## Covariance between X and Y

Let `$(x_i, y_i), i=1,2, \cdots , n$`

be $n$ pairs of observations.

Covariance measures the simultaneous variability between the two variables. It indicates how the two variables are related. A positive value of covariance indicate that the two variables moves in the same direction, whereas a negative value of covariance indicate that the two variables moves on opposite direction.

## Formula

The sample covariance between $x$ and $y$ is denoted by $Cov(x,y)$ or $s_{xy}$ and is defined as

`$Cov(x,y) =s_{xy}=\dfrac{1}{n-1}\sum_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})$`

OR

`$s_{xy} = \dfrac{1}{n-1}\bigg(\sum xy - \dfrac{(\sum x)(\sum y)}{n}\bigg)$`

where,

`$\overline{x}$`

sample mean of $x$,`$\overline{y}$`

sample mean of $y$

## Sample mean of $x$

`$\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$`

## Sample mean of $y$

`$\overline{y} =\dfrac{1}{n}\sum_{i=1}^{n}y_i$`

## 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 covariance 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 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 covariance between **advertising expenditure** and **sales** is $419.5$. Since the value of covariance is positive, there is a positive relationship between advertising expenditure and sales. That is the two variables moves together in the same direction.

## 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 covariance 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 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 covariance between **daily rainfall** and **particulate removed** is $-15.225$. Since the value of covariance is negative, there is a negative relationship between daily rainfall and particulate removed. That is the two variables moves together in the opposite direction.