With and without ties

The Pearson correlation coefficient, denoted \(r\) in the case of a sample, can be computed as follows²

\[ r = \frac{\sum^n_{i = 1} x_i y_i – n \bar{x} \bar{y}}{(n – 1) s_x s_y} \]

where:

• \(n\) is the sample size
• \(x_i\) and \(y_i\) are the observations
• \(\bar{x}\) and \(\bar{y}\) are the sample means of \(x\) and \(y\)
• \(s_x\) and \(s_y\) are the sample standard deviations of \(x\) and \(y\)

We show how to compute it step by step.

Step 1.

As you can see, before computing the correlation coefficient we first need to compute the mean and the standard deviation for each of the two variables.

The mean of \(x\) is computed as follows:

\[\bar{x} = \frac{1}{n}\sum^n_{i = 1} x_i\]

The standard deviation of \(x\) is computed as follows:

\[s_x = \sqrt{\frac{1}{n – 1} \sum^n_{i = 1}(x_i – \bar{x})^2}\]

The formulas can be used analogously for the variable \(y\).

We start by computing the means of the two variables:

• \(\bar{x} = \frac{-1 + 3 + 5 + 5 + 2}{5} = 2.8\)
• \(\bar{y} = \frac{-3 + 1 + 0 + 2 – 1}{5} = -0.2\)

Step 2.

We now need to compute the standard deviation of each variable. To ease the computations, it is best to use a table, starting with the observations:

To compute the standard deviations, we need the sum of the squared differences between each observation and its mean, that is, \(\sum^n_{i = 1}(x_i – \bar{x})^2\) and \(\sum^n_{i = 1}(y_i – \bar{y})^2\). We start by creating two new columns in the table, denoted x-xbar and y-ybar, corresponding to \((x_i – \bar{x})\) and \((y_i – \bar{y})\):

We take the square of these two new columns to have \((x_i – \bar{x})^2\) and \((y_i – \bar{y})^2\), denoted (x-xbar)^2 and (y-ybar)^2:

We then sum these two columns, which gives:

• for \(x\): \(\sum^n_{i = 1}(x_i – \bar{x})^2 =\) 24.8
• for \(y\): \(\sum^n_{i = 1}(y_i – \bar{y})^2 =\) 14.8

The standard deviations are thus:

• \(s_x = \sqrt{\frac{24.8}{5-1}} = 2.49\)
• \(s_y = \sqrt{\frac{14.8}{5-1}} = 1.92\)

Step 3.

We now need \(\sum^n_{i = 1} x_i y_i\), so we add a new column in the table, corresponding to \(x_iy_i\) and denoted x*y:

Step 4.

We take the sum of that last column, which gives \(\sum^n_{i = 1} x_i y_i =\) 14.

Step 5.

Finally, the Pearson correlation coefficient can be computed by plugging values found above in the initial formula:

\[ \begin{split} r &= \frac{\sum^n_{i = 1} x_i y_i – n \bar{x} \bar{y}}{(n – 1) s_x s_y} \\ &= \frac{14 – (5\times2.8\times-0.2)}{(5-1) \times 2.49\times1.92} \\ &= 0.88 \end{split} \]

Note that there are other formulas to compute the Pearson correlation coefficient. For instance,

\[ \begin{split} r &= \frac{1}{n – 1} \sum^n_{i = 1} \left(\frac{x_i – \bar{x}}{s_x}\right)\left(\frac{y_i – \bar{x}}{s_y}\right) \\ &= \frac{\sum^n_{i = 1}(x_i – \bar{x})(y_i – \bar{y})}{\sqrt{\sum^n_{i = 1}(x_i – \bar{x})^2} \sqrt{\sum^n_{i = 1}(y_i – \bar{y})^2}} \end{split} \]

All formulas will of course give the exact same results.

With ties

The Spearman correlation coefficient (with ties), denoted \(r_s\), is defined as follows

\[ r_s = \frac{\sum^n_{i = 1} Rx_i Ry_i – n \overline{Rx} \overline{Ry}}{(n – 1) s_{Rx} s_{Ry}} \]

where:

• \(n\) is the sample size
• \(Rx_i\) and \(Ry_i\) are the ranks for the two variables
• \(\overline{Rx}\) and \(\overline{Ry}\) are the sample means of the ranks for the two variables
• \(s_{Rx}\) and \(s_{Ry}\) are the sample standard deviations of the ranks for the two variables

Here is how to compute it by hand.

Step 1.

As mentioned before, the Spearman coefficient is based on the ranks. So we first need to add the ranks of the observations (from lowest to highest), separately for each of the two variables.

