33 DietmarPfeiferPaper
33 DietmarPfeiferPaper
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              New copulas based on general partitions-of-unity
                and their applications to risk management
1. Introduction
The theory of copulas and their applications has gained much interest in the recent
years, especially in the field of quantitative risk management, insurance and finance
(see e.g. MCNEIL, FREY AND EMBRECHTS (2005) or RANK (2006)). While classical
approaches like elliptically contoured copulas and Archimedean copulas are
widely explored, more modern approaches concentrate on non-standard, non-
symmetric or data-driven copula constructions (see e.g. LAUTERBACH AND PFEIFER
(2015), LAUTERBACH (2014), COTTIN AND PFEIFER (2014) or JAWORSKI, DURANTE AND HÄRDLE
(2013) and the papers therein for a survey). Statistical and computational aspects of
copulas have also been investigated in more detail recently (see e.g. BLUMENTRITT
(2012) and MAI AND SCHERER (2012)). In this paper, we want to focus on a particular
class of copulas and their generalizations, the so called partition-of-unity copulas
(see e.g. LI, MIKUSIŃSKI AND TAYLOR (1998) or KULPA (1999)). Whereas in the usual
approach, only finite partitions-of-unity are considered, which do not allow for a
modelling of tail-dependence, we extend this concept to infinite partitions-of-unity,
which allows for tail-dependence as well as for asymmetry, and which can also be
used to fit given data to a more realistic copula model. Our investigations resemble
in some sense more recent approaches such as YANG ET AL. (2015), GONZÁLEZ-BARRIOS
AND HERNÁNDEZ-CEDILLO (2013), ZHENG ET AL. (2011), HUMMEL AND MÄRKERT (2011), or
GHOSH AND HENDERSON (2009). Whereas in these papers, local modifications of known
standard copulas are considered in order to obtain tail dependence or
asymmetries, we focus on a closed form representation of completely new copula
densities which allows for easy Monte Carlo simulations as well as a data driven
modelling of tail dependence and asymmetries. This approach is not restricted to
two dimensions in general, but can likewise be used in arbitrary dimensions.
However, in order to illustrate our results, we will give examples in the bivariate case
only.
Let + = {0,1,2,3,} denote the set of non-negative integers and suppose that
{ji (u)}iÎ   +   and        {y (v)}
                                 j         i Î+
                                                     are non-negative maps defined on the interval (0,1) each
such that
¥ ¥
                                                                      å ji (u) = å yj (v) = 1
                                                                       i =0                   j =0
                                                                                                                                            (2.1)
and
                                      1                                       1
The maps ji (u) and yj (v) can be thought of as representing discrete distributions
over the non-negative integers + with parameters u and v, resp. The sequences
{ai }iÎ+ and {b j } + then represent the probabilities of the corresponding mixed
                                jÎ
distributions each.
                                                                 ¥     ¥         pij
                                               c(u, v):= å å                               ji (u)yj (v), u, v Î (0,1)                       (2.3)
                                                                i =0 j = 0    ai b j
        1                                                         1
                                     ¥     ¥        pij                                        ¥     ¥         pij
       ò      c(u, v)dv = å å                               ji (u) ò yj (v)dv = å å                                    b j ji (u)
        0                            i =0 j = 0    ai b j         0                           i =0 j =0       ai b j
                                       ¥     ¥      pij               ¥
                                                                              ji (u) ¥       ¥
                                                                                                ji (u)        ¥
                              = åå                          ji (u) = å              å  p = å           a  = å    ji (u) = 1,                (2.4)
                                      i =0 j = 0    ai                i =0     ai j=0 ij   i =0  ai     i
                                                                                                            i =0
                                                                             ¥     ¥
                                                            c(u, v) = åå pij fi (u) gj (v), u, v Î (0,1)                                    (2.5)
                                                                           i =0 j =0
                     ji (  )              yj (  )
where fi (  ) =              , gj (  ) =          , i, j Î + denote the Lebesgue densities induced by
                      ai                    bj
{ji (u)}iÎ   +   and   {y (v)}j    iÎ+
                                           . This means that the copula density c can also be seen as
an appropriate mixture of product densities, which possibly allows for a simple way
for a stochastic simulation.
                                                              åj
                                                               i =0
                                                                             ki   (u) = 1 for u Î (0,1)                      (2.6)
and
                                                        1
Let further        {pi }iÎ   +d   represent the distribution of an arbitrary discrete d-dimensional
random vector Z over +d where, for simplicity, we write i = (i1, , id ) , i.e.
P (Z = i) = pi , i Î +d . (2.8)
Then
                                                            pi              d
                                                                                                                        d
                                   c(u):=     å         d                         jk ,ik (uk ), u = (u1, , ud ) Î (0,1)   (2.10)
                                             i Î + d
                                                        a
                                                        k=1
                                                                   k , ik
                                                                            k=1
defines the density of a d-variate copula, which is also called generalized partition-
of-unity copula.
