_{f}\left(C))=\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({C}_{n},C)=0.Setting U≔0,14∪34,1U:= \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]and considering property (3) of Theorem 4.1 in [13] (see also Theorem 4.4(iii) in the sequel) we directly obtain that Sf(Cn){{\mathcal{S}}}_{f}\left({C}_{n})and Sf(C){{\mathcal{S}}}_{f}\left(C)are maximal asymmetric w.r.t. D1{D}_{1}, which shows that Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is not closed w.r.t. the metric D1{D}_{1}.Figure 3The support of the copulas μSf(Cn){\mu }_{{{\mathcal{S}}}_{f}\left({C}_{n})}(black) for n=4n=4(left panel) and n=8n=8(right panel) as well as the copula μSf(C){\mu }_{{{\mathcal{S}}}_{f}\left(C)}(magenta) as considered in Example 4.2.Leaving the subclass of mutually completely dependent copulas we will now derive novel and handy characterizations of copulas with maximal D1{D}_{1}-asymmetry and then show some topological properties. The following lemma, showing that the ∗\ast -product cannot increase the Dp{D}_{p}-distance, will be useful in the sequel. The result has already been stated for D1{D}_{1}in a slightly different context in [28].Lemma 4.3For every A,B,C∈CA,B,C\in {\mathcal{C}}, the following inequality holds for every p∈[1,∞)p\in {[}1,\infty ): Dpp(A∗B,A∗C)≤Dpp(B,C).{D}_{p}^{p}\left(A\ast B,A\ast C)\le {D}_{p}^{p}\left(B,C).ProofApplying Lemma 2.2, Jensen’s inequality, disintegration and using the fact that μA{\mu }_{A}is doubly stochastic we obtain Dpp(A∗B,A∗C)=∫[0,1]∫[0,1]∫[0,1]KB(t,[0,y])KA(x,dt)−∫[0,1]KC(t,[0,y])KA(x,dt)pdλ(x)dλ(y)≤∫[0,1]∫[0,1]∫[0,1]∣KB(t,[0,y])−KC(t,[0,y])∣pKA(x,dt)dλ(x)dλ(y)=∫[0,1]∫[0,1]2∣KB(t,[0,y])−KC(t,[0,y])∣pdμA(x,t)dλ(y)=∫[0,1]∫[0,1]∣KB(t,[0,y])−KC(t,[0,y])∣pdλ(t)dλ(y)=Dpp(B,C),\begin{array}{rcl}{D}_{p}^{p}\left(A\ast B,A\ast C)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{\left|\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{B}\left(t,\left[0,y]){K}_{A}\left(x,{\rm{d}}t)-\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{C}\left(t,\left[0,y]){K}_{A}\left(x,{\rm{d}}t)\right|}^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{K}_{A}\left(x,{\rm{d}}t){\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{{\left[0,1]}^{2}}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{\rm{d}}{\mu }_{A}\left(x,t){\rm{d}}\lambda (y)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{B}\left(t,\left[0,y])-{K}_{C}\left(t,\left[0,y])| }^{p}{\rm{d}}\lambda \left(t){\rm{d}}\lambda (y)={D}_{p}^{p}\left(B,C),\end{array}which completes the proof.□The next theorem gathers several equivalent characterizations of copulas having maximal D1{D}_{1}-asymmetry (see [13]), and the novel ones established here are (v) and (vi).Theorem 4.4For every A∈CA\in {\mathcal{C}}the following statements are equivalent: (i)κ(A)=1\kappa \left(A)=1,(ii)ΦA,At(12)=1{\Phi }_{A,{A}^{t}}\left(\frac{1}{2})=1(or equivalently, AAhas maximal D∞{D}_{\infty }-asymmetry),(iii)there exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with the following properties: λU∩0,12=λU∩12,1=14,μAU×0,12=12,μA0,12×U=0,\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right)=\lambda \left(U\cap \left[\frac{1}{2},1\right]\right)=\frac{1}{4},\hspace{1em}{\mu }_{A}\left(U\times \left[0,\frac{1}{2}\right]\right)=\frac{1}{2},\hspace{1em}{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=0,(iv)there exist sets U1,U2∈ℬ([0,1]){U}_{1},{U}_{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with U1⊆0,12{U}_{1}\subseteq \left[0,\frac{1}{2}\right], U2⊆12,1{U}_{2}\subseteq \left(\frac{1}{2},1\right], λ(U1)=λ(U2)=14\lambda \left({U}_{1})=\lambda \left({U}_{2})=\frac{1}{4}and V1≔0,12⧹U1{V}_{1}:= \left[0,\frac{1}{2}\right]\setminus {U}_{1}and V2≔12,1⧹U2{V}_{2}:= \left(\frac{1}{2},1\right]\setminus {U}_{2}, and copulas C1,C2,C3,C4∈C{C}_{1},{C}_{2},{C}_{3},{C}_{4}\in {\mathcal{C}}such that the following identityA(x,y)=14[C1(F1(x),G1(y))+C2(G1(x),G2(y))+C3(F2(x),F1(y))+C4(G2(x),F2(y))]A\left(x,y)=\frac{1}{4}{[}{C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))]holds, whereby Fi(x)≔4λ(Ui∩[0,x]){F}_{i}\left(x):= 4\lambda \left({U}_{i}\cap \left[0,x])and Gi(x)≔4λ(Vi∩[0,x]){G}_{i}\left(x):= 4\lambda \left({V}_{i}\cap \left[0,x])for i=1,2i=1,2.(v)(A∗A)12,12=0\left(A\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=0,(vi)D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}.ProofThe equivalences of (i), (ii), (iii), and (iv) have already been proved in [13]. Note that the equivalence in property (ii) directly follows from the facts that ΦA,At{\Phi }_{A,{A}^{t}}is Lipschitz continuous with Lipschitz constant 2 and the function ΦA,At:[0,1]→[0,1]{\Phi }_{A,{A}^{t}}:\left[0,1]\to \left[0,1]fulfills ΦA,At(y)≤min{2y,2(1−y)}{\Phi }_{A,{A}^{t}}(y)\le \min \left\{2y,2\left(1-y)\right\}for every y∈[0,1]y\in \left[0,1](see Lemma 5 in [27]). To show that (i) and (v) are equivalent we may proceed as follows: Suppose that κ(A)=1\kappa \left(A)=1. Then according to property (iii) there exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(U)=12\lambda \left(U)=\frac{1}{2}such that KAx,0,12=1{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=1for every x∈Ux\in U. Applying equation (2) and disintegration yields another Borel set V⊆UcV\subseteq {U}^{c}with λ(V)=12\lambda \left(V)=\frac{1}{2}and KAx,0,12=0{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0for every x∈Vx\in V. Setting V˜≔Uc⧹V\tilde{V}:= {U}^{c}\setminus V, then obviously λ(V˜)=0\lambda \left(\tilde{V})=0holds, and applying Lemma 2.2 we obtain μA∗A0,12×0,12=∫[0,12]∫[0,1]KA(s,0,12)KA(x,ds)dλ(x)=∫[0,12]∫U1KA(x,ds)+∫V0KA(x,ds)+∫V˜KA(s,0,12)KA(x,ds)dλ(x)≤∫[0,12]KA(x,U)dλ(x)+∫[0,12]KA(x,V˜)dλ(x)≤μA0,12×U+μA([0,1]×V˜)=0+λ(V˜)=0.