For \(y\), we see that:

• -3 is the smallest value, so we assign it the rank 1
• -1 is the second smallest value, so we assign it the rank 2
• then comes 0, so we assign it the rank 3
• then comes 1, so we assign it the rank 4
• finally, 2 is the largest value, so we assign it the rank 5

The same goes for \(x\), except that here we have two observations with the value 5 (so there will be ties in the ranks). In this case, we take the mean rank:

• -1 is the smallest value, so we assign it the rank 1
• 2 is the second smallest value, so we assign it the rank 2
• then comes 3, so we assign it the rank 3
• finally, the two largest values belongs to rank 4 and 5, so we assign both of them the rank 4.5

We include the ranks in the table, denoted Rx and Ry:

Step 2.

From there, it is similar than the Pearson coefficient except that we work on the ranks and not on the initial observations anymore. To avoid any confusion in the remaining steps, we remove the initial observations from the table and we keep only the ranks:

We start with the means of the ranks:

• \(\overline{Rx} = \frac{1+3+4.5+4.5+2}{5} = 3\)
• \(\overline{Ry} = \frac{1+4+3+5+2}{5} = 3\)

For the standard deviations, we use the table as we did for the Pearson coefficient:

We sum the last two columns, which gives:

• for \(x\): \(\sum^n_{i = 1}(Rx_i – \overline{Rx})^2 =\) 9.5
• for \(y\): \(\sum^n_{i = 1}(Ry_i – \overline{Ry})^2 =\) 10

The standard deviations are thus:

• \(s_{Rx} = \sqrt{\frac{9.5}{5-1}} = 1.54\)
• \(s_{Ry} = \sqrt{\frac{10}{5-1}} = 1.58\)

Step 3.

We now need \(\sum^n_{i = 1} Rx_i Ry_i\), so we add a new column in the table, corresponding to \(Rx_iRy_i\) and denoted Rx*Ry:

Step 4.

We take the sum of that last column, which gives \(\sum^n_{i = 1} Rx_i Ry_i =\) 53.

Step 5.

Finally, the Spearman correlation coefficient can be computed by plugging all values in the initial formula:

\[ \begin{split} r_s &= \frac{\sum^n_{i = 1} Rx_i Ry_i – n \overline{Rx} \overline{Ry}}{(n – 1) s_{Rx} s_{Ry}} \\ &= \frac{53 – (5\times3\times3)}{(5-1) \times 1.54\times1.58} \\ &= 0.82 \end{split} \]

Without ties

When all initial values are different within each variable, meaning that all ranks are distinct integers, there are no ties. In that specific case, the Spearman coefficient can be computed with the following shortened formula:

\[ r_s = 1 – \frac{6 \sum^n_{i = 1}(Rx_i – Ry_i)^2}{n (n^2 – 1)} \]

For example, suppose the sample without ties introduced at the beginning of the post:

Step 1.

All observations within each variable are different, so all ranks are distinct integers and there are no ties:

Step 2.

We only need to compute the difference between the two ranks of each row, denoted Rx-Ry:

Take the square of these differences, denoted (Rx-Ry)^2:

And then take the sum of that last column, which gives \(\sum^n_{i = 1}(Rx_i – Ry_i)^2 =\) 2.

Step 3.

Finally, we can fill in the initial formula to find the Spearman coefficient:

\[ \begin{split} r_s &= 1 – \frac{6 \sum^n_{i = 1}(Rx_i – Ry_i)^2}{n (n^2 – 1)} \\ &= 1 – \frac{6 \times2}{3 (3^2 – 1)} \\ &= 1 – \frac{12}{24} \\ &= 0.5 \end{split} \]

Without ties

When there are no ties, the Kendall coefficient, denoted \(\tau_a\), is defined as follows

\[ \tau_a = \frac{C – D}{C + D} \]

where:

• \(C\) is the number of concordant pairs
• \(D\) is the number of discordant pairs

This coefficient, also referred as Kendall tau-a, does not make any adjustment for ties.

Let’s see what are concordant and discordant pairs using the data without ties (the same data than for Spearman correlation without ties):

Step 1.

We start by computing the ranks for each variable (Rx and Ry), as we did for the Spearman coefficient:

Step 2.

We then arbitrarily choose a reference variable among the two, Rx or Ry. Suppose we take Rx as the reference variable here.

We sort the dataset by this reference variable:

From now, we only look at the ranks of the second variable (the one which is not the reference level, here Ry), so to avoid any confusion in the remaining steps we keep only the Ry column:

Step 3.

We now take each row of Ry one by one and check whether the rows below it in the table are smaller or larger.

In our table, the first row of Ry is 1. We see that the value just below it is 3, which is larger than 1. Since 3 is larger than 1, this is called a concordant pair. We write it in the table:

The next row, 2, is also larger than 1, so it is also a concordant pair. We also write it in the table:

We start again with the second row of Ry, which is 3. Again, we look at the row below it and check whether it is larger or smaller. Here, the row below it is 2, which is smaller than 3, so we have a discordant pair. We add this information into the table:

Step 4.

