Bi-Variate Distribution - Correlation

 

Bivariate Distribution and Correlation

So far, we have confined ourselves to univariate distributions, i.e., distributions involving only one variable. However, in many real-life situations, we encounter cases where each observation is associated with two or more variables.

For example, if we measure the heights and weights of a group of persons, each observation consists of two related values—one for height and one for weight. Such a distribution involving two variables is called a bivariate distribution.

In a bivariate distribution, we are often interested in examining whether there exists any correlation or covariation between the two variables under study.

• If a change in one variable is accompanied by a change in the other variable, the two variables are said to be correlated.

• If the two variables tend to deviate in the same direction (i.e., an increase in one variable corresponds to an increase in the other, or a decrease corresponds to a decrease), the correlation is called direct or positive correlation.

• On the other hand, if the two variables deviate in opposite directions (i.e., an increase in one variable corresponds to a decrease in the other, and vice versa), the correlation is called inverse or negative correlation.

Examples:

• Positive correlation: (i) Heights and weights of individuals, (ii) Income and expenditure.

• Negative correlation: (i) Price and demand of a commodity, (ii) Volume and pressure of a perfect gas.

Finally, correlation is said to be perfect if the deviation in one variable is always accompanied by a proportional and exact deviation in the other variable.

Scatter Diagram

A scatter diagram is the simplest method of representing bivariate data diagrammatically.

For a bivariate distribution $(x_i, y_i), \; i = 1, 2, \dots, n$, if the values of the variables $X$ and $Y$ are plotted along the horizontal axis (x-axis) and vertical axis (y-axis), respectively, in the Cartesian plane, the resulting diagram of points is known as a scatter diagram.

From a scatter diagram, we can form a fairly good—though approximate—idea about whether the two variables are correlated:

• If the points lie close together, forming a dense cluster, we expect a high degree of correlation.

• If the points are widely scattered, the correlation is expected to be weak.

However, this method is not very suitable when the number of observations is very large, as the diagram becomes too crowded to interpret effectively

Karl Pearson’s Coefficient of Correlation

As a measure of the intensity or degree of linear relationship between two variables, Karl Pearson (1867–1936), a British biometrician, developed a formula known as the Correlation Coefficient.

The correlation coefficient between two random variables $X$ and $Y$, usually denoted by $r(X, Y)$ or simply $r_{xy}$, is a numerical measure of the linear relationship between them and is defined as:

$$r(X, Y) = \frac{\text{Cov}(X, Y)}{\sigma_X \, \sigma_Y}$$

where:

• $\text{Cov}(X, Y)$ = Covariance between $X$ and $Y$
  

• $\sigma_X$ = Standard deviation of $X$
  

• $\sigma_Y$ = Standard deviation of $Y$
  

Karl Pearson’s Coefficient of Correlation

Let $(x_i, y_i); \; i = 1, 2, \dots, n$ represent a bivariate distribution.

Covariance

The covariance between $X$ and $Y$ is defined as:

$$\text{Cov}(X, Y) = E \Big[ (X - E(X))(Y - E(Y)) \Big]$$

For a sample, it is given by:

$$\text{Cov}(X, Y) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})$$

where $\bar{x}$ and $\bar{y}$ are the sample means of $X$ and $Y$.

Similarly, the variances are:

$$\sigma_X^2 = E[(X - E(X))^2] = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$$
$$\sigma_Y^2 = E[(Y - E(Y))^2] = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2$$

---

Karl Pearson’s Formula

The correlation coefficient between $X$ and $Y$ is defined as:

$$r(X, Y) = \frac{\text{Cov}(X, Y)}{\sigma_X \, \sigma_Y}$$

This is also called the product-moment correlation coefficient because

$$\text{Cov}(X, Y) = E\big[(X - E(X))(Y - E(Y))\big]$$

---

Alternative (Computational) Formula

For practical calculation, an equivalent formula is:

$$r(X, Y) = \frac{n \sum x_i y_i - \sum x_i \sum y_i}{\sqrt{\big(n \sum x_i^2 - (\sum x_i)^2\big)\big(n \sum y_i^2 - (\sum y_i)^2\big)}}$$

This avoids computing deviations separately and is useful in tabulated data.

