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This is Clayton's copula:

$C(u_1,u_2)=[u_1^{-\alpha} + u_2^{-\alpha} - 1]^{\frac{-1}{\alpha}}$

where $ (u_1,u_2) \in ]0,1]$ and $\alpha>0$

How do you prove the following limit to infinity ?

$lim_{\alpha \to \infty}C(u_1,u_2)=min(u_1,u_2) $

What about the other limit, to zero ?

$lim_{\alpha \to 0}C(u_1,u_2)=u_1u_2 $

I'm stuck here, help would be appreciated.

Pane
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1 Answers1

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Limit $\alpha\to\infty$: let $u\leqslant v$ in $[0,1]$, then $u^{-\alpha}\leqslant u^{-\alpha}+v^{-\alpha}-1\leqslant2u^{-\alpha}$ hence $2^{-1/\alpha}u\leqslant C(u,v)\leqslant u$. Now, consider $\alpha\to\infty$.

Limit $\alpha\to0$: let $u$ and $v$ in $[0,1]$, then $u^{-\alpha}=\mathrm e^{-\alpha\log u}=1-\alpha\log u+o(\alpha)$ when $\alpha\to0$, likewise $v^{-\alpha}=1-\alpha\log v+o(\alpha)$ when $\alpha\to0$, hence $\log C(u,v)=-\frac1{\alpha}\log(1-\alpha\log(uv)+o(\alpha))\to\log(uv)$ when $\alpha\to0$.

Did
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  • Perfect. Probably should have remembered the first method. The second I would have never thought of. Thanks! – Pane Nov 30 '13 at 22:08
  • You are welcome. Why such a pessimism about the second question? – Did Nov 30 '13 at 22:29
  • I've rarely seen Taylor series used in my field and wouldn't have thought of them. I now realize that L'Hôpital's rule could be an option too, although yours is more elegant. – Pane Nov 30 '13 at 23:06
  • @Did could I use an approach like this to at least show the first boundary condition: $C(0, u_{2}) = C(u_{1},0) = 0$ for this copula, like I'm asking about in this question here? https://math.stackexchange.com/questions/2537170/prove-that-cu-v-left-max-u-theta-v-theta-1-0-right-1-th –  Nov 25 '17 at 21:45