We now compute the total number of concordant and discordant pairs. There are:

• 2 concordant pairs so \(C = 2\), and
• 1 discordant pair so \(D = 1\).

Step 5.

Finally, we plug the values we have just found in the initial formula:

\[ \begin{split} \tau_a &= \frac{C – D}{C + D}\\ &= \frac{2 – 1}{2 + 1}\\ &= \frac{1}{3}\\ &= 0.33 \end{split} \]

Alternatively, we can also use the following formulas

\[ \begin{split} \tau_a &= 1 – \frac{4D}{n(n – 1)}\\ &= \frac{4C}{n(n – 1)} – 1 \end{split} \]

where \(n\) is still the sample size, and which both give the exact same results.

With ties

I must admit that the process with ties is slightly more complex than without ties.

The Kendall tau-b coefficient, which makes adjustments for ties, is defined as follows

\[ \tau_b = \frac{C – D}{\sqrt{(C^2_n – n_x)(C^2_n – n_y)}} \]

where:

• \(C\) and \(D\) are still the number of concordant and discordant pairs, respectively
• \(C^2_n\) is the total number of possible pairs
• \(n_x\) is the number of possible pairs with a tie among the \(x\) values
• \(n_y\) is the number of possible pairs with a tie among the \(y\) values

Note that here, the letter \(C\) in \(C^2_n\) denotes “combination” and not “concordant”.

Let’s illustrate that scenario and the formula with the dataset used for the Pearson correlation and the Spearman correlation with ties, that is:

Step 1.

Similarly to with no ties, the process with ties is based on the ranks so we start by adding the ranks for each variable, denoted as usual as Rx and Ry:

Step 2.

One of the two variables (\(x\) or \(y\)) must be selected as the reference variable. This time, we do not choose it arbitrarily, but we choose the one which does not have any ties. In our case, variable \(x\) has ties, while variable \(y\) does not have any ties. So \(y\) will be our reference variable.

We then:

• order the dataset by the reference variable, here Ry, and
• we keep only the necessary columns to avoid any confusion in the remaining steps:

Step 3.

We check whether it is a concordant or discordant pair the same way we did without ties, but this time we do not count ties.

The first row in column Rx is 1. We compare all rows below it in the table with that value. All rows below 1 in the table are larger, so we write that they are concordant pairs:

We repeat the same process for each row.

For instance, for the second row of the column Rx, we have the value 2. Again, all rows below in the table are larger so we write that they are concordant pairs:

For the third row of the column Rx, we have the value 4.5. The row just below in the table (= 3) is smaller than 4, so we write D for discordant pair:

Now, as you can see, the last row is also 4.5. Therefore, since it is equal to the value we are comparing to, it is neither a concordant nor a discordant pair, so we write “T” in the table for “ties”:

Last, the fourth row in the column Rx is 3, which we compare to 4.5 to see that 4.5 is larger so it is a concordant pair:

Step 4.

We then sum the number of concordant and discordant pairs:

• \(C = 8\)
• \(D = 1\)

Step 5.

We now have all the information required to compute the numerator of \(\tau_b\), but we still need to find \(C^2_n\), \(n_x\) and \(n_y\) to compute the denominator.

As mentioned above, \(C^2_n\) is the total number of possible pairs, thus corresponding to the number of combinations of two values. This number of pairs can be found with the formula of the combination, that is, \(C^2_n = \frac{n!}{2!(n – 2)!}\) where \(n\) is the sample size.

In our case, we have a sample size of 5, so we have:³

\[ \begin{split} C^2_n &= \frac{n!}{2!(n – 2)!}\\ &= \frac{5!}{2!(5-2)!}\\ &= 10 \end{split} \]

Furthermore, \(n_x\) and \(n_y\) are the number of possible pairs with a tie among the \(x\) and \(y\) variables, respectively.

When looking at Ry and Rx:

We see that there are:

• 2 identical values for the variable \(x\)
• 0 identical value for the variable \(y\)

This means that the number of possible pairs with a tie is equal to:⁴

• 1 for the variable \(x\) (the only possible pair with a tie is the pair {4.5, 4.5})
• 0 for the variable \(y\) (since all ranks are distinct, there exists no pair with a tie)

Therefore,

\[ \begin{split} n_x &= 1\\ n_y &= 0 \end{split} \]

And we now have all the necessary information to compute the denominator of \(\tau_b\)!

Step 6.

By plugging the values found above in the initial formula, we have

\[ \begin{split} \tau_b &= \frac{C – D}{\sqrt{(C^2_n – n_x)(C^2_n – n_y)}}\\ &= \frac{8-1}{\sqrt{(10 – 1)(10 – 0)}}\\ &= \frac{7}{\sqrt{90}}\\ &= 0.74 \end{split} \]