Clayton copula and Kendall's tau I'm currently preparing for an exam in Risk Management (mathematics) by doing exercises from old exams. One of these exercises proved to be too difficult because of the following:
Given Kendall's tau $ \tau = 1/3 $ and the Clayton copula 
$$
C(u,v) = (u^{-\theta} + v^{-\theta} - 1)^{-1/\theta}
$$
we can calculate the parameter $ \theta $ by
$$
\tau = \theta / (\theta + 2),
$$
which we see is $ \theta = 1 $.
My problem is that I can't see why this is so, how does this work?
After some more reading I have arrived at this:
$$
\begin{align}
\tau & = 4 \cdot \mathbf{E} [C(U,V)] - 1 \\ 
     & = 4 \int_{0}^{1} \int_{0}^{1} (u^{-\theta} + v^{-\theta} - 1)^{(-1/\theta)} dudv - 1,
\end{align}
$$
but I'm not able to solve this double integral. I've checked on the internet, tried Wolfram Alpha integral calculator as well as using a mathematics handbook.
Any help as to how I should proceed in order to solve this integral would be much appreciated!
 A: You have translated incorrectly the expected value of the copula in the integral representation -you forgot to include in the integrand the density function of the copula viewed as a random variable.
But in any case, straightforward integration is not the way to go here. 
Since you want to consider the expected value of the copula, you treat it as a univariate random variable (which in turn is a function of two random variables):
$$T = C(U,V)$$
In univariate level, when a distribution function is viewed as a random variable it follows a $U(0,1)$ distribution. An analogous probability integral transform result holds in the copula case: The copula viewed as a random variable has a distribution function called "Kendall distribution function", and it is equal to
$$K_C(t) = t- \frac {\varphi (t)}{\varphi'(t)}$$
where $\varphi(t)$ is the copula's generator function, and the prime denotes the first derivative.
This means that
$$E[C(U,V)] = E(T) = \int_0^1tdK_C(t)$$
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
$$\tau = 3 - 4\int_0^1\left[t- \frac {\varphi (t)}{\varphi'(t)}\right]dt = 1 + 4\int_0^1\frac {\varphi (t)}{\varphi'(t)}dt$$
The generator function for Clayton's Copula is (for $\theta \neq 0$)
$$\varphi(t) = \frac1{\theta}\left(t^{-\theta}-1\right)$$
It is now straightforward to complete the calculations and arrive at $\tau = \theta / (\theta + 2)$.