\begin{array}{rcl}{\mu }_{A\ast A}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\left(\mathop{\displaystyle \int }\limits_{U}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{V}0\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{\tilde{V}}{K}_{A}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s)\right){\rm{d}}\lambda \left(x)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,\tilde{V}){\rm{d}}\lambda \left(x)\\ & \le & {\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)+{\mu }_{A}(\left[0,1]\times \tilde{V})=0+\lambda \left(\tilde{V})=0.\end{array}Suppose now that (A∗A)12,12=0\left(A\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=0holds. Then according to equation (5) we have ∫[0,1]KAtx,0,12KAx,0,12dλ(x)=0,\mathop{\int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)=0,so there exists a set Λ∈ℬ([0,1])\Lambda \in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(Λ)=1\lambda \left(\Lambda )=1such that KAtx,0,12KAx,0,12=0{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0holds for all x∈Λx\in \Lambda . Considering min{a,b}=12(a+b−∣a−b∣)\min \left\{a,b\right\}=\frac{1}{2}(a+b-| a-b| )therefore yields ΦA,At12=∫[0,1]∣KAx,0,12−KAtx,0,12∣dλ(x)=∫[0,1]KAx,0,12dλ(x)+∫[0,1]KAtx,0,12dλ(x)−2∫[0,1]min{KAx,0,12,KAtx,0,12}dλ(x)=12+12−2∫Λmin{KAx,0,12,KAtx,0,12}dλ(x)=12+12−2⋅0=1.\begin{array}{rcl}{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)-{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)| {\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)-2\mathop{\displaystyle \int }\limits_{\left[0,1]}\min \left\{{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right),{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)\right\}{\rm{d}}\lambda \left(x)\\ & =& \frac{1}{2}+\frac{1}{2}-2\mathop{\displaystyle \int }\limits_{\Lambda }\min \left\{{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right),{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)\right\}{\rm{d}}\lambda \left(x)=\frac{1}{2}+\frac{1}{2}-2\cdot 0=1.\end{array}To show the equivalence of (i) and (vi) first assume that D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}. Then applying Lemma 4.3 directly yields D1(A,At)≥12{D}_{1}\left(A,{A}^{t})\ge \frac{1}{2}, hence κ(A)=1\kappa \left(A)=1. On the other hand, if κ(A)=1\kappa \left(A)=1holds we may proceed as follows: There exists a Borel set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])with λ(U)=12\lambda \left(U)=\frac{1}{2}and KAx,0,12=1{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=1as well as KAtx,0,12=0{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)=0for every x∈Ux\in U. Using disintegration and equation (2) there exists a Borel set V⊆UcV\subseteq {U}^{c}with λ(V)=12\lambda \left(V)=\frac{1}{2}and KAx,0,12=0{K}_{A}\left(x,\left[0,\frac{1}{2}\right]\right)=0and KAtx,0,12=1{K}_{{A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)=1for every x∈Vx\in V. As before, set V˜≔Uc⧹V\tilde{V}:= {U}^{c}\setminus V. Applying Lemma 2.2 yields A∗At12,12=μA∗At0,12×0,12=∫[0,12]∫[0,1]KAts,0,12KA(x,ds)dλ(x)≤∫[0,12]∫Uc1KA(x,ds)dλ(x)=∫[0,12]KA(x,Uc)dλ(x)=μA0,12×Uc=12−μA0,12×U=12\begin{array}{rcl}A\ast {A}^{t}\left(\frac{1}{2},\frac{1}{2}\right)& =& {\mu }_{A\ast {A}^{t}}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)=\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{{A}^{t}}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\mathop{\displaystyle \int }\limits_{{U}^{c}}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,{U}^{c}){\rm{d}}\lambda \left(x)\\ & =& {\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times {U}^{c}\right)=\frac{1}{2}-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\frac{1}{2}\end{array}as well as μA∗At0,12×0,12=∫[0,12]∫U0KA(x,ds)+∫V1KA(x,ds)+∫V˜KAts,0,12KA(x,ds)dλ(x)≥∫[0,12]KA(x,V)dλ(x)=μA0,12×V=12−μA0,12×Vc=12−μA0,12×U+μA0,12×V˜≥12.\begin{array}{rcl}{\mu }_{A\ast {A}^{t}}\left(\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\right)& =& \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}\left(\mathop{\displaystyle \int }\limits_{U}0\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{V}1\hspace{0.33em}{K}_{A}\left(x,{\rm{d}}s)+\mathop{\displaystyle \int }\limits_{\tilde{V}}{K}_{{A}^{t}}\left(s,\left[0,\frac{1}{2}\right]\right){K}_{A}\left(x,{\rm{d}}s)\right){\rm{d}}\lambda \left(x)\\ & \ge & \mathop{\displaystyle \int }\limits_{\left[0,\frac{1}{2}]}{K}_{A}\left(x,V){\rm{d}}\lambda \left(x)={\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times V\right)=\frac{1}{2}-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times {V}^{c}\right)\\ & =& \frac{1}{2}-\left({\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)+{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times \tilde{V}\right)\right)\ge \frac{1}{2}.\end{array}Together with property (v) there exist Borel sets Δ1⊆0,12{\Delta }_{1}\subseteq \left[0,\frac{1}{2}\right]and Δ2⊆12,1{\Delta }_{2}\subseteq \left(\frac{1}{2},1\right]with λ(Δ1)=λ(Δ2)=12\lambda \left({\Delta }_{1})=\lambda \left({\Delta }_{2})=\frac{1}{2}such that KA∗Ax1,0,12=0,KA∗Ax2,0,12=1andKA∗Atx1,0,12=1,KA∗Atx2,0,12=0{K}_{A\ast A}\left({x}_{1},\left[0,\frac{1}{2}\right]\right)=0,\hspace{1em}{K}_{A\ast A}\left({x}_{2},\left[0,\frac{1}{2}\right]\right)=1\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}{K}_{A\ast {A}^{t}}\left({x}_{1},\left[0,\frac{1}{2}\right]\right)=1,\hspace{1em}{K}_{A\ast {A}^{t}}\left({x}_{2},\left[0,\frac{1}{2}\right]\right)=0for every x1∈Δ1{x}_{1}\in {\Delta }_{1}and x2∈Δ2{x}_{2}\in {\Delta }_{2}, which gives ΦA∗A,A∗At12=∫[0,1]∣KA∗Ax,0,12−KA∗Atx,0,12∣dλ(x)=λ(Δ1)+λ(Δ2)=1.