                                                                      d
                                                                                                                   d
                                      c(u) =       å        pi  fk ,ik (uk ), u = (u1, , ud ) Î (0,1)                     (2.11)
                                                  iÎ+ d            k=1
                              jki (  )
where the fki (  ) =                   , i Î + , k = 1,, d denote the Lebesgue densities induced by
                               aki
the {jki (u)}iÎ+ .
3. The symmetric case (diagonal dominance)
For simplicity, we restrict ourselves to the two-dimensional case in the sequel. The
generalization to higher dimensions is obvious.
                                                          ìïa , if i = j
                                                  pij := ïí i                               (3.1)
                                                           ïïî0, otherwise.
Then
                                    ¥
                                           ji (u)ji (v)     ¥
                         c(u, v):= å                    = å ai fi (u)fi (v), u, v Î (0,1)   (3.2)
                                    i =0        ai        i =0
                                       ìïæm - 1ö i
                                        ïç
                                        ïç     ÷÷ u (1- u)m-1-i , i = 0, , m - 1
                            jm,i (u) = íè i ø÷÷
                                        ï   ç                                               (3.3)
                                        ïï
                                         ïïî0,                    i ³ m.
                1
                                 æm - 1ö÷ 1 i
         am,i = ò   jm,i (u)du = çç
                                  çè i ø÷÷ ò
                                        ÷ u (1- u)m-1-i du
                0                                 0
and hence
                                              2
                               æm - 1ö÷
                              m-1
              cm(u, v) = må çç
                                                             m-1-i
                                      ÷÷ (uv)i ((1- u)(1- v))      , u, v Î (0,1)           (3.5)
                               ç
                          i =0 è  i ø  ÷
which corresponds to the density of a particular Bernstein copula (see e.g. COTTIN
AND PFEIFER (2014), Theorem 2.1). Especially, for m = 2, we obtain
m= 2 m= 3 m= 4 m= 5
Clearly, all those densities are bounded by the constant m, hence the coefficients
lU and lL of upper and lower tail dependence are zero:
1 1
Example 2 (negative binomial distributions). Consider, for fixed b > 0, the family of
negative binomial distributions given by their point masses
                                              æb + i - 1ö÷
                                  jb ,i (u) = çç         ÷÷(1- u)b ui , i Î  + .                             (3.9)
                                               çè i       ÷ø
        1
                           æb + i - 1÷ö 1 i              G(b + i) G(i + 1)G(b + 1)           b
ab ,i = ò                 ç
            jb ,i (u)du = ç          ÷÷ ò u (1- u)b du =           ⋅               =                    (3.10)
                          çè   i      ÷
                                      ø0                  i ! G(b ) G(b + i + 2)     (b + i)(b + i + 1)
       0
and hence
                                       b
                      ((1- u)(1- v))        ¥                    æ b + i - 1÷ö
                                                                                     2
For integer choices of b, this expression can be explicitly evaluated as a finite sum,
as can be seen from the following result.
                                                           b
                                       ((1- u)(1- v))          b -1   æb - 1÷öæb + 1÷ö
                 cb (u, v) = (b + 1)
                                           (1- uv)2 b +1
                                                               å çççè
                                                               i =0
                                                                            ÷÷çç
                                                                         i øè i ÷ø
                                                                             ÷ ç
                                                                                    ÷÷(uv)i , u, v Î (0,1).           (3.12)
Proof. We will show by induction the equality of the following two expressions:
               ¥ æ
                    b + i - 1÷öæçb + i + 1ö÷ i                   1        b -1 æ      öæ       ö
                                                                               ççb - 1÷÷ççb + 1÷÷ z i for 0 < z < 1.
  K(b , z):= å çç            ÷÷ç           ÷÷ z = k(b , z):=              å
                  ç
             i =0 è    i      ÷ç
                              øè    i       ÷ø               (1- z)2 b +1
                                                                          i=0 è
                                                                                ç i ÷øèç i ø÷÷
                                                                                       ÷
                                                                                                                (3.13)
                ¥ æ
   ¶K(b , z)          b + i - 1öæ
                               ÷÷ççb + i + 1ö÷÷ z i-1 and ¶ K(b , z) =
                                                           2             ¥          æb + i - 1öæ
                                                                                              ÷÷ççb + i + 1ö÷÷ z i-2
             = å i çç           ÷÷øèç         ÷÷ø                      å    i(i - 1)çç
                                                                                               ÷÷øèç         ÷÷ø
     ¶z             ç
               i =1 è    i            i                      ¶z 2      i =2
                                                                                     çè i            i
(3.14)
                                                                                             ¶ 2 K(b , z) ¶K(b , z)
                                                                                         z               +
      2
     ¶ K(b , z)     1 æç                       ¶K(b , z)ö÷                                       ¶z 2       ¶z .
