Stats with Python: Rank Correlation

February 06, 2021 | 8 min read | 988 views

The correlation coefficient is a familiar statistic that we see everywhere from news articles to scientific papers, but there are several variations whose differences should be noted. This post aims to recap the definitions of those common correlation coefficients, with the derivation of the equation and experiment regarding Spearman rank correlation coefficient.

Pearson Correlation Coefficient

For measuring the linear correlation between two sets of data, it is common to use Pearson product-moment correlation coefficient. Pearson correlation coefficient is the most well-known measure for correlation. When the term “correlation coefficient” is used without further information, it usually refers to this type of definition. Given paired data $\{(x_i,y_i)\}_{i=1}^n$ , Pearson’s r is defined as:

r = \frac{{\displaystyle \sum_{i = 1}^n (x_i - \overline{x}) (y_i - \overline{y})}}{\sqrt{{\displaystyle \sum_{i = 1}^n (x_i - \overline{x})^2}} \sqrt{{\displaystyle \sum_{i = 1}^n (y_i - \overline{y})^2}}} ,

where $\bar{x}$ and $\bar{y}$ are the sample means. The numerator is the covariance between $x_i$ and $y_i$ and the denominator is the product of their standard deviations.

The correlation coefficient ranges from $−1$ to $1$ . $r=\plusmn1$ is observed if and only if all the data points lie on a line (perfect correlation).

Spearman Rank Correlation Coefficient

When the data is ordinal variable, you should consider rank correlation. One of the common measures for rank correlation is Spearman rank correlation coefficient, which is simply the Pearson correlation coefficient between the two rank variables. For the $n$ paired ranks $\{(a_i,b_i)\}_{i=1}^n$ for the raw scores $\{(x_i,y_i)\}_{i=1}^n$ , the Spearman’s ρ is defined as:

\rho = \frac{{\displaystyle \sum_{i = 1}^n (a_i - \overline{a}) (b_i - \overline{b})}}{\sqrt{{\displaystyle \sum_{i = 1}^n (a_i - \overline{a})^2}} \sqrt{{\displaystyle \sum_{i = 1}^n (b_i - \overline{b})^2}}} ,

where $\bar{a}$ and $\bar{b}$ are the sample means. This definition can be simplfied to:

\rho = 1 - \frac{6}{n(n^2 - 1)}\sum_{i = 1}^n (a_i-b_i)^2.

It takes the value 1 if the order of the raw scores all match, and the value -1 if the order is completely reversed.

Proof for the Simplified Form

Since $a_i$ and $b_i$ are ranks of $n$ scores, following equations hold.

\begin{gathered} \sum_{i = 1}^n a_i = \sum_{i = 1}^n b_i=\sum_{i = 1}^n i = \frac{n(n+1)}{2}\\ \sum_{i = 1}^n a_i^2 = \sum_{i = 1}^n b_i^2 = \sum_{i = 1}^n i^2= \frac{n(n+1)(2n+1)}{6}\\ \overline{a} = \overline{b} = \frac{1}{n} \sum_{i = 1}^n i= \frac{n+1}{2} \end{gathered}

Using the above equations, we have:

