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A copula is a function $C:[0,1]^2\to[0,1]$ such that $C(x,0)=C(0,x)=0$ for all $x\in[0,1]$, $C(x,1)=C(1,x)=x$ for all $x\in[0,1]$, and \begin{equation}\label{ineq} C(x_2,y_2)-C(x_1,y_2)-C(x_2,y_1)+C(x_1,y_1)\ge0\tag{*} \end{equation} for all $(x_1,y_1), (x_2,y_2)\in[0,1]^2$ with $x_1\le x_2$ and $y_1\le y_2$. I am trying to show that a copula is Lipschitz continuous in the following sense: $$ |C(x_2,y_2)-C(x_1,y_1)| \le|x_2-x_1|+|y_2-y_1| $$ for all $(x_1,y_1),(x_2,y_2)\in[0,1]^2$.

Suppose that $x_2\ge x_1$ and $y_2\ge y_1$. Using the definition of a copula, we have that $$ x_2-x_1-C(x_2,y_1)+C(x_1,y_1)\ge0 $$ and $$ y_2-y_1-C(x_2,y_2)+C(x_2,y_1)\ge0. $$ Adding these two inequalities, we obtain $$ C(x_2,y_2)-C(x_1,y_1) \le x_2-x_1+y_2-y_1. $$ Since copulas are increasing in each argument, $ C(x_2,y_2) \ge C(x_2,y_1) \ge C(x_1,y_1) $ so that $ C(x_2,y_2)-C(x_1,y_1)\ge0 $ and hence $$ |C(x_2,y_2)-C(x_1,y_1)|\le x_2-x_1+y_2-y_1, $$ when $x_1\le x_2$ and $y_1\le y_2$.

We also need to consider the case when $x_1\le x_2$ but $y_1\ge y_2$. If I understand correctly, inequality \eqref{ineq} is not valid in this case and I am not sure how to proceed. How can we proceed with the proof when $x_1\le x_2$ but $y_1\ge y_2$?

Any help is much appreciated!

Cm7F7Bb
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  • well, if you assume $x_2\geq x_1$ and $y_2\geq y_1$ then trivially $C(x2,y2)−C(x1,y1)\geq 0\geq -(x2−x1+y2−y1)$. – Albert Paradek Apr 01 '22 at 10:57
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    However, sligthly more complicated case is when $x_2\geq x_1, y_1\geq y_2$. Then, you need to divide the $[0,1]^2$ into two parts. – Albert Paradek Apr 01 '22 at 10:59
  • @AlbertParadek Thanks for your comment! How does it follow that $C(x_2,y_2)-C(x_1,y_1)\ge0$? A copula is non-decreasing in each argument, right? Can we conclude that $C(x_2,y_2)-C(x_1,y_1)\ge0$ if $x_2\ge x_1$ and $y_2\ge y_1$? – Cm7F7Bb Apr 01 '22 at 11:01
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    As you said, A copula is non-decreasing in each argument .Therefore, $C(x2, y2)\geq C(x2, y1)\geq C(x1, y1)$. Hence, $C(x2, y2)-C(x1, y1)\geq 0$ – Albert Paradek Apr 01 '22 at 11:20
  • @AlbertParadek Thanks for drawing my attention to the case when $x_2\ge x_1$ but $y_2\le y_1$ (I edited the question accordingly). If I understand correctly, the inequality in the definition of a copula is no longer applicable in this case. You mentioned that $[0,1]^2$ needs to be divided into two parts. Could you please elaborate on that? – Cm7F7Bb Apr 01 '22 at 11:57

1 Answers1

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We have $x_1\le x_2$ but $y_1\ge y_2$. Then either $C(x_2, y_2)\geq C(x_1, y_1)$ or $C(x_2, y_2)\leq C(x_1, y_1)$. Take the case when $C(x_2, y_2)\geq C(x_1, y_1)$. Then, you want to prove that

$$ C(x_2, y_2)- C(x_1, y_1)\leq x_2-x_1 + y_1-y_2. $$ From the marginal monotonicity you get

$$ C(x_2, y_2)- C(x_1, y_1)\leq C(x_2, y_2)- C(x_1, y_2). $$

Now, you are in the previous setting, i.e. holds that

$$ C(x_2, y_2)- C(x_1, y_2)\leq x_2-x_1. $$ Hence, $C(x_2, y_2)- C(x_1, y_1)\leq x_2-x_1. $ In the other case when $C(x_2, y_2)\leq C(x_1, y_1)$, we obtain the same $$ C(x_2, y_2)- C(x_1, y_1)\leq y_1-y_2. $$

Hence, together, $$ C(x_2, y_2)- C(x_1, y_1) \leq \max(x_2-x_1,y_1-y_2) \leq x_2-x_1+y_1-y_2. $$

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