2.1. General Results
To begin, let us recall the exact mathematical definition of an absolutely continuous two-dimensional copula.
Definition 1 (absolutely continuous two-dimensional copulas [
5]).
A two-dimensional function $C(x,y)$, $(x,y)\in {[0,1]}^{2}$ is said to be an absolutely continuous two-dimensional copula if and only if the following assumptions are met for any $(x,y)\in {[0,1]}^{2}$:- (I)
$C(x,0)=0$ and $C(0,y)=0$;
- (II)
$C(x,1)=x$ and $C(1,y)=y$;
- (III)
${\partial}_{x,y}C(x,y)\ge 0$, where ${\partial}_{x,y}={\partial}^{2}/(\partial x\partial y)$ represents the standard (mixed-second-order) partial derivatives according to x and y.
This definition is at the heart of most of our proofs. In the rest of the paper, we omit the sentence “absolutely continuous two-dimensional” to lighten the text.
The following result presents the general assumptions made on $f\left(x\right)$ to make copulas of the form $x{y}^{f\left(x\right)}$ valid.
Theorem 1. Let $f\left(x\right)$, $x\in [0,1]$ be a function satisfying the following assumptions:
- A1:
$f\left(1\right)=1$ (or ${lim}_{x\to 1}f\left(x\right)=1$);
- A2:
${lim}_{x\to 0}f\left(x\right)=\ell $ with $\ell \in [1,+\infty )\cup \{+\infty \}$;
- A3:
$f\left(x\right)$ differentiable for $x\in [0,1]$ with ${f}^{\prime}\left(x\right)\le 0$ (so that $f\left(x\right)$ is non-increasing);
- A4:
$x{f}^{\prime}\left(x\right)+f\left(x\right)\ge 0$.
Then, the following two-dimensional OVP function is a copula:with the usual convention that ${0}^{0}$ exists, i.e., equal to a finite value. Proof. The proof consists in demonstrating that $C(x,y)$ satisfies the points (I), (II) and (III) of Definition 1.
- Proof
for (I) To begin, let us consider
$C(x,0)$. For any
$x\in [0,1]$, since
$f\left(x\right)$ is non-increasing by
A3 and
$f\left(1\right)=1$ (or
${lim}_{x\to 1}f\left(x\right)=1$) by
A1, we have
$f\left(x\right)\ge f\left(1\right)=1>0$. Therefore, we obtain
(For the case $x=0$, owing to A2, the case $\ell <+\infty $ is included, and if $\ell =+\infty $, we use the fact that ${0}^{+\infty}=0$).
Let us now consider
$C(0,y)$ by distinguishing
$\ell <+\infty $ and
$\ell =+\infty $ based on
A2. For the case
$\ell <+\infty $, for any
$y\in [0,1]$, we have
For the case
$\ell =+\infty $, we can remark that
${y}^{f\left(x\right)}={e}^{f\left(x\right)log\left(y\right)},$ with
$log\left(y\right)<0$ for
$y\in [0,1)$, implying that
${lim}_{x\to 0}{y}^{f\left(x\right)}(={e}^{\ell log\left(y\right)})=0$. Therefore, we obtain
Note that, for the case $y=0$, with the convention that ${0}^{0}$ exists, we have used $C(0,0)=0\times {0}^{0}=0$. On the other hand, it is clear that $C(0,1)=0$. This ends this portion of the proof.
- Proof
for (II) For any
$x\in [0,1]$, we have
Similarly, for any
$y\in [0,1]$, since
$f\left(1\right)=1$ (or
${lim}_{x\to 1}f\left(x\right)=1$) by
A1, it is immediate that
This ends this portion of the proof.
- Proof
for (III) After differentiation with the usual power-exponential rules and several simplifications and factorizations, for any
$(x,y)\in {[0,1]}^{2}$, we find
Since $(x,y)\in {[0,1]}^{2}$, it is clear that ${y}^{f\left(x\right)-1}\ge 0$. Furthermore, since $log\left(y\right)\le 0$, ${f}^{\prime}\left(x\right)\le 0$ by A3, and $f\left(x\right)\ge f\left(1\right)=1>0$ by A1, we have $xf\left(x\right)log\left(y\right){f}^{\prime}\left(x\right)\ge 0$. These results and the inequality $x{f}^{\prime}\left(x\right)+f\left(x\right)\ge 0$ by A4 imply that ${\partial}_{x,y}C(x,y)\ge 0$.
