A New Family of Archimedean Copulas: The Truncated-Poisson Family of Copulas

Abstract

Copulas are multivariate distribution functions which their margins are distributed uniformly. Therefore, copulas are pretty useful for modeling several types of data. As they allow different dependence patterns. A numerous number of new classes of copulas have been suggested in the literature. Each granted different characteristics that make it compatible with certain type of data. In this paper, we introduce a new family of Archimedean copulas. The multiplicative Archimedean generator of this copula is the inverse of the probability generating function of a truncated-Poisson distribution. The properties of this copula are studied in detail. Three applications are provided for the sake of comparison between this copula and well-known ones.

Appendices

Proofs

1.1 Proof of Theorem 2

• i To prove that $\lim _{\theta \rightarrow 0}{C_T\left( u,v;\theta \right) }=uv$, we only need to prove that

  $$\begin{aligned} \lim _{\theta \rightarrow 0}{\frac{\psi _\theta (t)\ln \left[ \psi _\theta (s)\right] }{\psi _\theta ^{\prime }(t)}} =t\ln \left[ s\right] , \end{aligned}$$

  (see Nelsen [50], Pages 139–140). Therefore,

  $$\begin{aligned} \lim _{\theta \rightarrow 0}\frac{\psi _\theta (t)\ln \left[ \psi _\theta (s)\right] }{\psi _\theta ^{\prime }(t)}&=\lim _{\theta \rightarrow 0}\frac{(1+\theta t)\ln \left[ 1+\theta t\right] }{\theta }\ln \left[ {\frac{\ln \left[ 1+\theta s\right] }{\ln \left[ 1+\theta \right] }}\right] \\&=\lim _{\theta \rightarrow 0}\frac{(1+\theta t)\ln \left[ 1+\theta t\right] }{\theta }\ln \left[ \lim _{\theta \rightarrow 0}{\frac{\ln \left[ 1+\theta s\right] }{\ln \left[ 1+\theta \right] }}\right] . \end{aligned}$$

  Using L’Hôpital’s rule for both limits, we get

  $$\begin{aligned} \lim _{\theta \rightarrow 0}\frac{\psi _\theta (t)\ln \left[ \psi _\theta (s)\right] }{\psi _\theta ^{\prime }(t)} =&\lim _{\theta \rightarrow 0}\left[ t\ln \left[ 1+\theta t\right] +t\right] \ln \left[ \lim _{\theta \rightarrow 0}{\frac{\left( 1+\theta \right) s}{1+\theta s}}\right] =t\ln \left[ s\right] . \end{aligned}$$

  Thus, $\lim _{\theta \rightarrow 0}{C_T\left( u,v\right) }=uv$.

• ii For $\theta \rightarrow \infty$, we have$^*$¹

  $$\begin{aligned} \lim _{\theta \rightarrow \infty }\frac{\psi _\theta (t)\ln \left[ \psi _\theta (s)\right] }{\psi _\theta ^{\prime }(t)}&=\lim _{\theta \rightarrow \infty }\frac{\left( 1+\theta t\right) \ln \left[ 1+\theta t\right] \ln \left[ {\frac{\ln \left[ 1+\theta s\right] }{\ln \left[ 1+\theta \right] }}\right] }{\theta }\\&=\lim _{\theta \rightarrow \infty }{\left( \frac{1}{\theta }+t\right) } \lim _{\theta \rightarrow \infty }{\ln \left[ 1+\theta t\right] \ln {\left[ \frac{\ln \left[ \frac{\left( 1+\theta s\right) }{\left( 1+\theta \right) }\right] }{\ln \left[ 1+\theta \right] }+1\right] }}. \end{aligned}$$

