Basic statistics · Lesson 6 of 10

Correlation: covariance, Pearson's r and rank correlation

Do households with more years of schooling earn more? Does demand rise when income does? Correlation puts a number between −1 and +1 on how closely two variables move together in a straight line. This lesson builds it from the scatter plot, through covariance, to Pearson’s r and Spearman’s rank version, and spends equal time on where the number misleads.

← Basic statistics moduleLesson 6 of 10 · about 50 minutes

A question to hold on to

A random sample of size 10 has a covariance of 100 between X and Y, a variance of 25 for X and a variance of 400 for Y. What is the sample correlation coefficient?

CUET 2023Solved at the end ↓

New here?What is Ashanagar?The invented district behind this page's examples: 60 households, with incomes you can inspect.

Ashanagar is an invented district. The statistics lessons use it for every worked example, so you meet the same households again as the ideas build: first as a list of numbers, then as a spread, then as a population to sample from.

It has 60 households, each with a monthly income in rupees. Unless a page says otherwise, those 60 are the whole population, and a “sample” means a few of them drawn at random. Where a page needs two variables, it uses a smaller set of 12 households with the head’s years of schooling beside income.

The figures are illustrative, made up once and fixed. They are not official statistics.

Households: 60
Mean income: ₹28,783 a month
Median income: ₹25,000 a month
Standard deviation: ₹13,366
Lowest and highest: ₹10,500 and ₹72,000

Meet the full data set →

The idea

Do two things move together?

Until now every lesson looked at one variable. Economics is mostly about pairs: schooling and income, rainfall and crop yield, the price of a good and the quantity bought. When you have two measurements for each household, district or year, you have bivariate data, and the first thing to do with it is draw it.

A scatter plot puts one variable on each axis and a dot for every observation. It says a good deal without any arithmetic: whether the dots rise, fall or form a shapeless cloud, how tightly they cluster, and whether a few stray far from the rest. A rising band of dots is a positive relation (Hyderabad 2020 shows one and the right description is a positive statistical relation, not that Y is directly proportional to X). Note two things a scatter plot does not claim: it does not require a regression line to be drawn, and it does not show which variable causes the other, whichever one is on the horizontal axis (JNU 2015).

Correlation analysis then measures the degree of linear association between the two variables. It does not measure causation (CUET 2023, Hyderabad 2015), and it treats the two variables symmetrically: there is no “dependent” and “explanatory” one, and both can be thought of as random (CUET 2025, Hyderabad 2012). That last difference is what separates correlation from regression, which is the next lesson.

Interactive

Covariance: the quadrant idea

Take twelve Ashanagar households. For each, plot the schooling of the household head (years) against monthly income (₹ thousand). Mark the two means. The mean schooling is 10.4 years and the mean income ₹23 thousand, and the two dashed lines through those points cut the plot into four quadrants.

A household in the upper right has schooling above average and income above average. A household in the lower left has both below. In both cases the two deviations, x − x̄ and y − ȳ, have the same sign, so their product is positive. In the other two quadrants the deviations have opposite signs and the product is negative. The average of all those products is the covariance:

Covariance

Cov (X, Y) = \frac{1}{n} i = 1 \sum n (x_{i} - x ˉ) (y_{i} - y ˉ) = \frac{\sum x _{i} y _{i}}{n} - x ˉ y ˉ

Population form, dividing by n. Dividing by n − 1 gives the sample form. Its units are the units of x times the units of y.

Most of the dots fall in the two positive quadrants, so the covariance is positive: 21.83 years × ₹ thousand. Drag a point across the cross-hair into a red quadrant and watch the covariance and r fall. The blue points push the covariance up and the red points pull it down; r is positive when the blue win.

adds to the covariance (10 points)subtracts from it (2 points)

Mean schooling, mean income10.4 yrs, ₹23k

Covariance21.83

Pearson r0.869

Spearman ρ (ranks)0.882

A strong positive association: 10 points push r up, 2 pull it down. Move a single point well away from the rest and watch how far one household can drag r, while Spearman, which sees only ranks, moves less.

