Kendall's τ as a particular case

If ${\displaystyle r_{i}}$, ${\displaystyle s_{i}}$ are the ranks of the ${\displaystyle i}$-member according to the ${\displaystyle x}$-quality and ${\displaystyle y}$-quality respectively, then we can define

${\displaystyle a_{ij}=\operatorname {sgn}(r_{j}-r_{i}),\quad b_{ij}=\operatorname {sgn}(s_{j}-s_{i}).}$

The sum ${\displaystyle \sum a_{ij}b_{ij}}$ is the number of concordant pairs minus the number of discordant pairs (see Kendall tau rank correlation coefficient). The sum ${\displaystyle \sum a_{ij}^{2}}$ is just ${\displaystyle n(n-1)/2}$, the number of terms ${\displaystyle a_{ij}}$, as is ${\displaystyle \sum b_{ij}^{2}}$. Thus in this case,

${\displaystyle \Gamma ={\frac {2\,(({\text{number of concordant pairs}})-({\text{number of discordant pairs}}))}{n(n-1)}}={\text{Kendall's }}\tau }$

Spearman’s ρ as a particular case

If ${\displaystyle r_{i}}$, ${\displaystyle s_{i}}$ are the ranks of the ${\displaystyle i}$-member according to the ${\displaystyle x}$ and the ${\displaystyle y}$-quality respectively, we may consider the matrices ${\displaystyle a,b\in M(n\times n;\mathbb {R} )}$ defined by

${\displaystyle a_{ij}:=r_{j}-r_{i}}$

${\displaystyle b_{ij}:=s_{j}-s_{i}}$

The sums ${\displaystyle \sum a_{ij}^{2}}$ and ${\displaystyle \sum b_{ij}^{2}}$ are equal, since both ${\displaystyle r_{i}}$ and ${\displaystyle s_{i}}$ range from ${\displaystyle 1}$ to ${\displaystyle n}$. Hence

${\displaystyle \Gamma ={\frac {\sum (r_{j}-r_{i})(s_{j}-s_{i})}{\sum (r_{j}-r_{i})^{2}}}}$

To simplify this expression, let ${\displaystyle d_{i}:=r_{i}-s_{i}}$ denote the difference in the ranks for each ${\displaystyle i}$. Further, let ${\displaystyle U}$ be a uniformly distributed discrete random variables on ${\displaystyle \{1,2,\ldots ,n\}}$. Since the ranks ${\displaystyle r,s}$ are just permutations of ${\displaystyle 1,2,\ldots ,n}$, we can view both as being random variables distributed like ${\displaystyle U}$. Using basic summation results from discrete mathematics, it is easy to see that for the uniformly distributed random variable, ${\displaystyle U}$, we have ${\displaystyle \mathbb {E} [U]=\textstyle {\frac {n+1}{2}}}$ and ${\displaystyle \mathbb {E} [U^{2}]=\textstyle {\frac {(n+1)(2n+1)}{6}}}$ and thus ${\displaystyle \mathrm {Var} (U)=\textstyle {\frac {(n+1)(2n+1)}{6}}-\textstyle {\frac {(n+1)(n+1)}{4}}=\textstyle {\frac {n^{2}-1}{12}}}$. Now, observing symmetries allows us to compute the parts of ${\displaystyle \Gamma }$ as follows:

