# Karl Pearson’s Coefficient Method (Correlation)

## 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}}$$\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)}{{{\left[ \left\{ \frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}} \right\}\left\{ \frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$Other equivalent forms of Coefficient of correlation formulas are:$Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)$$\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

Solution:

$\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

While the correct values were

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

Solution:

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

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

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