I found the relation below while studying the measure of dependence. How we can prove this relation about Kendall's tau?
$$ \tau = 4\int\int H(x, y)\, h(x,y)\, dx\, dy - 1 $$

  • 2
    $\begingroup$ Please explain what $H$ and $h$ refer to. $\endgroup$
    – whuber
    Jul 25, 2019 at 11:00
  • $\begingroup$ Unfortunately it looks similar but not the same ,here the relation in general and for kendalls tau for copula it is specific for the archimedean family of copula $\endgroup$ Jul 26, 2019 at 4:56
  • $\begingroup$ H: the joint distribution function and h : the p.d.f of H $\endgroup$ Jul 26, 2019 at 5:03
  • $\begingroup$ I have noticed this relationship in the book "Continuous Bivariate Distributions" Of the authors " N. Balakrishnan · Chin-Diew Lai " Chapter 4 page 155 @whuber $\endgroup$ Jul 26, 2019 at 5:23
  • $\begingroup$ I think we could use this relation 4 ∫∫ C(x, y) dC(x, y) dx dy − 1 and by Skalar's Thm we can prove that Kendall's tau τ = 4 ∫ ∫ H(x, y) h(x, y) dx dy – 1 @Alecos Papadopoulos $\endgroup$ Jul 26, 2019 at 5:38

1 Answer 1


Let $(X_1,Y_1)$ be a bivariate continuous random vector, independently and identically distributed with the vector $(X_2, Y_2)$. $X_1$ may be dependent with $Y_1$, and $X_2$ may be dependent with $Y_2$. Kendall's tau is defined as

$$\tau_{X,Y} = \Pr[(X_1-X_2)(Y_1-Y_2) > 0] - \Pr[(X_1-X_2)(Y_1-Y_2) < 0]$$

Now, we have

$$\Pr[(X_1-X_2)(Y_1-Y_2) < 0] = 1-\Pr[(X_1-X_2)(Y_1-Y_2) > 0] $$

$$\implies \tau_{X,Y} =2\Pr[(X_1-X_2)(Y_1-Y_2) > 0] -1$$

The probability of the event $\{(X_1-X_2)(Y_1-Y_2) > 0\}$ is equal to the sum of the probabilities of the joint events $\{X_1>X_2, Y_1>Y_2\}$ and $\{X_1<X_2, Y_1<Y_2\}$ ,

$$\Pr[(X_1-X_2)(Y_1-Y_2) > 0] = \Pr[X_1>X_2, Y_1>Y_2] + \Pr[X_1<X_2, Y_1<Y_2]$$

Note also that because the two $X$'s are i.i.d and the two $Y$'s likewise, we have

$$\Pr[X_1>X_2, Y_1>Y_2] = \Pr[X_1<X_2, Y_1<Y_2]$$

Combining all we have arrived at

$$\tau_{X,Y} = 4\Pr[X_1<X_2, Y_1<Y_2] -1$$

Now, let's translate $\Pr[X_1<X_2, Y_1<Y_2]$. I will initially compact the bivariate vector into $(X_i,Y_i)=\mathbf w_i$. Let $H(x_i,y_i) \equiv H(w_i)$ be the joint distribution function of $(X_i,Y_i)$ (for index $1$ as well as for index $2$), and $h(x_i,y_i) \equiv h(w_i)$ the corresponding density.

Then (remembering that $\mathbf w_1$ is independent from $\mathbf w_2$, so their joint distribution is the product of their bivariate marginals) we have $$\Pr[X_1<X_2, Y_1<Y_2] = \Pr[\mathbf w_1 < \mathbf w_2]$$ $$=\int h(\mathbf w_2) \int^{\mathbf w_2} h(\mathbf t_1) d\mathbf t_1 d\mathbf w_2,$$

where the integrals are bi-dimensional. Then $$\Pr[X_1<X_2, Y_1<Y_2] = \int h(w_2) H(w_2) dw_2$$

and reverting back to writing the bivariate vector explicitly the RHS integral is

$$\Pr[X_1<X_2, Y_1<Y_2] = \int \int _{S_{X,Y}}h(x,y) H(x,y)dx dy.$$

Losing the "2" index is inconsequential because these are dummy variables of integration, the juice is in the functions comprising the integrand, as well as in the range of integration: $S_{X,Y}$ is the joint support.

So we have arrived at

$$\tau_{X,Y} = 4\int \int _{S_{X,Y}}h(x,y) H(x,y)dx dy -1 =4E[H(X,Y)] -1$$

In the last expression, $H(X,Y)$ is not treated as the distribution function of $X,Y$ but as a single-valued bivariate function of these two random variables.

  • 1
    $\begingroup$ Could you explain what "$S_{X,Y}$" refers to and what these "basic principles" might be? $\endgroup$
    – whuber
    Apr 5, 2020 at 22:42
  • $\begingroup$ @whuber Will take care of it in a few hours $\endgroup$ Apr 6, 2020 at 9:03
  • $\begingroup$ been a minute, but would be helpful to me if you could answer whuber's question $\endgroup$ May 30, 2022 at 22:57
  • 1
    $\begingroup$ @RobertHickman Yeah, I got distracted. Please see the re-written answer. $\endgroup$ Jun 3, 2022 at 1:34

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