I want to show that the bivariate Fréchet-Hoefdding lower bound is indeed a copula.
The bivariate function is defined by: $W(x,y)=\max\{x+y-1,0\}, \ x,y\in[0,1] $
Definition "Copula": A two-dimensional copula is a function $A:\left[ 0,1\right]^2\rightarrow \left[ 0,1\right]$ with the following two properties:
- $\forall x\in \left[ 0,1 \right]:A(x,1)=A(1,x)=x$, $A(x,0)=A(0,x)=0$.
- For $0\leq x_1 \leq x_2 \leq 1$ and $ 0 \leq y_1 \leq y_2 \leq 1$ the following inequality holds: $$ A(x_2,y_2)-A(x_1,y_2)-A(x_2,y_1)+A(x_1,y_1) \geq 0 $$
I want to show that $W(x,y)$ is a copula by verifying the properties in the definition. I already managed to show the first property, but I'm struggling with the second one.
Ad property 1: For all $x,y\in [0,1]$ the following holds: $$ W(x,1)=\max\{x+1-1,0\}=x=\max\{1+x-1,0\}= W(1,x) $$ $$ W(x,0)=\max\{x+0-1,0\}=0=\max\{0+x-1,0\}=W(0,x). $$ Ad property 2: I think I have to differ between several cases, but I don't know exactly how to describe the cases.
HINT
Consider the upper triangle defined by
\begin{align*} U := \{(x,y)\in[0,1]^{2}\mid y \geq 1 - x\} \end{align*}
For every $(x,y)\in U$, we have that $\max\{x + y - 1,0\} = x + y - 1$.
Similarly, consider the lower triangle defined by \begin{align*} L := \{(x,y)\in[0,1]^{2} \mid y \leq 1 - x\} \end{align*}
For every $(x,y)\in L$, we have that $\max\{x + y - 1,0\} = 0$.
Now study every possible configuration of the box $B = [x_{1},x_{2}]\times[y_{1},y_{2}]$.
Can you take it from here?