                =      ççb(b + 2)K(b + 1, z) -           ÷ or K(b + 1, z) =                                           (3.15)
       ¶z 2         zè                           ¶z ÷ø                                             b(b + 2)
A similar, but more elaborate calculation shows that the latter equality remains valid
if K(b , z) is replaced by k(b , z):
                                                               ¶2 k(b , z) ¶k(b , z)
                                                                        z +
                                                 k(b + 1, z) =    ¶z 2       ¶z .                                     (3.16)
                                                                    b(b + 2)
                          ¥
                              (i + 1)(i + 2) i   ¥
                                                    j( j - 1) i-2 1           1
               K(1, z) = å                  z =å             z = h ''(z) =         = k(1, z)                          (3.17)
                         i =0        2         j =2     2         2        (1- z)3
               ¥
                    1
with h(z):= å z i =     for z <1. For the second step, assume that relation (3.13)
            i=0   1- z
holds for some b Î . Then it follows by (3.15) and (3.16) that
          (1- u)(1- v)
1     2            3
           (1- uv)
          (1+ 3uv)(1- u)2 (1- v)2
2     3
                (1- uv)5
3     4
                        (1- uv)7
4
          (1+ 15uv + 30u v   2   2
                                         + 10u3 v3 )(1- u)4(1- v)4
      5
                                 (1- uv)9
5
          (1+ 24uv + 90u v   2       2
                                         + 80u3 v3 + 15u4 v 4 )(1- u)5(1- v)5
      6
                                            (1- uv)11
b
     Cb (x, y), x, y Î (0,1)
          (2 - x - y)
1    xy
            1- xy
2
          (3 - 3 x - 3 y + x     2
                                     + y 2 + 3 x2 y 2 - x2 y 3 - x3 y 2 )
     xy
                                         (1- xy)3
        xy
              (4 - 6x - 6y + 4 x2 + 4 xy + 4y 2 + 4 x3 y + 24 x2 y 2 + 4 xy 3 - x3 - 6 x2 y - 6 xy 2 - y 3 -
     (1- xy)5
3
      - x 5 y 4 - x 4 y 5 + 4 x 4 y 4 - x 4 y 3 - x3 y 4 + 4 x4 y 2 + 4 x 3 y 3 + 4 x 2 y 4 - x4 y - 16 x3 y 2 - 16 x 2 y 3 -
The following graphs show the negative binomial copula densities cb for b = 1, ,4.
  b         1      2             3            4       5            6           7          8           9        10
            1     5          11               93     193        793        1619        26333         53381   215955
  lU(b )
            2     8          16              128     256       1024        2048        32768         65536   262144
A closed formula for the tail dependence coefficients for integer values of b as a
finite sum is given in the following result.
1 1
                             ò ò c (u, v)du dv
                                         b
                                                           2G(2b )
                                                                       1   1
                                                                                  xb y b
                                                            G (b ) ò0      ò
           lU(b ) = lim      t       t
                                                          = 2      ⋅                         dx dy
                       t1                   1- t                          0
                                                                               (x + y)2 b +1
                    2G(2b ) çæ b æçb ÷ö (-1)k b +k-2 æçb + k÷ö (-1)j +1 æç   1 ö÷÷ö
                   = 2      ⋅ ççå ç ÷÷         å èç j + 2 ÷ø÷ j + 1 èç 2 j+1 ø÷÷÷ø÷÷.
                                                      ç     ÷            ç1-                                    (3.19)
                     G (b ) çè çè k ÷ø b + k k=0            j =0
                                 b -1 æb - 1÷öæb + 1ö÷ b -1 æ b - 1 öæ
                                                                    ÷÷ççb + 1÷÷ö = ççæ 2b ÷÷ö
                                 å ççèç
                                 i =0
                                            ÷çç
                                            ÷       ÷÷ = å çç       ÷
                                                                    ÷ç i ÷÷ø çèb - 1÷÷ø
                                         i ÷øèç i ÷ø i=0 çèb - 1- iøè
                                                                                                                (3.20)
                                     b -1 æ
                                            b - 1÷öæçb + 1÷ö           æ 2b ÷ö 2G(2b )
                              (b + 1)å çç        ÷÷ç      ÷÷ = (b + 1)çç      ÷= 2     .                        (3.21)
                                     i=0 è
                                          ç   i øè ç
                                                  ÷ i ø÷              èb - 1ø÷÷
                                                                       ç        G (b )
Now, in the light of Lemma 1, we obtain
                    1     1                                                                        1   1
                                                                                                       (1- u)b (1- v)b
                    òò        cb (u, v)du dv                b -1 æ
                                                                   b - 1÷öæçb + 1ö÷
                                                                                               òò        (1- uv) 2 b +1
                                                                                                                        (uv)i du dv
lU(b ) = lim 1-h 1-h                                      = å çç        ÷÷ç       ÷ lim        1-h 1-h
                                                                                                                                          .   (3.22)
              h0                     h                          ç i øè
                                                            i =0 è
                                                                         ÷ç i ø÷÷ h0                                   h
                      1   1                                                         h   h
                            (1- u)b (1- v)b                                                       sb wb
  I(b , h, i):= ò        ò (1- uv)2b +1 (uv) du dv = ò
                                            i
                                                                                        ò
                                                                                                                     i
                                                                                                               (1- s)(1- w)i ds dw.           (3.23)
                    1-h 1-h                          0                                  0
                                                                                            (s + w - sw)2 b +1
                                                  1   1
                                                                xb y b
                          I(b , h, i) = hò            ò
                                                                                     i
                                                                              (1- hx)(1- hy)i dx dy ,                                         (3.24)
                                                  0   0
                                                          (x + y - hxy)2 b +1
