To calculate the strength and direction of the linear relationship of two variables of bivariate distribution, the quantity (*r*) used is known as linear correlation coefficient. This is the often called as the Pearson product as it was developed by Karl Pearson.

When the linear relationship of bivariate variables x and y is needed to be measured then the coefficient of correlation will be *r _{xy }*=Cov (x,y) /σ

_{x}. σ

_{y}

In the equation, Cov is Covariance between bivariate variables, σ_{x }is S.D of x and σ_{y }is S.D of y and n is the number of pairs of data.

_{ } 1/n ∑ (x_{i }_{ }– x̅) (y_{i }– y̅)

r_{xy =} _________________________

√1/n ∑(x_{i }– x̅)^{2 }. √1/n ∑(y_{i }– y̅)^{2}

∑ (x_{i }_{ }– x̅) (y_{i }– y̅)

= ____________________

√∑(x_{i} -x̅)^{2} √∑(y_{i} -y̅)^{2}

There are some drawbacks found when measuring correlations:

- As it is already mentioned, that by coefficient correlation one can measures the linear relationship between bivariate variables, so this correlation fails to provide any calculation when there is neither linear relationship nor other types of relationship. This is exactly why scatter diagrams are scrutinized thoroughly.
- In case data is picked from a different source, different coefficient of correlation will come as a result.
- It is not necessary that two bivariate variables are influenced specifically by each other. There may be other effects. If those effects are removed then it may be possible to have a zero coefficient correlation.

**Features:**

You can find some specific characteristics of this correlation:

- This correlation coefficient is completely independent of their source and those referential data.
- -1 ≤ r
_{xy}≤ 1

Let us measure:

u_{i } = __x _{i} – x__

__̅__and v

_{i }=

__y__

_{i}– y__̅__

σ_{x} σ_{y}

Then we have,

1/n ∑u_{i}^{2 }= 1, 1/n ∑v_{i}^{2 }=1, 1/n ∑u_{i }v_{i }= r_{xy}

1/n ∑(u_{i} -v_{i})^{2 }≥ 0

1/n ∑u_{i}^{2 }+ 1/n ∑v_{i}^{2 }– 2/n ∑u_{i }v_{i} ≥ 0

2( 1- r_{xy }) ≥ 0

r_{xy} ≤ 1

1/n [∑(*ui + vi)*^{2}] ^{ }≥ 0

1/n ∑u_{i}^{2 }+ 1/n ∑v_{i}^{2 }+ 2/n ∑u_{i }v_{i} ≥ 0

2 ( 1 + r_{xy })≥ 0

r_{xy }≥ -1

So we get -1 ≤ r _{xy } ≤ 1.

Here, two uncorrelated variables are found. They are independent but the converse isn’t supposedly true always. If r_{xy }is correlation coefficient and n is data of pairs then the standard error is probable error which you can find from:

P.E. (r_{xy}) = 0.6745 __(1 – r _{xy}^{2})__

√n

**By Step Deviation Method:**

Let *d _{x} *=

*x*– A,

*d*=

_{y}*y*– B which are the deviations and A, B are assumed values, then

∑d_{x} d_{y } __–____∑____d _{x }. __

__∑__

__d__

_{y}n

r_{xy} = _________________________________

√[∑d_{x}^{2 }– __(∑d _{x})^{2}__] . √[∑d

_{y}

^{2 }–

__(∑d__] .

_{y})^{2}n n

Here n = number of observations.

The coefficient of correlation given by Karl Pearson is

∑XY

r_{xy}= __________ = 85/ √90*112 = 0.85

∑X^{2 }∑Y^{2 }

So let us calculate the coefficient correlation of this price and supply related grouped data:

Price (Rs.) | 60 | 65 | 70 | 75 | 80 | 85 | 90 | 95 | 100 |

Demand (Qts) | 35 | 30 | 25 | 25 | 23 | 21 | 20 | 20 | 18 |

Solution:

Price (x) | d_{x} = x – 80
<d |
d_{x}^{2} |
Demand (y) | D_{y} = y – 25
<d |
d_{y}^{2} |
d_{x}d_{y} |

60
65 70 75 80 85 90 95 100 |
-20
-15 -10 -5 0 5 10 15 20 |
400
225 100 25 0 25 100 225 400 |
35
30 25 25 23 21 20 20 18 |
10
5 0 0 -2 -4 -5 -5 -7 |
100
25 0 0 4 16 25 25 49 |
-200
-75 0 0 0 -20 -50 -75 -140 |

∑ | 0 | 1500 | – | -8 | 244 | -560 |

∑d_{x} d_{y } – __∑____d _{x }. __

__∑__

__d__

_{y}9

r_{xy} = _________________________________

√[∑d_{x}^{2 }– __(∑d _{x})^{2}__] . √[∑d

_{y}

^{2 }–

__(∑d__] .

_{y})^{2}- 9

__=________-560______________ = -0.94.

√1500 √244 – 64/9

__ __