Skip to content
All library documents

Showing Nonnegative Derivatives for an Extreme-Value Copula

Article Quant Q&A · Author: wrb98
Read the original

Summary

The document shows how to verify that the partial derivative of a bivariate extreme-value copula with respect to one argument is nonnegative. After differentiating, it reduces the task to proving that the Pickands dependence function satisfies A(t) minus t times its derivative being nonnegative for t between zero and one.

The proof uses the convexity of A and its endpoint value at one to bound its slope above by one. Since A(t) is at least t, this slope bound implies the required inequality. The argument also invokes standard properties of the extreme-value dependence function, including convexity and its endpoint conditions. These conditions matter: the proof is tied to the stated class of Pickands functions and does not establish the claim for an arbitrary differentiable function.

Key ideas

  • The derivative’s sign can be reduced to showing that A(t) minus tA'(t) is nonnegative.
  • Convexity bounds the derivative of A above by one using the endpoint at one.
  • The lower bound A(t) ≥ t then establishes the needed inequality.
  • The proof depends on the standard convexity and boundary properties of the Pickands dependence function.

Tags

StatisticsRisk management
Full text
# Verifying that the extreme value copula is indeed a copula


# Verifying that the extreme value copula is indeed a copula












Given the extreme value copula as defined in Schölzel/Friederichs (2008), how does one verify that $\frac{\partial C(u_1, u_2)}{\partial u_1} \geq 0?$ For the LHS, I have $$\exp\left[\log(u_1u_2)A\left(\frac{\log(u_2)}{\log(u_1u_2)}\right)\right]\left[\frac{1}{u_1}A\left(\frac{\log(u_2)}{\log(u_1u_2)}\right)-\frac{\log(u_2)}{u_1\log(u_1u_2)}A^{\prime}\left(\frac{\log(u_2)}{\log(u_1u_2)}\right)\right]$$

The derivative is causing difficulty. Any help on progressing would be much appreciated.

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/41997

Note that, you only need to show that \begin{align*} A\left(\frac{\log(u_2)}{\log(u_1u_2)}\right)-\frac{\log(u_2)}{\log(u_1u_2)}A'\left(\frac{\log(u_2)}{\log(u_1u_2)}\right) \ge 0, \end{align*} or, for any $t \in (0, 1)$, \begin{align*} A(t) - t A'(t) \ge 0. \end{align*} Recall that $A$ is a convex function from $[0,\, 1]$ to $[1/2,\, 1]$, $A(0)=A(1)=1$, and $A(t) \ge \max(t, 1-t)$. From the convexity, the path from the function is always above the tangent line at any point. That is, for any $\xi, t \in (0, 1)$, \begin{align*} A(\xi) \ge A(t) + A'(t) (\xi -t). \end{align*} Let $\xi\rightarrow 1$, \begin{align*} 1 \ge A(t) + A'(t) (1 -t). \end{align*} In other words, \begin{align*} A'(t) (1 -t) &\le 1-A(t)\\ &\le 1-t. \end{align*} Consequently, $A'(t) \le 1$. Then, \begin{align*} A(t) - t A'(t) \ge A(t) - t \ge 0. \end{align*}

## Answer by Magic is in the chain (score 0)

https://quant.stackexchange.com/a/42002

Attached contains a very detailed account (also see appendix A for derivation ofnyhe derivative):

https://mediatum.ub.tum.de/doc/1145695/1145695.pdf

Hope this helps.

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.