{\Phi }_{A\ast A,A\ast {A}^{t}}\left(\frac{1}{2}\right)=\mathop{\int }\limits_{\left[0,1]}| {K}_{A\ast A}\left(x,\left[0,\frac{1}{2}\right]\right)-{K}_{A\ast {A}^{t}}\left(x,\left[0,\frac{1}{2}\right]\right)| {\rm{d}}\lambda \left(x)=\lambda \left({\Delta }_{1})+\lambda \left({\Delta }_{2})=1.Since y↦ΦA,B(y)y\mapsto {\Phi }_{A,B}(y)is Lipschitz continuous with Lipschitz constant 2 (see [27] [Lemma 5]), the property that D1(A∗A,A∗At)=12{D}_{1}\left(A\ast A,A\ast {A}^{t})=\frac{1}{2}follows immediately and the proof is complete.□Remark 4.5For mutually completely dependent copulas Ah∈Cmcd{A}_{h}\in {{\mathcal{C}}}_{mcd}, property (vi) of Theorem 4.4 simplifies to κ(Ah)=1if and only ifD1(Ah∗Ah,M)=D1(Ah2,M)=‖h2−id‖1=12,\kappa \left({A}_{h})=1\hspace{1em}\hspace{0.1em}\text{if and only if}\hspace{0.1em}\hspace{0.33em}{D}_{1}\left({A}_{h}\ast {A}_{h},M)={D}_{1}\left({A}_{{h}^{2}},M)=\Vert {h}^{2}-id{\Vert }_{1}=\frac{1}{2},where the second equality of the right-hand side directly follows from [27] [Proposition 15 (ii)]. Furthermore, considering property (v) of Theorem 4.4 one might conjecture that (A∗At)12,12=12\left(A\ast {A}^{t})\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}also implies κ(A)=1\kappa \left(A)=1. For the symmetric copula MM, however, it is clear that M12,12=12M\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}as well as M∗Mt=M∗M=MM\ast {M}^{t}=M\ast M=Mholds.Not surprisingly, the following result holds.Proposition 4.6The set Cκ=1{{\mathcal{C}}}^{\kappa =1}is closed in (C,D∂)\left({\mathcal{C}},{D}_{\partial }).ProofLet (An)n∈N{({A}_{n})}_{n\in {\mathbb{N}}}be a sequence of maximal D1{D}_{1}-asymmetric copulas fulfilling limn→∞D∂(An,A)=0{\mathrm{lim}}_{n\to \infty }{D}_{\partial }\left({A}_{n},A)=0for some A∈CA\in {\mathcal{C}}. Then applying Theorem 4.4, the triangle inequality and the fact that D∂{D}_{\partial }-convergence implies both limn→∞ΦAn,A12=0{\mathrm{lim}}_{n\to \infty }{\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)=0and limn→∞ΦAnt,At12=0{\mathrm{lim}}_{n\to \infty }{\Phi }_{{A}_{n}^{t},{A}^{t}}\left(\frac{1}{2}\right)=0we obtain 1=ΦAn,Ant12≤ΦAn,A12+ΦA,At12+ΦAt,Ant121={\Phi }_{{A}_{n},{A}_{n}^{t}}\left(\frac{1}{2}\right)\le {\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)+{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)+{\Phi }_{{A}^{t},{A}_{n}^{t}}\left(\frac{1}{2}\right)and hence ΦA,At12≥1−limn→∞ΦAn,A12−limn→∞ΦAt,Ant12=1.□\hspace{11em}{\Phi }_{A,{A}^{t}}\left(\frac{1}{2}\right)\ge 1-\mathop{\mathrm{lim}}\limits_{n\to \infty }{\Phi }_{{A}_{n},A}\left(\frac{1}{2}\right)-\mathop{\mathrm{lim}}\limits_{n\to \infty }{\Phi }_{{A}^{t},{A}_{n}^{t}}\left(\frac{1}{2}\right)=1.\hspace{12em}\square Remark 4.7Proposition 4.6 certainly is not surprising, however, the following result is. Key for proving the statement is property (v) of Theorem 4.4.Theorem 4.8The set Cκ=1{{\mathcal{C}}}^{\kappa =1}is closed in (C,D1)\left({\mathcal{C}},{D}_{1}).ProofSuppose that A,A1,A2,…A,{A}_{1},{A}_{2},\ldots are copulas, that κ(An)=1\kappa \left({A}_{n})=1for every n∈Nn\in {\mathbb{N}}, and that limn→∞D1(An,A)=0{\mathrm{lim}}_{n\to \infty }{D}_{1}\left({A}_{n},A)=0. Since the ∗\ast -product is jointly continuous w.r.t. D1{D}_{1}(see [29]) we have limn→∞D1(An∗An,A∗A)=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }{D}_{1}\left({A}_{n}\ast {A}_{n},A\ast A)=0.Considering that D1{D}_{1}-convergence implies d∞{d}_{\infty }-convergence, limn→∞(An∗An)12,12=A∗A12,12{\mathrm{lim}}_{n\to \infty }\left({A}_{n}\ast {A}_{n})\left(\frac{1}{2},\frac{1}{2}\right)=A\ast A\left(\frac{1}{2},\frac{1}{2}\right)follows, and applying Theorem 4.4 the proof is complete.□Analogous to the fact that shuffles are dense in (C,d∞)\left({\mathcal{C}},{d}_{\infty })the set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is dense in (Cκ=1,d∞)\left({{\mathcal{C}}}^{\kappa =1},{d}_{\infty }).Theorem 4.9The set Cmcdκ=1{{\mathcal{C}}}_{mcd}^{\kappa =1}is dense in (Cκ=1,d∞)\left({{\mathcal{C}}}^{\kappa =1},{d}_{\infty }).ProofFix ε>0\varepsilon \gt 0and let A∈Cκ=1A\in {{\mathcal{C}}}^{\kappa =1}be a copula with maximal D1{D}_{1}-asymmetry. According to property (iv) in Theorem 4.4 there exist sets U1,U2,V1,V2∈ℬ([0,1]){U}_{1},{U}_{2},{V}_{1},{V}_{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])and copulas C1,C2,C3,C4∈C{C}_{1},{C}_{2},{C}_{3},{C}_{4}\in {\mathcal{C}}such that A(x,y)=14[C1(F1(x),G1(y))+C2(G1(x),G2(y))+C3(F2(x),F1(y))+C4(G2(x),F2(y))],A\left(x,y)=\frac{1}{4}{[}{C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))],whereby Fi(x)≔4λ(Ui∩[0,x])=4∫[0,x]1Ui(s)λ(s){F}_{i}\left(x):= 4\lambda \left({U}_{i}\cap \left[0,x])=4{\int }_{\left[0,x]}{{\mathbb{1}}}_{{U}_{i}}\left(s)\lambda \left(s)and Gi(x)≔4λ(Vi∩[0,x])=4∫[0,x]1Vi(s)dλ(s){G}_{i}\left(x):= 4\lambda \left({V}_{i}\cap \left[0,x])=4{\int }_{\left[0,x]}{{\mathbb{1}}}_{{V}_{i}}\left(s){\rm{d}}\lambda \left(s)for i∈{1,2}i\in \left\{1,2\right\}. It is well-known that Cmcd{{\mathcal{C}}}_{mcd}(in fact even the family of straight shuffles) is dense in (C,d∞)\left({\mathcal{C}},{d}_{\infty })(see, e.g., [6][Corollary 4.1.16]), hence, we can find mutually completely dependent copulas Ch1,Ch2,Ch3,Ch4∈Cmcd{C}_{{h}_{1}},{C}_{{h}_{2}},{C}_{{h}_{3}},{C}_{{h}_{4}}\in {{\mathcal{C}}}_{mcd}with d∞(Ci,Chi)<ε{d}_{\infty }\left({C}_{i},{C}_{{h}_{i}})\lt \varepsilon for every i∈{1,2,3,4}i\in \left\{1,2,3,4\right\}. Defining A˜\tilde{A}by A˜(x,y)≔14[Ch1(F1(x),G1(y))+Ch2(G1(x),G2(y))+Ch3(F2(x),F1(y))+Ch4(G2(x),F2(y))],\tilde{A}\left(x,y):= \frac{1}{4}{[}{C}_{{h}_{1}}\left({F}_{1}\left(x),{G}_{1}(y))+{C}_{{h}_{2}}\left({G}_{1}\left(x),{G}_{2}(y))+{C}_{{h}_{3}}\left({F}_{2}\left(x),{F}_{1}(y))+{C}_{{h}_{4}}\left({G}_{2}\left(x),{F}_{2}(y))],and applying Theorem 4.4 we conclude that A˜\tilde{A}has maximal D1{D}_{1}-asymmetry too. Furthermore, using the triangle inequality we obtain supx,y∈[0,1]∣A(x,y)−A˜(x,y)∣≤14supx,y∈[0,1]∣C1(F1(x),G1(y))−Ch1(F1(x),G1(y))∣+14supx,y∈[0,1]∣C2(G1(x),G2(y))−Ch2(G1(x),G2(y))∣+14supx,y∈[0,1]∣C3(F2(x),F1(y))−Ch3(F2(x),F1(y))∣+14supx,y∈[0,1]∣C4(G2(x),F2(y))−Ch4(G2(x),F2(y))∣≤14∑i=14d∞(Ci,Chi)<ε4+ε4+ε4+ε4=ε.