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?

  • 1
    $\begingroup$ 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). $\endgroup$
    – Glen_b
    Commented May 26, 2017 at 2:29
  • $\begingroup$ 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. $\endgroup$
    – ducks17
    Commented May 26, 2017 at 7:47

1 Answer 1


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 $\alpha = 1/3, \beta =2/3$.


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