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.


2 Answers 2


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*}

  • $\begingroup$ I don’t know why this hasn’t been accepted. $\endgroup$
    – Theodore
    Commented Sep 30, 2018 at 16:08

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


Hope this helps.


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