\begin{array}{rcl}\mathop{\sup }\limits_{x,y\in \left[0,1]}| A\left(x,y)-\tilde{A}\left(x,y)| & \le & \frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{1}\left({F}_{1}\left(x),{G}_{1}(y))-{C}_{{h}_{1}}\left({F}_{1}\left(x),{G}_{1}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{2}\left({G}_{1}\left(x),{G}_{2}(y))-{C}_{{h}_{2}}\left({G}_{1}\left(x),{G}_{2}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{3}\left({F}_{2}\left(x),{F}_{1}(y))-{C}_{{h}_{3}}\left({F}_{2}\left(x),{F}_{1}(y))| \\ & & +\frac{1}{4}\mathop{\sup }\limits_{x,y\in \left[0,1]}| {C}_{4}\left({G}_{2}\left(x),{F}_{2}(y))-{C}_{{h}_{4}}\left({G}_{2}\left(x),{F}_{2}(y))| \\ & \le & \frac{1}{4}\mathop{\displaystyle \sum }\limits_{i=1}^{4}{d}_{\infty }\left({C}_{i},{C}_{{h}_{i}})\lt \frac{\varepsilon }{4}+\frac{\varepsilon }{4}+\frac{\varepsilon }{4}+\frac{\varepsilon }{4}=\varepsilon .\end{array}As final step we have to show A˜∈Cmcd\tilde{A}\in {{\mathcal{C}}}_{mcd}, which can be done as follows: Fix y∈[0,1]y\in \left[0,1], then by applying Lemma 1 in [20] using the fact that KChi(x,[0,y]){K}_{{C}_{{h}_{i}}}\left(x,\left[0,y])is given by KChi(x,[0,y])=1[0,y](hi(x)){K}_{{C}_{{h}_{i}}}\left(x,\left[0,y])={{\mathbb{1}}}_{\left[0,y]}\left({h}_{i}\left(x))for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]and every i∈{1,…,4}i\in \left\{1,\ldots ,4\right\}, the Markov kernel KA˜(x,[0,y]){K}_{\tilde{A}}\left(x,\left[0,y])of A˜\tilde{A}can be expressed by KA˜(x,[0,y])=1[0,G1(y)](h1∘F1(x))forx∈U11[0,G2(y)](h2∘G1(x))forx∈V11[0,F1(y)](h3∘F2(x))forx∈U21[0,F2(y)](h4∘G2(x))forx∈V2{K}_{\tilde{A}}\left(x,\left[0,y])=\left\{\begin{array}{ll}{{\mathbb{1}}}_{\left[0,{G}_{1}(y)]}\hspace{0.33em}({h}_{1}\circ {F}_{1}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {U}_{1}\\ {{\mathbb{1}}}_{\left[0,{G}_{2}(y)]}\hspace{0.33em}({h}_{2}\circ {G}_{1}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {V}_{1}\\ {{\mathbb{1}}}_{\left[0,{F}_{1}(y)]}\hspace{0.33em}({h}_{3}\circ {F}_{2}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {U}_{2}\\ {{\mathbb{1}}}_{\left[0,{F}_{2}(y)]}\hspace{0.33em}({h}_{4}\circ {G}_{2}\left(x))\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in {V}_{2}\end{array}\right.for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]. Since {U1,U2,V1,V2}\left\{{U}_{1},{U}_{2},{V}_{1},{V}_{2}\right\}form a partition of [0,1]\left[0,1], for each y∈[0,1]y\in \left[0,1]we have that KA˜(x,[0,y])∈{0,1}{K}_{\tilde{A}}\left(x,\left[0,y])\in \left\{0,1\right\}for λ\lambda -a.e. x∈[0,1]x\in \left[0,1], which is equivalent to A˜\tilde{A}being completely dependent (see [3]). Using the same arguments we also obtain for every y∈[0,1]y\in \left[0,1]that KA˜t(x,[0,y])=(∂1A˜t)(x,y)=(∂2A˜)(y,x)∈{0,1}{K}_{{\tilde{A}}^{t}}\left(x,\left[0,y])=\left({\partial }_{1}{\tilde{A}}^{t})\left(x,y)=\left({\partial }_{2}\tilde{A})(y,x)\in \left\{0,1\right\}for λ\lambda -a.e. x∈[0,1]x\in \left[0,1], i.e., A˜t{\tilde{A}}^{t}is completely dependent too. Altogether, we have shown that A˜∈Cmcd\tilde{A}\in {{\mathcal{C}}}_{mcd}, which completes the proof.□5Maximal Dp{D}_{p}-asymmetrySince the metrics Dp{D}_{p}, p∈[1,∞]p\in \left[1,\infty ]induce the same topology on C{\mathcal{C}}one could conjecture that maximal Dp{D}_{p}-asymmetry might be the same as maximal D1{D}_{1}-asymmetry. We will falsify this idea and start with three simple lemmata.Lemma 5.1[27] Suppose that h1,h2{h}_{1},{h}_{2}are λ\lambda -preserving transformations on [0,1]\left[0,1]and let Ah1{A}_{{h}_{1}}, Ah2{A}_{{h}_{2}}denote the corresponding completely dependent copulas. ThenDpp(Ah1,Ah2)=D1(Ah1,Ah2)=‖h1−h2‖1{D}_{p}^{p}\left({A}_{{h}_{1}},{A}_{{h}_{2}})={D}_{1}\left({A}_{{h}_{1}},{A}_{{h}_{2}})=\Vert {h}_{1}-{h}_{2}{\Vert }_{1}holds for every p∈(1,∞)p\in \left(1,\infty ).Lemma 5.2The metric space (C,Dp)\left({\mathcal{C}},{D}_{p})has the following diameter: (1)diamDp(C)=2−1p{{\rm{diam}}}_{{D}_{p}}\left({\mathcal{C}})={2}^{-\tfrac{1}{p}}for p∈[1,∞)p\in {[}1,\infty ).(2)diamD∞(C)=1{{\rm{diam}}}_{{D}_{\infty }}\left({\mathcal{C}})=1.ProofAccording to Lemma 5 in [27] we have diamD1(C)=∫[0,1]min{2y,2(1−y)}dλ(y)=12.{{\rm{diam}}}_{{D}_{1}}\left({\mathcal{C}})=\mathop{\int }\limits_{\left[0,1]}\min \left\{2y,2\left(1-y)\right\}{\rm{d}}\lambda (y)=\frac{1}{2}.Since ∣KA(x,[0,y])−KB(x,[0,y])∣∈[0,1]| {K}_{A}\left(x,\left[0,y])-{K}_{B}\left(x,\left[0,y])| \in \left[0,1]it is straightforward to verify that (8)Dpp(A,B)≤D1(A,B)≤Dp(A,B){D}_{p}^{p}\left(A,B)\le {D}_{1}\left(A,B)\le {D}_{p}\left(A,B)holds for every A,B∈CA,B\in {\mathcal{C}}and p∈[1,∞)p\in {[}1,\infty ). As a direct consequence, we obtain Dp(A,B)≤D1(A,B)1p≤2−1p{D}_{p}\left(A,B)\le {D}_{1}{\left(A,B)}^{\tfrac{1}{p}}\le {2}^{-\tfrac{1}{p}}. On the other hand, there exist copulas A,B∈CA,B\in {\mathcal{C}}with Dp(A,B)=2−1p{D}_{p}\left(A,B)={2}^{-\tfrac{1}{p}}. Considering A=MA=Mand B=WB=Wand applying Lemma 5.1 yield Dpp(M,W)=D1(M,W)=12.{D}_{p}^{p}\left(M,W)={D}_{1}\left(M,W)=\frac{1}{2}.The assertion for p=∞p=\infty is a direct consequence of Lemma 5 in [27].