This ends this portion of the proof.
The proof of Theorem 1 ends. □
The copula described in Equation (
3) is thus modulated by a one-dimensional function
$f\left(x\right)$ that only affects the power of
y. The interest of Theorem 1 is that the assumptions
A1,
A2,
A3, and
A4 on
$f\left(x\right)$ are fulfilled by functions of various nature. The most immediate example is
$f\left(x\right)=1$, for which the copula in Equation (
3) becomes the independence copula, i.e.,
$C(x,y)=xy$. Another simple example is
$f\left(x\right)=1/x$, for which the copula in Equation (
3) becomes
$C(x,y)=x{y}^{1/x}$. This intriguing copula seems poorly referenced in the literature. More sophisticated functions with several parameters are presented in
Section 2.2. This opens the door to the construction of new dependence models that can be used efficiently in diverse statistical scenarios. This claim is illustrated in
Section 3 with the complete study of a special OVP copula that unified the independence copula and the intriguing
$C(x,y)=x{y}^{1/x}$.
Of course, Theorem 1 and all the coming results can be applied to the copulas of the following form:
the roles of
x and
y being exchanged in comparison with Equation (
3), but the required assumptions on
$f\left(x\right)$ hold.
As one of the main important copula functions, the copula density related to the copula in Equation (
3) is given as
in such a way that
$C(x,y)={\int}_{0}^{x}{\int}_{0}^{y}c(s,t)dsdt$,
$(x,y)\in {[0,1]}^{2}$. It has the advantage of being simple and quite manageable provided a moderate complexity in the definition of
$f\left(x\right)$.
The next result shows how new OVP copulas can be immediately derived from Theorem 1 from the flipping (x-flipping and y-flipping) and survival approaches.
Proposition 1. Let $f\left(x\right)$, $x\in [0,1]$ be a function satisfying the assumptions A1, A2, A3, and A4 of Theorem 1. Then, the following two-dimensional OVP functions are copulas:
- 1.
$\overline{C}(x,y)=y-(1-x){y}^{f(1-x)}$, $(x,y)\in {[0,1]}^{2}$;
- 2.
$\tilde{C}(x,y)=x\left[1-{(1-y)}^{f\left(x\right)}\right]$, $(x,y)\in {[0,1]}^{2}$;
- 3.
$\widehat{C}(x,y)=x+y-1+(1-x){(1-y)}^{f(1-x)}$, $(x,y)\in {[0,1]}^{2}$.
Proof. According to [
5,
25], the
x-flipping,
y-flipping, and survival copulas of a given copula are valid copulas. By denoting the baseline copula as
$C(x,y)$, they are defined by
and
respectively. By considering Theorem 1 and the copula defined in Equation (
3), i.e.,
$C(x,y)=x{y}^{f\left(x\right)}$, we obtain the desired results. The proof of Proposition 1 ends. □
To the best of our knowledge, Proposition 1 is the first result describing such OVP copulas, along with the precise theoretical assumptions behind them.
The next result presents some new two-dimensional inequalities that are of separate interest.
Proposition 2. Let $f\left(x\right)$, $x\in [0,1]$ be a function satisfying the assumptions A1, A2, A3, and A4 of Theorem 1. Then, for any $(x,y)\in {[0,1]}^{2}$, the following two-dimensional OVP inequalities holds:
- 1.
$max(x+y-1,0)\le x{y}^{f\left(x\right)}\le min(x,y);$
- 2.
$max(x+y-1,0)\le y-(1-x){y}^{f(1-x)}\le min(x,y)$;
- 3.
$max(x+y-1,0)\le x\left[1-{(1-y)}^{f\left(x\right)}\right]\le min(x,y)$;
- 4.