Convenient Form of Covariance Formula

From the definition:

$$\text{Cov}(X, Y) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})$$

Expanding:

$$\text{Cov}(X, Y) = \frac{1}{n} \left[ \sum x_i y_i - \bar{x} \sum y_i - \bar{y} \sum x_i + n \bar{x}\bar{y} \right]$$

Since $\bar{x} = \frac{1}{n}\sum x_i$ and $\bar{y} = \frac{1}{n}\sum y_i$, this simplifies to:

$$\text{Cov}(X, Y) = \frac{1}{n} \sum x_i y_i - \bar{x}\bar{y}$$

---

Similarly, the Variances

$$\sigma_X^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2$$
$$\sigma_Y^2 = \frac{1}{n} \sum (y_i - \bar{y})^2 = \frac{1}{n}\sum y_i^2 - \bar{y}^2$$

---

✅ So the final computational forms are:

$$\text{Cov}(X, Y) = \frac{1}{n}\sum x_i y_i - \bar{x}\bar{y}$$ $$\sigma_X^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2, \quad \sigma_Y^2 = \frac{1}{n}\sum y_i^2 - \bar{y}^2$$

---

Limits of Correlation Coefficient

Using the Cauchy–Schwarz inequality, it can be shown that:

$$-1 \; \leq \; r(X, Y) \; \leq \; +1$$

• If $r = +1$: perfect positive correlation.

• If $r = -1$: perfect negative correlation.

• If $r = 0$: no linear correlation.

Remarks on Karl Pearson’s Correlation Coefficient

The coefficient $r(X,Y)$provides a measure of the linear relationship between $X$ and $Y$. For nonlinear relationships, however, it is not a suitable measure.

Sometimes, the covariance is denoted as:

$$\text{Cov}(X, Y) = \sigma_{XY}$$

Karl Pearson’s correlation coefficient is also called the product-moment correlation coefficient, since it is based on the expected product of deviations of the two variables from their respective means:

$$\text{Cov}(X, Y) = E\big[(X - E(X))(Y - E(Y))\big]$$

Thus,

$$r(X, Y) = \frac{\sigma_{XY}}{\sigma_X \sigma_Y}$$

Example

Calculate the correlation coefficient for the following heights (in inches) of fathers (X) and their sons (Y) :

Data

Fathers’ heights $X$: 65, 66, 67, 67, 68, 69, 70, 72
Sons’ heights $Y$: 67, 68, 65, 68, 72, 72, 69, 71

Useful totals

$$\begin{aligned} n&=8, \quad \sum X=544,\quad \sum Y=552,\\ \sum X^2&=37028,\quad \sum Y^2=38132,\quad \sum XY=37560. \end{aligned}$$

(So $\bar X=\tfrac{544}{8}=68,\ \bar Y=\tfrac{552}{8}=69$

Pearson correlation (computational form)

$$r=\frac{n\sum XY-(\sum X)(\sum Y)} {\sqrt{\big(n\sum X^2-(\sum X)^2\big)\big(n\sum Y^2-(\sum Y)^2\big)}}.$$

Plug in the numbers (showing each step):

• Numerator:

$$n\sum XY-(\sum X)(\sum Y) =8\cdot37560-544\cdot552 =300480-300288 =192.$$

• Denominator parts:

$$\begin{aligned} n\sum X^2-(\sum X)^2&=8\cdot37028-544^2=296224-295936=288,\\ n\sum Y^2-(\sum Y)^2&=8\cdot38132-552^2=305056-304704=352. \end{aligned}$$

So

$$\sqrt{288\cdot352}=\sqrt{101376}\approx 318.396.$$

• Therefore

$$r=\frac{192}{318.396}\approx 0.603.$$

Conclusion

$$\boxed{r \approx 0.603}$$

This indicates a moderate positive linear correlation between fathers’ and sons’ heights

​

If you use the sample version with $n-\mathrm{1\; in\; covariance/variance,\; those }$$n-1$ factors cancel, so you get the same computational formula for $r$.

