# Convolution copula?

Using copula formulation for the following probability:

$$\mathbb{P}(X\leq x,y_{1}\leq Y\leq y_{2})=\mathbb{P}(X\leq x,Y\leq y_{2})-\mathbb{P}(X\leq x,Y\leq y_{1})$$ $$=C(F_{X}(x),F_{Y}(y_{2}))-C(F_{X}(x),F_{Y}(y_{1}))$$

There is no need or requirement for the two copulas above to be the same. Is there a link between these two copulas?

I read somewhere that there should be a link of the type of convolution copula.

Does anyone know if there is a link or perhaps has some references of research concerning this topic?

• There is no need or requirement for the two copulas above to be the same. Do you mean here that $$\mathbb{P}(X\leq x,y_{1}\leq Y\leq y_{2})=C_1(F_{X}(x),F_{Y}(y_{2}))-C_2(F_{X}(x),F_{Y}(y_{1}))$$ with $C_1\neq C_2$ in general? – Ilya Apr 11 '14 at 13:16
• Yes. In general $C_1\neq C_2$. – Math Girl Apr 11 '14 at 17:04

Is'nt it true that your first line can be written as $$F_{X,Y}(x,y_2) - F_{X,Y}(x,y_1),$$ where $F_{X,Y}$ is the joint cdf of $(X,Y)$. If we assume that the distributions of $X$ and $Y$ are continuous without atoms (I have to check the exact formulation), then it is clear from Sklar's theorem that there is exactly one copula $C$ such that $$F_{X,Y}(x,y) = C(F_X(x),F_Y(y)).$$ So the answer is: no the "two" copulas are just one.