This post concerns Archimedean copulas, specifically as given in [1]. In this post, I ask a series of questions with the intent of clarifying what is written in [1].


In [1], it states,

"The copula,$ C$, is called Archimedean if it admits the representation $$ C ( u_1,\ldots, u_d ; \theta ) = \psi^{- 1}{( \psi ( u_1 ; \theta ) + \cdots + \psi( u_d ; \theta ) ; \theta )},$$ where ... $\psi$ is the so-called generator function."

Q.1 Is $\psi$ a generator function or not? (In other words, why use the words $\textit{so-called}$? )

Q.2 Is the $\textit{so-called}$ generator function related to the univariate marginal distribution functions of the multivariate joint distribution? If so, how?

Q.3 How can one derive an explicit representation of the the generator function for a copula? In particular, can you derive the $\textit{so-called}$ generator function of the independent copula (i.e., $\psi{(t)} = -\log(t)$ [1])?


[1] https://en.wikipedia.org/wiki/Copula_(probability_theory)#Most_important_Archimedean_copulas


The third question is the important and interesting one. (Surprisingly, it's not discussed in the standard textbook on the subject, Nelsen's An Introduction to Copulas.) Not only does it have the immediate application to copulas, but the same ideas apply to constructing multi-attribute valuation functions: see Keeney & Raiffa, Decisions with Multiple Objectives.

For completeness, let's first dispose of the first two questions.

  1. One can only speculate about the motivation of the authors of the Wikipedia article, but perhaps "so-called" is added because "generator" has so many completely different meanings in mathematics that its use here should be flagged lest the unwary reader be led to suppose the term suggests any more than that the copula $C$ is determined by $\psi$ and the axiomatic properties of copulas. For instance, Nelsen comes right out and plainly writes [at p. 112]

    The function $\phi$ [his term for $\psi$] is called the generator of the copula.

  2. By definition, the marginal distributions of any copula are uniform on the interval $[0,1].$ Thus, the generator $\psi$ adds no information to that fact.

  1. Given an Archimedean copula $C,$ how can one find a generator $\psi$?

I will answer this in the two-dimensional case $d=2$ because it illustrates the general technique. To avoid technical issues that could distract from the basic idea, I will assume $\psi$ is twice differentiable.

The problem, then, is this:

Given a (distribution) function $C:[0,1]\times [0,1]\to [0,1]$ that is symmetric and twice differentiable, with $C(u,0)=C(0,v)=0$ for all $u$ and $v$ and $C(u,1)=C(1,u)=u$ for all $u,$ find a decreasing function $\psi:[0,1]\to \mathbb{R}_{+}$ for which $\psi(1)=0$ and for all $u$ and $v,$ $$\psi(C(u,v)) = \psi(u) + \psi(v).\tag{*}$$

Let's begin by interpreting this relation: $\psi$ transforms or re-expresses numbers between $0$ and $1$ in such a way that the copula, when both its arguments and its values are re-expressed, is an additive function. In other words, $\psi$ exposes one of the simplest possible underlying structures for the copula.

Let me make a short motivational digression. Additive functions $f$ such as $(u,v)\to \psi(C(u,v))$ are easy to work out once you know a few values of $f,$ using the basic relations $$\pmatrix{f(u,v)\\f(u^\prime,v)\\f(u,v^\prime)\\f(u^\prime,v^\prime)} = \pmatrix{\psi(u)+\psi(v)\\ \psi(u^\prime)+\psi(v)\\ \psi(u)+\psi(v^\prime)\\ \psi(u^\prime)+\psi(v^\prime)}$$

for any $u,v,u^\prime,v^\prime$ since they imply

$$f(u^\prime,v^\prime) = f(u,v) - f(u^\prime,v) - f(u,v^\prime).\tag{**}$$

(This arises from taking two differences in the first variable--$f(u^\prime,v^\prime) - f(u,v^\prime)$ and $f(u^\prime,v) - f(u,v)$--and then differencing them.)

