Clockwise rotation

The most prominent copula modification is a rotation of a given copula by either 90, 180 or 270 degrees. Thereby one can rotate in two different directions: clockwise or counter clockwise. This page shows the derivation of pdf, cdf, h- and v-functions for clockwise rotation, which is the default setting of this package. In order to get the respective formulas for counter clockwise rotation, take a look at this page.

When deriving the individual formulas it helps to look at a concrete example, which shall be illustrated by the following contour line plots of a copula pdf:

pdf contour plot

When using new copulas obtained through modification of given copulas, properties like pdf or v-function are calculated by relating them to properties of the respective original copula.

pdf

In order to derive the pdf value of a given rotated copula, the rotation needs to be reverted and the original copula be evaluated at the a obtained point with new coordinates. For example, for a 90 degrees clockwise rotation and the given purple point shown on the left, the coordinates in terms of the original underlying copula without rotation can be seen on the right. Going from the left to the right figure, the rotation has been reverted.

Resolving 90° rotation

Hence, we get the pdf value of a clockwise 90° rotated copula with

$\begin{aligned} c^{90}(u_{1},u_{2})&=c(1-u_{2},u_{1}) \end{aligned}$

The same procedure also works for 180 and 270 degrees:

Resolving 90° rotation

To sum up, the following formulas for pdf values of clockwise rotated copulas hold:

$\begin{aligned} c^{90}(u_{1},u_{2})&=c(1-u_{2},u_{1}) \end{aligned}$ $\begin{aligned} c^{180}(u_{1},u_{2})&=c(1-u_{1},1-u_{2}) \end{aligned}$ $\begin{aligned} c^{270}(u_{1},u_{2})&=c(u_{2},1-u_{1}) \end{aligned}$

cdf

The logic for the cdf of a rotated copula is quite similar. Recalling the definition of the copula cdf, $C(u_{1},u_{2})$ is just an integration of the probability mass contained in a rectangular starting at the origin with upper right corner given by $(u_{1},u_{2})$ . Reverting the rotation, the same rectangular can be found in a different location of the original underlying copula. Therefore, we need to make use of the formula for integration of a general rectangular $[a_{1},a_{2}]\times[b_{1},b_{2}]$ for a copula, often called $H$ -volume [REFERENCE_KEY!]:

$C(a_{2}, b_{2})-C(a_{1},b_{2})-C(a_{2},b_{1})+C(a_{1},b_{1})$

Hence, for any given rectangular starting at the origin the location needs to be found after reversion of the rotation. In the following plots this is illustrated through the gray shaded rectangular.

Resolving 90° rotation

Once the coordinates are fixed for the original un-rotated copula, the $H$ -volume of the rectangular can be calculated through the standard formula:

$\begin{aligned} C^{90}(u_{1},u_{2})&=C(1,u_{1})-C(1-u_{2},u_{1})\\ &=u_{1} - C(1-u_{2},u_{1}) \end{aligned}$

The same also holds for rotations of 180 and 270 degrees:

Resolving 90° rotation

$\begin{aligned} C^{180}(u_{1},u_{2})&=1-C(1-u_{1},1)-C(1,1-u_{2})+C(1-u_{1},1-u_{2})\\ &=u_{1}+u_{2}+C(1-u_{1},1-u_{2})-1 \end{aligned}$

Resolving 90° rotation

$\begin{aligned} C^{270}(u_{1},u_{2})&=C(u_{2},1)-C(u_{2},1-u_{1})\\ &=u_{2}-C(u_{2},1-u_{1}) \end{aligned}$

h- and v-functions

As $h$ - and $v$ -functions are just partial derivatives of a copula (REFERENCE_KEY), they can easily be derived from the cdf formulas for rotated copulas. Here are the respective formulas as a lookup table:

$\begin{aligned} h^{90}(u_{1},u_{2})&=v(u_{1},1-u_{2})\\ h^{180}(u_{1},u_{2})&=1-h(1-u_{1},1-u_{2})\\ h^{270}(u_{1},u_{2})&=1-v(1-u_{1},u_{2})\\ \end{aligned}$ $\begin{aligned} v^{90}(u_{2},u_{1})&=1-h(1-u_{2},u1)\\ v^{180}(u_{2},u_{1})&=1-v(1-u_{2},1-u_{1})\\ v^{270}(u_{2},u_{1})&=h(u_{2},1-u_{1}) \end{aligned}$

Proof: 90° rotation

$\begin{aligned} h^{90}(u_{1},u_{2})&=\partial_{2}C^{90}(u_{1},u_{2})\\ &=\frac{\partial C^{90}(u_{1},u_{2})}{\partial u_{2}}\\ &=\frac{\partial(u_{1}-C(1-u_{2},u_{1}))}{\partial u_{2}}\\ &=-\partial_{1} C(1-u_{2},u_{1})(-1)\\ &=v(u_{1},1-u_{2}) \end{aligned}$ $\begin{aligned} v^{90}(u_{2},u_{1})&=\partial_{1}C^{90}(u_{1},u_{2})\\ &=\frac{\partial C^{90}(u_{1},u_{2})}{\partial u_{1}}\\ &=\frac{\partial(u_{1}-C(1-u_{2},u_{1}))}{\partial u_{1}}\\ &=1-\partial_{2}C(1-u_{2},u_{1})\\ &=1-h(1-u_{2},u_{1}) \end{aligned}$