giving
                  b -1 æ
                         b - 1öæ
                              ÷÷ çç b + 1ö÷÷ lim I(b , h, i) =
                                                               b -1 æ       öæ         ö1                      1
        lU (b ) = å çç                                              çç b - 1÷÷ çç b + 1÷÷                             xb y b
                  i=0 è
                       ç i ÷÷øèç i ÷÷ø h0           h
                                                               å
                                                               i=0 è
                                                                     ç i ÷øè ÷ ç i ÷ø÷ò                     ò      (x + y )2 b +1
                                                                                                                                  dx dy
                                                                                          0                    0
                                          1   1
                          2 G(2 b )                       x y b   b
                           G 2 (b ) ò0        ò (x + y )
                    =               ⋅                             2 b +1
                                                                           dx dy.                                                             (3.25)
                                              0
It remains to evaluate the integral term in the expression above. Using the
substitution z = x + y and the binomial theorem twice, we get
           1    1                                         1       y+1                          1       y+1 b
                    xb y b                   (z - y)b                                                          æb÷ö        k
                                                                                                               çç ÷(-1)k y dz dy
          ò    ò (x + y)2b+1 dx dy = ò ò z2b+1 dz dy = ò y ò
                                       y b                b
                                                                                                           å    ç ÷÷
                                                                                                           k=0 è kø     zb+k+1
           0   0                     0     y           0    y
                                    æbö÷     b +k 1 æ
                                  b
                                    ç    k y        çç 1 -       1 ö÷
                              = åç ÷÷(-1)
                                           b + k ò0 èç yb +k (y +1)b +k ø÷÷
                                                                        ÷ dy
                                    ç ÷
                                k=0 è kø
                                     b (-1)k æç çæ y ÷ö ÷÷ö                              b +k 1 æ
                                                                                                          1 ö÷ ö÷÷
                                 b æ ö         1            b +k         b æ ö                                b +k
                                                                             b÷      k y          çç æç
                              = åçç ÷÷÷          çç1-ç
                                              ò çç èç y +1÷ø ÷÷÷        å   ç
                                                                                       b + k ò0 çèç çè y +1÷ø ÷ø÷
                                                           ÷       dy =     ç   ÷(-1)               1-
                                                                                                   ç ç 1-    ÷     ÷ dy
                                    ç ÷
                                k=0 è kø b + k 0 è               ø          ç ÷÷
                                                                        k=0 è k ø
                                      b (-1)k çæ                                                                ö÷
                                  b æ ö                                 b +k æ                   1
                                                                                   b + k÷ö j+1        1
                              = åçç ÷÷÷           çç(b + k)ln(2) + åçç                  ÷÷(-1) ò            dy÷÷
                                     ç ÷
                                 k=0 è kø b + k ç
                                                                               ç j ÷ø              (y +1) j       ÷
                                                   è                    j =2 è                   0               ø÷
                                                                                æ                           1 ÷ö
                                                                                ç                     1-
                                         b æ ö
                                             b                   b (-1) ç çb + k÷ j+1 2 j-1 ÷÷÷
                                                             b æ ö           k ç b +k æ       ö
                              = ln(2)åçç ÷÷÷(-1)k + åçç ÷÷÷                     ççåç          ÷(-1)           ÷
                                            ç ÷
                                       k=0 è k ø
                                                                ç ÷              ç ç j ø÷÷
                                                            k=0 è k ø b + k ç j=2 è                     j -1 ÷÷÷
                                                                                 ççè                           ÷÷
                                                                                                                ø
                                  b æ ö         k b +k-2 æ         ö
                                                         ççb + k÷÷ (-1) æçç1- 1 ö÷÷
                                                                           j  1
                                      b (-1)                                 +
                              = åçç ÷÷÷             å              ÷÷                                                                         (3.26)
                                     ç ÷                  ç
                                 k=0 è k ø b + k j=0 è j + 2 ø j +1 è
                                                                                   ç 2 j+1 ø÷
          b    æb ö
since    å çççè k ø÷÷÷÷(-1)
         k=0
                              k
                                  = (1- 1)b = 0, which proves the statement above.                                                
Example 3 (Poisson distributions). Consider the family of Poisson distributions given by
their point masses
                                                                     g i L(u)i
                                               jg ,i (u) = (1- u)g             , i Î +                   (3.27)
                                                                          i!
where L(u) = - ln(1- u) > 0, u Î (0,1) and g > 0. Here we get, for i Î + , with the
substitutions z = L(u) and y = (1+ g )z,
                       1                  1                               ¥
                                                   g i L(u)i        g i z i -(1+g )z
               ag , i = ò   jg ,i (u)du = ò (1- u)      g
                                                             du = ò        e         dz
                       0                  0
                                                        i!        0
                                                                     i!
