# Karl Pearson’s Coefficient Method (Correlation)

Statistics / Saturday, October 6th, 2018

# Karl Pearson’s Coefficient Method

Karl Pearson’s method of coefficient of correlation is also known as Pearsonian coefficient or correlation or product correlation method.

Let X and Y be two random variables, then the correlation of coefficient between the variables X and Y is denoted by r(X, Y) or simply by rXY, and is defined as:
${{r}_{XY}}=r\left( X,Y \right)=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}$
$Where,Cov\left( X,Y \right)=E\left[ \left\{ X-E\left( X \right) \right\}\left\{ Y-E\left( Y \right) \right\} \right]$
$\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)={{\mu }_{11}}$
${{\sigma }_{X}}^{2}=E{{\left\{ X-E\left( X \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}}$
${{\sigma }_{Y}}^{2}=E{{\left\{ Y-E\left( Y \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}}$
$\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)}{{{\left[ \left\{ \frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}} \right\}\left\{ \frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$
Other equivalent form of coefficient of correlation formulas are:
$Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)$
$\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}-\bar{y}\frac{1}{n}\sum{{{x}_{i}}}}-\bar{x}\frac{1}{n}\sum{{{y}_{i}}}+\bar{x}\bar{y}$
$\therefore Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}$
${{\sigma }_{X}}^{2}=\frac{1}{n}\sum{{{x}_{i}}^{2}}-{{\bar{x}}^{2}}$
${{\sigma }_{Y}}^{2}=\frac{1}{n}\sum{{{y}_{i}}^{2}}-{{\bar{y}}^{2}}$
$\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{x}_{i}}^{2}}-{{{\bar{x}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{y}_{i}}^{2}}-{{{\bar{y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$

Note:

The values of correlation coefficient cannot exceed unity numerically. It always lies in between -1 and +1. That is -1 ≤ r(X, Y) ≤ 1. If r =+1, the correlation is perfect and positive and if r = -1, correlation is perfect and negative.

1. It is not affected by change of origin or change of scale.
2. It is a relative measure. It does not have any unit attached to it.

 Example 01

Calculate the coefficient of correlation by Karl Pearson’s method based on following values

 T1 75 60 45 30 15 T2 150 175 200 225 250

Solution:

 T1 X = T1/15 X2 T2 Y = T2/25 Y2 XY 75 5 25 150 6 36 30 60 4 16 175 7 49 28 45 3 9 200 8 64 24 30 2 4 225 9 81 18 15 1 1 250 10 100 10 Total 15 55 40 330 110

$\bar{X}=\frac{\sum{X}}{5}=\frac{15}{5}=3$
$\bar{Y}=\frac{\sum{Y}}{5}=\frac{40}{5}=8$
$\frac{1}{n}\sum{{{X}^{2}}=\frac{1}{5}\times 55=11}$
$\frac{1}{n}\sum{{{Y}^{2}}=\frac{1}{5}\times 330=66}$
$\frac{1}{n}\sum{XY=\frac{1}{5}\times 110=22}$
$\therefore {{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$
$\therefore {{r}_{XY}}=\frac{22-3\times 8}{{{\left[ \left\{ 11-{{3}^{2}} \right\}\left\{ 66-{{8}^{2}} \right\} \right]}^{\frac{1}{2}}}}=\frac{-2}{2}=-1$

 Example 02

A computer while calculating correlation coefficient between two variables X and Y from 25 pairs of observation obtained the following results: n = 25, ΣX = 125, ΣX2 = 650, ΣY = 100, ΣY2 = 460, ΣXY = 508. It was later noticed that there was some mistake while copied down two pairs as

 X 6 8 Y 14 6

While the correct values were

 X 8 6 Y 12 8

Obtain the correct value of the correlation coefficient.

Solution:

Corrected ΣX = 125 – 6 – 8 + 8 + 6 = 125
Corrected ΣY = 100 – 14 – 6 + 12 + 8 = 100
Corrected ΣX2 = 650 – 62 – 82 + 82 + 62 = 650
Corrected ΣY2 = 460 – 142 – 62 + 122 + 82 = 436
Corrected ΣXY = 508 – 6(14) – 8(6) + 8(12) + 6(8) = 520
$\bar{X}=\frac{\sum{X}}{25}=\frac{125}{25}=5$
$\bar{Y}=\frac{\sum{Y}}{25}=\frac{100}{25}=4$
$\therefore \text{Corrected}~~{{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$
$\therefore {{r}_{XY}}=\frac{\frac{1}{25}\times 520-5\times 4}{{{\left[ \left\{ \frac{1}{25}\times 650-{{5}^{2}} \right\}\left\{ \frac{1}{25}\times 436-{{4}^{2}} \right\} \right]}^{\frac{1}{2}}}}$
$\therefore {{r}_{XY}}=\frac{\frac{4}{5}}{1\times \frac{6}{5}}=\frac{2}{3}=0.67$

 Example 03

Calculate the correlation coefficient for the following

 Firm 1 2 3 4 5 6 7 8 9 10 Sales 50 50 55 60 65 65 65 60 60 60 Expenses 11 13 14 16 16 15 15 14 13 13

Solution:

Here,
$\bar{X}=\frac{\sum{X}}{N}=\frac{580}{10}=58$
$\bar{Y}=\frac{\sum{Y}}{N}=\frac{140}{10}=14$

 Firm Sale X $x=X-\bar{X}$ ${{x}^{2}}$ Expenses Y $y=Y-\bar{Y}$ ${{y}^{2}}$ $xy$ 1 50 -8 64 11 -3 9 24 2 50 -8 64 13 -1 1 8 3 55 -3 9 14 0 0 0 4 60 2 4 16 2 4 4 5 65 7 49 16 2 4 14 6 65 7 49 15 1 1 7 7 65 7 49 15 1 1 7 8 60 2 4 14 0 0 0 9 60 2 4 13 -1 1 -2 10 60 -8 64 13 -1 1 8 N = 10 $\sum{X=580}$ $\sum{x=0}$ $\sum{{{x}^{2}}=360}$ $\sum{Y=140}$ $\sum{y=0}$ $\sum{{{y}^{2}}=22}$ $\sum{xy=70}$

$\therefore r(x,y)=\frac{\sum{xy}}{\sqrt{\sum{{{x}^{2}}.\sum{{{y}^{2}}}}}}=\frac{70}{\sqrt{360\times 22}}=0.787$

<< Previous     Next>>

## One thought on “Karl Pearson’s Coefficient Method (Correlation)”

1. Monzu

Thanks a lot bro Rajib Kumar Saha for your nice article.