Kendall's Rank for Normal Distribution and Correlation

Is there a closed formula to compute the Kendall rank correlation coefficient from r?

The new book by Kahneman, Sibony and Sunstein, Noise a Flaw in Human Judgment, has a statistical metric Percent Concordant (PC), that I had not seen before (page 108). In the notes they state that it is related to Kendall's W, and after some searching it seems to me that it is actually Kendall's rank correlation coefficient. Because in the book they give a table showing a univocal relationship between correlation r and PC, I assume that the calculation was done for normal distributions. For instance for a correlation coefficient r = 0.4, PC = 63%, and for r = 0.60, PC = 71%.

I assume that for normal distributions there is a closed formula to compute the Kendall rank corr. coef. from r, but I have not been able to find it my search. Is there one such formula and what is it?

• There is no one-to-one relation between r and Kendall's $\tau$ (and neither Kendall's $W$). The wording "is related to" doesn't imply that there is one, at least in my view. Don't know about PC though. Commented Jun 8, 2021 at 14:32
• @Lewian It looks like there is one. See below the selected answer. Commented Jun 9, 2021 at 0:55
• In @Lewian's defence, the transformation provided by the formula technically only holds in the special case of bivariate normality (although this was specified in the original question). It seems reasonably robust when things are roughly normal, but to be fair there's all kinds of wacky Anscombe's Quartet-style relationships that could break it, so it's not always one-to-one. Commented Jun 9, 2021 at 1:39
• I had missed that normality should be assumed. (Not sure whether it was already there when I wrote my comment, but may well be that I missed it anyway.) Commented Jun 9, 2021 at 15:46

If you can assume bivariate normality, there is a formula for Kendall's $$\tau$$ from $$r$$ given in Rank Correlation Methods (5th Ed.) by Kendall & Gibbons (1990, p. 167):

$$E(\tau)=\frac{2}{\pi}\arcsin r$$

The Percent Concordant coefficient is unfamiliar to me. I can't see the connection to Kendall's $$W$$ personally, but I agree that the coefficient does appear to relate to $$\tau$$, at least based on how it's described in the book. Although $$\tau$$ ranges from $$-1$$ to $$1$$, if we do a simple rescale using $$(1 + \tau)/2$$ to shift this to between $$0$$ and $$1$$, the results seem pretty close to those in the book.

from numpy import pi, arcsin

# Convert r to tau
r_to_t = lambda r: (2 / pi) * arcsin(r)

# Values in Table 1, Kahneman et al.
rs = [0, 0.1, 0.2, 0.3, 0.4, 0.6, 0.8, 1]

# Convert example r's and rescale
[round((1 + r_to_t(r))/2, 3) for r in rs]
# [0.5, 0.532, 0.564, 0.597, 0.631, 0.705, 0.795, 1.0]

• It works as a charm after the rescaling, thanks! I'm calculating PC from a sample, and from the formula you provided, and they come very close. Commented Jun 9, 2021 at 0:54