I am working with the following copula, and have a few questions about it:
$C(x,y) = xy + \theta (1-x)(1-y)xy$
Here $\theta \in [-1,1]$ and $x,y \in [0,1]$
First, I am trying to show this copula is d-increasing. To do this, I took $\frac{\partial C}{\partial x \partial y}$ hoping $\frac{\partial C}{\partial x \partial y} \geq 0$
What I ended up with was $\frac{\partial C}{\partial x \partial y} = 1 + \theta - \theta (1-2x-2y+4xy)$. If I think of the case where $x=0, y=1, \theta = -1$ then this is equal to -1 so my condition isn't satisfied. Am I going about this the wrong way?
Second, I am trying to calculate the copula of $(x,y^2)$. My first thought was just to plug in $x=x, y=y^2$ into my original copula. However I thought I couldn't do this because it would violate the assumption of uniform margins (as $y^2$ would no longer be uniform). Any hints here?
Many thanks!
You can show that this bivariate copula is d-increasing (2-increasing in this case), directly, using the definition.
For a function $C:[0,1]^2\rightarrow \mathbb{R}$ the C-volume of any rectangle $R=[x_1,x_2]\times[y_1,y_2] \subset [0,1]^2$ is defined by
$$V_C(R)= C(x_2,y_2)-C(x_2,y_1)-C(x_1,y_2)+C(x_1,y_1),$$
and $C$ is said to be 2-increasing if for any such rectangle $V_C(R) \geq 0.$ Note that the property that $C$ is 2-increasing neither implies nor is implied by the property that $C$ is non-decreasing in each argument.
The 2-increasing property is one requirement that must be satisfied for $C$ to be a two-dimensional copula. This is related to the fact that a joint distribution function $F$ of random variables $X$ and $Y$ with joint density $f$ is easily seen to have the 2-increasing property, since
$$0\leq P(x_1\leq X \leq x_2,y_1\leq Y \leq y_2)=\int_{y_1}^{y_2}\int_{x_1}^{x_2}f(x,y)dxdy=F(x_2,y_2)-F(x_2,y_1)-F(x_1,y_2)+F(x_1,y_1) $$
For the given function $C(x,y)= xy + \theta(1-x)(1-y)xy$ we have
$$C(x_2,y_2)-C(x_2,y_1)-C(x_1,y_2)+C(x_1,y_1)= \\\ [x_2y_2+\theta(1-x_2)(1-y_2)x_2y_2-x_2y_1-\theta(1-x_2)(1-y_1)x_2y_1]-[x_1y_2+\theta(1-x_1)(1-y_2)x_1y_2-x_1y_1-\theta(1-x_1)(1-y_1)x_1y_1]= \\\ [x_2(y_2-y_1) +\theta x_2(1-x_2)(y_2-y_1)(1-y_1-y_2)] - [x_1(y_2-y_1) + \theta x_1(1-x_1)(y_2-y_1)(1-y_1-y_2)] = \\\ (x_2-x_1)(y_2-y_1)[1+\theta(1-x_1-x_2)(1-y_1-y_2)]. $$
For all $x_1,x_2,y_1,y_2$ in $[0,1]$ we have
$$-1 \leq (1-x_1-x_2)(1-y_1-y_2) \leq 1,$$
and for $-1 \leq \theta \leq 1$ it follows that
$$1 +\theta (1-x_1-x_2)(1-y_1-y_2) \geq 0,$$
Finally, since $x_2 \geq x_1$ and $y_2 \geq y_1$ we have
$$(x_2-x_1)(y_2-y_1)[1+\theta(1-x_1-x_2)(1-y_1-y_2)] \geq 0$$
and $C$ is 2-increasing.