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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!

Bergson
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    Interesting that this appears to have the inequality flipped from that linked question. – Glen_b Jul 01 '20 at 02:38
  • @Thomas - where did you find that first formula? – Glen_b Jul 01 '20 at 02:39
  • @Sycorax Actually, I believe that question posts the inequality the wrong way! – Bergson Jul 01 '20 at 03:16
  • @Glen_b In a course I took. After more research, the inequality appears in the book "An introduction to copulas" by Roger B. Nelsen (page 176 of the second edition). Though I must say that I did not understand the proof very well. – Bergson Jul 01 '20 at 03:23
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    With minus instead of plus signs it's on the last page of https://www.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 $\tau>0$ – Thomas Lumley Jul 01 '20 at 03:57
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    Related: https://stats.stackexchange.com/questions/473683. – whuber Jul 01 '20 at 13:40

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