# 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.

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*}

• I don’t know why this hasn’t been accepted. – Theodore Weld Sep 30 '18 at 16:08

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.