                                 ¥                                              i
                        gi         y i -y         gi          æ g ö÷ æ       g ö÷
                   =        i +1 ò
                                      e   dy =             = çç       ÷ çç1-    ÷,                        (3.28)
                     (1+ g ) 0 i !             (1+ g )i +1
                                                              èç1+ g ø÷ èç 1+ g ø÷÷
                                                                      ÷
The following graphs show some of these copula densities for different choices of g.
g =1 g=2 g=5
                            g = 10                           g = 20                             g = 30
The corresponding copula C cannot be calculated explicitly. However, in contrast
to the visual impression, the coefficient lU(g ) of upper tail dependence is zero here
for all g > 0, although we have a singularity in the point (1,1) in all cases.
                         ¥
                                   x i y i æç ¥ x i ÷ö æç ¥ y i ÷ö
               h(x, y):= å                £ ççå ÷÷ ⋅ ççå ÷÷ = exp(x + y) for all x, y ³ 0                                                    (3.30)
                        i =0        i !2     çè i=0 i ! ÷ø èç i=0 i ! ÷ø
                                                                  (
        cg (u, v) = (1+ g )(1- u)g (1- v)g h - g(1+ g )ln(1- u), - g(1+ g )ln(1- v)                                     )
                  £ 2(1- u)K (1- v)K , u, v Î (0,1).                                                                                         (3.31)
This implies
                                                     1       1
                                                 ò ò c (u, v)du dv g
                                                                                                  (1- t)2 K +1
                        lU(g ) = lim                 t       t
                                                                                     = 2lim                    = 0,     b                    (3.32)
                                          t 1                         1- t                t 1    (K + 1)2
Example 4 (log series distribution). Consider the family of log series distributions given
by their point masses
                                                                                 ui
                                                                  ji (u) =             , i Î +                                              (3.33)
                                                                              i ⋅ L(u)
                                          1
                                                                        1 i æç i ÷ö
                               ai = ò ji (u)du =                         å    ç ÷÷÷(-1) ln( j + 1) for i Î .
                                                                              ç
                                                                        i j=1 è jø
                                                                                       j +1
                                                                                                                                             (3.34)
                                          0
The proof of this relation requires some more sophisticated arguments, as is shown in
the sequel.
                                                 n
                        ¥
                             (1- e )      -x
                                                                 n æ ö
                                                                      n
                        ò                            e-cx dx = å çç ÷÷÷(-1)j +1 (ln( j + c) - ln(c)).                                        (3.35)
                                      x                             ç   ÷
                                                               j =0 è j ø
                        0
                                                         n                                                                           n
                             ¥
                                   (1- e )     -x
                                                                                                                      (1- e )   -x
                  ænö                      ¥     n æ ö
                                                      n         æ                           ¥    ö    n æ ö
                                                                      1                                  ççn÷÷(-1)j+1 1
               n
                                                                ç
           = å çç ÷÷÷(-1)j+1 ò e-( j +c)x dx = å çç ÷÷÷(-1)j +1 çç-                             ÷÷÷ =
                  ç   ÷                             ç   ÷
                                                                          e-( j+c)x               ÷÷  å       ÷÷         (3.36)
             j =0 è j ø      0                 j =0 è j ø
                                                                 ç
                                                                 çè j + c                   0      ø j=0 çè j ø      j+c
                                          n æ ö
                                               n
                                hn(c):= å çç ÷÷÷(-1)j +1 (ln( j + c) - ln(c)) for c > 0.                                           (3.37)
                                             ç   ÷
                                        j =0 è j ø
Then
                    n æ ö                                       n æ ö          æ 1
                        n           d                                n                  1ö
       hn '(c) = å çç ÷÷÷(-1)j+1       (ln( j + c) - ln(c)) = å çç ÷÷÷(-1)j+1 çç      - ÷÷÷
                       ç ÷
                  j =0 è j ø       dc                              ç ÷
                                                              j =0 è j ø
                                                                              çè j + c c ÷ø
                    n æ ö                            n æ ö
                        n       æ 1           1ö        n              1    1 n ænö
              = å çç ÷÷÷(-1)j+1 çç        - ÷÷÷ = å çç ÷÷÷(-1)j +1       + å çç ÷÷÷(-1)j
                       ç ÷
                 j =0 è j ø
                                 çè j + c c ø÷ j=0 çè j ø÷          j + c c j=0 èç j ø÷
                       n æ ö
                            n         1
                   = å çç ÷÷÷(-1)j+1                                                                                               (3.38)
                          ç   ÷
                     j =0 è j ø      j+c
                     n æ ö
                          n
since 0 = (1- 1)n = å çç ÷÷÷(-1)j . This implies gn ' = hn ' and hence gn(c) = hn(c) + Kn or
                        ç   ÷
                    j=0 è j ø
equivalently, Kn = gn(c) - hn(c) for all c > 0, for some constant Kn Î . But then also
                                     æ¥          n     ö
                                     çç (1- e-x ) -cx ÷÷           n æ ö
                                                                        n
    Kn = lim gn(c) - lim hn(c) = lim ç ò
                                      ç            e dx÷÷÷ - lim å çç ÷÷÷(-1)j+1 (ln( j + c) - ln(c))
         c¥         c¥         c¥ ç      x            ÷   c¥      ç   ÷
                                                                 j =0 è j ø
                                      èç 0             ø÷
                            n
       ¥
           (1- e ) -x                              n æ ö
                                                       n               æ æ     j öö
