Bivariate frequency tables and covariance – Bivariate Statistics (Senior 5 Mathematics) | Ganzaa

Bivariate Statistics

Lessons

Senior 5 · Mathematics

Bivariate Statistics

4 lessons

Lesson 2 of 4

Bivariate frequency tables and covariance

40 min

Forty households, two questions

A pupil surveying a village asks 40 households two questions: how many people live there, and how many mobile phones they own (illustrative data). Many households give the same pair of answers, so listing 40 pairs would waste space. A double-entry table counts each pair once. From it we can rebuild each variable's own distribution, and measure with one number, the covariance, whether the two tend to rise together.

Definition

A bivariate (double-entry) frequency table lists the values $x_1, \dots, x_p$ of $X$ down one side and $y_1, \dots, y_q$ of $Y$ across the top. The cell in row $i$ , column $j$ holds the joint frequency $f_{ij}$ , the number of individuals with $X = x_{i}$ and $Y = y_{j}$ . The total is $N = \sum_i \sum_j f_{ij}$ .

The row totals $f_{i\cdot} = \sum_j f_{ij}$ give the marginal distribution of $X$ ; the column totals $f_{\cdot j}$ give that of $Y$ .
One row (or column) on its own is a conditional distribution: for example the distribution of $Y$ among the individuals with $X = x_{i}$ .

Worked example 1: reading a double-entry table

$X$ = number of people in a household, $Y$ = number of mobile phones, for 40 households (illustrative):

$X$ \ $Y$	1	2	3	Total $f_{i\cdot}$
2	5	2	0	7
3	4	6	1	11
4	2	7	4	13
5	1	3	5	9
Total $f_{\cdot j}$	12	18	10	40

Find the marginal means and the mean number of phones in 4-person households.

Solution. From the margins: $\bar{x} = \frac{2(7) + 3(11) + 4(13) + 5(9)}{40} = \frac{144}{40} = 3.6$ people and $\bar{y} = \frac{1(12) + 2(18) + 3(10)}{40} = \frac{78}{40} = 1.95$ phones. The conditional distribution of $Y$ for $X = 4$ is the row $2, 7, 4$ (13 households), with mean $\frac{2 + 14 + 12}{13} = \frac{28}{13} \approx 2.15$ phones, higher than for 2-person households ( $\frac{9}{7} \approx 1.29$ ).

Definition

The covariance of $X$ and $Y$ is the mean of the products of their deviations from their means:

\operatorname{cov}(X, Y) = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y}) = \frac{1}{n}\sum_{i=1}^{n} x_iy_i - \bar{x}\,\bar{y} .

For a double-entry table, $\operatorname{cov}(X, Y) = \frac{1}{N}\sum_i\sum_j f_{ij}\,x_iy_j - \bar{x}\,\bar{y}$ . (For grouped data, use the class midpoints.)

Why the two forms agree. Expanding, $\sum (x_i - \bar{x})(y_i - \bar{y}) = \sum x_iy_i - \bar{y}\sum x_i - \bar{x}\sum y_i + n\bar{x}\bar{y} = \sum x_iy_i - n\bar{x}\bar{y}$ , because $\sum x_i = n\bar{x}$ and $\sum y_i = n\bar{y}$ .

What the sign of the covariance means

The dashed lines through the mean point cut the scatter diagram into four quadrants. A point up and to the right of $(\bar{x}, \bar{y})$ , or down and to the left, has both deviations of the same sign, so a positive product; the other two quadrants give negative products. The covariance is the average product: positive when most points lie in the "+" quadrants (positive correlation), negative when most lie in the "−" quadrants, and near 0 when they balance.

Worked example 2: covariance of the revision data

Solution. With $\bar{x} = 7$ , $\bar{y} = 30$ , the products $(x_{i} - 7) (y_{i} - 30)$ are $65, 36, 12, 0, 1, 6, 32, 80$ , with sum 232. So $\operatorname{cov}(X, Y) = \frac{232}{8} = 29$ . With the shortcut: $\sum x_iy_i = 1912$ , and $\frac{1912}{8} - 7 \times 30 = 239 - 210 = 29$ . ✓ No product is negative, which matches the figure.

Worked example 3: covariance from the table

Find $\operatorname{cov}(X, Y)$ for the 40 households.

Solution. Add $f_{ij}x_iy_j$ over all cells. Row $X = 2$ : $5 (2) (1) + 2 (2) (2) = 18$ . Row $X = 3$ : $12 + 36 + 9 = 57$ . Row $X = 4$ : $8 + 56 + 48 = 112$ . Row $X = 5$ : $5 + 30 + 75 = 110$ . Total 297. So

\operatorname{cov}(X, Y) = \frac{297}{40} - 3.6 \times 1.95 = 7.425 - 7.02 = 0.405 .

