Checking Validity and Transformations of a Symmetric Copula
Summary
The document examines a proposed bivariate copula, C(x,y) = xy + θ(1−x)(1−y)xy, with θ between −1 and 1. It addresses how to verify that the function is 2-increasing, a condition needed for it to define a valid copula. The response corrects the mixed partial derivative and bounds its factor between −1 and 1, supporting nonnegativity across the stated parameter range.
It also asks how to obtain the copula for a pair in which the second variable is squared. The response relies on invariance of copulas under strictly increasing transformations, so the dependence structure is retained when a variable is transformed monotonically. The discussion is concise and gives a mathematical argument rather than numerical or empirical evidence. Its transformation conclusion assumes the variables lie in a domain where squaring is strictly increasing, such as nonnegative values; squaring over all real values is not strictly monotonic.
Key ideas
- A copula must be 2-increasing, which can be checked through its mixed partial derivative when the function is sufficiently smooth.
- The corrected mixed derivative is nonnegative for the stated parameter interval.
- Copulas are invariant under strictly increasing transformations of their variables.
- Squaring preserves the copula only on a domain where the square function is strictly increasing.
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# Properties of a Symmetric Copula
# Properties of a Symmetric Copula
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!
## Answer by Gordon (score 1)
https://quant.stackexchange.com/a/14244
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 $C(x, y)$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.