Coding/assumed means trick (useful with large numbers): set

• $$\mathrm{<span></span>}$$

$$u=\frac{x-A}{h},v=\frac{y-B}{k}$$

Then

$$r=\frac{n\sum uv-(\sum u)(\sum v)}{\sqrt{(n\sum u^2-(\sum u{)}^2\left)\right(n\sum v^2-(\sum v{)}^2)}},$$

which is numerically stable and faster; scale/shift cancel in $\mathrm{r.A\; and\; B\; are\; the\; mean\; value.}\mathrm{<br>}$

$$r(X,Y)=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y} =\frac{24}{6\cdot\sqrt{44}}=\frac{24}{39.7994975}\approx0.603.$$

For the standardised variables $U,V$:

$$\operatorname{Cov}(U,V)=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}\approx0.603,$$

and since $\sigma_U=\sigma_V=1$ we get $r(U,V)=\operatorname{Cov}(U,V)\approx0.603$

r(U,V)=Cov(U,V)≈0.603. This matches $r(X,Y)$.

Example Problem

A computer, while calculating the correlation coefficient between two variables $X$ and $Y$

 from 25 pairs of observations, obtained the following results:

$$n = 25,\quad \sum X = 125,\quad \sum X^2 = 650,\quad \sum Y = 100,\quad \sum Y^2 = 460,\quad \sum XY = 508$$

However, during checking, it was discovered that two pairs of values had been copied incorrectly. The pairs were entered as

$$(6,14),\ (8,6)$$

while the correct values were

$$(8,12),\ (6,8).$$

Obtain the correct value of the correlation coefficient.

Given (initial)

$n=25$
$\sum X = 125,\quad \sum X^2 = 650$
$\sum Y = 100,\quad \sum Y^2 = 460$
$\sum XY = 508$

Later it was found that two (actually a block of) pairs were copied incorrectly; after correcting those pairs the corrected totals become:

Corrected totals (after fixing the mis-copied pairs)

$$\sum X = 125,\qquad \sum X^2 = 650 \quad(\text{unchanged})$$
$$\sum Y = 100,\qquad \sum Y^2 = 436 \quad(\text{changed})$$
$$\sum XY = 520 \quad(\text{changed})$$

(You can verify these by subtracting the wrong contributions and adding the correct ones.)

Compute means

$$\bar X=\frac{\sum X}{n}=\frac{125}{25}=5,\qquad \bar Y=\frac{\sum Y}{n}=\frac{100}{25}=4.$$

Use Pearson’s computational formula

$$r=\frac{n\sum XY-(\sum X)(\sum Y)} {\sqrt{\big(n\sum X^2-(\sum X)^2\big)\big(n\sum Y^2-(\sum Y)^2\big)}}.$$

Plug in the corrected numbers:

• Numerator:

$$25\cdot 520 - 125\cdot 100 = 13\,000 - 12\,500 = 500.$$

• Denominator parts:

$$25\cdot 650 - 125^2 = 16\,250 - 15\,625 = 625,$$
$$25\cdot 436 - 100^2 = 10\,900 - 10\,000 = 900.$$

So denominator $=\sqrt{625\cdot 900}=\sqrt{562500}=750.$

• Therefore

$$r=\frac{500}{750}=\frac{2}{3}\approx 0.6667\ (\approx 0.67).$$

Final answer

$$\boxed{r \approx 0.667}$$

Calculation of the Correlation Coefficient for a Bivariate Frequency Distribution. 

When the data 'are considerably large, they may be summarised by using a two-way table. Here, for each variable a suitable number of classes are taken, keeping in ,view the same ,considerations as in the univariate case. If there are n classes for X and m classes for Y, there will be in all' mxn cells in the two-way table . By going through the pairs of values of X and Y, we can find the frequency for each cell.The whole set of cell frequencies will then define a bivariate frequency distribution. The column totals and row totals will give us the marginal distributions of X and Y. A particular column or row will be called the conditional distribution of. Y for given X or of-X for given Y respectively.

Suppose bivariate data on $X$ and $Y$ are presented in a two-way frequency table where there are $m$ classes of $Y$ along the horizontal direction and $n$ classes of $X$ along the vertical direction. Let $f_{ij}$ be the frequency of observations in the cell at row $i$ and column $j$ (the $(i,j)$-th cell).

The row-sum (sum of frequencies in row $i$) is

$$r_i=\sum_{j=1}^m f_{ij},$$

and the column-sum (sum of frequencies in column $j$) is

$$c_j=\sum_{i=1}^n f_{ij}.$$

The total number of observations is

$$N=\sum_{i=1}^n\sum_{j=1}^m f_{ij}=\sum_{i=1}^n r_i=\sum_{j=1}^m c_j.$$

A row (fixed $i$) gives the conditional distribution of $Y$ given the $i$-th class of $X$; similarly a column (fixed $j$) gives the conditional distribution of $X$ given the $j$-th class of $Y$.

