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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!

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See my answer on StackExchange Mathematics for a proof that $C$ is d-increasing. – RRL Jun 1 '14 at 6:53
@RRL a link might help – Clark Henry Aug 1 '14 at 0:05
@nsw: This message was sent to OP who was asking on both sites at that time. math.stackexchange.com/questions/816321/… – RRL Aug 1 '14 at 0:17
Sorry. I didn't realize it was a cross-post. :) I thought you wanted OP (and lurkers) to go searching through your profile and post history. – Clark Henry Aug 1 '14 at 0:19
No, I should have deleted it. Are you interested in this problem? – RRL Aug 1 '14 at 0:20

For your first question, your derivative is incorrect. It instead is $\frac{\partial C^2}{\partial x \partial y} = 1+\theta(1-2x-2y+4xy)$. Note also that $x+y-2xy \geq x^2 + y^2 -2xy = (x-y)^2 \geq 0$. That is, $1-2x-2y+4xy \leq 1$. On the other hand, $1-2x-2y+4xy = 2(1-x)(1-y)+2xy - 1 \geq -1$. Then, $\frac{\partial C^2}{\partial x \partial y} \geq 0$, for $\theta \in [-1, 1]$.

As for the second question, note that the copula function is invariant of any monotonic transformations, then the copula for $(X, Y^2)$ is also given by $C(x, y)$.

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Note that $C$ is invariant to strictly increasing transformations only, and $Y^2$ is not monotonic on $\mathbb{R}$, but on $[0,1]$ here indeed. – emcor Jul 31 '14 at 21:25
Thanks for making it precise. – Gordon Aug 1 '14 at 1:19

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