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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.

user0102
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  • 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 06 '20 at 08:13

1 Answers1

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$C(\mathrm x)$ is a cdf of a vector $(U_1,\dots,U_n)$ with $U(0,1)$ marginals. That said, for any $\mathrm x = (x_1,\dots x_n)\in [0,1]^n$, $$ C(\mathrm x) = \mathrm{P} (U_1\le x_1,\dots, U_n\le x_n) = 1 - \mathrm{P} \left( \bigcup_{i=1}^n \{U_i> x_i\}\right)\\ \ge 1 - \sum_{i=1}^n \mathrm{P} (U_i> x_i) = 1 - \sum_{i=1}^n (1- x_i) = \sum_{i=1}^n x_i - n+ 1. $$

zhoraster
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