  Note that for small x, we have $\ln {\left[ 1+x\right] }\sim x$. Also, for large $\theta$, $\ln \left[ \frac{\left( 1+\theta s\right) }{1+\theta }\right]$ tends to $\ln {\left[ s\right] }$ and hence $\frac{\ln \left[ \frac{\left( 1+\theta s\right) }{\left( 1+\theta \right) }\right] }{\ln \left[ 1+\theta \right] }$ becomes small. Therefore, $\ln {\left[ \frac{\ln \left[ \frac{\left( 1+\theta s\right) }{\left( 1+\theta \right) }\right] }{\ln \left[ 1+\theta \right] }+1\right] }\sim \frac{\ln \left[ s\right] }{\ln \left[ 1+\theta \right] }$. This implies

  $$\begin{aligned}&\lim _{\theta \rightarrow \infty }{\left( \frac{1}{\theta }+t\right) }\lim _{\theta \rightarrow \infty }{\ln {\left[ 1+\theta t\right] }}\ln {\left[ \frac{\ln \left[ \frac{\left( 1+\theta s\right) }{\left( 1+\theta \right) }\right] }{\ln \left[ 1+\theta \right] }+1\right] } \\&\quad =t\lim _{\theta \rightarrow \infty }\frac{\ln \left[ 1+\theta t\right] \ln \left[ s\right] }{\ln \left[ 1+\theta \right] }=t\ln \left[ s\right] . \end{aligned}$$

  Hence, $\lim _{\theta \rightarrow \infty }{C_T\left( u,v;\theta \right) }=uv$. $\square$

1.2 Proof of Theorem 3

The density of the truncated-Poisson copula is given by

$$\begin{aligned} c_T(u,v)=\frac{\theta e^{\frac{\ln \left[ 1+\theta u\right] \ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }}\left[ 1+\frac{\ln \left[ 1+\theta u\right] \ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }\right] }{(1+\theta u)(1+\theta v)\ln \left[ 1+\theta \right] };\quad 0\le u,v\le 1,\theta \ge 0. \end{aligned}$$

This function is a product of the two non-negative functions

$$\begin{aligned} f_1\left( u,v\right) =e^{\frac{\ln \left[ 1+\theta u\right] \ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }}; \end{aligned}$$

and

$$\begin{aligned} f_2\left( u,v\right) =\theta \frac{\ln \left[ 1+\theta \right] +\ln \left[ 1+\theta u\right] \ln \left[ 1+\theta v\right] }{\left( 1+\theta u\right) \left( 1+\theta v\right) \ln ^{2}\left[ 1+\theta \right] }. \end{aligned}$$

Second derivatives for the natural logarithms of $f_1$ and $f_2$ are $\frac{\partial ^{2}\ln \left[ f_1\left( u,v\right) \right] }{\partial u \partial v}=\frac{\theta ^2}{\left( 1+\theta u\right) \left( 1+\theta v\right) \ln \left[ 1+\theta \right] }\ge 0$ and $\frac{\partial \ln \left[ f_2\left( u,v\right) \right] }{\partial u}=\frac{\theta \ln {\left[ 1+\theta v\right] }}{\left( 1+\theta u\right) \left[ \ln {\left[ 1+\theta \right] }+\ln {\left[ 1+\theta u\right] }\ln {\left[ 1+\theta v\right] }\right] }-\frac{\theta }{1+\theta u}$, respectively. Hence, $\ln {f_1}$ and $\ln {f_2}$ are both supermodulars. Thus, $f_1$ and $f_2$ are $TP_2$ functions. Therefore, the multiple of the two functions $c_T=f_1f_2$ is also $TP_2$ (see Karlin [32]). $\square$