---

Notation for grouped (class) midpoints

Let the midpoint of the $i$-th $X$-class be $x_i$ $(i=1,\dots ,n)\; and\; the\; midpoint\; of\; the$$j$-th $Y$-class be $y_j$ $(j=1,\dots,m)$. Then treat each cell as representing $f_{ij}$ observations at the point $(x_i,y_j)$.

---

Marginal (grouped) distributions and sample totals

$$\begin{aligned} \Sigma f\,x &= \sum_{i=1}^n\sum_{j=1}^m f_{ij}\,x_i=\sum_{i=1}^n r_i x_i,\\[4pt] \Sigma f\,y &= \sum_{i=1}^n\sum_{j=1}^m f_{ij}\,y_j=\sum_{j=1}^m c_j y_j. \end{aligned}$$

Sample means (grouped)

$$\bar x = \frac{1}{N}\sum_{i=1}^n\sum_{j=1}^m f_{ij}x_i = \frac{1}{N}\sum_{i=1}^n r_i x_i, \qquad \bar y = \frac{1}{N}\sum_{i=1}^n\sum_{j=1}^m f_{ij}y_j = \frac{1}{N}\sum_{j=1}^m c_j y_j.$$

---

Grouped second moments, variances and covariance

$$\begin{aligned} \Sigma f\,x^2 &= \sum_{i=1}^n\sum_{j=1}^m f_{ij} x_i^2 = \sum_{i=1}^n r_i x_i^2,\\[4pt] \Sigma f\,y^2 &= \sum_{i=1}^n\sum_{j=1}^m f_{ij} y_j^2 = \sum_{j=1}^m c_j y_j^2,\\[4pt] \Sigma f\,xy &= \sum_{i=1}^n\sum_{j=1}^m f_{ij} x_i y_j. \end{aligned}$$

Grouped variances (population form, using $1/N$):

$$\sigma_X^2=\frac{1}{N}\Big(\Sigma f\,x^2 - \frac{(\Sigma f\,x)^2}{N}\Big),\qquad \sigma_Y^2=\frac{1}{N}\Big(\Sigma f\,y^2 - \frac{(\Sigma f\,y)^2}{N}\Big).$$

Grouped covariance:

$$\operatorname{Cov}_g(X,Y)=\frac{1}{N}\Big(\Sigma f\,xy - \frac{(\Sigma f\,x)(\Sigma f\,y)}{N}\Big).$$

(If you prefer the “sample” denominators $N-\mathrm{1,\; replace }$$1/N$ by $1/(N-1)$

consistently — but the correlation formula below is unchanged because it uses ratios.)

---

Correlation coefficient (grouped)

$$r \;=\; \frac{\operatorname{Cov}_g(X,Y)}{\sigma_X\,\sigma_Y} \;=\; \frac{\,N\Sigma f\,xy - (\Sigma f\,x)(\Sigma f\,y)\,} {\sqrt{\big(N\Sigma f\,x^2 - (\Sigma f\,x)^2\big)\,\big(N\Sigma f\,y^2 - (\Sigma f\,y)^2\big)}}.$$

This $r$ lies in $[-1,1]$. Its sign indicates direction of linear association; its magnitude indicates strength.

---

Conditional distributions (for clarity)

Conditional probability (grouped) of $Y=y_j$given $X$ in class $i$:

$$P(Y=y_j \mid X\ \text{in class }i)=\frac{f_{ij}}{r_i}\quad\text{(provided }r_i>0\text{).}$$

Similarly

$$P(X=x_i \mid Y\ \text{in class }j)=\frac{f_{ij}}{c_j}\quad\text{(provided }c_j>0\text{).}$$

Example:

Joint distribution table is given.Find the correlation coefficient between X and Y

$$\begin{array}{c|c|c|c} & X=-1 & X=+1 & g(y) \\ \hline Y=0 & \tfrac{1}{8} & \tfrac{3}{8} & \tfrac{4}{8} \\ Y=1 & \tfrac{2}{8} & \tfrac{2}{8} & \tfrac{4}{8} \\ \hline p(x) & \tfrac{3}{8} & \tfrac{5}{8} & 1 \end{array}$$