Covariance has the right sign but the wrong scale: 21.83 depends on whether income is measured in rupees or thousands, so it can compare nothing with nothing, and it is not bounded (JNU 2019 contrasts it with the correlation coefficient, which is). The fix, in the next section, is to divide by the two standard deviations. Before that, covariance has an important job in sums of variables.

Variance of a sum, and covariance of combinations

Variances add only when the variables are uncorrelated. In general the covariance enters, and it enters with a factor of 2:

Variance of a combination

Var (a X + bY) = a^{2} σ_{X}^{2} + b^{2} σ_{Y}^{2} + 2 ab Cov (X, Y)

The same rule gives Var(X + Y) with a = b = 1 and Var(X − Y) with a = 1, b = −1.

Derive it

The variance of a combination of two variables

We want to show

Var (a X + bY) = a^{2} σ_{X}^{2} + b^{2} σ_{Y}^{2} + 2 ab Cov (X, Y)

Centre the combination.
Write U = X − μ_X and V = Y − μ_Y for the deviations from the means. The mean of aX + bY is aμ_X + bμ_Y, so the deviation of the combination from its own mean is aU + bV.
Square it.
$(a U + bV)^{2} = a^{2} U^{2} + 2 ab U V + b^{2} V^{2}$
This is the ordinary expansion of (p + q)². Nothing about statistics has happened yet.
Average each term.
The average of U² is σ_X² and of V² is σ_Y², by the definition of variance, and the average of UV is Cov(X, Y), by the definition of covariance. That gives the formula. The cross term is there because the squared bracket has a middle term, and it carries a factor of 2 for the same reason.
See it as areas.
Draw a square of side aσ_X and another of side bσ_Y, and put the two rectangles between them. Their areas are a²σ_X², b²σ_Y² and two copies of abσ_Xσ_Y. The covariance is ρσ_Xσ_Y, so each rectangle is filled in only to the fraction ρ. Move ρ to 1 and the pieces make one big square, (aσ_X + bσ_Y)². Move it to 0 and the rectangles vanish.
5
Read off the special cases.
- Independent variables have Cov = 0, so the variances just add: Var(X + Y) = σ_X² + σ_Y².
- A difference has b = −1, so the cross term changes sign but the variances still add: Var(X − Y) = σ_X² + σ_Y² − 2 Cov. Never subtract the variances themselves.
- Example (JNU 2013). With Var X = 2, Var Y = 4 and Cov = 2, Var(2X + 3Y) = 4 × 2 + 9 × 4 + 2 × 2 × 3 × 2 = 8 + 36 + 24 = 68.

Step 1 of 5

Want the covariance itself built from scratch, with the shortcut and the independence trap derived? See the full derivation of covariance.

The covariance itself is linear in each argument, which makes the combinations easy:

Var(2X + 3Y) with Var X = 2, Var Y = 4 and Cov = 2: 4 × 2 + 9 × 4 + 2 × 2 × 3 × 2 = 8 + 36 + 24 = 68 (JNU 2013).
Var(X₁ + X₂) with variances 4 and 9 and a correlation of 0.5: the covariance is 0.5 × 2 × 3 = 3, so the variance is 4 + 9 + 2 × 3 = 19 (JAM 2021). The same formula appears in the form Var(X − Y) = σ_x² + σ_y² − 2ρσ_xσ_y, one of the correct statements in JAM 2024.
If y = 2w + 4u, then Cov(w, y) = 2 Var(w) + 4 Cov(w, u) = 2 × 5 + 4 × 1 = 14 (JNU 2018).
X ~ Uniform(8, 20) and Z ~ Uniform(0, 6) independent, Y = X + Z, W = X − Z: Cov(Y, W) = Var X − Var Z = 12 − 3 = 9 (JAM 2023). The cross terms vanish because X and Z are independent, and the uniform variance is (b − a)² ÷ 12.

DSE 2007 · Delhi School level

E[X] = 3, E[Y] = 0, V[X] = 4, V[Y] = 25 and E[XY] = 5. Find Cov(X, Y), the Pearson correlation and V[X + Y].

Covariance.
E[XY] − E[X]E[Y] = 5 − 3 × 0 = 5.
Correlation.
5 ÷ (√4 × √25) = 5 ÷ 10 = 0.5.
Variance of the sum.
4 + 25 + 2 × 5 = 39.