$max(x+y-1,0)\le x+y-1+(1-x){(1-y)}^{f(1-x)}\le min(x,y)$.
Proof. The proof follows immediately from the four copulas described in Theorem 1 and Proposition 1, and the Fréchet–Hoeffding theorem that is fulfilled for each of these copulas (as for any other copulas; see [
5]). More precisely, this theorem states that for any copula
$C(x,y)$, we have
$max(x+y-1,0)\le C(x,y)\le min(x,y)$ for any
$(x,y)\in {[0,1]}^{2}$. The proof of Proposition 2 ends. □
Proposition 2 can be picked up independently of the copula theory; new two-dimensional OVP inequalities are proved and can be used in various two-dimensional analysis settings.
The next results show how the proposed copulas and two baseline CDFs can be used to generate new two-dimensional distributions.
Proposition 3. Let $f\left(x\right)$, $x\in [0,1]$ be a function satisfying the assumptions A1, A2, A3, and A4 of Theorem 1. Let $U\left(x\right)$ and $V\left(x\right)$ be two CDFs of absolutely continuous distributions. Then, for any $(x,y)\in {\mathbb{R}}^{2}$, the following two-dimensional OVP function are CDFs:
- 1.
$F(x,y)=U\left(x\right)V{\left(y\right)}^{f\left[U\right(x\left)\right]}$;
- 2.
$F(x,y)=V\left(y\right)-[1-U\left(x\right)]V{\left(y\right)}^{f[1-U(x\left)\right]}$;
- 3.
$F(x,y)=U\left(x\right)\left\{1-{[1-V\left(y\right)]}^{f\left[U\right(x\left)\right]}\right\}$;
- 4.
$F(x,y)=U\left(x\right)+V\left(y\right)-1+[1-U\left(x\right)]{[1-V\left(y\right)]}^{f[1-U(x\left)\right]}$.
Proof. The proof follows immediately from the four copulas described in Theorem 1 and Proposition 1, and the original definition of a copula (see Definition 1). This ends the proof of Proposition 3. □
Proposition 3 can be the starting point for various two-dimensional data analyses. For a panel of choices of lifetime baseline CDFs, we may suggest the overview in [
26] and the references therein.
Theorem 1 opens the horizon for the creation of new copulas with an original definition beyond the OVP form. The following proposition illustrates this claim with a general ratio-form copula.
Proposition 4. Let $f\left(x\right)$, $x\in [0,1]$ be a function satisfying the assumptions A1, A2, A3, and A4 of Theorem 1. Then, the following two-dimensional ratio-form function is a copula: Proof. The proof is based on the so-called copula product. For two-dimensional copulas, say
${C}_{1}(x,y)$ and
${C}_{2}(x,y)$, this product function is defined by
A fundamental result is that
$({C}_{1}\u2605{C}_{2})(x,y)$ is a copula (see [
5]). Let us apply it to
${C}_{1}(x,y)=x{y}^{f\left(x\right)}$ and
${C}_{2}(x,y)=y{x}^{f\left(y\right)}$, which are both copulas under the assumptions
A1,
A2,
A3, and
A4 of Theorem 1. We have
${\partial}_{y}{C}_{1}(x,y)=xf\left(x\right){y}^{f\left(x\right)-1}$ and
${\partial}_{x}{C}_{2}(x,y)=yf\left(y\right){x}^{f\left(y\right)-1}$. Hence, since
$f\left(x\right)\ge f\left(1\right)=1$, we obtain
This ends the proof of Proposition 4. □
It is worth noting that the copula in Equation (
4) is connected with the Ali–Mikhail–Haq copula (with parameter
$\theta =1$). Indeed, by taking
$f\left(x\right)=1/x$, we obtain
which is a special case of the Ali–Mikhail–Haq copula. Thus, Proposition 4 offers a new perspective on dependence ratio-copula models.
The following section includes numerous examples to illustrate the above theory.
2.2. Examples
We now aim to exemplify Theorem 1. More precisely, some examples of parametric functions $f\left(x\right)$ satisfying the main assumptions of this theorem and the related OVP copulas are described. The lemma below is the first result in this regard.