                                                                                        ¥          n
    =ò                          lim (e-cx ) dx - å çç ÷÷÷(-1)j +1 lim ççlnçç1+ ÷÷÷÷÷÷ = ò 0 dx - å 0 = 0                           (3.39)
               x                c¥                   ç ÷
                                                 j =0 è j ø           è èç cø÷ø 0
                                                                  c¥ ç
                                                                                                 j =0
       0
                                                                  n
                        1
                                    un
                                                ¥
                                                  (1- e-x ) -x      n æ ö
                                                                         n
           bn := ò                         du = ò          e dx = å çç ÷÷÷(-1)j+1 ln( j + 1).                                      (3.40)
                                - ln(1- u)            x                ç   ÷
                                                                  j =1 è j ø
                        0                       0
                                             ui
Hence with ji (u) = -                               for i Î , this means
                                       i ⋅ ln(1- u)
                             1
                                          bi 1 i æç i ÷ö
                   ai = ò ji (u)du =         = å ç ÷÷(-1)j+1 ln( j + 1) for i Î .       (3.41)
                             0
                                           i  i j=1 çè j÷ø
                   ¥
                         1                        1          ¥
                                                               (uv)i
         c(u, v) = å        ji (u)ji (v) =                 å         for 0 < u, v < 1.   (3.42)
                  i =1   ai                ln(1- u)ln(1- v) i=1 ibi
plot of c(u, v)
The log series copula does not have a positive tail dependence either, as in the
case of the Poisson copula.
The proof of this statement again requires some more sophisticated arguments. We
proceed in the following steps.
                                                      1
                                                  1     L(ut)
                                         lim
                                          t 1       ò
                                                 1- t t L(u)
                                                              du = 1.                    (3.43)
                                                 1 ln(w + s - ws)
                                                    s
                                                 s ò0
                                         lim                      dw = 1.                (3.45)
                                          s0          ln(w)
Define
                                                         ln(w + s)             ln(w + s(1- s))
                                        F(w, s):=                  , G(w, s):=                 .                                  (3.46)
                                                           ln(w)                   ln(w)
Then
                                                             ln(w + s - ws)
                                        F(w, s) £                           £ G(w, s) for 0 < w £ s                               (3.47)
                                                                 ln(w)
(note that ln(w) < 0 for 0 < w < 1). Now for 0 < s < 1,
                                                         æ     s2 ÷ö
                                                    - ln ççç1-     ÷÷
                                                           è w + s ÷ø
           s                                   s                                      s
                                                                                           1         s ln(1- s)
 0 £ ò G(w, s) - F(w, s)dw £ò                                         dw £ - ln(1- s)ò          dw £                              (3.48)
           0                                   0
                                                        - ln(w)                      0
                                                                                        - ln(w)         ln(s)
                                                         s
                                                    1                              ln(1- s)
                                       0 £ lim
                                              s0     ò
                                                    s 0
                                                        G(w, s) - F(w, s)dw £ lim
                                                                               s0   ln(s)
                                                                                            = 0.                                  (3.49)
                                                    s                          s
                                              1                      1 - ln(w + s)    !
                                              s ò0               s0 s ò
                                        lim        F(w, s) dw = lim                dw = 1.                                        (3.50)
                                        s0
                                                                       0
                                                                         - ln(w)
                                                              1
                                                                 ¥
                                                                        - ln (e- x + s) - x !
                                                        lim
                                                        s0      ò
                                                              s -ln( s)        x
                                                                                       e dx =1.                                   (3.51)
Note that
  ¥
          - ln (e- x + s)                  ¥
                                                   - ln(e- x (1+ se- x ))                  ¥
                                                                                                   x - ln (1+ se- x )
 ò              x
                            e- x dx =     ò                      x
                                                                            e- x dx =     ò                  x
                                                                                                                        e- x dx
-ln( s)                                  -ln( s)                                         -ln( s)
                            ¥                      ¥
                                                         ln(1+ se- x )                     ¥
                                                                                                   ln (1+ se- x )
                     =      ò      e- x dx -       ò             x
                                                                         e- x dx = s -    ò              x
                                                                                                                    e- x dx.      (3.52)
                         -ln( s)               -ln( s)                                   -ln( s)
                                                         1
                                                            ¥
                                                                   ln (1+ se- x ) - x !
