# How do we prove that $\max\{x_1 + x_2+ \ldots + x_n - n + 1,0\} \leq C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}$?

I am interested in proving the generalized version of the Fréchet-Hoeffding inequality. Precisely speaking, given a $$n$$-copula $$C:[0,1]^{n}\rightarrow[0,1]$$, how do we demonstrate that

$$\max\{x_1 + x_2 + \ldots + x_n - n + 1, 0\} \leq C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}$$

MY ATTEMPT

Since $$\textbf{x} = (x_1,x_2,\ldots,x_n) \leq (1,1,\ldots,1)$$, I have been able to prove the upper bound inequality as next \begin{align*} C(\textbf{x}) & \leq C(x_1,x_2,\ldots,x_{n-1},1)\\ & \leq C(x_1,x_2,\ldots,1,1) \leq \ldots\\ & \leq C(x_1,1,\ldots,1,1) = x_1 \end{align*} because copulas are non-decreasing in each argument and have uniform margins. Once the same reasoning applies to each coordinate, the result $$C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}$$ follows.

But what about the first inequality? Any help is appreciated.

• It may not help, but if each $x_i \in [0,1]$ then $x_1 + x_2 + \ldots + x_n - n + 1= x_1+(x_2-1)+ \ldots + (x_n-1)$ so $\max\{x_1 + x_2 + \ldots + x_n - n + 1, 0\} \le x_1$ – Henry Feb 6 '20 at 8:13

$$C(\mathrm x)$$ is a cdf of a vector $$(U_1,\dots,U_n)$$ with $$U(0,1)$$ marginals. That said, for any $$\mathrm x = (x_1,\dots x_n)\in [0,1]^n$$, $$C(\mathrm x) = \mathrm{P} (U_1\le x_1,\dots, U_n\le x_n) = 1 - \mathrm{P} \left( \bigcup_{i=1}^n \{U_i> x_i\}\right)\\ \ge 1 - \sum_{i=1}^n \mathrm{P} (U_i> x_i) = 1 - \sum_{i=1}^n (1- x_i) = \sum_{i=1}^n x_i - n+ 1.$$
• In the first place, thanks for the contribution. Could you also provide an answer based on $C$-volumes as well? – APCorreia Feb 6 '20 at 20:31
• @user1337, I've no idea about $C$-volumes. – zhoraster Feb 6 '20 at 20:42