1.3 Proof of Theorem 4

In terms of copulas, Spearman’s Rho correlation is given by

$$\begin{aligned} \rho _{C_T}&=12 \int _{0}^{1}\int _{0}^{1}C_T\left( u,v;\theta \right) \times dudv-3\\&=12\int _{0}^{1}\int _{0}^{1}\frac{1}{\theta }\left[ e^{\frac{\ln \left[ 1+\theta u\right] \ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }}-1\right] dudv -3\\&=12\int _{0}^{1}\int _{0}^{1}\frac{1}{\theta }\left[ \left( 1+\theta u\right) ^\frac{\ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }-1\right] dudv-3\\&=12\int _{0}^{1}\frac{1}{\theta } \left[ \frac{\left( 1+\theta \right) ^{\frac{\ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }+1}}{\theta \left[ \frac{\ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }+1\right] }-\frac{1}{\theta \left[ \frac{\ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }+1\right] }-1 \right] dv-3\\&=12\int _{0}^{1} \left[ \frac{\ln \left[ 1+\theta \right] \left[ \left( 1+\theta \right) ^{\frac{\ln \left[ 1+\theta v\right] }{\ln \left[ 1+\theta \right] }+1}-1\right] }{\theta ^2 \left[ \ln \left[ 1+\theta v\right] +\ln \left[ 1+\theta \right] \right] }-\frac{1}{\theta } \right] dv-3\\&=12\frac{\ln \left[ 1+\theta \right] }{\theta ^2}\left[ \int _{0}^{1} \frac{\left( 1+\theta \right) \left( 1+\theta v\right) }{\ln \left[ \left( 1+\theta \right) \left( 1+\theta v\right) \right] } dv-\int _{0}^{1} \frac{1}{\ln \left[ \left( 1+\theta \right) \left( 1+\theta v\right) \right] } dv\right] \\&\qquad -\frac{12}{\theta }-3. \end{aligned}$$

For $\theta \ge 0$, take $t=2\ln \left[ \left( 1+\theta \right) \left( 1+\theta v\right) \right]$ for the first integral, and $s=\left( 1+\theta \right) \left( 1+\theta v\right)$ for the second one. Hence,

$$\begin{aligned} \rho _{C_T}=&12\frac{\ln \left[ 1+\theta \right] }{\theta ^2}\left[ \frac{1}{\theta \left( 1+\theta \right) } \int _{2\ln \left[ 1+\theta \right] }^{4\ln \left[ 1+\theta \right] } \left[ \frac{e^{t}}{t}\right] dt-\frac{1}{\theta \left( 1+\theta \right) } \int _{\left( 1+\theta \right) }^{\left( 1+\theta \right) ^2} \frac{ds}{\ln \left[ s\right] } \right] \\&\qquad -\frac{12}{\theta }-3\\ =&12\frac{\ln \left[ 1+\theta \right] }{\theta ^3\left( 1+\theta \right) }\biggl [ \int _{-\infty }^{4\ln \left[ 1+\theta \right] } \left[ \frac{e^{t}}{t}\right] dt-\int _{-\infty }^{2\ln \left[ 1+\theta \right] } \left[ \frac{e^{t}}{t}\right] dt- \int _{0}^{\left( 1+\theta \right) ^2} \frac{ds}{\ln \left[ s\right] }\\&+\int _{0}^{\left( 1+\theta \right) } \frac{ds}{\ln \left[ s\right] } \biggl ]-\frac{12}{\theta }-3\\ =&12\frac{\ln \left[ 1+\theta \right] }{\theta ^3\left( 1+\theta \right) }\bigl [ Ei\left( 4\ln \left[ 1+\theta \right] \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) - li\left( \left( 1+\theta \right) ^2\right) +li\left( 1+\theta \right) \bigl ]\\&-\frac{12}{\theta }-3\\ =&12\frac{\ln \left[ 1+\theta \right] }{\theta ^3\left( 1+\theta \right) }\bigl [ Ei\left( 4\ln \left[ 1+\theta \right] \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) -Ei\left( \ln \left[ \left( 1+\theta \right) ^2\right] \right) \\&+li\left( 1+\theta \right) \bigl ]-\frac{12}{\theta }-3\\ =&12\frac{\ln \left[ 1+\theta \right] }{\theta ^3\left( 1+\theta \right) }\left[ Ei\left( 4\ln \left[ 1+\theta \right] \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) +li\left( 1+\theta \right) \right] \\&-\frac{12}{\theta }-3\\ =&12\frac{\ln \left[ 1+\theta \right] }{\theta ^3\left( 1+\theta \right) }\left[ Ei\left( 4\ln \left[ 1+\theta \right] \right) -2 Ei\left( 2\ln \left[ 1+\theta \right] \right) +li\left( 1+\theta \right) \right] -\frac{12}{\theta }-3. \end{aligned}$$