□Slightly adapting the notation of the previous section we will now focus on the family Cκp=1{{\mathcal{C}}}^{{\kappa }_{p}=1}of all bivariate copulas with maximal Dp{D}_{p}-asymmetry, i.e., Cκp=1≔{A∈C:κp(A)≔21pDp(A,At)=1}{{\mathcal{C}}}^{{\kappa }_{p}=1}:= \left\{A\in {\mathcal{C}}:{\kappa }_{p}\left(A):= {2}^{\tfrac{1}{p}}{D}_{p}\left(A,{A}^{t})=1\right\}. Building upon Lemma 5.1 and Theorem 3.5 in [13] there are mutually completely dependent copulas A∈CmcdA\in {{\mathcal{C}}}_{mcd}such that κp(A)=1{\kappa }_{p}\left(A)=1is attained for every p∈[1,∞]p\in \left[1,\infty ]. In fact, the copula Ah{A}_{h}defined in Example 3.4 in [13] has maximal Dp{D}_{p}-asymmetry for every p∈[1,∞]p\in \left[1,\infty ]. The following lemma shows that a copula with maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty )also has maximal D1{D}_{1}-asymmetry.Lemma 5.3If A∈CA\in {\mathcal{C}}satisfies κp(A)=1{\kappa }_{p}\left(A)=1for some p∈(1,∞)p\in \left(1,\infty ), then κ1(A)=1{\kappa }_{1}\left(A)=1holds.ProofUsing the inequality Dpp(A,B)≤D1(A,B){D}_{p}^{p}\left(A,B)\le {D}_{1}\left(A,B)as well as the fact that D1(A,B)≤12{D}_{1}\left(A,B)\le \frac{1}{2}holds for all A,B∈CA,B\in {\mathcal{C}}we obtain 1=21pDp(A,At)≤(2D1(A,At))1p≤1,1={2}^{\tfrac{1}{p}}{D}_{p}\left(A,{A}^{t})\le {(2{D}_{1}\left(A,{A}^{t}))}^{\tfrac{1}{p}}\le 1,which yields D1(A,At)=12{D}_{1}\left(A,{A}^{t})=\frac{1}{2}.□The following example, however, shows that the reverse implication does not hold in general.Example 5.4Suppose that A∈CA\in {\mathcal{C}}corresponds to the uniform distribution on the union of the four squares (Figure 4) 0,14×14,24,14,24×24,34,24,34×34,1,34,1×0,14.\left(0,\frac{1}{4}\right)\times \left(\frac{1}{4},\frac{2}{4}\right),\left(\frac{1}{4},\frac{2}{4}\right)\times \left(\frac{2}{4},\frac{3}{4}\right),\left(\frac{2}{4},\frac{3}{4}\right)\times \left(\frac{3}{4},1\right),\left(\frac{3}{4},1\right)\times \left(0,\frac{1}{4}\right).Since AA(and At{A}^{t}) is a checkerboard copula (see, for instance, [11,18]) a version of the Markov kernel of AAis piecewise linear in yyfor fixed x∈[0,1]x\in \left[0,1]and does not depend on the choice of the point x∈i−14,i4x\in \left(\frac{i-1}{4},\frac{i}{4}\right)for every i∈{1,…,4}i\in \left\{1,\ldots ,4\right\}, (a version of) the corresponding Markov kernel is given by KA(x,[0,y])=(4y−1)114,24(y)+124,1(y)forx∈0,14(4y−2)124,34(y)+134,1(y)forx∈14,24(4y−3)134,1(y)forx∈24,34(4y)10,14(y)+114,1(y)forx∈34,1.{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}\left(4y-1){{\mathbb{1}}}_{\left(\tfrac{1}{4},\tfrac{2}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{2}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(0,\frac{1}{4}\right)\\ \left(4y-2){{\mathbb{1}}}_{\left(\tfrac{2}{4},\tfrac{3}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{3}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{1}{4},\frac{2}{4}\right)\\ \left(4y-3){{\mathbb{1}}}_{\left(\tfrac{3}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{2}{4},\frac{3}{4}\right)\\ \left(4y){{\mathbb{1}}}_{\left[0,\tfrac{1}{4}\right]}(y)+{{\mathbb{1}}}_{\left(\tfrac{1}{4},1\right]}(y)\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\in \left(\frac{3}{4},1\right).\end{array}\right.It is straightforward to verify κ1(A)=1{\kappa }_{1}\left(A)=1(e.g., by using property (iv) or property (v) in Theorem 4.4). On the other hand, simple calculations (see Appendix A) yield Dpp(A,At)=14+2∫[0,14](4x)pdλ(x)=14+24p+4{D}_{p}^{p}\left(A,{A}^{t})=\frac{1}{4}+2\mathop{\int }\limits_{\left[0,\frac{1}{4}]}{\left(4x)}^{p}{\rm{d}}\lambda \left(x)=\frac{1}{4}+\frac{2}{4p+4}for every p∈[1,∞)p\in {[}1,\infty ). As a direct consequence we obtain Dpp(A,At)<2−1{D}_{p}^{p}\left(A,{A}^{t})\lt {2}^{-1}for every p∈(1,∞)p\in \left(1,\infty ), i.e., although AAhas maximal D1{D}_{1}-asymmetry, it fails to have maximal Dp{D}_{p}-asymmetry.Figure 4Density of the copula AA(left panel) and the copula At{A}^{t}(right panel) considered in Example 5.4. The copula AAhas maximal D1{D}_{1}-asymmetry, i.e., κ1(A)=1{\kappa }_{1}\left(A)=1, nevertheless κp(A)<1{\kappa }_{p}\left(A)\lt 1holds for p∈(1,∞)p\in \left(1,\infty ).Contrary to D1{D}_{1}, the class Cκp=1{{\mathcal{C}}}^{{\kappa }_{p}=1}, p∈(1,∞)p\in \left(1,\infty )only contains mutually completely dependent copulas.Theorem 5.5If A∈CA\in {\mathcal{C}}has maximal Dp{D}_{p}-asymmetry for p∈(1,∞)p\in \left(1,\infty ), then A is a mutually completely dependent copula.ProofIf κp(A)=1{\kappa }_{p}\left(A)=1we have κ1(A)=1{\kappa }_{1}\left(A)=1and Dp(A,At)=2−1p{D}_{p}\left(A,{A}^{t})={2}^{-\tfrac{1}{p}}, which implies 12=∫[0,1]∫[0,1]∣KA(x,[0,y])−KAt(x,[0,y])∣pdλ(x)dλ(y)≤∫[0,1]∫[0,1]∣KA(x,[0,y])−KAt(x,[0,y])∣dλ(x)dλ(y)=12.\begin{array}{rcl}\frac{1}{2}& =& \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}{| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| }^{p}{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)\\ & \le & \mathop{\displaystyle \int }\limits_{\left[0,1]}\mathop{\displaystyle \int }\limits_{\left[0,1]}| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| {\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)=\frac{1}{2}.\end{array}Therefore, we obtain ∣KA(x,[0,y])−KAt(x,[0,y])∣p=∣KA(x,[0,y])−KAt(x,[0,y])∣,{| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| }^{p}=| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| ,or equivalently, that (9)∣KA(x,[0,y])−KAt(x,[0,y])∣∈{0,1}| {K}_{A}\left(x,\left[0,y])-{K}_{{A}^{t}}\left(x,\left[0,y])| \in \left\{0,1\right\}holds for λ2{\lambda }^{2}-a.e. (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}. According to Lemma 5.3 and property (iii) in Theorem 4.4 there exist sets U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])and V∈ℬ([0,1])V\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])such that U∩V=∅U\cap V=\varnothing , λ(U)=λ(V)=12\lambda \left(U)=\lambda \left(V)=\frac{1}{2}, and KA(x,[0,y])=≤1for everyy∈0,121for everyy∈12,1KAt(x,[0,y])=0for everyy∈0,12≤1for everyy∈12,1,{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}\le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.