Answer

Cov = 5, ρ = 0.5 and V[X + Y] = 39. The shortcut formula for covariance, the mean of the products minus the product of the means, is the same trick as the shortcut for variance.

The measure

Pearson's correlation coefficient

Divide the covariance by the product of the two standard deviations and the units cancel:

Karl Pearson's coefficient of correlation

r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}} = \frac{n \sum x y - \sum x \sum y}{[ n \sum x ^{2} - ( \sum x ) ^{2} ] [ n \sum y ^{2} - ( \sum y ) ^{2} ]}

Whether you divide by n or n − 1 in all three terms, r comes out the same.

For the twelve households, the covariance is 21.83, the standard deviations are 3.33 years and ₹7.55 thousand, and r = 21.83 ÷ (3.33 × 7.55) = 0.87. That is a strong positive linear association, in a sample with no claim about cause.

Facts about r the examiners lean on

It lies between −1 and +1 (CUET 2018, CUET 2019, JNU 2019). It can equal either end: −1 ≤ r ≤ 1, with the end points included (CUET 2018). The regression coefficient has no such limit (CUET 2023).
It is a pure number. Heights in centimetres or inches, weights in kilos or pounds, give the same r. JNU 2021 asks which statement is untrue, and it is the one that says r is not independent of units.
It is unchanged by a change of origin and scale. Let U = (X − A) ÷ h and V = (Y − B) ÷ K with h, K positive: then r_uv = r_xy (UGC NET 2025, Hyderabad 2016). Converting temperature from Fahrenheit to Celsius, which is a shift and a positive multiplication, leaves r with electricity use the same (Hyderabad 2020). If z = 2x + 5 and r(x, y) = 0.36, then r(y, z) = 0.36 as well (Hyderabad 2013).
A negative multiplier flips the sign. If r(X, Y) = 0.4 then r(2X, −Y) = −0.4 (Hyderabad 2020). If Y = aX + b with a < 0, then r_xz = −r_yzfor any third variable Z (Hyderabad 2018).
It is ±1 exactly when the points lie on a straight line. For Y = 25 + 0.65X, r = 1, not 0.65 and not 25: the slope has nothing to do with r when the line is exact (Hyderabad 2010). With Y = 3X + 6 and W = −3X + 6, the correlation of W and Y is −1 (JNU 2017); the covariance is (−3)(3) × 9 = −81.
The two variables are treated symmetrically. r(X, Y) = r(Y, X).

Why origin and scale do not matter

r is built from z-scores. Subtracting a constant and dividing by a positive number changes nobody’s z-score, and r is the average product of the two z-scores. A negative divisor reverses one variable, so the z-scores change sign and r follows.

Computing r

You will meet three ways to be given the inputs. If the covariance and the standard deviations are given, divide: a covariance of 6 with standard deviations of 2.45 and 2.61 gives 6 ÷ (2.45 × 2.61) = 0.938 (Hyderabad 2020), and a covariance of 0.80 with standard deviations of 2.5 and 1.25 gives 0.80 ÷ 3.125 = 0.256, for which the nearest option, 0.25, is the answer (CUET 2018). If the variances are given, take the square roots first, which is the trap in the opening puzzle. And if raw sums are given, use the sums formula.

JAM 2025 · standard

For n = 100, Σx = 280, Σy = 60, Σx² = 2,384, Σy² = 117 and Σxy = 438. Find the correlation coefficient to two decimal places.

Means.
x̄ = 2.8 and ȳ = 0.6.
Covariance.
Σxy ÷ n − x̄ȳ = 4.38 − 1.68 = 2.70.
Variances.
Var x = 23.84 − 7.84 = 16, so σₓ = 4. Var y = 1.17 − 0.36 = 0.81, so σᵧ = 0.9.
Correlation.
2.70 ÷ (4 × 0.9) = 2.70 ÷ 3.6 = 0.75.