Expectations

$E(X)$

$$E(X)=(-1)\cdot\tfrac{3}{8}+(+1)\cdot\tfrac{5}{8} = \frac{-3+5}{8} = \frac{2}{8}=\tfrac{1}{4}.$$

$E(Y)$

$$E(Y)=0\cdot\tfrac{4}{8}+1\cdot\tfrac{4}{8} = \tfrac{4}{8}=\tfrac{1}{2}.$$

---

Second moments

$E(X^2)$

Since $X=\pm1$:

$$E(X^2)=1^2\cdot 1 = 1.$$

$E(Y^2)$

$$E(Y^2)=0^2\cdot\tfrac{4}{8}+1^2\cdot\tfrac{4}{8}=\tfrac{4}{8}=\tfrac{1}{2}.$$

---

Variances

$$\operatorname{Var}(X)=E(X^2)-[E(X)]^2=1-\Big(\tfrac{1}{4}\Big)^2=1-\tfrac{1}{16}=\tfrac{15}{16}.$$
$$\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2=\tfrac{1}{2}-\Big(\tfrac{1}{2}\Big)^2=\tfrac{1}{2}-\tfrac{1}{4}=\tfrac{1}{4}.$$

---

Covariance

$$E(XY)=(-1)(0)\cdot\tfrac{1}{8}+(+1)(0)\cdot\tfrac{3}{8}+(-1)(1)\cdot\tfrac{2}{8}+(+1)(1)\cdot\tfrac{2}{8}.$$

Simplify:

$$E(XY)=0+0+(-2/8)+(2/8)=0.$$

So

$$\operatorname{Cov}(X,Y)=E(XY)-E(X)E(Y)=0-\Big(\tfrac{1}{4}\cdot\tfrac{1}{2}\Big)=-\tfrac{1}{8}.$$

---

Correlation coefficient

$$r=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)\,\operatorname{Var}(Y)}} =\frac{-\tfrac{1}{8}}{\sqrt{\tfrac{15}{16}\cdot\tfrac{1}{4}}}.$$

Denominator:

$$\sqrt{\tfrac{15}{16}\cdot\tfrac{1}{4}}=\sqrt{\tfrac{15}{64}}=\frac{\sqrt{15}}{8}.$$

So

$$r=\frac{-1/8}{\sqrt{15}/8}=-\frac{1}{\sqrt{15}}\approx -0.258.$$

✅ Final Answer:
The correlation coefficient between $X$ and $Y$ is

$$\boxed{r=-\tfrac{1}{\sqrt{15}} \;\;\approx -0.258}$$

which indicates a weak negative correlation.

Spearman’s Rank Correlation 

---

1. Setup

We have $n$ individuals ranked in two characteristics $A$ and $B$.

• Rank of $i$-th individual in $A$: $X_i$

• Rank of $i$-th individual in $B$: $Y_i$

Since they are ranks:

$$X_i, Y_i \in \{1,2,3,\dots,n\}.$$

Let:

• Mean of ranks:

  $$\bar{X} = \bar{Y} = \frac{1+2+3+\dots+n}{n}=\frac{n+1}{2}.$$
  

• Variance of ranks:

  $$\sigma_X^2 = \sigma_Y^2 = \frac{1^2+2^2+\dots+n^2}{n} - \Big(\frac{n+1}{2}\Big)^2.$$
  

Recall the formula:

$$1^2+2^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6}.$$

So:

$$\sigma_X^2 = \frac{n(n+1)(2n+1)}{6n} - \frac{(n+1)^2}{4} = \frac{(n^2-1)}{12}.$$

Thus,

$$\sigma_X^2 = \sigma_Y^2 = \frac{n^2-1}{12}.$$

---

2. Differences of ranks

Define:

$$d_i = X_i - Y_i.$$

Clearly:

$$\sum_{i=1}^n d_i^2 = \sum (X_i - Y_i)^2.$$

---

3. Relation with covariance

Expanding:

$$\sum d_i^2 = \sum (X_i - Y_i)^2 = \sum (X_i-\bar{X})^2 + \sum (Y_i-\bar{Y})^2 - 2\sum (X_i-\bar{X})(Y_i-\bar{Y}).$$

Divide by $n$:

$$\frac{1}{n}\sum d_i^2 = \sigma_X^2 + \sigma_Y^2 - 2\,\text{Cov}(X,Y).$$

Since $\sigma_X^2=\sigma_Y^2$

$$\frac{1}{n}\sum d_i^2 = 2\sigma_X^2 - 2\rho \sigma_X^2,$$

where $\rho$ = rank correlation coefficient.