Thus, the three values on the right hand side (which are the values of $f$ at three vertices of a square) determine the fourth value on the left hand side (which is the value of $f$ at the remaining vertex). By starting with the values of $f$ on the upper and right sides of the domain $[0,1]\times[0,1]$ we can gradually work out close approximations to $f$ on values just inside those sides, and then repeat.

This ends the motivation: let's return to the analysis.

The "close approximation" part can be made exact by working infinitesimally: in other words, we will differentiat instead of using the finite differences in $(**).$ To this end, begin with the defining relation $(*)$ and differentiate twice, doing so once in each variable to perform the infinitesimal version of the finite differences. The Chain Rule and Product Rule of Calculus give

$$\psi^{\prime\prime}(C(u,v))\, C^\prime_1(u,v)C^\prime_2(u,v) + \psi^\prime(C(u,v))\,C^{\prime\prime}_{12}(u,v) = 0\tag{***}$$

where the subscripts on the $C$'s indicate which variable is differentiated.

Because $C$ is Archimedean, it is (obviously) symmetric, so $C^\prime_1(u,v) = C^\prime_2(v,u).$ This suggests examining values where $u=v.$ So, define $f:[0,1]\to \mathbb{R}_+$ (this is a new $f,$ differing from the previous one!) as

$$f(t) = \psi^\prime(C(t,t)).$$

Equation $(***)$ becomes a linear first order ordinary differential equation for $f.$ It can be solved (via integration) for $f$ up to an additive constant $A.$ The function $t\to \psi(C(t,t))$ is then found with a second integration, introducing another constant $B.$ $B$ will be determined by the initial condition

$$\psi(C(1,1)) = \psi(1) = 0$$

and $A$ will remain indeterminate (but must be positive to guarantee $\psi$ decreases). This is no surprise, because a positive multiple of any generator of $C$ also is a generator (see Nelsen Theorem 4.1.5(3), which is an (easy) exercise).

Let's do an example. The independence copula is

$$C(u,v) = uv,\quad 0 \le u, v\le 1.$$


$$C(t,t) = t^2;\quad C^\prime_1(t,t) = C^\prime_2(t,t) = t;\quad C^{\prime\prime}_{12}(t,t) = 1.$$

(Remember, in these expressions the derivatives are taken as partial derivatives of $C(u,v)$ and then $u$ and $v$ are both set to $t.$)

The differential equation $(***)$ is

$$\psi^{\prime\prime}(t^2) (t)(t) + \psi^\prime(t^2)(1) = 0.$$

In terms of $f$ it can be written

$$f^\prime(t^2) t^2 = -f(t^2).$$

Letting $x=t^2$ and dividing by $xf(x)$ this becomes

$$\frac{f^\prime(x)}{f(x)} = -\frac{1}{x}.$$

Integrating both sides (bearing in mind $f(x)\lt 0$ because $\psi$ must be decreasing) yields

$$\log(-f(x)) = A - \log(1/x)$$

and exponentiating produces

$$f(x) = -e^A \frac{1}{x}.$$

Integrating once more gives us the generator,

$$\psi(x) = \int^x f(x)\mathrm{d}x = B - e^A \log(x).$$

The initial condition $\psi(1) = 0$ shows $B=0,$ so the general solution for the independence copula is the independence generator family

$$\psi(x) = -e^A\log(x).$$

These are all the positive multiples of $-\log(x).$ Equivalently, they are $-\log(x)$ to any base of logarithms you care to use.

The process for any other Archimedean copula is the same as this one, but the two integrations tend to be more complicated.

  • $\begingroup$ Of relevance to this answer are functional equations [en.wikipedia.org/wiki/Functional_equation]. "For instance, properties of functions can be determined by considering the types of functional equations they satisfy." In particular, "f(xy)=f(x)+f(y), [is] satisfied by all logarithmic functions" $\endgroup$ Jan 21 '20 at 17:11

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