$\square$

1.4 Proof of Theorem 5

As our copula belongs to the Archimedean class, we will use the following formula to derive Kendall’s tau correlation coefficient

$$\begin{aligned} \tau _{C_T}=&1+4\int _{0}^{1}\frac{\psi _\theta (s)\ln \left[ \psi _\theta (s)\right] }{\psi _\theta ^{\prime }(s)}ds =1+4\int _{0}^{1}\frac{\left( 1+\theta s\right) \ln \left[ 1+\theta s\right] }{\theta }\ln \left[ \frac{\ln \left[ 1+\theta s\right] }{\ln \left[ 1+\theta \right] }\right] ds \end{aligned}$$

By setting $t=\frac{\ln \left[ 1+\theta s\right] }{\ln \left[ 1+\theta \right] }$, we get

$$\begin{aligned} \tau _{C_T}=&1+4\ln ^2\left[ 1+\theta \right] \int _{0}^{1}\frac{e^{2\ln \left[ 1+\theta \right] t}t\ln \left[ t\right] }{\theta ^2}dt\\ =&1+\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\int _{0}^{1}\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] t^n}{n!}t\ln \left[ t\right] dt\\ =&1+\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{n!}\int _{0}^{1}t^{n+1}\ln \left[ t\right] dt\\ =&1+\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{n!}\left[ \frac{t^{n+2}}{n+2}\ln \left[ t\right] |_{0}^{1}-\int _{0}^{1}\frac{t^{n+1}}{n+2}dt\right] \\ =&1+\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{n!}\left[ -\frac{t^{n+2}}{(n+2)^2}|_{0}^{1}\right] \\ =&1-\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{(n+2)^2 n!}\\ =&1-\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\frac{n+1}{n+2}\frac{2^n\ln ^n\left[ 1+\theta \right] }{(n+2)!}\\ =&1-\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\sum _{n=0}^{\infty }\left[ 1-\frac{1}{n+2}\right] \frac{2^n\ln ^n\left[ 1+\theta \right] }{(n+2)!}\\ =&1-\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\left[ \sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{(n+2)!}-\sum _{n=0}^{\infty }\frac{2^n\ln ^n\left[ 1+\theta \right] }{(n+2)(n+2)!}\right] \\ =&1-\frac{4\ln ^2\left[ 1+\theta \right] }{\theta ^2}\left[ \sum _{m=2}^{\infty }\frac{2^{m-2}\ln ^{m-2}\left[ 1+\theta \right] }{m!}-\sum _{m=2}^{\infty }\frac{2^{m-2}\ln ^{m-2}\left[ 1+\theta \right] }{mm!}\right] \\ =&1-\frac{1}{\theta ^2}\left[ \sum _{m=2}^{\infty }\frac{2^{m}\ln ^{m}\left[ 1+\theta \right] }{m!}-\sum _{m=2}^{\infty }\frac{2^{m}\ln ^{m}\left[ 1+\theta \right] }{mm!}\right] \\ =&1-\frac{1}{\theta ^2}\left[ e^{2\ln \left[ 1+\theta \right] }-2\ln \left[ 1+\theta \right] -1-\sum _{m=1}^{\infty }\frac{2^{m}\ln ^{m}\left[ 1+\theta \right] }{mm!}+2\ln \left[ 1+\theta \right] \right] \\ =&1-\frac{1}{\theta ^2}\left[ \left( 1+\theta \right) ^2-1-\sum _{m=1}^{\infty }\frac{\left( 2\ln \left[ 1+\theta \right] \right) ^m}{mm!}\right] \\ =&1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -\sum _{m=1}^{\infty }\frac{\left( 2\ln \left[ 1+\theta \right] \right) ^m}{mm!}\right] \\ =&{\left\{ \begin{array}{ll} 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -\sum _{m=1}^{\infty }\frac{\left( 2\ln \left[ 1+\theta \right] \right) ^m}{mm!}\right] ;\theta \ge 0\\ 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) +\sum _{m=1}^{\infty }\frac{(-1)^{m+1}\left( -2\ln \left[ 1+\theta \right] \right) ^m}{mm!}\right] ;-0.63\le \theta<0 \end{array}\right. }\\ =&{\left\{ \begin{array}{ll} 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) \right] +\gamma +\ln \left[ 2\ln \left[ 1+\theta \right] \right] ;\theta \ge 0\\ 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) +\gamma +\ln \left[ -2\ln \left[ 1+\theta \right] \right] \right] ;-0.63\le \theta<0 \end{array}\right. }\\ =&{\left\{ \begin{array}{ll} 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) +\ln \left[ \ln \left[ 1+\theta \right] \right] +\gamma +\ln \left[ 2\right] \right] ;\theta \ge 0\\ 1-\frac{1}{\theta ^2}\left[ \theta \left( 2+\theta \right) -Ei\left( 2\ln \left[ 1+\theta \right] \right) +\ln \left[ -\ln \left[ 1+\theta \right] \right] +\gamma +\ln \left[ 2\right] \right] ;\\ \qquad -0.63\le \theta <0 \end{array}\right. } \end{aligned}$$