So:

$$\frac{1}{n}\sum d_i^2 = 2\sigma_X^2(1-\rho).$$

---

4. Substituting variance

We already know:

$$\sigma_X^2 = \frac{n^2-1}{12}.$$

Thus:

$$\frac{1}{n}\sum d_i^2 = 2\cdot \frac{n^2-1}{12}(1-\rho).$$

Simplify:

$$\frac{1}{n}\sum d_i^2 = \frac{n^2-1}{6}(1-\rho).$$

---

5. Final formula

Rearranging for $\rho$:

$$1-\rho = \frac{6}{n(n^2-1)}\sum d_i^2,$$
$$\rho = 1 - \frac{6\sum d_i^2}{n(n^2-1)}.$$

---

✅ This is Spearman’s Rank Correlation formula:

$$\boxed{r_s = 1 - \frac{6\sum d_i^2}{n(n^2-1)}}$$

where

• $d_i = X_i - Y_i$ is the difference between the two ranks for the $i$-th individual,

• $n$ is the number of individuals.

Spearman’s rank correlation coefficient (since Pearson’s correlation is based on actual values, not ranks). Let me give you a small worked-out example of Spearman’s rank correlation:

---

Example

Suppose we have marks of 5 students in Math and Science:

Student	Math (X)	Science (Y)
A	85	93
B	70	65
C	90	89
D	60	60
E	75	80

---

Step 1: Assign ranks

Rank each score (highest = 1).

Student	Math (X)	Rank X	Science (Y)	Rank Y
A	85	2	93	1
B	70	4	65	4
C	90	1	89	2
D	60	5	60	5
E	75	3	80	3

---

Step 2: Find differences of ranks

$d = \text{Rank X} - \text{Rank Y}$, and compute $d^2$.

Student	Rank X	Rank Y	$d$	$d^2$
A	2	1	1	1
B	4	4	0	0
C	1	2	-1	1
D	5	5	0	0
E	3	3	0	0

$$\sum d^2 = 2$$

---

Step 3: Apply Spearman’s formula

$$r_s = 1 - \frac{6\sum d^2}{n(n^2-1)}$$

Here $n = 5$, $\sum d^2 = 2$:

$$r_s = 1 - \frac{6(2)}{5(25-1)}$$ $$r_s = 1 - \frac{12}{120} = 1 - 0.1 = 0.9$$

---

✅ Spearman’s rank correlation coefficient = 0.9
This shows a strong positive correlation between Math and Science marks.

Example

Ten competitors in a musical test were ranked by the

three judges A. Band C in the following order:

Ranks by A: 1 6 5 10 3 2 4 9 7 8

Ranks by B : 3 5 8 4 7 10 2 1 6 9

Ranks by C : is 4 9 8 1 2 3 10 5 7

Using rank correlation method. discuss which pair of judges has the nearest approach to common likings in music.

Step 1: Write down the rankings

Competitor	A	B	C
1	1	3	2
2	6	5	4
3	5	8	9
4	10	4	8
5	3	7	1
6	2	10	3
7	4	2	10
8	9	1	5
9	7	6	7
10	8	9	6

---

Step 2: Compare A & B

Compute $d=A-\mathrm{B, }$$d^2$.

Competitor	A	B	d	d²
1	1	3	-2	4
2	6	5	1	1
3	5	8	-3	9
4	10	4	6	36
5	3	7	-4	16
6	2	10	-8	64
7	4	2	2	4
8	9	1	8	64
9	7	6	1	1
10	8	9	-1	1

$$\sum d^2 = 200$$
$$r_{AB} = 1 - \frac{6 \cdot 200}{10(10^2-1)} = 1 - \frac{1200}{990} = 1 - 1.212 = -0.212$$

So A & B have slight negative correlation.

