## Karl Pearson's Correlation Coefficient

Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations then the Karl Pearson's coefficient of correlation between two variables $X$ and $Y$ is denoted by $r_{xy}$ or $r$ and is given by

$r = \dfrac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}}$

where,

• the sample covariance between $x$ and $y$ is

\begin{aligned} Cov(x,y) =s_{xy}&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})(y_i-\overline{y})\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_iy_i - \frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}\bigg) \end{aligned}

• the sample variance of $x$ is

\begin{aligned} V(x) =s_{x}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_i^2 - \frac{(\sum_{i=1}^n x_i)^2}{n}\bigg) \end{aligned}

• the sample variance of $y$ is

\begin{aligned} V(y) =s_{y}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(y_i -\overline{y})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n y_i^2 - \frac{(\sum_{i=1}^n y_i)^2}{n}\bigg) \end{aligned}

• the sample mean of $x$ is

\begin{aligned} \overline{x}&=\frac{1}{n}\sum_{i=1}^n x_i \end{aligned}

• the sample mean of $y$ is

\begin{aligned} \overline{y}&=\frac{1}{n}\sum_{i=1}^n y_i \end{aligned}

The correlation coefficient $r$ can not exceed unity numerically. i.e. $|r|\leq 1 \implies -1 \leq r \leq +1$.

Two independent variables are uncorrelated. But the converse is not necessarily true.

## Interpretation

• If $r =0$, then there is no correlation between the ranks.
• If $r >0$, then there is a positive correlation between the ranks.
• If $r = 1$, then there is a perfect positive correlation between the ranks.
• If $0 <r < 1$, then there is a partially positive correlation between the ranks.
• If $r <0$, then there is a negative correlation between the ranks.
• If $r = -1$, then there is a perfect negative correlation between the ranks.
• If $-1 <r < 0$, then there is a partially negative correlation between the ranks.

## Example 1

A study was conducted to analyze the relationship between advertising expenditure and sales. The following data were recorded:

X Advertising (in $) 20 24 30 32 35 Y Sales (in$) 310 340 400 420 490

Compute the correlation coefficient between advertising expenditure and sales.

### Solution

Let $x$ denote the advertising expenditure and $y$ denote the sales.

$x$ $y$ $x^2$ $y^2$ $xy$
1 20 310 400 96100 6200
2 24 340 576 115600 8160
3 30 400 900 160000 12000
4 32 420 1024 176400 13440
5 35 490 1225 240100 17150
Total 141 1960 4125 788200 56950

The sample variance of $x$ is

\begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(4125-\frac{(141)^2}{5}\bigg)\\ &= \frac{1}{4}\bigg(4125-\frac{19881}{5}\bigg)\\ &= \frac{1}{4}\bigg(4125-3976.2\bigg)\\ &= \frac{148.8}{4}\\ &= 37.2. \end{aligned} The sample variance of $x$ is

\begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(788200-\frac{(1960)^2}{5}\bigg)\\ &= \frac{1}{4}\bigg(788200-\frac{3841600}{5}\bigg)\\ &= \frac{1}{4}\bigg(788200-768320\bigg)\\ &= \frac{19880}{4}\\ &= 4970. \end{aligned}

The sample covariance between $x$ and $y$ is

\begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(56950-\frac{(141)(1960)}{5}\bigg)\\ &= \frac{1}{4}\bigg(56950-\frac{276360}{5}\bigg)\\ &= \frac{1}{4}\bigg(56950-55272\bigg)\\ &= \frac{1678}{4}\\ &= 419.5. \end{aligned} The Karl Pearson's sample correlation coefficient between advertising expenditure and sales is

\begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{419.5}{\sqrt{37.2\times 4970}}\\ &=\frac{419.5}{\sqrt{184884}}\\ &=0.9756. \end{aligned} The correlation coefficient between advertising expenditure and sales is $0.9756$. Since the value of correlation coefficient is positive, there is a strong positive relationship between advertising expenditure and sales.

## Example 2

A study of the amount of rainfall and the quantity of air pollution removed produced the following data:

Daily Rainfall (0.01cm) 4.3 4.5 5.9 5.6 6.1 5.2 3.8 2.1 7.5
Particulate Removed ($\mu g/m^3$) 126 121 116 118 114 118 132 141 108

Calculate correlation coefficient between daily rainfall and particulate removed.

### Solution

Let $x$ denote the daily rainfall (0.01 cm) and $y$ denote the particulate removed ($\mu g/m^3$).

$x$ $y$ $x^2$ $y^2$ $xy$
1 4.3 126 18.49 15876 541.8
2 4.5 121 20.25 14641 544.5
3 5.9 116 34.81 13456 684.4
4 5.6 118 31.36 13924 660.8
5 6.1 114 37.21 12996 695.4
6 5.2 118 27.04 13924 613.6
7 3.8 132 14.44 17424 501.6
8 2.1 141 4.41 19881 296.1
9 7.5 108 56.25 11664 810.0
Total 45.0 1094 244.26 133786 5348.2

The sample variance of $x$ is

\begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(244.26-\frac{(45)^2}{9}\bigg)\\ &= \frac{1}{8}\bigg(244.26-\frac{2025}{9}\bigg)\\ &= \frac{1}{8}\bigg(244.26-225\bigg)\\ &= \frac{19.26}{8}\\ &= 2.4075. \end{aligned} The sample variance of $x$ is

\begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(133786-\frac{(1094)^2}{9}\bigg)\\ &= \frac{1}{8}\bigg(133786-\frac{1196836}{9}\bigg)\\ &= \frac{1}{8}\bigg(133786-132981.7778\bigg)\\ &= \frac{804.2222}{8}\\ &= 100.5278. \end{aligned}

The sample covariance between $x$ and $y$ is

\begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(5348.2-\frac{(45)(1094)}{9}\bigg)\\ &= \frac{1}{8}\bigg(5348.2-\frac{49230}{9}\bigg)\\ &= \frac{1}{8}\bigg(5348.2-5470\bigg)\\ &= \frac{-121.8}{8}\\ &= -15.225. \end{aligned} The Karl Pearson's sample correlation coefficient between daily rainfall and particulate removed is

\begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{-15.225}{\sqrt{2.4075\times 100.5278}}\\ &=\frac{-15.225}{\sqrt{242.0207}}\\ &=-0.9787. \end{aligned} The correlation coefficient between daily rainfall and particulate removed is $-0.9787$. Since the value of correlation coefficient is negative, there is a strong negative relationship between daily rainfall and particulate removed.