$\square$

1.5 Proof of Theorem 6

The upper tail dependence is given by

$$\begin{aligned} \uplambda _U=2-\lim _{t\rightarrow 1^{-}}\frac{1-\frac{e^{\frac{\ln ^{2}\left[ 1+\theta t\right] }{\ln \left[ 1+\theta \right] }}-1}{\theta }}{1-t}. \end{aligned}$$

Using L’Hôpital’s rule, we get

$$\begin{aligned} \uplambda _U=2-\lim _{t\rightarrow 1^{-}}\frac{2\ln \left[ 1+\theta t\right] }{\left( 1+\theta t\right) \ln \left[ 1+\theta \right] }e^{\frac{\ln ^{2}\left[ 1+\theta t\right] }{\ln \left[ 1+\theta \right] }}=2-2=0. \end{aligned}$$

Also, the lower tail dependence is given by

$$\begin{aligned} \uplambda _L=\lim _{t\rightarrow 0^{+}}\frac{e^{\frac{\ln ^{2}\left[ 1+\theta t\right] }{\ln \left[ 1+\theta \right] }}-1}{\theta t}. \end{aligned}$$

Using L’Hôpital’s rule, we get

$$\begin{aligned} \uplambda _L=\lim _{t\rightarrow 0^{+}}\frac{2\ln \left[ 1+\theta t\right] }{\left( 1+\theta t\right) \ln \left[ 1+\theta \right] }e^{\frac{\ln ^{2}\left[ 1+\theta t\right] }{\ln \left[ 1+\theta \right] }}=0. \end{aligned}$$

$\square$

1.6 Proof of Theorem 7

An Archimedean copula could be generalized to the multivariate case if the inverse of its Archimedean additive generator is completely monotone (see Nelsen [50], Theorem 4.6.2). The inverse of the additive generator of the truncated-Poisson copula takes the form $\phi _{\theta }^{-1}(t)=\psi _{\theta }^{-1}\left( e^{-t}\right) =\frac{e^{e^{-t}\ln \left[ 1+\theta \right] }-1}{\theta }$. As the function $f_1\left( t\right) =e^{-t}$ is completely monotone and $f_2\left( t\right) =\frac{e^{t\ln \left[ 1+\theta \right] }-1}{\theta }$ is absolutely monotone for $\theta \ge 0$. Then, we have $f_2\circ f_1=\phi _{\theta }^{-1}\left( t\right)$ is completely monotone, which completes the proof.

Applications’ Datasets

Table 7 Kidney Infection Data

Table 8 Aircrafts Data