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For a function $H(x,y)$ to be a copula, it has to be increasing in $x$ and in $y$. But, instead of this condition, other authors say that the function has to be grounded. Are these properties equivalent?. I cannot prove that if $H(x,y)$ is increasing in $x$ and $y$, then it is grounded. Any help will be appreciated.

Luis
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  • Here's a related question: http://math.stackexchange.com/questions/1257160/why-it-is-necessary-for-copula-functions-to-be-grounded – Michael Hardy Dec 19 '15 at 19:31
  • Many thanks, I had already seen this related question. But mine is different. My problem is how to prove that H(x,y) is grounded when:1) it is 2-increasing.- 2) it is increasing for x and y.- 3) it is such that H(1,y)= y and H(x,1)=x. – Luis Dec 20 '15 at 15:17
  • "Grounded" should mean $H(0,y)=0=H(x,0)$ for all $x,y\in[0,1]$. If $H(0,1/2)$ were positive, that would mean there is a positive probability that the first of the two component random variables is $0$. And that would mean it's not uniformly distributed on $[0,1]$. ${}\qquad{}$ – Michael Hardy Dec 20 '15 at 16:16
  • Ok, Michael but your reasoning needs to make use of hypothesis or properties of H(x,y) other than the three formerly stated: 1) 2-increasing.- 2) increasing in x and y.- 3) H(x,1) = x and H(1,y)= y. – Luis Dec 20 '15 at 20:28
  • The fact that $H(x,1)=x$ and $H(1,y)=y$ is the same as saying the marginals are uniformly distributed. ${}\qquad{}$ – Michael Hardy Dec 20 '15 at 21:27
  • $H(0,y) \le H(0,1)$ since $H$ is nondecreasing in each variable separately. And $H(0,1) = 0$ because $H(x,1)=x$ for all $x\in[0,1]$. ${}\qquad{}$ – Michael Hardy Dec 20 '15 at 21:29
  • Thank you for your help. I will think about your ideas. – Luis Dec 20 '15 at 22:14

2 Answers2

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If $H(x, y)= x + y -1$ we have a function that fulfills our three properties or requirements, but it is not grounded.

Harish Chandra Rajpoot
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Luis
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The properties are not equivalent and a copula function has to satisfy both of them. What happens is that some authors define the copula as a joint distribution with cumulative distribution function as arguments. This definition is not the formal definition of copulas as stated by Nelsen.

As defined in Nelsen's book, a copula is a function $C(u,v): I^2 \rightarrow I$ that is 2-increasing, grounded and $C(u,1) = u, C(1,v) = v$. A function that is grounded and 2-increasing is non-decreasing in each argument. The 2-increasing definition comes from the definition of the C-volume over a rectangle with vertices on $I^2$:

Given a rectangle $B = [x_1, x_2] \times [y_1, y_2]$, the C-Volume over B is:

$$V_C(B) = C(x_2, y_2) - C(x_2, y_1) - C(x_1, y_2) + C(x_1, y_1)$$

And C is 2-increasing if and only if $V_C(B) \geq 0$ for every B that has vertices on $I^2$ (domain of C).

When some authors define a copula function as a function of the cumulative distribution function of two variables x and y $C(F(x), G(y)) = H(x,y)$, it already satisfies all those properties (non-decreasing of arguments, grounded and 2-increasing) because of the probability properties of F, G and H. By this definition. the copula can be seen as a volume (the result of the double integral):

$$C(F(x), G(y)) = H(x,y) = \int_0^x\int_0^y h(x,y)dxdy$$

It is easy to see that whenever $x = 0$ or $y = 0$, the integral is also zero (grounded) and because $F \geq 0$ and $G \geq 0$, the volume is always non-negative (2-increasing).

Note that in Nelsen's book the definition of a copula function is not related to the cumulative distribution function of random variables, and the link is made through Sklar's Theorem (Probability Integral Transform). By this theorem, every joint distribution function is a copula and vice versa.

But it is not true that every function $H: I^2 \rightarrow I$ is a copula. For example $H(x,y) = max\{x,y\}$ for $x,y \in I$ is not a copula because it is not 2-increasing:

$$V_H([0, 1] \times [0, 1]) = H(1,1) - H(1,0) - H(0,1) + H(0,0) = 1 - 1- 1- 0 = -1 < 0$$

David
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