                                                    lim ò                        e dx = 0.                                        (3.53)
                                                     s0 s               x
                                                           -ln( s)
                                     ¥
                                             ln(1+ ey )                                    ¥
                                                                                               ln(1+ ey )             !
                                     ò                                     dy = lim ò
                                 T                             -(y + T )
                         lim e                            e                                                   e-y dy = 0.               (3.55)
                          T ¥
                                     0
                                                  y+T                               T ¥
                                                                                           0
                                                                                                   y+T
                   ¥
                       ln (1+ ey )                               ¥
                                                                      y + 1 -y
                                                                                  ¥
                                                                                          æ y + 1 -y ö÷
      0 £ lim ò                          e-y dy £ lim ò                    e dy = ò lim çç       e ÷÷ dy = 0                            (3.56)
            T ¥         y+T                             T ¥         y+T            T ¥ ç
                                                                                          èy + T      ø
                   0                                              0               0
Lemma 5. With L(u) = - ln(1- u), the ai given in (3.34) and the copula density given in
(3.42), it holds that
                                         1    1                                            1
                             1                                                      t      L(ut)
                            1- t òt          ò                                     1- t òt L(u)
                   K(t):=                         c(u, v)du dv £ 1-                              du for 0 < t < 1,                      (3.57)
                                              t
which in turn implies that the log series copula has no tail dependence.
                                                                æ 1 ö÷      1         u
Proof. First notice that by the relation L(u) = - ln(1- u) = ln çç      ÷£      - 1=      for
                                                                 çè1- u ø÷ 1- u      1- u
0 < u < 1, we obtain
                   1                              1                            1
                          ui        1 ui         1                     1
          ai = ò                du ³ ò (1- u)du = ò ui-1(1- u)du = 2         for all i Î .                                             (3.58)
                   0
                       i ⋅ L(u)     i 0 u        i 0               i (i + 1)
Now
                         1
                                 1   1       ¥
                                                      (uv)i              1
                                                                                               1   ì
                                                                                                   ï  ¥
                                                                                                               vi
                                                                                                                      1
                                                                                                                         ui      ü
                                                                                                                                 ï
                                                                                                   ï                             ï dv
            K(t):=
                        1- t òt      ò       å        2
                                             i =1 ai i L(u)L(v)
                                                                du dv =
                                                                        1- t òt
                                                                                                   íå 2
                                                                                                   ï      a  i  L (v) ò L(u )
                                                                                                                              du ý
                                                                                                                                 ï
                                                                                                                                        (3.59)
                                     t                                                             ï
                                                                                                   î i =1  i          t          ï
                                                                                                                                 þ
      ¥
               (uv)i        ¥
                                         ui   vi   ¥
                                                                        2 + (uv)2 - 3uv      3
      å        2
      i =1 ai i L(u)L(v)
                         £ å
                           i =1
                                (i + 1)     ⋅    £ å
                                        L(u) L(v) i=1
                                                      (i + 1)(uv)i -1
                                                                      =
                                                                           (1- uv) 3
                                                                                        £
                                                                                          (1- uv)3
                                                                                                   ,                                    (3.60)
                                                                        1
                                                                              3                  3(2 - v)
the r.h.s being integrable w.r.t. u with value                         ò (1- uv)      3
                                                                                          du =
                                                                                                 2(1- v)2
                                                                                                          . Now for 0 < t < 1,
                                                                        0
we have
                                      t                          1                1
                                           ui            vi ti                  vi
                                   ò      L(u)
                                               du = t ò
                                                        L(vt   )
                                                                 dv ³ t i +1 ò
                                                                               L(v )
                                                                                     dv                                 (3.61)
                                      0               0                      0
and hence
                   1              1                          t                        1
                        ui          ui          ui                      vi
                   ò        du = ò      du - ò      du £ (1- t i +1) ò      dv = (1- t i +1) iai                       (3.62)
                   t
                       L(u)      0
                                   L(u)      0
                                               L(u)                  0
                                                                       L(v)
          1  ì
             ï 1
                        vi
                               1
                                  ui       ü
                                           ï                1 ¥
                                                                     vi
                                           ï dv £ 1
                ¥
             ï
K(t) =       íå 2
         1- t òt
             ï     a  i  L (v) ò L(u  )
                                        du ý
                                           ï          1- t ò  å    iL(v  )
                                                                           (1- ti+1) dv
             ï
             î i=1 i           t           ï
                                           þ                t i =1
which proves relation (3.57). From Lemma 4 we thus obtain the final result
indicating that the log series copula has no upper tail dependence. 