Answer

0.75. Each step reuses the shortcut formula from Lesson 4: the mean of the squares minus the square of the mean.

A related idea: two combinations of the same variables can be correlated through what they share. If X₁, X₂ and X₃ are uncorrelated, each with variance 3, then X₁ + X₂ and X₁ + X₃ have a covariance of 3 (the shared X₁) and variances of 6 each, so their correlation is 3 ÷ 6 = 0.5 (DSE 2021).

Pause and predict

The correlation between people's heights in inches and weights in pounds is 0.7. Both are converted to centimetres and kilograms. The new correlation is

Derive it

Why r can never leave −1 to 1

We want to show

For any data,

- 1 \leq r \leq 1

, and r is exactly 1 or −1 only when every point lies on a straight line.

Put both variables on the same ruler.
Standardise each variable: $z_{x} = \frac{x - x ˉ}{σ _{X}}$ and $z_{y} = \frac{y - y ˉ}{σ _{Y}}$ . Two facts follow at once. By the definition of the standard deviation the average of z_x² is 1, and so is the average of z_y². And the correlation is the average product of the two z scores:
$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}} = \frac{1}{n} \sum z_{x} z_{y}$
Squares are never negative, so look at a gap.
For each household the gap z_x − z_y squared is at least 0, so its average is too. Expand it:
$0 \leq (z_{x} - z_{y})^{2} = z_{x}^{2} - 2 z_{x} z_{y} + z_{y}^{2} = 1 - 2 r + 1 = 2 - 2 r$
So 2 − 2r ≥ 0, which says r ≤ 1. For the twelve households r = 0.869 and the average squared gap is 2 − 2 × 0.869 = 0.262.
Do the same with the sum.
$0 \leq (z_{x} + z_{y})^{2} = 1 + 2 r + 1 = 2 + 2 r$
So 2 + 2r ≥ 0, which says r ≥ −1. Both bounds come from nothing more than the fact that a square cannot be negative. The picture shows what the two averages measure: how far the standardised points stray from the line y = x and from the line y = −x.
When is the bound reached?
An average of squares is 0 only if every square is 0. So r = 1 means z_x = z_y for every household, which is the straight line y = ȳ + (σ_Y ÷ σ_X)(x − x̄) with a positive slope. In the same way r = −1 forces z_x = −z_y, a straight line sloping down. Any scatter that is not exactly a straight line leaves some gap, and then r is strictly inside the interval.

Step 1 of 4

Interactive

Reading r from a picture

A number is only useful if you can picture what it looks like. A rough guide: below about 0.3 in size the cloud looks shapeless, around 0.5 there is a visible tilt with plenty of scatter, 0.7 is a clear band, and above 0.9 the points hug a line. But many people find it hard to judge, especially in the middle of the range: a cloud with r of only 0.3 can look as if it has a clear tilt. The game below shows you forty points built to have an exact correlation. Guess it, then see how close you came.

Round 1 of 8

Your guess for r0

Each cloud has exactly the r it was built with, so there is always a right answer. Look at how far the points scatter around an imagined line, not at the slope of that line.

A related quantity, r², is the share of the variation in one variable that a straight line on the other accounts for. An r of 0.87, as in the schooling example, gives r² of about 0.76. Lesson 7 on regression uses it properly.

Interactive

Where r misleads

Pearson’s r measures one thing: how close the points are to a straight line. Everything it can mislead you about follows from that.

Four data sets, one r

In 1973 the statistician Francis Anscombe published four small data sets that have exactly the same mean of x, mean of y, variance of x, variance of y, correlation and line of best fit. The table cannot tell them apart. The pictures are nothing alike.

Summary	I	II	III	IV
Mean of x	9	9	9	9
Mean of y	7.5	7.5	7.5	7.5
Variance of x (n − 1)	11	11	11	11
Variance of y (n − 1)	4.1	4.1	4.1	4.1
Correlation r	0.82	0.82	0.82	0.82

Every set has the same line, y = 3 + 0.5x, and the same r of about 0.82, but only set I looks like what r seems to describe. In II the relationship is a curve, so a straight-line measure misses its shape. In III a single outlier bends the line away from an otherwise perfect fit. In IV every x but one is 8, and that lone point at 19 creates the whole correlation. Always draw the picture before trusting the number.

