Help with proving an inequality

Question

I need help proving the following inequality that relates spearman and kendall correlations to be true.

$$\frac{1+\rho}{2}\le\left(\frac{1+\tau}{2}\right)^2$$

where:

$\rho$ is the spearman correlation

$\tau$ is the kendall correlation

Any insight would be helpful Edit: How do i show this inequality apply for x and y ordinal vectors

@Adrian Presumably the inequality is intended to hold for any (finite) bivariate dataset. — whuber, Jun 23 '20 at 21:41
i added that i need to show how this inequality holds for two ordinal vectors x and y — Zaros, Jun 23 '20 at 22:18
If you need to show that the inequality holds, then the word "solve" is not the right word. Use the word "show", instead. — Adrian Keister, Jun 23 '20 at 22:23
Do you know any material or reference that could give a hint in showing this? @AdrianKeister — Zaros, Jun 24 '20 at 02:19
i took an introductory course of applied statistics(mostly learning to use R) last semester and in one of the homeworks this was mark as challenge for the student, since i couldn´t solve it then, i thought i could solve it know but it was futile, i don´t know where to start @StubbornAtom — Zaros, Jun 25 '20 at 02:57

score 5 · Answer 1 · answered Jun 24 '20 at 13:23

The inequality is not generally true.

As counterexample, let the ranks of the $x$ values be $1,2,3,4$ and those of their associated $y$ values be $3,1,4,2.$ You can check that $\rho=\tau=0,$ but the statement "$1/2 = (1+\rho)/2 \le ((1+\tau)/2)^2 = 1/4$" is false.

How can one come up with such examples? In this case I compared the plots of $y=(1+x)/2$ and $y=((1+x)/2)^2,$ for $-1\le x \le 1,$ in my imagination, and noted that the latter lies below the former and diverges from it the most at $x=0.$ Thus, for the inequality to be true, $\tau$ would have to exceed $\rho$ greatly, especially when $\rho\approx 0.$

We can visualize this using crayons, as in my description of (the usual) covariance at https://stats.stackexchange.com/a/18200/919 in terms of "net amount of color" used to color in all the data rectangles in the scatterplot. When the point locations in the scatterplot are determined by the data ranks, we obtain the Spearman analog of the covariance. When, in addition, we use a fixed (unit) amount of ink to color in each data rectangle, we obtain the Kendall tau analog of the covariance. Here, then, is the chain of intuition:

When Spearman's $\rho = 0,$ Kendall's $\tau$ needs to be fairly large and positive for the inequality to hold.
$\rho=0$ means the area of the positive ("concordant") data rectangles equals the area of the negative ("discordant") data rectangles in the scatterplot of ranks (because covariance and correlation are proportional).
$\tau \gg 0$ means there are many more positive data rectangles than negative data rectangles (because $\tau$ is the average number of data rectangles counted with their signs).

The last statement doesn't seem like a necessary consequence of what precedes it. With this as an intuitive guide, it's easy to construct examples with equal numbers of positive and negative rectangles (implying $\tau=0$) whose areas balance out (implying $\rho=0$). The counterexample I offered at the outset is the simplest such situation.

wow, i didn´t expect that. But i have a doubt what would happen if i use vectors that are part of oredered vector space or one of total order — Zaros, Jun 25 '20 at 02:54
Could you elaborate on that? As you likely know, it's not possible to create a total order for a real vector space of dimension greater than 1 and, regardless, it's unclear what the relevance of any kind of ordering would be to computing correlation coefficients. — whuber, Jun 25 '20 at 13:32

Help with proving an inequality

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