So I'm trying to verify the first two properties of a copula for the Clayton model. The first two properties being:

  1. $C(u_1,…,u_d)$ is non-decreasing in each component, $u_i$

  2. The $i^{th}$ marginal distribution is obtained by setting $u_j=1$ for $j≠i$ and since it is uniformly distributed, $C(1,…,1,u_i,1,…,1)=u_i$.

The Clayton copula is defined as: $$C(u,v,\theta)=(u^{-\theta}+v^{-\theta}-1)^{- \frac{1}{\theta} }, \theta>0$$

I might be going about this completely wrong, but are these two properties the same as saying:

  1. $C(u,0,\theta)=C(0,v,\theta)=0$

  2. $C(u,1,\theta)=u$ and $C(1,v,\theta)=v$

In which case,

$$C(u,0,\theta)=(u^{-\theta}+1-1)^{-\frac{1}{\theta}}=(1+v^{-\theta}-1)^{- \frac{1}{\theta} }=C(0,v,\theta)$$

$$\implies C(u,0,\theta)=(u^{-\theta})^{-\frac{1}{\theta}}=(v^{-\theta})^{- \frac{1}{\theta} }=C(0,v,\theta)$$

$$\implies C(u,0,\theta)=u=v=C(0,v,\theta)$$

And for the second property:

$$C(u,1,\theta)=(u^{-\theta}+1^{-\theta}-1)^{-\frac{1}{\theta}}, C(1,v,\theta)= (1^{-\theta}+v^{-\theta}-1)^{- \frac{1}{\theta} }$$

$$\implies C(u,1,\theta)=(u^{-\theta}+1-1)^{-\frac{1}{\theta}}, C(1,v,\theta)=(1+v^{-\theta}-1)^{- \frac{1}{\theta} }$$

$$\implies C(u,1,\theta)=(u^{-\theta})^{-\frac{1}{\theta}}, C(1,v,\theta)=(v^{-\theta})^{- \frac{1}{\theta} }$$

$$\implies C(u,1,\theta)=u, C(1,v,\theta)=v$$

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    $\begingroup$ I guess the answer by g g will suffice $\endgroup$ – MarissaB Apr 19 '19 at 9:28
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    $\begingroup$ Thank you for the bounty also, Emma $\endgroup$ – MarissaB Apr 19 '19 at 9:29

Your reasoning for the first property does not look correct or at least I do not understand it. Your arguments for the second property seem sound. But your wording of the second property is a bit fuzzy. You should state this more clearly, for example: $C(1,\ldots,1,u_j,1,\ldots,1) = u_j$ for all $u_j\in [0,1]$ and $j\in 1,\ldots, d.$

You don't mention it but in addition to the two properties you state you would need to show two more properties to make sure Clayton is a proper copula. First that $C$ is a well defined function from the unit cube to $[0,1]$ and then the rectangle inequality, which is slightly more involved.

For your property 1. you need to show that $C(u_1,v) \le C(u_2,v)$ for all $v\in[0,1]$ and $0\le u_1\le u_2\le 1$ and the analogous statement for $v$.

Once you established well definedness, you can argue as follows: $-\theta <0$ hence $u_1^{-\theta}\ge u_2^{-\theta}$, which means $u_1^{-\theta} + v^{-\theta} -1 \ge u_2^{-\theta} + v^{-\theta} -1$. Now the exponential with $-\frac{1}{\theta}$ flips the inequality again and you conclude $$(u_1^{-\theta} + v^{-\theta} -1)^{-\frac{1}{\theta}} \le (u_2^{-\theta} + v^{-\theta} -1)^{-\frac{1}{\theta}}.$$

  1. Fix u, obtain derivative of v. And do it again for fixing v.
  2. To get marginal density of v, one has to do the integration w.r.t. u

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