Zero correlation is not independence

If r = 0 then there is no linear relationship (JNU 2017), but the variables can be tightly linked in another way. Let X take the values −1, 0 and 1 with equal probability and let Y = X². Y is completely determined by X, so they are as dependent as variables can be. Yet Cov(X, Y) = E[X³] − E[X]E[X²] = 0 − 0 = 0, so they are uncorrelated but dependent (DSE 2010). The same holds for X ~ Uniform[−1, 1] and X², whose covariance is 0 (DSE 2012). Independence does imply zero covariance, but the implication does not run the other way. JAM 2024 includes exactly this trap: “ρ = 0 implies X and Y are independent” is false. A strongly curved relationship is another case where r underestimates the link (Hyderabad 2016).

One point can make or break r

In set III one outlier bends the line away from an otherwise perfect straight pattern. In set IV a single point at x = 19 supplies the whole correlation: take it away and every other x is 8, so there is no linear pattern to measure. Try it in the first lab by moving one household far from the rest. Because r uses squared deviations, it is sensitive to such points.

Correlation is not causation

Months with high fan sales also have high cold-drink sales. The correlation is strong and neither causes the other: hot weather drives both. A third factor behind both variables, a confounder, is one common source of a correlation that means nothing for policy. Others are reverse causation, where the effect drives the supposed cause, and plain coincidence in short series. Correlation measures linear association and nothing more (CUET 2023, Hyderabad 2015), and the JNU 2013 question has a related warning: a negative correlation coefficient indicates an inverse relationship, but a scatter plot with points in the lower left and upper right quadrants indicates a positive one, and a negative “p-value” does not exist.

Watch out for

The mistakes examiners are fishing for

Always plot first. The same r can describe four different pictures.
r = 0 means no linear relation, not no relation. A curve can have r = 0.
Independent implies uncorrelated; uncorrelated does not imply independent.
Strong r is not causation. Think about a confounder, reverse causation and coincidence.
Slope is not r. A perfect line with a tiny slope has r = 1, and a steep, noisy cloud can have a small r.
Covariance is not bounded. Only the correlation coefficient is.

Ranks and reliability

Rank correlation and the probable error

Sometimes you have no measurements, only an ordering: judges ranking contestants, or districts ranked by literacy and by income. Or the data are measured but have outliers and a curved but steadily rising relationship. For these, Spearman’s rank correlation is Pearson’s r computed on the ranks. It is 1 when the two rankings agree exactly, −1 when one is the reverse of the other, and near 0 when they bear no relation to each other.

When no ranks are tied, it reduces to a short formula in the differences d between each item’s two ranks:

Spearman's rank correlation

ρ = 1 - \frac{6 \sum d _{i}^{2}}{n ( n ^{2} - 1 )}

For tied values give each the average of the ranks they occupy: values 10, 20, 20, 30 have ranks 1, 2.5, 2.5, 4. Then compute Pearson's r on the ranks.

District	A	B	C	D	E	F
Rank in literacy	1	2	3	4	5	6
Rank in income	2	1	4	3	6	5
Difference d	−1	1	−1	1	−1	1
d²	1	1	1	1	1	1

Six invented districts. Σd² = 6 and n = 6, so ρ = 1 − 6 × 6 ÷ (6 × 35) = 1 − 36 ÷ 210 = 0.829. Each district is only ever one place out of line, but there are six of them, and it is enough to take the correlation noticeably below 1.

The probable error

A correlation computed from a sample is itself uncertain. The older textbooks measure that with the probable error:

Probable error of r

P E = 0.6745 \frac{1 - r ^{2}}{n}

Its use today has largely given way to the t-test, which later modules cover. Exams still ask it.

The rule of thumb is that r is significant when it exceeds six times its probable error (JNU 2020), and is not worth reading if it is less than the probable error. For n = 16 and r = 0.9544 (JNU 2021), PE = 0.6745 × (1 − 0.9109) ÷ 4 = 0.015, so 6 PE = 0.09, and r is very significant.

Solving the opening question

The opening question, solved

CUET 2023 · standard

A sample of 10 has Cov(X, Y) = 100, Var(X) = 25 and Var(Y) = 400. Find r.