${\displaystyle {\begin{aligned}{\frac {1}{n^{2}}}\sum _{i,j=1}^{n}(r_{j}-r_{i})(s_{j}-s_{i})&=2\left({\frac {1}{n^{2}}}\cdot n\sum _{i=1}^{n}r_{i}s_{i}-({\frac {1}{n}}\sum _{i=1}^{n}r_{i})({\frac {1}{n}}\sum _{j=1}^{n}s_{j})\right)\\&={\frac {1}{n}}\sum _{i=1}^{n}(r_{i}^{2}+s_{i}^{2}-d_{i}^{2})-2(\mathbb {E} [U])^{2}\\&={\frac {1}{n}}\sum _{i=1}^{n}r_{i}^{2}+{\frac {1}{n}}\sum _{i=1}^{n}s_{i}^{2}-{\frac {1}{n}}\sum _{i=1}^{n}d_{i}^{2}-2(\mathbb {E} [U])^{2}\\&=2(\mathbb {E} [U^{2}]-(\mathbb {E} [U])^{2})-{\frac {1}{n}}\sum _{i=1}^{n}d_{i}^{2}\\\end{aligned}}}$

and

${\displaystyle {\begin{aligned}{\frac {1}{n^{2}}}\sum _{i,j=1}^{n}(r_{j}-r_{i})^{2}&={\frac {1}{n^{2}}}\cdot n\sum _{i,j=1}^{n}(r_{i}^{2}+r_{j}^{2}-2r_{i}r_{j})\\&=2{\frac {1}{n}}\sum _{i=1}^{n}r_{i}^{2}-2({\frac {1}{n}}\sum _{i=1}^{n}r_{i})({\frac {1}{n}}\sum _{j=1}^{n}r_{j})\\&=2(\mathbb {E} [U^{2}]-(\mathbb {E} [U])^{2})\\\end{aligned}}}$

Hence

${\displaystyle \Gamma =1-{\frac {\sum _{i=1}^{n}d_{i}^{2}}{2n\mathrm {Var} (U)}}=1-{\frac {6\sum _{i=1}^{n}d_{i}^{2}}{n(n^{2}-1)}}}$

where ${\displaystyle d_{i}=r_{i}-s_{i}}$ is the difference between ranks, which is exactly Spearman's rank correlation coefficient ${\displaystyle \rho }$.

Kerby simple difference formula

Dave Kerby (2014) recommended the rank-biserial as the measure to introduce students to rank correlation, because the general logic can be explained at an introductory level. The rank-biserial is the correlation used with the Mann–Whitney U test, a method commonly covered in introductory college courses on statistics. The data for this test consists of two groups; and for each member of the groups, the outcome is ranked for the study as a whole.

Kerby showed that this rank correlation can be expressed in terms of two concepts: the percent of data that support a stated hypothesis, and the percent of data that do not support it. The Kerby simple difference formula states that the rank correlation can be expressed as the difference between the proportion of favorable evidence (f) minus the proportion of unfavorable evidence (u).

${\displaystyle r=f-u}$

Example and interpretation

To illustrate the computation, suppose a coach trains long-distance runners for one month using two methods. Group A has 5 runners, and Group B has 4 runners. The stated hypothesis is that method A produces faster runners. The race to assess the results finds that the runners from Group A do indeed run faster, with the following ranks: 1, 2, 3, 4, and 6. The slower runners from Group B thus have ranks of 5, 7, 8, and 9.

The analysis is conducted on pairs, defined as a member of one group compared to a member of the other group. For example, the fastest runner in the study is a member of four pairs: (1,5), (1,7), (1,8), and (1,9). All four of these pairs support the hypothesis, because in each pair the runner from Group A is faster than the runner from Group B. There are a total of 20 pairs, and 19 pairs support the hypothesis. The only pair that does not support the hypothesis are the two runners with ranks 5 and 6, because in this pair, the runner from Group B had the faster time. By the Kerby simple difference formula, 95% of the data support the hypothesis (19 of 20 pairs), and 5% do not support (1 of 20 pairs), so the rank correlation is r = .95 − .05 = .90.

The maximum value for the correlation is r = 1, which means that 100% of the pairs favor the hypothesis. A correlation of r = 0 indicates that half the pairs favor the hypothesis and half do not; in other words, the sample groups do not differ in ranks, so there is no evidence that they come from two different populations. An effect size of r = 0 can be said to describe no relationship between group membership and the members' ranks.