Lemma 1. Let $(\alpha ,\beta ,\gamma )\in {\mathbb{R}}^{3}$. The following functions satisfy the assumptions A1, A2, A3, and A4 of Theorem 1:
for $\alpha >0$, $\gamma \ge 1$ and $\beta \in [0,1/(\alpha \gamma \left)\right]$.
- Function 2:
Exponential-power:
for $\alpha >0$ and $\beta \in [0,1/\alpha ]$.
for $\beta \in [0,\pi /2]$, $\alpha >0$, and $\lambda \in [0,1/(\alpha \beta \left)\right]$.
for $\beta \in [0,1]$.
- Function 5:
Inverse-power:
for $\alpha \in (0,1]$ and $\beta \in [0,1/\alpha ]$.
Proof. Let us prove that each of the proposed functions $f\left(x\right)$ satisfies the assumptions A1, A2, A3, and A4 of Theorem 1.
- Proof
for Function 1: We have the following:
Since $\gamma \ge 1$, $f\left(1\right)=1+\beta {(1-{1}^{\alpha})}^{\gamma}=1$, A1 is satisfied;
Since $\alpha >0$, ${lim}_{x\to 0}f\left(x\right)=1+\beta {(1-{0}^{\alpha})}^{\gamma}=1+\beta $ (so $\ell =1+\beta $), A2 is satisfied;
Since $\alpha >0$, $\gamma \ge 1$ and $\beta \in [0,1/(\alpha \gamma \left)\right]$, ${f}^{\prime}\left(x\right)=-\alpha \beta \gamma {x}^{\alpha -1}{(1-{x}^{\alpha})}^{\gamma -1}\le 0$, A3 is satisfied;
Since
$\alpha >0$,
$\gamma \ge 1$,
$\beta \in [0,1/(\alpha \gamma \left)\right]$,
${x}^{\alpha}\le 1$ and
${(1-{x}^{\alpha})}^{\gamma -1}\le 1$ for
$x\in [0,1]$,
so
A4 is satisfied. Let us mentioned that the above inequality can be refined for values of
$\alpha $,
$\beta $, and
$\gamma $ such that
$\alpha \beta \gamma {sup}_{x\in [0,1]}{x}^{\alpha}{(1-{x}^{\alpha})}^{\gamma -1}\le 1$.
- Proof
for Function 2: We have the following:
$f\left(1\right)={e}^{\beta (1-{1}^{\alpha})}={e}^{0}=1$, so A1 is satisfied;
Since $\alpha >0$, ${lim}_{x\to 0}f\left(x\right)={e}^{\beta (1-{0}^{\alpha})}={e}^{\beta}$ (so $\ell ={e}^{\beta}$), A2 is satisfied;
Since $\alpha >0$ and $\beta \in [0,1/\alpha ]$, ${f}^{\prime}\left(x\right)=-\alpha \beta {x}^{\alpha -1}{e}^{\beta (1-{x}^{\alpha})}\le 0$, A3 is satisfied;
Since
$\alpha >0$,
$\beta \in [0,1/\alpha ]$ and
${x}^{\alpha}\le 1$ for
$x\in [0,1]$,
so
A4 is satisfied.
- Proof
for Function 3: We have:
$f\left(1\right)=1+\lambda sin\left[\beta (1-{1}^{\alpha})\right]=1+sin\left(0\right)=1$, so A1 is satisfied;
Since $\beta \in [0,\pi /2]$, $\alpha >0$, and $\lambda \in [0,1/(\alpha \beta \left)\right]$,
${lim}_{x\to 0}f\left(x\right)=1+\lambda sin\left[\beta (1-{0}^{\alpha})\right]=1+\lambda sin\left(\beta \right)$ (so $\ell =1+\lambda sin\left(\beta \right)$), so A2 is satisfied;
Since $\beta \in [0,\pi /2]$, $\alpha >0$, and $\lambda \in [0,1/(\alpha \beta \left)\right]$,
${f}^{\prime}\left(x\right)=-\lambda \alpha \beta {x}^{\alpha -1}cos\left[\beta (1-{x}^{\alpha})\right]\le 0$, A3 is satisfied;
Using basic differentiation rules, we obtain
Since
$\beta \in [0,\pi /2]$,
$\alpha >0$, and
$\lambda \in [0,1/(\alpha \beta \left)\right]$,
${x}^{\alpha}\le 1$ and
$cos\left[\beta (1-{x}^{\alpha})\right]\le 1$ for
$x\in [0,1]$,
and it is clear that
$\lambda sin\left[\beta (1-{x}^{\alpha})\right]\ge 0$. Hence,
$x{f}^{\prime}\left(x\right)+f\left(x\right)\ge 0$; so
A4 is satisfied.