\begin{aligned} \rho &= \frac{{\displaystyle \sum_{i = 1}^n (a_i - \overline{a}) (b_i - \overline{b})}}{\sqrt{{\displaystyle \sum_{i = 1}^n (a_i - \overline{a})^2}} \sqrt{{\displaystyle \sum_{i = 1}^n (b_i - \overline{b})^2}}} \\ &= \frac{{\displaystyle \sum_{i = 1}^n \Bigl(a_i - \frac{n+1}{2}\Bigr) \Bigl(b_i - \frac{n+1}{2}\Bigr)}}{\sqrt{{\displaystyle \sum_{i = 1}^n \Bigl(a_i - \frac{n+1}{2}\Bigr)^2}} \sqrt{{\displaystyle \sum_{i = 1}^n \Bigl(b_i - \frac{n+1}{2}\Bigr)^2}}} \\ &= \frac{ {\displaystyle \sum_{i = 1}^n \Bigl(a_ib_i - \frac{n+1}{2}(a_i+b_i) + \frac{(n+1)^2}{4}\Bigr)}} { \sqrt{{\displaystyle \sum_{i = 1}^n \Bigl(a_i^2 - (n+1)a_i + \frac{(n+1)^2}{4}\Bigr)}} \sqrt{{\displaystyle \sum_{i = 1}^n \Bigl(b_i^2 - (n+1)b_i + \frac{(n+1)^2}{4}\Bigr)}} } \\ &= \frac{ {\displaystyle \sum_{i = 1}^n a_ib_i - \frac{n(n+1)^2}{2} + \frac{n(n+1)^2}{4}} }{ {\displaystyle \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)^2}{2} + \frac{n(n+1)^2}{4}} }\\ &= \frac{ 12{\displaystyle \sum_{i = 1}^n a_ib_i - 3n(n+1)^2} }{ (n-1)n(n+1) }. \end{aligned}

Here, let’s consider the sum of the squared difference between $a_i$ and $b_i$ :

\sum_{i = 1}^n (a_i-b_i)^2 = -2\sum_{i = 1}^na_ib_i +\frac{n(n+1)(2n+1)}{3}.

Therefore,

\begin{aligned} \rho &= \frac{ -6{\displaystyle \sum_{i = 1}^n (a_i-b_i)^2 +2n(n+1)(2n+1)- 3n(n+1)^2} }{ (n-1)n(n+1) }\\ &= 1 - \frac{6}{n(n^2 - 1)}\sum_{i = 1}^n (a_i-b_i)^2 ~~~\blacksquare \end{aligned}

Experiment

Here, I conducted a quick experiment to confirm that the Pearson’s r of the ranks is equivalent to Spearman’s ρ. I generated 100 pairs of random samples (x and y) and calculated several types of correlation coefficients.

import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns

sns.set_style("darkgrid")

n = 100
x = np.random.rand(n)
y = x + 0.5*np.random.rand(n)

from scipy import stats
print(stats.pearsonr(x, y))
# >> (0.8863388430290433, 1.5374433582768292e-34)

print(stats.spearmanr(x, y))
# >> SpearmanrResult(correlation=0.8835643564356436, pvalue=4.677602781530847e-34)

print(stats.kendalltau(x, y))
# >> KendalltauResult(correlation=0.6981818181818182, pvalue=7.62741751146521e-25)

By converting x and y to ranks, it is confirmed that stats.spearmanr(x, y) is equal to stats.pearsonr(a, b).

a = len(x) - stats.rankdata(x) + 1
b = len(y) - stats.rankdata(y) + 1
print(stats.pearsonr(a, b))
# >> (0.8835643564356437, 4.677602781530673e-34)

Kendall Rank Correlation Coefficient

Kendall rank correlation coefficient is another common type of rank correlation efficients. Among $N$ pairs of indices $\{(i,j)\}_{i<j}$ , it considers the number of concordant pairs $P$ , the number of discordant pairs $Q$ , and ties $T_x$ and $T_y$ .

\begin{gathered} P = |\{ (i,j)~|~ 0\leq i <j<n, ~ (x_j-x_i)(y_j-y_i)>0 ) \}|\\ Q = |\{ (i,j)~|~ 0\leq i <j<n, ~ (x_j-x_i)(y_j-y_i)<0 ) \}|\\ T_x= |\{ (i,j)~|~ 0\leq i <j<n, ~ x_i=x_j ) \}|\\ T_y= |\{ (i,j)~|~ 0\leq i <j<n, ~ y_i=y_j ) \}|\\ N = \frac{n(n-1)}{2} \end{gathered}

Given these quantities, Kendall’s τ (a) is defined as:

\tau_a = \frac{P - Q}{N} .

As well as Spearman’s ρ, Kendall’s τ (a) takes the value 1 if the order of the raw scores all match, and the value -1 if the order is completely reversed.

Kendall’s τ (b) cares about the case where ties are around.