Calculation of 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 $(X_1, X_2)$ a r.v. vector with $X_1$ ~ Exp$(1)$, $X_2$ ~ $N(0,1)$ and the dependecne structure given by the copula

$$C(u_1, u_2) = \frac{1}{3}W(u_1,u_2)+\frac{2}{3}\Pi(u_1,u_2), \text{ } u \in [0,1]^2$$

where the contra-montone copula is $W(u_1,u_2) = (u_1+u_2+1)_+$ and the indepence copula is $\Pi(u_1,u_2) = u_1u_2$. Calculate Kendall's Tau.

So I know the follwing formula for Kendall's Tau:

$$\rho_{\tau}(X_1,X_2) = 4 \int_0^1 \int_0^1 C(u_1,u_2)dC(u_1,u_2)-1.$$

But I really don't know how to start here, should I evaluate the integral

$$4 \int_0^1 \int_0^1 (\frac{1}{3}(u_1+u_2-1)_++\frac{2}{3}u_1u_2)du_1du_2 -1$$ That would result $-\frac{1}{9}$, but I don't think thats true. Can someone help me?

• Why don't you think $-\frac19$ could be right? (I'm not saying it is right, because I haven't checked, but I'd certainly expect it to be negative). – Glen_b May 26 '17 at 2:29
• I'm just not sure, if the integration is right. Because in the formula there is $dC(u_1, u_2)$ and I've basically just substituted it by $du_1du_2$. Furthermore in the calculation I've basically didn't use the distributions of $X_1$ and $X_2$ which kind of makes me uncertain. – ducks17 May 26 '17 at 7:47

The Lebesgue decomposition of $\mathbb{P}_C$, the measure induced by $C$, has a singular part (continuous singular). https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem. So one should find the decomposition in its absolutely continuous part plus singular continuous part (there's no pure point part) in order to calculate that double integral.
One can prove that $\mathbb{P}_C = \frac{1}{3}\mathbb{P}_W + \frac{2}{3}\mathbb{P}_\Pi$ and also that \begin{align} \int\int {C(u,v)} dC(u,v) & := \int {C(u,v)} \mathbb{P_C}(du,dv)\\ & = \frac{1}{3}\int C(u,v)\mathbb{P}_W(du,dv) + \frac{2}{3} \int C(u,v)\mathbb{P}_\Pi(du,dv)\\ &= \frac{1}{3}\int_0^1 C(t,1-t)dt + \frac{2}{3}\int_0^1\int_0^1{C(u,v)}dudv \end{align}
Using for the last line that $\mathbb{P}_W$ has support on the line $\left\{(u,v)\in [0,1]^2: u+v=1\right\}$ and that its mass is uniformly spread along that same line.
I got $\tau = \beta(1 + \alpha/3)-1$, where $\alpha = 1/3, \beta =2/3$.