I came across the following inequality:

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

where $\rho$ denotes Spearman's correlation coefficient and $\tau$ denotes Kendall's rank correlation coefficient. How does one derive this inequality?

For example, in this article they cite the following inequality:

$$ -1 \leqslant \frac{3(n+2)}{(n-2)} \tau-\frac{2(n+1)}{(n-2)} \rho \leqslant+1 $$

and, considering the right hand side inequality, I could reduce the expression to $$(n+2)\tau \leq n \rho.$$

However, how does one derive the inequality in (1)? Starting from the basic definitions of $\rho$ and $\tau$ would be highly appreciated so as to understand the complete process.

Any help is appreciated!

  • 1
    $\begingroup$ Interesting that this appears to have the inequality flipped from that linked question. $\endgroup$
    – Glen_b
    Commented Jul 1, 2020 at 2:38
  • $\begingroup$ @Thomas - where did you find that first formula? $\endgroup$
    – Glen_b
    Commented Jul 1, 2020 at 2:39
  • $\begingroup$ @Sycorax Actually, I believe that question posts the inequality the wrong way! $\endgroup$
    – Bergson
    Commented Jul 1, 2020 at 3:16
  • 1
    $\begingroup$ With minus instead of plus signs it's on the last page of jstor.org/stable/2984072, and that turns out to be where Nelsen attributes the proof. Looking at the corollary Nelsen gives just below it, the inequality seems to only be useful for $\tau\leq 0$, since it's weaker than the inequality $\rho>(3\tau-1)/2$ when $\tau>0$ $\endgroup$ Commented Jul 1, 2020 at 3:57
  • 1
    $\begingroup$ Related: stats.stackexchange.com/questions/473683. $\endgroup$
    – whuber
    Commented Jul 1, 2020 at 13:40


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