# How we can prove this relation for Kendall's tau

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

• Please explain what $H$ and $h$ refer to.
– whuber
Jul 25, 2019 at 11:00
• 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 Jul 26, 2019 at 4:56
• H: the joint distribution function and h : the p.d.f of H Jul 26, 2019 at 5:03
• I have noticed this relationship in the book "Continuous Bivariate Distributions" Of the authors " N. Balakrishnan · Chin-Diew Lai " Chapter 4 page 155 @whuber Jul 26, 2019 at 5:23
• 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 Jul 26, 2019 at 5:38

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 ,

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

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

Combining all we have arrived at

$$\tau_{X,Y} = 4\Pr[X_1

Now, let's translate $$\Pr[X_1. I will initially compact the bivariate vector into $$(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 $$\Pr[X_1 $$=\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

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

$$\Pr[X_1

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.

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