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In which cases should one prefer the one over the other?

I found someone who claims an advantage for Kendall, for pedagogical reasons, are there other reasons?

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I found that Spearman correlation is mostly used in place of usual linear correlation when working with integer valued scores on a measurement scale, when it has a moderate number of possible scores or when we don't want to make rely on assumptions about the bivariate relationships. As compared to Pearson coefficient, the interpretation of Kendall's tau seems to me less direct than that of Spearman's rho, in the sense that it quantifies the difference between the % of concordant and discordant pairs among all possible pairwise events. In my understanding, Kendall's tau more closely resembles Goodman-Kruskal Gamma.

I just browsed an article from Larry Winner in the J. Statistics Educ. (2006) which discusses the use of both measures, NASCAR Winston Cup Race Results for 1975-2003.

I also found @onestop answer about Pearson's or Spearman's correlation with non-normal data interesting in this respect.

Of note, Kendall's tau (the a version) has connection to Somers' D (and Harrell's C) used for predictive modelling (see e.g., Interpretation of Somers’ D under four simple models by RB Newson and reference 6 therein, and articles by Newson published in the Stata Journal 2006). An overview of rank-sum tests is provided in Efficient Calculation of Jackknife Confidence Intervals for Rank Statistics, that was published in the JSS (2006).

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  • $\begingroup$ Thanks chl for the answer, I accepted it for the sheer scope of it. Best, Tal $\endgroup$
    – Tal Galili
    Commented Oct 31, 2010 at 13:34
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    $\begingroup$ Spearman used on two integer variables regularly compalins about ties, which seem to be handled better by Kendall's tau. $\endgroup$
    – vinnief
    Commented Nov 18, 2016 at 7:18
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I refer the honorable gentleman to my previous answer: "...confidence intervals for Spearman’s rS are less reliable and less interpretable than confidence intervals for Kendall’s τ-parameters", according to Kendall & Gibbons (1990).

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    $\begingroup$ I think the thanks are due to Roger Newson, as I'm just quoting from his article. $\endgroup$
    – onestop
    Commented Oct 31, 2010 at 15:52
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Again somewhat philosophical answer; the basic difference is that Spearman's Rho is an attempt to extend R^2 (="variance explained") idea over nonlinear interactions, while Kendall's Tau is rather intended to be a test statistic for nonlinear correlation test. So, Tau should be used for testing nonlinear correlations, Rho as R extension (or for people familiar with R^2 -- explaining Tau to unsuspecting audience in limited time is painful).

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    $\begingroup$ could you please explain "non-linear interactions". The Spearman Rho it seems reflects a measure of validity coefficient in terms of psychometry. I do not know about the nature of Tau. $\endgroup$
    – user10619
    Commented Dec 20, 2013 at 0:57
  • $\begingroup$ I do not understand your comment's psychometry thing. $\endgroup$
    – Léo Léopold Hertz 준영
    Commented Nov 10, 2016 at 13:44
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    $\begingroup$ "non-linear interactions" because all that matters is the ordering, not the linear correlation. For example, $x$ and $x^2$ have a Pearson correlation of 0 while the Kendall's tau or the Spearman's rho will have a score of 1. $\endgroup$
    – Yohan Obadia
    Commented Nov 20, 2017 at 17:01
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    $\begingroup$ That is only true when x is non-negative. $\endgroup$
    – alan ocallaghan
    Commented Jan 3, 2018 at 12:45
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    $\begingroup$ @YohanObadia You need to specify the range of the data for that statement to make any sense. Even if you did, I don't think there's any range for which that is correct. $\endgroup$
    – Denziloe
    Commented Apr 5, 2022 at 13:44
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Here's a quote from Andrew Gilpin (1993) advocating Kendall's τ over Spearman's ρ for theoretical reasons:

"[Kendall's $τ$] approaches a normal distribution more rapidly than $ρ$, as $N$, the sample size, increases; and $τ$ is also more tractable mathematically, particularly when ties are present."

Reference

Gilpin, A. R. (1993). Table for conversion of Kendall's Tau to Spearman's Rho within the context measures of magnitude of effect for meta-analysis. Educational and Psychological Measurement, 53(1), 87-92.

