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I know $C(u,v)=uv$ is a copula corresponding to the independent case. How about $C(u,v)=u^2v^2$?

ndor
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  • This is not a copula. You can find a discussion of copulas with quadratic section in Nelsen, An introduction to copulas, second edition, 2006, pp. 77-80. – mlc Mar 25 '17 at 07:11

1 Answers1

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A bivariate copula $C:[0,1]^2 \to [0,1]$ must satisfy

$$\tag{1} C(u,0) = C(0,v) = 0,$$ $$\tag{2} C(u,1) = C(1,u) = u,$$

and, if $0 \leqslant u_1 \leqslant u_2 \leqslant 1$ and $0 \leqslant v_1 \leqslant v_2 \leqslant 1,$

$$\tag{3} C(u_2,v_2) - C(u_2,v_1) - C(u_1,v_2) + C(u_1,v_1) \geqslant 0 $$

In this case condition (2) is violated.

Condition (2) is required because given random variables $X_1, X_2$ with marginal distributions $F_1, F_2$ we must have, for example,

$$\begin{align}C(u,1) &= P(X_1 \leqslant F_1^{-1}(u),X_2 \leqslant F_2^{-1}(1)) \\ &= P(X_1 \leqslant F_1^{-1}(u)) \\ &= F_1(F_1^{-1}(u)) \\ &= u \end{align}$$

RRL
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