Positive: bigger households tend to own more phones.

Important

Properties. $\operatorname{cov}(X, Y) = \operatorname{cov}(Y, X)$ ; $\operatorname{cov}(X, X) = \operatorname{Var}(X)$ ; and for constants $a, b, c, d$ ,

\operatorname{cov}(aX + b,\ cY + d) = ac\,\operatorname{cov}(X, Y) .

Adding constants shifts the means but not the deviations; multiplying scales each deviation. So the covariance depends on the units: measuring revision in minutes instead of hours multiplies it by 60. Its size alone cannot say whether a correlation is strong. Lesson 3 fixes this by dividing by the standard deviations.

Worked example 4: changing units

The marks in the revision data are converted to percentages ( $Y^{'} = 2 Y$ ), and revision time to minutes ( $X^{'} = 60 X$ ). Find $\operatorname{cov}(X', Y')$ .

Solution. $\operatorname{cov}(60X, 2Y) = 60 \times 2 \times 29 = 3480$ . The relationship has not changed at all, yet the covariance is 120 times bigger.

Warning

Multiply by the frequency. In a table, each product $x_{i} y_{j}$ counts $f_{ij}$ times.
Subtract $\bar{x}\bar{y}$ once, after dividing the sum by $n$ : $\frac{1}{n}\sum x_iy_i - \bar{x}\bar{y}$ , not $\frac{1}{n}\left(\sum x_iy_i - \bar{x}\bar{y}\right)$ .
A large covariance does not mean a strong correlation; it may just mean large units.

Tip

Organise the work in a table with columns $x$ , $y$ , $x^{2}$ , $y^{2}$ , $x y$ and a row of totals. The same five totals give the means, variances, covariance, correlation coefficient and regression lines of lessons 3 and 4.

StatQuest with Josh Starmer: Covariance, Clearly Explained (about 22 min). Watch why the sign of the covariance shows the trend while its size depends on the scale of the data.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Find the covariance of $(1, 2), (2, 4), (3, 5), (4, 4), (5, 7)$ .
For the 40 households, write the marginal distribution of $Y$ and the conditional distribution of $Y$ for 5-person households.
Find the mean number of phones in 5-person households.
Using question 1, find $\operatorname{cov}(2X + 1,\ 3Y - 4)$ .
In a class of 10, $X = 1$ if a pupil walks to school and 0 otherwise, $Y = 1$ if the pupil is early and 0 otherwise. The table has $f_{00} = 3$ , $f_{01} = 1$ , $f_{10} = 1$ , $f_{11} = 5$ . Find $\operatorname{cov}(X, Y)$ .
Find $\operatorname{Var}(X)$ and $\operatorname{Var}(Y)$ for the revision data from $\sum x^2 = 484$ and $\sum y^2 = 7816$ .
Explain, using the four quadrants, why data with no trend have a covariance close to 0.

Show answersHide answers

$\bar{x} = 3$ , $\bar{y} = 4.4$ , $\sum xy = 76$ : $\operatorname{cov} = 15.2 - 13.2 = 2$ .
Marginal: $Y = 1, 2, 3$ with frequencies $12, 18, 10$ . Conditional for $X = 5$ : $1, 3, 5$ (9 households).
$\frac{1 + 6 + 15}{9} = \frac{22}{9} \approx 2.44$ phones.
$2 \times 3 \times 2 = 12$ .
$\bar{x} = 0.6$ , $\bar{y} = 0.6$ , $\frac{1}{10}\sum f xy = 0.5$ : $\operatorname{cov} = 0.5 - 0.36 = 0.14$ .
$\operatorname{Var}(X) = \frac{484}{8} - 49 = 11.5$ ; $\operatorname{Var}(Y) = \frac{7816}{8} - 900 = 77$ .
With no trend, points are spread evenly over the four quadrants, so the positive and negative products roughly cancel and their mean is close to 0.

Sources and credits

Written by Ganzaa: the explanations, the quadrant diagram (plotted from the data), the worked examples, the practice items and the quiz questions. The household and revision data are illustrative, not real survey results. Every sum and covariance was checked by computer.
Video: Covariance, Clearly Explained!!!, StatQuest with Josh Starmer, embedded from YouTube.

1Forty households, two questions

2What the sign of the covariance means

Forty households, two questions

What the sign of the covariance means