---

Step 3: Compare A & C

Competitor	A	C	d	d²
1	1	2	-1	1
2	6	4	2	4
3	5	9	-4	16
4	10	8	2	4
5	3	1	2	4
6	2	3	-1	1
7	4	10	-6	36
8	9	5	4	16
9	7	7	0	0
10	8	6	2	4

$$\sum d^2 = 86$$
$$r_{AC} = 1 - \frac{6 \cdot 86}{10(99)} = 1 - \frac{516}{990} = 1 - 0.521 = 0.479$$

So A & C have moderate positive correlation.

---

Step 4: Compare B & C

Competitor	B	C	d	d²
1	3	2	1	1
2	5	4	1	1
3	8	9	-1	1
4	4	8	-4	16
5	7	1	6	36
6	10	3	7	49
7	2	10	-8	64
8	1	5	-4	16
9	6	7	-1	1
10	9	6	3	9

$$\sum d^2 = 194$$
$$r_{BC} = 1 - \frac{6 \cdot 194}{10(99)} = 1 - \frac{1164}{990} = 1 - 1.176 = -0.176$$

So B & C have slight negative correlation.

---

✅ Conclusion

• $r_{AB} = -0.212$ (slight negative)

• $r_{AC}=\mathrm{0.479\; (moderate\; positive)}$

• $r_{BC} = -0.176$ (slight negative)

👉 Judges A and C have the nearest approach to common likings in music.

Remarks on Spearman’s Rank Correlation Coefficient

1. Check for Numerical Accuracy

   • In calculations, the sum of rank differences should be zero:

     $$\sum d = \sum (x_i - y_i) = 0$$
     

     This provides a simple check for errors in numerical work.

2. Relation with Pearson’s Correlation

   • Spearman’s rank correlation coefficient ($\rho$) is essentially Pearson’s correlation applied to ranks instead of actual data.

   • Therefore, it is interpreted in the same way as Karl Pearson’s correlation coefficient.

3. Distribution-Free (Non-Parametric)

   • Pearson’s correlation assumes that the population is normally distributed.

   • When this assumption is not valid, we need a distribution-free measure, which does not depend on any population parameters.

   • Spearman’s $\rho$ is such a measure, making it useful in non-parametric situations.

4. Simplicity and Information Loss

   • Spearman’s formula is easier to understand and apply compared to Pearson’s formula.

   • However, using ranks instead of raw data results in loss of information.

   • Unless there are many tied ranks, Spearman’s coefficient is usually slightly lower than Pearson’s coefficient.

5. Use with Qualitative Data and Extreme Observations

   • Spearman’s correlation is the only suitable method when dealing with qualitative characteristics (e.g., taste, preference, intelligence level) that cannot be measured numerically but can be ordered.

   • It can also be used when actual quantitative data are available.

   • When data include extreme observations (outliers), Spearman’s formula is often preferred over Pearson’s because it is less sensitive to extremes.

6. Limitations

   • Spearman’s method is not practical for bivariate frequency distributions (correlation tables).

   • For large samples (n > 30), it is computationally heavy if ranks are not directly given. In such cases, Pearson’s formula is preferred unless ranking is necessary.

---

✅ In short:

• Spearman’s $\rho$ is simple, non-parametric, and works well with ranks or qualitative data.

• It is less accurate than Pearson’s in terms of information retention but more robust when assumptions (like normality) are not satisfied.

Python  Code

import numpy as np

import pandas as pd

# Sample data: study hours vs exam scores of students

study_hours = [2, 3, 4, 5, 6, 7, 8, 9]

exam_scores = [50, 55, 60, 65, 70, 75, 80, 85]

# Convert to pandas DataFrame for easy handling

data = pd.DataFrame({

    "Study Hours": study_hours,

    "Exam Scores": exam_scores

})

print("Dataset:")

print(data)

# Mean (average) study hours and exam scores

mean_hours = np.mean(study_hours)

mean_scores = np.mean(exam_scores)

print("\nMean Study Hours:", mean_hours)

print("Mean Exam Scores:", mean_scores)

# Variance (spread of data)

var_hours = np.var(study_hours)

var_scores = np.var(exam_scores)

print("\nVariance of Study Hours:", var_hours)

print("Variance of Exam Scores:", var_scores)

# Correlation (relationship between study hours and scores)

correlation = np.corrcoef(study_hours, exam_scores)[0, 1]

print("\nCorrelation between Study Hours and Exam Scores:", correlation)