- Proof
for Function 4: We have:
$f\left(1\right)=1-\beta log\left(1\right)=1$, so A1 is satisfied;
${lim}_{x\to 0}f\left(x\right)=1-\beta {lim}_{x\to 0}log\left(x\right)=+\infty $ (so $\ell =+\infty $), so A2 is satisfied;
Since $\beta \in [0,1]$, ${f}^{\prime}\left(x\right)=-\beta {x}^{-1}\le 0$, A3 is satisfied;
Since
$\beta \in [0,1]$, and
$-log\left(x\right)\ge 0$ for
$x\in [0,1]$,
so
A4 is satisfied.
- Proof
for Function 5: We have:
$f\left(1\right)=\beta ({1}^{-\alpha}-1)+1=1$, so A1 is satisfied;
Since $\alpha \in (0,1]$, ${lim}_{x\to 0}f\left(x\right)=\beta ({lim}_{x\to 0}{x}^{-\alpha}-1)+1=+\infty $ (so $\ell =+\infty $), A2 is satisfied;
Since $\alpha \in (0,1]$ and $\beta \in [0,1/\alpha ]$, ${f}^{\prime}\left(x\right)=-\alpha \beta {x}^{-\alpha -1}\le 0$, A3 is satisfied;
Since
$\alpha \in (0,1]$,
$\beta \in [0,1/\alpha ]$, and
${x}^{-\alpha}\ge 1$ for
$x\in [0,1]$,
so
A4 is satisfied.
The proof of Lemma 1 comes to an end. □
Of course, Lemma 1 contains only a few examples, so much more can be found. This claim is illustrated in a coming result with the notion of CDF.
The functions in Lemma 1 combined with the result in Theorem 1 give the copulas collected in
Table 1.
By taking
$\beta =0$, all of the suggested copulas are reduced to the independence copula. By taking
$\alpha =\gamma =1$, the P copula is reduced to the copula in Equation (
1), so exemplified in ([
18] Example 2.3). The L copula as described in
Table 1 is not new. Indeed, we can write
and we obtain the so-called Gumbel–Barnett copula (see [
5]). To the best of our knowledge, the other copulas presented in the table are totally new and original. Because of its particular interest, the IP copula has a special treatment in
Section 4 for reasons explained later.
In a similar manner, based on the specific functions of Lemma 1, we may express all the OVP copulas presented in Proposition 1; this is, however, omitted for space reasons. The same is true for the ratio copulas presented in Proposition 4, which are less relevant to the OVP copula topic but are of a certain interest.
The next result shows that the assumptions A1, A2, A3, and A4 of Theorem 1 may be connected with the notion of CDF; they are satisfied by a simple ratio transformation of a CDF under suitable assumptions.
Proposition 5. Let $G\left(x\right)$ be a CDF of an absolutely continuous distribution with support $[0,1]$ and $g\left(x\right)$ be the corresponding probability density function. Suppose the following:
- B1:
${lim}_{x\to 0}xg\left(x\right)=0$;
- B2:
${g}^{\prime}\left(x\right)\le 0$ (so that $g\left(x\right)$ is non-increasing).