\hspace{1.0em}{K}_{{A}^{t}}\left(x,\left[0,y])=\left\{\begin{array}{ll}0\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ \le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.,for every x∈Ux\in Uas well as KA(x,[0,y])=0for everyy∈0,12≤1for everyy∈12,1KAt(x,[0,y])=≤1for everyy∈0,121for everyy∈12,1,{K}_{A}\left(x,\left[0,y])=\left\{\begin{array}{ll}0\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ \le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.\hspace{1.0em}{K}_{{A}^{t}}\left(x,\left[0,y])=\left\{\begin{array}{ll}\le 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left[0,\frac{1}{2}\right]\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for every}\hspace{0.1em}\hspace{0.33em}y\in \left(\frac{1}{2},1\right]\end{array}\right.,for every x∈Vx\in V. Fix x∈Ux\in Usuch that equation (9) holds and suppose that KA(x,[0,y])=y0∈(0,1){K}_{A}\left(x,\left[0,y])={y}_{0}\in \left(0,1)for some y∈0,12y\in \left[0,\frac{1}{2}\right]. Then due to equation (9) the Markov kernel of At{A}^{t}must satisfy KAt(x,[0,y])=y0{K}_{{A}^{t}}\left(x,\left[0,y])={y}_{0}, which is a contradiction to the fact that KAt(x,[0,y])=0{K}_{{A}^{t}}\left(x,\left[0,y])=0for every y∈0,12y\in \left[0,\frac{1}{2}\right]. Hence, we obtain that KA(x,[0,y])∈{0,1}{K}_{A}\left(x,\left[0,y])\in \left\{0,1\right\}for every y∈[0,1]y\in \left[0,1]. In an analogous way, we obtain that KAt(x,[0,y])∈{0,1}{K}_{{A}^{t}}\left(x,\left[0,y])\in \left\{0,1\right\}holds for every y∈[0,1]y\in \left[0,1]. Proceeding in the exactly same manner for x∈Vx\in Vwe obtain that for λ\lambda -a.e. x∈[0,1]x\in \left[0,1]and every y∈[0,1]y\in \left[0,1]the Markov kernels of AAand At{A}^{t}satisfy KA(x,[0,y])∈{0,1}{K}_{A}\left(x,\left[0,y])\in \left\{0,1\right\}and KAt(x,[0,y])∈{0,1}{K}_{{A}^{t}}\left(x,\left[0,y])\in \left\{0,1\right\}. By Theorem 7.1 in [3] and Lemma 3.4 in [10], it follows that AAand At{A}^{t}are completely dependent, implying that AAis mutually completely dependent.□Altogether we have shown the following results:Corollary 5.6The following properties hold: (1)Cκ1=1=Cκ∞=1{{\mathcal{C}}}^{{\kappa }_{1}=1}={{\mathcal{C}}}^{{\kappa }_{\infty }=1}.(2)Cκ1=1⊋Cκp=1{{\mathcal{C}}}^{{\kappa }_{1}=1}\hspace{0.33em} \supsetneq \hspace{0.33em}{{\mathcal{C}}}^{{\kappa }_{p}=1}for every p∈(1,∞)p\in \left(1,\infty ).(3)Cmcdκ1=1=Cκp=1{{\mathcal{C}}}_{mcd}^{{\kappa }_{1}=1}={{\mathcal{C}}}^{{\kappa }_{p}=1}for every p∈(1,∞)p\in \left(1,\infty ).6Maximal Dp{D}_{p}-asymmetric copulas and their values for ζ1{\zeta }_{1}and ξ\xi In Section 3, we have shown that copulas A∈CA\in {\mathcal{C}}with maximal d∞{d}_{\infty }-asymmetry have very high dependence scores with respect to ζ1{\zeta }_{1}and ξ\xi . Here we now focus on the range of these dependence measures to maximal Dp{D}_{p}-asymmetric copulas.Theorem 6.1If A∈CA\in {\mathcal{C}}satisfies κp(A)=1{\kappa }_{p}\left(A)=1for some p∈(1,∞)p\in \left(1,\infty ), then ζ1(A)=ξ(A)=1{\zeta }_{1}\left(A)=\xi \left(A)=1.ProofSince ζ1(A){\zeta }_{1}\left(A)and ξ(A)\xi \left(A)are 1 if and only if AAis completely dependent, the assertion directly follows from Theorem 5.5.□For the case p=1p=1different values for ξ\xi and ζ1{\zeta }_{1}are possible.Theorem 6.2If A∈CA\in {\mathcal{C}}satisfies κ1(A)=1{\kappa }_{1}\left(A)=1, then ξ(A)∈12,1\xi \left(A)\in \left(\frac{1}{2},1\right]holds.ProofProceeding analogously to the proof of Theorem 4.4 we obtain (At∗A)12,12=12\left({A}^{t}\ast A)\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}if κ1(A)=1{\kappa }_{1}\left(A)=1(see Appendix A). Since At∗A{A}^{t}\ast Ais a copula, we find copulas A1,A2∈C{A}_{1},{A}_{2}\in {\mathcal{C}}with (At∗A)=i−12,i2,Aii∈{1,2}≕A˜\left({A}^{t}\ast A)={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2\right\}}\hspace{0.33em}=: \hspace{0.33em}\tilde{A}. Setting CΠ≔i−12,i2,Πi∈{1,2}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{2},\frac{i}{2},\Pi \right\rangle \right)}_{i\in \left\{1,2\right\}}, μAt∗A≠μCΠ{\mu }_{{A}^{t}\ast A}\ne {\mu }_{{C}_{\Pi }}follows. In fact, according to Theorem 4.4 there exists a set U∈ℬ([0,1])U\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])such that λU∩0,12=λU∩12,1=14\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right)=\lambda \left(U\cap \left[\frac{1}{2},1\right]\right)=\frac{1}{4}and 0=μA0,12×U=∫0,12KA(x,U)dλ(x)0={\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\mathop{\int }\limits_{\left[0,\frac{1}{2}\right]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x). Hence, we can find Borel sets Λ1⊆0,12{\Lambda }_{1}\subseteq \left[0,\frac{1}{2}\right], Λ2⊆12,1{\Lambda }_{2}\subseteq \left(\frac{1}{2},1\right]such that λ(Λ1)=λ(Λ2)=12\lambda \left({\Lambda }_{1})=\lambda \left({\Lambda }_{2})=\frac{1}{2}and KA(x,U)=0{K}_{A}\left(x,U)=0for all x∈Λ1x\in {\Lambda }_{1}and KA(x,U)=1{K}_{A}\left(x,U)=1for all x∈Λ2x\in {\Lambda }_{2}. The set Λ3{\Lambda }_{3}defined by [0,1]⧹(Λ1∪Λ2)\left[0,1]\setminus \left({\Lambda }_{1}\cup {\Lambda }_{2})obviously fulfills λ(Λ3)=0\lambda \left({\Lambda }_{3})=0. Hence, we have μAt∗A(U×U)=∫U∫[0,1]KA(s,U)KAt(x,ds)dλ(x)=∫UKAt(x,Λ2)dλ(x)+∫U∫Λ3KA(s,U)KAt(x,ds)dλ(x)≥∫UKAt(x,Λ2)dλ(x)=μAt(U×Λ2)=μA(Λ2×U)=∫Λ2KA(x,U)dλ(x)=∫12,1KA(x,U)dλ(x)=μA12,1×U=λ(U)−μA0,12×U=λ(U)=12.\begin{array}{rcl}{\mu }_{{A}^{t}\ast A}\left(U\times U)& =& \mathop{\displaystyle \int }\limits_{U}\mathop{\displaystyle \int }\limits_{\left[0,1]}{K}_{A}\left(s,U){K}_{{A}^{t}}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{U}{K}_{{A}^{t}}\left(x,{\Lambda }_{2}){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U}\mathop{\displaystyle \int }\limits_{{\Lambda }_{3}}{K}_{A}\left(s,U){K}_{{A}^{t}}\left(x,{\rm{d}}s){\rm{d}}\lambda \left(x)\\ & \ge & \mathop{\displaystyle \int }\limits_{U}{K}_{{A}^{t}}\left(x,{\Lambda }_{2}){\rm{d}}\lambda \left(x)={\mu }_{{A}^{t}}\left(U\times {\Lambda }_{2})={\mu }_{A}\left({\Lambda }_{2}\times U)=\mathop{\displaystyle \int }\limits_{{\Lambda }_{2}}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{\left(\frac{1}{2},1\right]}{K}_{A}\left(x,U){\rm{d}}\lambda \left(x)={\mu }_{A}\left(\left(\frac{1}{2},1\right]\times U\right)=\lambda \left(U)-{\mu }_{A}\left(\left[0,\frac{1}{2}\right]\times U\right)=\lambda \left(U)=\frac{1}{2}.