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.
