How to prove that a function is 2-increasing (copula) There are three conditions to prove that a function is a copula:

*

*$C(u,0)=0=C(0,v)$ grounded.

*$C(u,1)= u, C(1,v)= v$.

*$C(u,v)$ 2-increasing function.

Here I am concerning in the last condition how to prove that a function is 2-increasing
as example $H(x,y)= (2x-1)(2y-1)$.
Is it correct if the second derivative of $H(x,y)$ is greater than zero then the $H(x,y)$ is 2-increasing?
 A: The 3rd condition is


*For every $u_1,u_2,v_1,v_2 \in [0,1]$ such that $ u_1 \leq u_2 $ and $v_1 \leq v_2$ we have $C(u_1,v_1) \leq C(u_2, v_2)$
That means it is non-decreasing (which is less strict than increasing). We can rewrite this as

*

*$C(u+x,v+y) - C(u, v) \geq 0$ for every $u,v \in [0,1]$, $x \in [0,1-u]$ and $y \in [0,1-v]$
Or we can use two conditions

*

*$C(u+x,v) - C(u, v) \geq 0$ for every $u,v \in [0,1]$, and $x \in [0,1-u]$


*$C(u,v+y) - C(u, v) \geq 0$ for every $u,v \in [0,1]$, and $y \in [0,1-v]$
With those two conditions we can see that your example doesn't work. For example
$$\begin{array}{}
C(u,v+y) - C(u, v)& =& (2u-1)(2(v+y)-1) -(2u-1)(2(v)-1)\\
& = &(2u-1)y\end{array}$$
And that is negative for $u <0.5$ and $y>0$. So you do not have a non-decreasing function.
This shows that a positive second partial derivative $\frac{\partial^2}{\partial u \partial v} C(u,v)$ is not a sufficient condition.

Whuber comments about the rectangle inequality
$$P[a \leq x\leq b, c \leq y \leq d] =  C(b,d)-C(a,d)-C(b,c)+C(a,c)\ge 0$$
This resembles a second derivative. In your example case this quantity is positive, but still your function is a decreasing function.
I guess that the reason for this discrepancy is in the boundary conditions in the 1st and 2nd conditions.
In the special case that the 1st and 2nd conditions are fulfilled, then the rectangle inequality can be used to show that the function is non-decreasing.