\end{array}On the other hand, μCΠ(U×U)=∫UKCΠ(x,U)dλ(x)=∫U∩[0,12]KCΠ(x,U)dλ(x)+∫U∩(12,1]KCΠ(x,U)dλ(x)=∫U∩[0,12]2λU∩0,12dλ(x)+∫U∩(12,1]2λ(U∩12,1)dλ(x)=14\begin{array}{rcl}{\mu }_{{C}_{\Pi }}\left(U\times U)& =& \mathop{\displaystyle \int }\limits_{U}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)=\mathop{\displaystyle \int }\limits_{U\cap \left[0,\frac{1}{2}]}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U\cap \left(\frac{1}{2},1]}{K}_{{C}_{\Pi }}\left(x,U){\rm{d}}\lambda \left(x)\\ & =& \mathop{\displaystyle \int }\limits_{U\cap \left[0,\frac{1}{2}]}2\lambda \left(U\cap \left[0,\frac{1}{2}\right]\right){\rm{d}}\lambda \left(x)+\mathop{\displaystyle \int }\limits_{U\cap \left(\frac{1}{2},1]}2\lambda \left(U\cap \left(\frac{1}{2},1\right]){\rm{d}}\lambda \left(x)=\frac{1}{4}\end{array}holds, implying Ai≠Π{A}_{i}\ne \Pi for i=1,2i=1,2, hence, considering that according to Lemma 4.3 we have D22(A,Π)≥D22(At∗A,At∗Π)=D22(At∗A,Π){D}_{2}^{2}\left(A,\Pi )\ge {D}_{2}^{2}\left({A}^{t}\ast A,{A}^{t}\ast \Pi )={D}_{2}^{2}\left({A}^{t}\ast A,\Pi )and applying Corollary 3.2 finally yields 1≥ξ(A)≥ξ(A˜)=14ξ(A1)+14ξ(A2)+12>12.□\hspace{10.25em}1\ge \xi \left(A)\ge \xi \left(\tilde{A})=\frac{1}{4}\xi \left({A}_{1})+\frac{1}{4}\xi \left({A}_{2})+\frac{1}{2}\gt \frac{1}{2}.\hspace{15em}\square Theorem 6.3If A∈CA\in {\mathcal{C}}satisfies κ1(A)=1{\kappa }_{1}\left(A)=1, then ζ1(A)∈34,1{\zeta }_{1}\left(A)\in \left[\frac{3}{4},1\right]holds.ProofUsing the same arguments as in the proof of Theorem 6.2 we find copulas A1,A2∈C{A}_{1},{A}_{2}\in {\mathcal{C}}such that (At∗A)=i−12,i2,Aii∈{1,2}≕A˜\left({A}^{t}\ast A)={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}\right\rangle \right)}_{i\in \left\{1,2\right\}}\hspace{0.33em}=: \hspace{0.33em}\tilde{A}holds. Since A˜\tilde{A}is an ordinal sum it is clear that the (SI)-rearrangement A˜↑{\tilde{A}}^{\uparrow }of A˜\tilde{A}satisfies A˜↑=i−12,i2,Ai↑i∈{1,2}{\tilde{A}}^{\uparrow }={\left(\left\langle \frac{i-1}{2},\frac{i}{2},{A}_{i}^{\uparrow }\right\rangle \right)}_{i\in \left\{1,2\right\}}. As SI copula, Ai↑{A}_{i}^{\uparrow }fulfills Ai↑(x,y)≥Π(x,y){A}_{i}^{\uparrow }\left(x,y)\ge \Pi \left(x,y)for all (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}and i∈{1,2}i\in \left\{1,2\right\}, implying A˜↑(x,y)≥CΠ(x,y){\tilde{A}}^{\uparrow }\left(x,y)\ge {C}_{\Pi }\left(x,y)for every (x,y)∈[0,1]2\left(x,y)\in {\left[0,1]}^{2}, whereby CΠ{C}_{\Pi }is defined as CΠ≔i−12,i2,Πi∈{1,2}{C}_{\Pi }:= {\left(\left\langle \frac{i-1}{2},\frac{i}{2},\Pi \right\rangle \right)}_{i\in \left\{1,2\right\}}. Hence, using Lemma 4.3 we obtain 13≥D1(A,Π)≥D1(At∗A,At∗Π)=D1(At∗A,Π)=D1(A˜↑,Π)≥D1(CΠ,Π),\frac{1}{3}\ge {D}_{1}\left(A,\Pi )\ge {D}_{1}\left({A}^{t}\ast A,{A}^{t}\ast \Pi )={D}_{1}\left({A}^{t}\ast A,\Pi )={D}_{1}\left({\tilde{A}}^{\uparrow },\Pi )\ge {D}_{1}\left({C}_{\Pi },\Pi ),whereby we used the fact that D1(A,Π){D}_{1}\left(A,\Pi )is monotone with respect to the pointwise order in C↑{{\mathcal{C}}}^{\uparrow }(see [26]). Using Lemma 3.1 we obtain ζ1(CΠ)=3D1(CΠ,Π)=34∫[0,1]∫[0,1]y−y2dλ(x)dλ(y)+34∫[0,1]∫[0,1]y−12−y2dλ(x)dλ(y)+38=316+38−316+38=68=34,{\zeta }_{1}\left({C}_{\Pi })=3{D}_{1}\left({C}_{\Pi },\Pi )=\frac{3}{4}\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|y-\frac{y}{2}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\frac{3}{4}\mathop{\int }\limits_{\left[0,1]}\mathop{\int }\limits_{\left[0,1]}\left|y-\frac{1}{2}-\frac{y}{2}\right|{\rm{d}}\lambda \left(x){\rm{d}}\lambda (y)+\frac{3}{8}=\frac{3}{16}+\frac{3}{8}-\frac{3}{16}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4},which completes the proof.□The following example demonstrates that it is possible to find copulas A∈Cκ1=1A\in {{\mathcal{C}}}^{{\kappa }_{1}=1}such that ζ1(A){\zeta }_{1}\left(A)(or ξ(A)\xi \left(A), respectively) is arbitrarily close to the lower bound derived in Theorems 6.2 and 6.3.Example 6.4Let n∈Nn\in {\mathbb{N}}be a natural number with n≥3n\ge 3, set N≔2nN:= {2}^{n}and define the sets UN1,UN2,VN1,VN2∈ℬ([0,1]){U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\in {\mathcal{ {\mathcal B} }}\left(\left[0,1])by UN1=⋃j=1N42j−2N,2j−1N,VN1=⋃j=1N42j−1N,2jN,UN2=⋃j=1N412+2j−2N,12+2j−1N,VN2=⋃j=1N412+2j−2N,12+2j−1N.\begin{array}{rcl}{U}_{N}^{1}& =& \mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{2j-2}{N},\frac{2j-1}{N}\right),\hspace{1em}{V}_{N}^{1}=\mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{2j-1}{N},\frac{2j}{N}\right),\\ {U}_{N}^{2}& =& \mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{1}{2}+\frac{2j-2}{N},\frac{1}{2}+\frac{2j-1}{N}\right),{V}_{N}^{2}=\mathop{\bigcup }\limits_{j=1}^{\frac{N}{4}}\left(\frac{1}{2}+\frac{2j-2}{N},\frac{1}{2}+\frac{2j-1}{N}\right).\end{array}Obviously, we have λ(UN1)=λ(UN2)=λ(VN1)=λ(VN2)=14\lambda \left({U}_{N}^{1})=\lambda \left({U}_{N}^{2})=\lambda \left({V}_{N}^{1})=\lambda \left({V}_{N}^{2})=\frac{1}{4}and λ(UN1∪UN2∪VN1∪VN2)=1\lambda \left({U}_{N}^{1}\cup {U}_{N}^{2}\cup {V}_{N}^{1}\cup {V}_{N}^{2})=1. Letting AN{A}_{N}denote the copula corresponding to the uniform distribution on the union of the four sets UN1×VN1{U}_{N}^{1}\times {V}_{N}^{1}, UN2×UN1{U}_{N}^{2}\times {U}_{N}^{1}, VN1×VN2{V}_{N}^{1}\times {V}_{N}^{2}and VN2×UN2{V}_{N}^{2}\times {U}_{N}^{2}(Figure 5), then by Theorem 4.4 AN{A}_{N}has maximal D1{D}_{1}-asymmetry. As next step we calculate the dependence measure ζ1(AN){\zeta }_{1}\left({A}_{N}).Applying Lemma 6.3 in [10] and using the fact that for every y∈[0,1]y\in \left[0,1]the identity of KAN(x1,[0,y])=KAN(x2,[0,y]){K}_{{A}_{N}}\left({x}_{1},\left[0,y])={K}_{{A}_{N}}\left({x}_{2},\left[0,y])holds for λ\lambda -a.e. x1,x2∈X{x}_{1},{x}_{2}\in X, whereby X∈{UN1,UN2,VN1,VN2}X\in \left\{{U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\right\}, we obtain ζ1(AN)3=D1(AN,Π)≤2N+1N∑j=1N∫[0,1]KANx,0,jN−jNdλ(x)=2N+∑X∈{UN1,UN2,VN1,VN2}1N∑j=1N∫XKANx,0,jN−jNdλ(x)︸≕m(X).