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FWIW, a quote from Myers & Well (research design and statistical analyses, second edition, 2003, p. 510). If you still care about the p-values;

Seigel and Castellan (1988, nonparametric statistics for the behavioral sciences) point out that, although $\tau$ and Spearman $\rho$ will generally have different values when calculated for the same data set, when significance tests for $\tau$ and Spearman $\rho$ are based on their sampling distributions, they will yield the same p-values.

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    $\begingroup$ Do you know if they offer any support for this claim? I don't see how it can actually be true in general (they may fairly often be similar, but I really don't see how the assertion that they will be the same can hold up). [I wonder if Siegel and Castellan really said exactly that, or something slightly different.] $\endgroup$
    – Glen_b
    Commented Apr 12, 2016 at 2:09
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    $\begingroup$ I have checked Siegel&Castellan (2ed p253). They do say something slightly different ... but it's actually slightly worse than the above paraphrase, even with the addition of "approximately" (worse since they restrict it to be the case under the null, but since they're conditioning on the data that won't help. Anyway, for a fixed order of $x$, all possible rank orders of $y$ are equally likely under H0.) The fact that they think conditioning on the null after conditioning on the data matters is a worry. I wonder if they meant to say something else or if they really misunderstand $\endgroup$
    – Glen_b
    Commented Apr 13, 2016 at 4:03
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    $\begingroup$ As a counterexample, take n=7 and exact p-values. Let x=1,2,3,4,5,6,7 and let y = 2,1,4,3,7,6,5 ... spearman gives p=0.048, Kendall gives 0.136 ... which are not at all alike. A different arrangement gives the same value for kendall but spearman has p=0.302. There are many such examples and various sample sizes $\endgroup$
    – Glen_b
    Commented Apr 13, 2016 at 4:03
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    $\begingroup$ Here's a plot for the n=8 case. As you see there's a lot of variation between the p-values for the two measures of correlation: i.sstatic.net/5JMbj.png ... I may write up a Q&A on this $\endgroup$
    – Glen_b
    Commented Apr 13, 2016 at 7:19
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    $\begingroup$ Here are two example data sets (after ranking) that show two cases (this time with n=9) where the Spearman correlation p-values are the same, but the Kendall correlation p-values are quite different: i.sstatic.net/3ILD8.png $\endgroup$
    – Glen_b
    Commented Apr 13, 2016 at 12:18
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To me it seems, that Kendall is conceptually more easy to grasp, as it only relates to the permutations of ranks that occur under random shuffles (this could be the "peadagogical reason"). For understanding $\rho$ one needs to learn first the theory of linear correlations, and then the interpretation is actually different (loosely speaking: $\tau$ measures how non-random are my data rank pairs, whereas $\rho$ measures how much variance can a linear model explain on transformed data; for the mapping that is used in $\rho$ see this answer).

In most cases the inferences from both will be the same - edge cases are hardly something that you need or want to deal with; when really in doubt, and an inference is required at all cost, then take the one with smaller absolute value.

Also note the following relation between Kendall $\tau$ and and Spearman $\rho$: $-1 \leq 3\times\tau -2 \times \rho \leq +1$ (left as exercise to the reader).

Reference

Puka, L. (2011). Kendall’s Tau. In International Encyclopedia of Statistical Science (pp. 713–715). Springer Berlin Heidelberg. https://link.springer.com/referenceworkentry/10.1007/978-3-642-04898-2_324

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    $\begingroup$ This needs to be downvoted. Spearman's rho is well known as a non-parametric measure since it only looks at the ranks. Here's a good description of generally the difference between parametric and non-parametric tests. $\endgroup$
    – blehblehblecksheep
    Commented Jun 6 at 18:11
  • $\begingroup$ I removed the first sentence - this was indeed a wrong oversimplification. But correct me if I am wrong: The interpretation of Spearman requires a linear model, whereas this is not needed for Kendall. $\endgroup$
    – Christian Herenz
    Commented Jun 7 at 11:31

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