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When $\theta =-\frac{1}{2}$ the Clayton copula is given by $C(x,y)=(\sqrt{x}+\sqrt{y}-1)^2$. I've been asked to show in this case that Kendall's tau is $-\frac{1}{3}$. Using $$\rho_{\tau}=4 \int_0^1 \int_0^1 C(x,y)dC(x,y)-1=4 \int_0^1 \int_0^1(\sqrt{x}+\sqrt{y}-1)^2\frac{1}{2 \sqrt{xy}}dxdy-1$$ I find that $\rho_{\tau}=\frac{1}{3}$. I've done the integration twice by hand and once using Mathematica and always get $\frac{1}{3}$. Am I making some kind of mistake or is the question stated wrong?

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    $\begingroup$ Your $C(x,y)$ is not a valid copula. Try $C(x,y)=\max(0, \sqrt{x}+\sqrt{y}-1)^2$ instead. $\endgroup$
    – whuber
    Commented Jan 1, 2015 at 20:44
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    $\begingroup$ $-\frac{1}{3}$ is correct; the question hasn't made an error there. $\endgroup$
    – Glen_b
    Commented Jan 1, 2015 at 22:29

1 Answer 1

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As pointed out by @whuber, the expression for the Copula is

$$C(u,v) = \big[\max \{u^{-\theta}+v^{-\theta}-1,\;0\}\big]^{-1/\theta} , \;\;\theta \in [-1,\infty), \theta\neq 0 $$

When $\theta >0 \Rightarrow u^{-\theta}+v^{-\theta}-1>0$ for the whole joint support, and we can "ignore" the $\max$ operator. But when $\theta<0$ the $\max$ operator becomes effective, and this must be taken into account when performing the integration, since for subsets of the joint support the value of the copula is zero.

An alternative approach could be the following:

Treat the Copula itself as a (univariate) random variable (which in turn is a function of two other random variables):

$$T = C(U,V)$$

Then $T$ has a distribution function called "Kendall distribution function", and it is equal to

$$K_C(t) = t- \frac {\varphi (t)}{\varphi'(t)} \tag{1}$$

where $\varphi(t)$ is the copula's generator function, and the prime denotes the first derivative.

The relation between Kendall's tau and a copula is

$$\tau = 4 \cdot E[C(U,V)] - 1 \tag{2}$$

and from $(1)$ we have

$$E[C(U,V)] = E(T) = \int_0^1tdK_C(t)$$ Integrating by parts we have

$$\int_0^1tdK_C(t) = tK_C(t)\Big|_0^1 - \int_0^1K_C(t)dt = 1 -\int_0^1K_C(t)dt$$

So Kendall's tau now is

$$\tau = 4\cdot \left(1 -\int_0^1K_C(t)dt\right) - 1 = 3 - 4\int_0^1K_C(t)dt$$ Inserting the expression for $K_C(t)$ we have

$$\tau = 3 - 4\int_0^1\left[t- \frac {\varphi (t)}{\varphi'(t)}\right]dt = 1 + 4\int_0^1\frac {\varphi (t)}{\varphi'(t)}dt$$

which is a general expression for the calculation of Kendall's tau related to a copula.

For Clayton's Copula, its generator function is (for $\theta \neq 0$)

$$\varphi(t) = \frac1{\theta}\left(t^{-\theta}-1\right)$$

Completing the calculations, one arrives at $\tau = \theta / (\theta + 2)$.

Then , for $\theta = -1/2$ we have indeed that $\tau = -1/3$.

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    $\begingroup$ +1 I was considering pointing to a recent answer of yours on math.SE, but I think the present adaptation of it to directly address the question is probably for the better. $\endgroup$
    – Glen_b
    Commented Jan 2, 2015 at 2:51

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