\begin{array}{rcl}\frac{{\zeta }_{1}\left({A}_{N})}{3}& =& {D}_{1}\left({A}_{N},\Pi )\le \frac{2}{N}+\frac{1}{N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\mathop{\displaystyle \int }\limits_{\left[0,1]}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|{\rm{d}}\lambda \left(x)\\ & =& \frac{2}{N}+\displaystyle \sum _{X\in \left\{{U}_{N}^{1},{U}_{N}^{2},{V}_{N}^{1},{V}_{N}^{2}\right\}}\mathop{\underbrace{\frac{1}{N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\mathop{\displaystyle \int }\limits_{X}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|{\rm{d}}\lambda \left(x)}}\limits_{=: m\left(X)}.\end{array}Considering X=UN1X={U}_{N}^{1}and x∈UN1x\in {U}_{N}^{1}, a version of the Markov kernel KAnx,0,jN{K}_{{A}_{n}}\left(x,\left[0,\frac{j}{N}\right]\right)is given by KANx,0,jN=2jNforj∈2,4,…,N22(j−1)Nforj∈1,3,…,N2−11forj>N2,{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)=\left\{\begin{array}{ll}\frac{2j}{N}\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\in \left\{2,4,\ldots ,\frac{N}{2}\right\}\\ \frac{2\left(j-1)}{N}\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\in \left\{1,3,\ldots ,\frac{N}{2}-1\right\}\\ 1\hspace{1.0em}& \hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}j\gt \frac{N}{2},\end{array}\right.which yields m(UN1)=14N∑j=1NKANx,0,jN−jN=14N∑j∈{2,4,…,N∕2}2jN−jN+∑j∈{1,3,…,N∕2−1}2(j−1)N−jN+∑j=N2+1N1−jN=14N∑j=1N44jN−2jN+∑j=1N44(j−1)N−2j−1N+∑j=1N212−jN=14NN4+2N−12≤116+12N2.\begin{array}{rcl}m\left({U}_{N}^{1})& =& \frac{1}{4N}\mathop{\displaystyle \sum }\limits_{j=1}^{N}\left|{K}_{{A}_{N}}\left(x,\left[0,\frac{j}{N}\right]\right)-\frac{j}{N}\right|\\ & =& \frac{1}{4N}\left(\displaystyle \sum _{j\in \left\{2,4,\ldots ,N/2\right\}}\left|\frac{2j}{N}-\frac{j}{N}\right|+\displaystyle \sum _{j\in \left\{1,3,\ldots ,N/2-1\right\}}\left|\frac{2\left(j-1)}{N}-\frac{j}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=\frac{N}{2}+1}^{N}\left(1-\frac{j}{N}\right)\right)\\ & =& \frac{1}{4N}\left(\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{4}}\left|\frac{4j}{N}-\frac{2j}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{4}}\left|\frac{4\left(j-1)}{N}-\frac{2j-1}{N}\right|+\mathop{\displaystyle \sum }\limits_{j=1}^{\frac{N}{2}}\left(\frac{1}{2}-\frac{j}{N}\right)\right)\\ & =& \frac{1}{4N}\left(\frac{N}{4}+\frac{2}{N}-\frac{1}{2}\right)\le \frac{1}{16}+\frac{1}{2{N}^{2}}.\end{array}In a similar manner, we obtain m(UN2)=116+18Nm\left({U}_{N}^{2})=\frac{1}{16}+\frac{1}{8N}, m(VN1)=116+18Nm\left({V}_{N}^{1})=\frac{1}{16}+\frac{1}{8N}, and m(VN2)≤116+12N2m\left({V}_{N}^{2})\le \frac{1}{16}+\frac{1}{2{N}^{2}}. Together with Theorem 6.3 it follows that 34≤ζ1(AN)≤34+274N+3N2,\frac{3}{4}\le {\zeta }_{1}\left({A}_{N})\le \frac{3}{4}+\frac{27}{4N}+\frac{3}{{N}^{2}},which shows that for sufficiently large N∈NN\in {\mathbb{N}}the dependence value ζ1(AN){\zeta }_{1}\left({A}_{N})is arbitrarily close to 34\frac{3}{4}.Using similar calculations (see Appendix A) yields 12<ξ(AN)≤12+aN,\frac{1}{2}\lt \xi \left({A}_{N})\le \frac{1}{2}+{a}_{N},whereby limN→∞aN=0{\mathrm{lim}}_{N\to \infty }{a}_{N}=0.Figure 5Density of the copula AN{A}_{N}(gray) as considered in Example 6.4 and support of the mutually completely dependent copula BN{B}_{N}(magenta) according to Remark 6.5 for N=8N=8(left panel) and N=32N=32(right panel).Remark 6.5Slightly modifying the construction from Example 6.4 (which corresponds to copying shrunk versions of the product copula Π\Pi in the small squares) we now construct the copula BN{B}_{N}by copying shrunk versions of MMin every square of the “diagonal” of each of the four sets UN1×VN1{U}_{N}^{1}\times {V}_{N}^{1}, UN2×UN1{U}_{N}^{2}\times {U}_{N}^{1}, VN1×VN2{V}_{N}^{1}\times {V}_{N}^{2}, and VN2×UN2{V}_{N}^{2}\times {U}_{N}^{2}as depicted in Figure 5 (magenta lines). The shuffle BN{B}_{N}is obviously maximal D1{D}_{1}-asymmetric and, being completely dependent, fulfills ζ1(BN)=1=ξ(BN){\zeta }_{1}\left({B}_{N})=1=\xi \left({B}_{N}). Hence, setting CNα≔αAN+(1−α)BN{C}_{N}^{\alpha }:= \alpha {A}_{N}+\left(1-\alpha ){B}_{N}for every α∈[0,1]\alpha \in \left[0,1](with AN{A}_{N}according to Example 6.4) obviously yields a maximal D1{D}_{1}-asymmetric copula CNα{C}_{N}^{\alpha }. Due to the fact that ζ1(CNα){\zeta }_{1}\left({C}_{N}^{\alpha })and ξ(CNα)\xi \left({C}_{N}^{\alpha })are continuous in α\alpha the intermediate value theorem implies that for every s∈[ζ1(AN),1]s\in \left[{\zeta }_{1}\left({A}_{N}),1]we can find a copula CNα{C}_{N}^{\alpha }with ζ1(CNα)=s{\zeta }_{1}\left({C}_{N}^{\alpha })=sand the same result holds for ζ1{\zeta }_{1}replaced by ξ\xi . In other words, each point in the intervals mentioned in Theorems 6.2 and 6.3 is attained.Remark 6.6We have been able neither to find a copula A∈Cκ1=1A\in {{\mathcal{C}}}^{{\kappa }_{1}=1}fulfilling ζ1(A)=34{\zeta }_{1}\left(A)=\frac{3}{4}, nor to prove that such a copula does not exist.

Journal

Dependence Modelingde Gruyter

Published: Jan 1, 2022

Keywords: asymmetry; copula; dependence measure; exchangeability; Markov kernel; 62H05; 62H20; 60E05; 54E52

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