n n
                                      åp
                                      k=0
                                                    ik   = å pki = ai for i = 0, , n
                                                           k=0
                                                                                                                         (4.1)
and
                                              ìïa , if i = j
                                      pij := ïí i                           for i, j > n.                                (4.2)
                                               ïïî0, otherwise
Example 5 (negative binomial distributions, asymmetric case). We consider the
negative  binomial   distributions  from    Example   2   with   b = 1.  Then
      1
                               1
ai = ò j1,i (u)du =                    for i Î  + . With n = 4 and
      0
                         (1+ i)(2 + i)
                                                     é18             5 5 0 2ù
                                                     ê                        ú
                                                     ê10             0 0 0 0úú
                                                  1 êê
                                           M4 :=        0            5 0 0 0úú                                              (4.3)
                                                 60 êê
                                                     ê0              0 0 3 0úú
                                                     ê2              0 0 0 0úúû
                                                     êë
the conditions above are fulfilled, giving the copula density, according to (2.3),
             ¥    ¥      pij                   n     n      pij                        ¥
                                                                                             1
  c(u, v) = å å                  ji (u)j j (v) = å å                ji (u)j j (v) +   å         jk (u)jk (v), u, v Î (0,1) , (4.4)
            i =0 j =0   ai a j                i = 0 j =0   ai a j                     k= n+1 ak
                                                            ìïb j   if j = 2i
                                                             ïï
                                                              ï
                                                      pij = íb j +1 if j = 2i + 1 for i, j Î + ,                                              (4.7)
                                                              ïï
                                                               ïï0  otherwise
                                                                î
i.e.
                                                    é b0     b1       ù
                                                    ê                          ú
                                                    ê        b2 b3     úú
                                                    ê
                              ép ù +
                              ëê ij ûú i , jÎ   = êê          b4 b5   úú                                                               (4.8)
                                                    ê            b6 b7 úú
                                                    ê
                                                    ê                  ú
                                                    êë                         úû
                                                                      ¥                               ¥
where  stands for zero. Then pi  = å pij = ai and p j = å pij = b j for i, j Î + since
                                                                      j =0                            i =0
                                     2                  2                1
           b2 i + b2 i +1 =                    +                  =              = ai for i Î  + .                                            (4.9)
                              (2 + 2i)(3 + 2 i) (3 + 2 i)(4 + 2 i) (1+ i)(2 + i)
                                                                     ¥       ¥
                                                           c(u, v) = åå pij fi (u) gj (v), u, v Î (0,1)                                       (4.10)
                                                                     i =0 j =0
                       xy
        C(x, y) =               4   (2 - x - 2 xy   3
                                                        + xy 4 + x 2 y 3 - 2y 2 + y 3 ) , x, y Î (0,1).   (4.12)
                    (1- xy )2
                                                                        5
                                                                 lU =                                     (4.13)
                                                                        9
which is between the coefficients of upper tail dependence for the symmetric case
with b = 1 and b = 2, cf. the final table in Example 2.
empirical copula, for instance as was proposed in PFEIFER, STRASSBURGER AND PHILIPPS
(2009). In the particular case of Bernstein copulas (see Example 1) such a procedure
can be very easily implemented, even in higher dimensions (cf. COTTIN AND PFEIFER
(2014)).
As a practical exercise, we refer to Example 4.2 in COTTIN AND PFEIFER (2014) where the
empirical copula from an original data set was fitted to a Bernstein copula. The
following two graphs show the scatter plot from the empirical copula (big red dots)
superimposed by 1000 simulated points of that Bernstein copula (left) and of a
negative binomial copula of type (3.11), with b = 5.
As can be nicely seen, the Bernstein copula represents the local asymmetry of the
empirical copula better, but shows no tail dependence, as does the negative
binomial copula.
The fit to the negative binomial copula was, for the sake of simplicity, performed by
a numerical match between the theoretical correlation for the negative binomial
copula and the correlation of the empirical copula, which is 0.815. Note that the
theoretical correlation r(b ) for the negative binomial copula of type (3.11) can be
explicitly calculated as
                   æ¥                (i + 1)2           ö÷
     r(b ) = 12b çççå                                  2÷
                                                         ÷ - 3 = 3b (2(b + 1)2 Y(1, b + 2) - 2b - 1)       (4.14)
                   çè i=0 (b + i)(b + i + 1)(b + i + 2) ø÷
               b            1          2             3           4          5          6               7
             r(b )     0.4784      0.6529     0.7410     0.7937       0.8288    0.8537       0.8723
For the sake of completeness, we finally show a comparison between the Bernstein
copula fit and a Poisson copula fit with parameter g = 6. The empirical correlation
for the Poisson copula here is 0.814.
             Bernstein copula fit                     Poisson copula fit
Note that although the empirical plot for the Poisson copula might suggest some tail
dependence here this is actually not true in the light of (3.32).
It should be finally pointed out that copula constructions as presented in this paper
will have a major impact in the construction of Internal Models under the new
Solvency II insurance supervising regime in Europe (see e.g. HUMMEL AND MÄRKERT
(2011) or SANDSTRǾM (2011), Chapter 13).
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