I am learning about copula's, using Nelsen's book, and more specifically about the geometric method of constructing copula's. The problem is replicated in the following link: http://www.stat.ubc.ca/lib/FCKuserfiles/file/huacopula.pdf (Slides 37 - 42).

My question (referring to the posted link above) is related to Slide 39, 40, and 41. In this problem, we have defined the support of the copula to be the triangle shown in the diagram on Slide 39,40, and 41. The line on the left can be shown to be $u=a v$ and the line on the right to be $u = 1 - (1-a)v$. From a general definition, the support of a function is the subset of the domain of the copula where the value of the copula is not 0. So now onto my questions:

  1. In Slide 39, $C_a(u,v) = V_{C_a}([0,u]\times [0,v])$. Because $$V_{C_a}([0,u]\times [0,v]) = C(u,v) - C(u,0) - C(0,v) + C(0,0)$$ (definition of C-volume) and by the grounded property that $C(u,0)= C(0,v)= C(0,0)=0$, we get that $$V_{C_a}([0,u]\times [0,v]) = C(u,v)$$ Because we are considering the region where $u\leq av$ (i.e. u is restricted by the left line segment, v can go from 0 to 1), the C-volume calculation becomes $$V_{C_a}([0,u]\times [0,1]) => C(u,v) = u$$ Is my reasoning sound here (mainly the part where I say u is restricted by the line and v can go from 0 to 1)?
  2. In Slide 40, we consider the region where $1−1−avuav$. This equation is clear to me, as it is the region bounded by the equations I gave above for the support of the copula. However, Nelsen's reasoning for saying that $$C_a(u,v) = C_a(av,v) = av$$ is that the C-volume of of the rectangle $[av,u]\times[0,v]$ is 0. This is the confusing part to me. I understand that we defined the support of the copula, so that leads me to think that every term in the calculation of the C-volume will be 0, because the rectangles in this region do not fall on the support of the copula. But this seems like circular logic to justify the statement that the C-volume of the rectangle is 0. More specifically, it is my understanding that Nelsen arrives at the answer as follows: $$V_{C_a}([av,u]\times[0,v]) = C(u,v)-C(av,v)-C(u,0)+C(av,0) = 0$$ which implies $$C(u,v) = C(av,v)$$ Then using the result from Slide 39, we make the leap that $$C(u,v) = C(av,v) = av$$ However, my lack of understanding in this is as follows: no points (u,v) as defined in this region will be on the support of the copula, and as such the conclusion that I am wanting to draw (and most likely incorrectly) is that $$C(u,v)=C(av,v)=C(u,0)=C(av,0)=0$$ So my question is, why is this incorrect?
  3. In proving slide 41, to me it seems like the same logic is applied as in question 2. Again, what am I missing?

1 Answer 1


although this questions has been asked for quite some time, I'll try to share the information I obtained while studying the construction of copula using the geometric method.

I believe this question is asking how the copula $C$ can be found when the specified support is known to be as in the diagram below. enter image description here

Indeed the explanation given in Nelsen(2006) might create some confusion to the reader. However, I found some chapter that clearly explained how the geometric method is conducted. You can follow this link for more details (use other source to read the chapter).

I think the basic idea on how to use geometric method is to know the fact that all drawn rectangles that do not intersect the support line will have volume 0.

Thus, for all three cases, we can find the copula $C(u,v)$ as follows, enter image description here enter image description here enter image description here

Hence, the obtained copula is written as $$ C(u,v)=\begin{cases} u & \text{if $u<\alpha v$} \\ \alpha v & \text{if $u>\alpha v$ and $u<1-(1-\alpha )v$} \\ u+v-1 & \text{if $u>1-(1-\alpha)v$} \end{cases}$$


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