I have recently come accross the star product of copulas, that is if $A$ and $B$ are 2-copulas and $\{C\_t\}\_{t\in[0,1]}$ is a family of copulas, then $C(x,y,z) = \int\_0^y C\_t(\frac{\partial}{\partial t} A(x,t),\frac{\partial}{\partial t} B(t,z))dt$ is the star product of $A$ and $B$, and $C$ itself is a copula.


Actually, I was wondering if we take two other 2-copulas $H$ and $G$, with $A\leq H$ and $B\leq G$ if then 


$\int\_0^y C\_t(\frac{\partial}{\partial t} A(x,t),\frac{\partial}{\partial t} B(t,z))dt\leq \int\_0^y C\_t(\frac{\partial}{\partial t} H(x,t),\frac{\partial}{\partial t} G(t,z))dt\ \forall x,y,z\in[0,1]$ 


holds for all possible families $\{C\_t\}\_{t\in[0,1]}$. I am not too familiar with all the features of copulas so I am not sure if it follows directly from them.



star-product of copulas
Below is an example where the inequality does not hold:


For each $t$ we choose $C\_t$ to be the independent copula $C\_t(u,v)=u\cdot v$ and denote the corresponding star-product of $A$ and $B$ by 
$A\star B$. 


For copulas $A$ and $B$ with probability density functions (pdf) $a$ and $b$, resp., we obtain from 
$$
A\star B(x,y,z)=\int\_0^y \frac{\partial A}{\partial t}(x,t)\frac{\partial B}{\partial t}(t,z)\,dt
$$
that the pdf of $A\star B$ is given by 
$$
a\star b(x,y,z) = a(x,y)\cdot b(y,z)\,.
$$
**Interpretation:** Suppose that $A$ is a copula for random variables $X,Y$ and $B$ for $Y,Z$. If, conditioned on $Y$, the variables $X$ and $Z$ are independent, then $A\star B$ is a copula for $X,Y,Z$. 


**Example:** Define $A,B,H,G$ by their pdfs $a,b,h,g$ on $[0,1]^2$ as follows:


* $a(x,y) = 1$ for all $x,y$,
* $b(y,z) = 0$ for $(y,z)\in [0,1/2]^2\cup(1/2,1]^2$ and $b(y,z) = 2$ otherwise,
* $h = 2 -b$,
* $g = b$.


Straightforward calculations yield that $A,B,H,G$ are indeed copulas and that the conditions $A\leq H$ and $B\leq G$ hold. 


Observe now that $a\star b(x,y,z) = b(y,z)$ which implies that $A\star B(1/2,1,1/2)=1/4>0$. On the other hand, $h\star g$ is constant and equal to zero on the set $[0,1/2]\times[0,1]\times[0,1/2]$ and therefore $0=H\star G(1/2,1,1/2)<A\star B(1/2,1,1/2)$.


