# Verifying two properties of the Clayton Copula

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:

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

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

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

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

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

## 2 Answers

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