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On all Pickands Dependence Functions whose corresponding Extreme-Value-Copulas have Spearman ρ (Kendall τ) identical to some value v ∈ [0,1]

We answer an open question posed by the second author at the Salzburg workshop on Dependence Models and Copulas in 2016 concerning the size of the family A^ρ_v (A^τ_v) of all Pickands dependence functions A whose corresponding Extreme-Value-Copulas have Spearman ρ (Kendall τ) equal to some arbitrary, fixed value v ∈ [0,1]. After determining compact sets Ω^ρ_v, Ω^τ_v ⊆ [0,1] × [1/2,1] containing the graphs of all Pickands dependence functions from the classes A^ρ_v and A^τ_v respectively, we then show that both sets are best possible.

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1 Notation and preliminaries

Let denote the family of all two-dimensional copulas, and the class of all extreme-value copulas. As usual, denotes the uniform metric on . will denote the set of all Pickands dependence functions (see [2]). Arzela-Ascoli thereom ([1]) implies that is a compact (and convex) subset of the Banach space . For every the corresponding EVC will be denoted by , i.e.

It is well-known that Spearman’s and Kendall’s can be calculated as

(1)
(2)

For every we will let () denote the family of all Pickands dependence functions fulfilling (). Given we will write if for all with strict inequality in at least one point (and hence on an interval).

In the sequel we will work with the following five types of piecewise linear Pickands dependence functions, see Figure 1.
(i) Consider and and let be defined by

(ii) For every define by

(iii) For every and define by

(iv) For every , and let be defined by

(v) For every and define by

Figure 1: The five types of auxiliary functions considered.

2 Spearman’s

A straightforward calculation yields . Hence, defining by

we get

(3)

for every with , i.e. does not depend on . Obviously is a strictly decreasing homeomorphism mapping onto . Furthermore, letting denote its inverse,

(4)

holds for every .

A straightforward calculation yields , implying

(5)

Considering we get

and defining by

we get

(6)

for every . Notice that for obviously , hence , so is a strictly decreasing homeomorphism mapping onto . Letting denote its inverse,

(7)

holds for every .

For every and every obviously . Additionally, for , fixed and we have , implying that the mapping is an increasing homeomorphism mapping onto .

Again consider . Then for every there exists exactly one with . It is easy to verify that this is given by

and that as well as holds. Since the derivative of is given by

we get . As direct consequence, is strictly increasing and for

(8)

This non-contractivity property will be key in the proof of Lemma 2.2.

Straightforward (but tedious) calculations (see Appendix) show that the upper envelope of the family is given by

(9)

The function is symmetric w.r.t. , convex and continuously differentiable on .

We now show that holds for every and and start with the lower bound.

Lemma 2.1.

Suppose that . Then every fulfills .

Proof.

The assertion is trivial for , so we may consider . Suppose that and that there exists some with . Convexity of together with implies , from which we immediately get

a contradiction to . ∎

Lemma 2.2.

Suppose that . Then every fulfills .

Proof.

For the extreme cases and we get and respectively, so the result obviously holds. For the rest of the proof we consider . Suppose that and that for some . The case yields a contradiction immediately, so we assume . Symmetry of implies that it suffices to consider , continuity of yields .
Define by (increasing green with positive slope in Figure 2). Convexity of yields for and for . Let denote the unique point in the interval fulfilling . The following observation is key for the rest of the proof: For every , setting , defining by , and letting denote the unique point in fulfilling

(10)

holds.
Step 1: Define by (green line with negative slope in Figure 2). Convexity of yields for and for . Let denote the unique point fulfilling and set . Then and for . Case 1: If , then , so , contradiction.
Case 2: If and for all , then , contradiction.
Case 3: If and that holds for some we proceed with Step 2.
Step 2: Define the function by (blue line starting from in Figure 2). Then for and convexity of implies for . Let denote the unique point with and set . Considering and using ineq. (8) we get . In case of jump to the final step, if not, continue in the same manner to construct , where denote the first integer fulfilling and (notice that can be reached in finally many steps since holds, Figure 2 depicts the case ).
Final step: Since implies , which directly yields a contradiction, assume that there exists some with and set . Convexity of implies for as well as for . Considering we get and , contradiction. ∎

Figure 2: The construction used in the proof of Lemma 2.2.
Figure 3: All lie in the shaded region . The graphic depicts the cases (upper left panel), (upper right panel), for (lower left panel), and (lower right panel).

Summing up, we have proved the following result, where is defined by

Theorem 2.3.

Let be arbitrary but fixed. Then every fulfills . Additionally, is best-possible in the sense that for each point there exists some with .

Proof.

The first assertion has already been proved, the second one is a direct consequence of the construction via the functions and . ∎

Figure 3 depicts the set for some choices of .

3 Kendall

It is well-known that implies . In the sequel, however, we need strict inequality as stated in the following lemma:

Lemma 3.1.

For with we have .

Proof.

If then holds and we get

(11)

According to [2], setting

for every the support of coincides with the set , whereby is defined as for , and as for .
Suppose now that holds and that does not coincide with (in which case is trivial). Obviously as well as and we can find some fulfilling . By continuity there exists some such that holds for every . Considering

and

shows and . Hence

follows, and applying eq. (3) yields . ∎

Defining by we get

(12)

for every with , i.e. does not depend on . Obviously is a strictly decreasing homeomorphism mapping onto . Furthermore, letting denote its inverse,

(13)

holds for every .
A straightforward calculation yields

(14)

for and for . Considering and defining by

we get

(15)

for every . Obviously is a strictly decreasing homeomorphism mapping onto . Letting denote its inverse,

holds for every .

For every and every obviously . Additionally, for fixed and we have , implying that the mapping is an increasing homeomorphism mapping onto .

Again consider . Then for every there exists exactly one with . It is easy to verify that this is given by

and that as well as holds. Since the derivative of is given by

we get . As direct consequence, is strictly increasing and for

(16)

holds. This non-contractivity property will be key in the proof of Lemma 3.3.

For every , , so the upper envelope of the family is given by