Proof: 180° rotation

$\begin{aligned} h^{180}(u_{1},u_{2})&=\frac{\partial C^{180}(u_{1},u_{2})}{\partial u_{2}}\\ &=\frac{\partial (u_{1}+u_{2}+C(1-u_{1},1-u_{2})-1)}{\partial u_{2}}\\ &=1+ \partial_{2}C(1-u_{1},1-u_{2})(-1)\\ &=1-h(1-u_{1},1-u_{2}) \end{aligned}$ $\begin{aligned} v^{180}(u_{2},u_{1})&=\frac{\partial C^{180}(u_{1},u_{2})}{\partial u_{1}}\\ &=\frac{\partial (u_{1}+u_{2}+C(1-u_{1},1-u_{2})-1)}{\partial u_{1}}\\ &=1+ \partial_{1}C(1-u_{1},1-u_{2})\cdot \frac{\partial (1-u_{1})}{\partial u_{1}}\\ &=1-v(1-u_{2},1-u_{1}) \end{aligned}$

Proof: 270° rotation

$\begin{aligned} h^{270}(u_{1},u_{2})&=\frac{\partial C^{270}(u_{1},u_{2})}{\partial u_{2}}\\ &=\frac{\partial(u_{2}-C(u_{2},1-u_{1}))}{\partial u_{2}}\\ &=1- \partial_{1} C(u_{2},1-u_{1})\\ &=1-v(1-u_{1},u_{2}) \end{aligned}$ $\begin{aligned} v^{270}(u_{2},u_{1})&=\frac{\partial C^{270}(u_{1},u_{2})}{\partial u_{1}}\\ &=\frac{\partial (u_{2}-C(u_{2},1-u_{1}))}{\partial u_{1}}\\ &=-\partial_{2}C(u_{2},1-u_{1})\cdot \frac{\partial(1-u_{1})}{\partial u_{1}}\\ &=h(u_{2},1-u_{1}) \end{aligned}$

inverse h- and v-functions

With given $h$ - and $v$ -functions the following inverses can be derived:

$\begin{aligned} (h^{90})^{-1}(x,u_{2})&=v^{-1}(x,1-u_{2})\\ (h^{180})^{-1}(x, u_{2})&=1-h^{-1}(1-x,1-u_{2})\\ (h^{270})^{-1}(x, u_{2})&=1-v^{-1}(1-x,u_{2}) \end{aligned}$ $\begin{aligned} (v^{90})^{-1}(x,u_{1})&=1-h^{-1}(1-x,u_{1})\\ (v^{180})^{-1}(x, u_{1})&=1-v^{-1}(1-x,1-u_{1})\\ (v^{270})^{-1}(x, u_{1})&=h^{-1}(x,1-u_{1}) \end{aligned}$

Proof: 90° rotation

$\begin{aligned} h^{90}(y,u_{2})&=x&\Leftrightarrow \\ v(y,1-u_{2})&=x&\Leftrightarrow \\ y &= v^{-1}(x,1-u_{2}) \end{aligned}$ $\begin{aligned} v^{90}(y,u_{1})&=x&\Leftrightarrow \\ 1-h(1-y,u_{1})&=x&\Leftrightarrow \\ 1-x&=h(1-y,u_{1})&\Leftrightarrow \\ h^{-1}(1-x,u_{1})&=1-y&\Leftrightarrow \\ y&=1-h^{-1}(1-x,u_{1}) \end{aligned}$

Proof: 180° rotation

$\begin{aligned} h^{180}(y,u_{2})&=x&\Leftrightarrow \\ 1-h(1-y,1-u_{2})&=x&\Leftrightarrow \\ 1-x &= h(1-y,1-u_{2})&\Leftrightarrow \\ h^{-1}(1-x,1-u_{2})&=1-y&\Leftrightarrow \\ y&=1-h^{-1}(1-x,1-u_{2}) \end{aligned}$ $\begin{aligned} v^{180}(y,u_{1})&=x&\Leftrightarrow \\ 1-v(1-y,1-u_{1})&=x&\Leftrightarrow \\ v^{-1}(1-x,1-u_{1})&=1-y&\Leftrightarrow \\ y&=1-v^{-1}(1-x,1-u_{1}) \end{aligned}$

Proof: 270° rotation

$\begin{aligned} h^{270}(y,u_{2})&=x&\Leftrightarrow \\ 1-v(1-y,u_{2})&=x&\Leftrightarrow \\ 1-x&= v(1-y,u_{2})&\Leftrightarrow \\ v^{-1}(1-x,u_{2})&=1-y&\Leftrightarrow \\ y&=1-v^{-1}(1-x,u_{2}) \end{aligned}$ $\begin{aligned} v^{270}(y,u_{1})&=x&\Leftrightarrow \\ h(y,1-u_{1})&=x&\Leftrightarrow \\ y&=h^{-1}(x,1-u_{1}) \end{aligned}$

Copula theory

Notes on copula theory

Clockwise rotation

pdf

cdf

h- and v-functions

Proof: 90° rotation

Proof: 180° rotation

Proof: 270° rotation

inverse h- and v-functions

Proof: 90° rotation

Proof: 180° rotation

Proof: 270° rotation