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

T17560453015
T2150175200225250

Solution:

T1X = T1/15X2T2Y = T2/25Y2XY
7552515063630
6041617574928
453920086424
302422598118
15112501010010
Total155540330110

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

 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

X68
Y146

While the correct values were

X86
Y128

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

Firm12345678910
Sales50505560656565606060
Expenses11131416161515141313

Solution:

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

FirmSale X\[x=X-\bar{X}\]\[{{x}^{2}}\]Expenses Y\[y=Y-\bar{Y}\]\[{{y}^{2}}\]\[xy\]
150-86411-3924
250-86413-118
355-3914000
4602416244
565749162414
66574915117
76574915117
8602414000
9602413-11-2
1060-86413-118
N = 10\[\sum{X=580}\]\[\sum{x=0}\]\[\sum{{{x}^{2}}=360}\]\[\sum{Y=140}\]\[\sum{y=0}\]\[\sum{{{y}^{2}}=22}\]\[\sum{xy=70}\]

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

 

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