Then, the following ratio CDF satisfies the assumptions A1, A2, A3, and A4 of Theorem 1: Proof. Since $G\left(x\right)$ is the CDF of an absolutely continuous distribution with support $[0,1]$, we recall that $G\left(0\right)=0$, $G\left(1\right)=1$, $G\left(x\right)={\int}_{0}^{x}g\left(t\right)dt$ and ${G}^{\prime}\left(x\right)=g\left(x\right)\ge 0$. As a result:
$f\left(1\right)=1/G\left(1\right)=1$, so A1 is satisfied;
${lim}_{x\to 0}f\left(x\right)=1/{lim}_{x\to 0}G\left(x\right)=+\infty $ (implying that $\ell =+\infty $), so A2 is satisfied;
${f}^{\prime}\left(x\right)=-g\left(x\right)/G{\left(x\right)}^{2}\le 0$, so A3 is satisfied;
Since
${G}^{\prime}\left(x\right)=g\left(x\right)$, we have
It is clear that
$1/G{\left(x\right)}^{2}\ge 0$. Let us now prove that
$-xg\left(x\right)+G\left(x\right)\ge 0$. Using an integration by part,
${lim}_{x\to 0}xg\left(x\right)=0$ by
B1, and
${g}^{\prime}\left(x\right)\le 0$ by
B2, we obtain
Hence, $-xg\left(x\right)+G\left(x\right)\ge 0$, implying that $x{f}^{\prime}\left(x\right)+f\left(x\right)\ge 0$, so A4 is satisfied.
The proof of Proposition 5 ends. □
Thus, based on Proposition 5, if
$G\left(x\right)$ satisfies the assumptions
B1 and
B2, then the following two-dimensional OVP function is a valid copula:
Some examples of well-established parametric CDFs $G\left(x\right)$ satisfying the main assumptions in Proposition 5 are described in the lemma below. They are mainly based on truncated CDFs of recognized lifetime distributions (exponential distribution, Lomax distribution, etc.).
Lemma 2. Let $(\alpha ,\beta )\in {\mathbb{R}}^{2}$. The following CDFs satisfy the assumptions B1 and B2 of Proposition 5:
- CDF 1:
Truncated exponential:
for $\alpha >0$.
- CDF 2:
Truncated exponential-logarithmic:
for $\alpha >0$ and $\beta \in (0,1)$.
for $\alpha >0$ and $\beta >0$.
- CDF 4:
Truncated half-normal:
where $erf\left(x\right)=(2/\sqrt{\pi}){\int}_{0}^{x}{e}^{-{t}^{2}}dt$ for $\beta >0$.
Proof. Let us prove that each of the proposed CDFs satisfies the assumptions B1 and B2 of Proposition 5.
- Proof
for CDF 1: We have the following:
so it is immediate that
${lim}_{x\to 0}xg\left(x\right)=0$, so
B1 is satisfied;
For
$\alpha >0$,
so
B2 is satisfied.
- Proof
for CDF 2: We have the following:
so it is clear that
${lim}_{x\to 0}xg\left(x\right)=0$, so
B1 is satisfied;
For
$\alpha >0$ and
$\beta \in (0,1)$,
all the main terms being positive, the minus in factor gives the final sign, so
B2 is satisfied.
- Proof
for CDF 3: We have the following:
so it is immediate that
${lim}_{x\to 0}xg\left(x\right)=0$, so
B1 is satisfied;
For
$\alpha >0$ and
$\beta >0$,
so
B2 is satisfied.
- Proof
for CDF 4: We have
so it is immediate that
${lim}_{x\to 0}xg\left(x\right)=0$, so
B1 is satisfied;
For
$\beta >0$,
so
B2 is satisfied.
The proof of Lemma 2 ends. □
The interest of the functions in Lemma 2 is that the assumptions on the parameters are relaxed; when $\alpha $ and $\beta $ are involved, they can be chosen independently. This aspect contrasts with the joint parameter restrictions imposed on the functions proposed in Lemma 1.
The functions in Lemma 2 combined with the result in Theorem 1 give the copulas collected in
Table 2.
As a quick illustration of the findings,
Figure 1 shows the TE copula for various values of the parameters
$\alpha $.
This figure illustrates the mathematical validity of the TE copula, its flexibility in terms of shape deformations, and the ease of its implementation; all the graphical and numerical analyses of this article are realized by using the R software (see [
27]).