Take the square roots.
σₓ = √25 = 5 and σᵧ = √400 = 20.
Divide.
r = 100 ÷ (5 × 20) = 100 ÷ 100 = 1.

Answer

r = 1: a perfect positive linear relationship. The option 0.01 comes from dividing by the variances, 100 ÷ (25 × 400), instead of by their square roots. The sample size of 10 is irrelevant because the divisor, n or n − 1, cancels from top and bottom.

Revise

The lesson on one screen

Covariance: Cov = Σ(x − x̄)(y − ȳ) ÷ n = mean(xy) − x̄ȳ. Positive when the variables rise together. Unbounded, with units.
Pearson's r: r = Cov ÷ (σₓσᵧ), between −1 and +1. Pure number. ±1 only for an exact straight line.
Invariance: Unchanged by shifting either variable and by positive rescaling. A negative multiplier flips the sign.
Sums: Var(aX + bY) = a²σₓ² + b²σᵧ² + 2ab Cov. Var(X + Y) = σₓ² + σᵧ² + 2Cov. Cov(aX + b, cY + d) = ac Cov.
Limits: r = 0 means no linear relation, not independence. Plot the data first. Outliers can create or destroy r. Correlation is not causation.
Spearman: ρ = 1 − 6Σd² ÷ n(n² − 1). Pearson's r on ranks; average ranks for ties.
Probable error: PE = 0.6745(1 − r²) ÷ √n. r is significant if r > 6 PE.
Correlation vs regression: Correlation treats the two variables symmetrically. Regression does not (Lesson 7).

Practice

Check yourself

0 of 13 correct so far

Compute

From covariance

σ_X = 4, σ_Y = 5 and Cov(X, Y) = 15. Find r.

Back to the covariance

Var(X) = 9, Var(Y) = 16 and r = −0.5. Find Cov(X, Y).

Cov

Variance of a sum

Var(X) = 4, Var(Y) = 9 and Cov(X, Y) = 3. Find Var(X + Y).

Var

Variance of a combination

Var(X) = 5, Var(Y) = 8 and Cov(X, Y) = 1. Find Var(2X − Y).

Var

From raw sums

n = 5, Σx = 15, Σy = 20, Σx² = 55, Σy² = 90 and Σxy = 68. Find r.

A negative multiplier

r(X, Y) = 0.6. Find r(−3X + 2, 4Y + 1).

Spearman

Ten items are ranked two ways, with Σd² = 33. Find Spearman’s rank correlation.

Probable error

r = 0.8 from n = 25 pairs. Find the probable error, to three decimal places.

Quick questions

1The correlation between two variables is zero. We can conclude that

2The correlation coefficient is independent of

3Which of these is NOT bounded between −1 and +1?

4A strong positive correlation between fan sales and cold-drink sales shows that

5Spearman's rank correlation is

Real questions

Past papers, by level

Foundation

Definitions, one-step arithmetic and which-measure questions. Most CUET and Hyderabad questions sit here.

18 questionsPractise →

Standard

Grouped data, combined means, effects of shifting and scaling, and statement questions, up to JNU level.

17 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 10: Correlation and covariance.

5 questionsLesson 10 →

All 35 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 5Shape: skewness, kurtosis and boxplots Lesson 7 →A first look at regression: the line of best fit

Correlation: covariance, Pearson's r and rank correlation

Do two things move together?

Covariance: the quadrant idea

Variance of a sum, and covariance of combinations

Pearson's correlation coefficient

Facts about r the examiners lean on

Computing r

Reading r from a picture

Where r misleads

Four data sets, one r

Zero correlation is not independence

One point can make or break r

Correlation is not causation

The mistakes examiners are fishing for

Rank correlation and the probable error

The probable error

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Correlation: covariance, Pearson's r and rank correlation

Do two things move together?

Covariance: the quadrant idea

Variance of a sum, and covariance of combinations

Pearson's correlation coefficient

Facts about r the examiners lean on

Computing r

Reading r from a picture

Where r misleads

Four data sets, one r

Zero correlation is not independence

One point can make or break r

Correlation is not causation

The mistakes examiners are fishing for

Rank correlation and the probable error

The probable error

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next