Did you know? Two variables can follow a perfectly predictable relationship while their Pearson correlation coefficient is exactly zero. Take X equal to −1, 0 or 1 with equal probabilities, and define Y=X². Knowing X determines Y completely. Yet the positive and negative sides of the U-shaped graph cancel in the linear covariance calculation, giving correlation r=0. Zero correlation is not the same statement as independence.
This JC2 H2 Mathematics Tutor Bukit Timah guide answers one exact A-Level statistics question: what does a correlation coefficient of zero mean, and why can it coexist with a nonlinear relationship? We start with a concise interpretation, explain a discrete counterexample and a simple four-point scatter dataset, then compare it with genuine independence. Thirty-six original numerical clinics let learners verify the cancellation and interpret a scatter diagram without making claims unsupported by a single statistic.
The 2027 SEAB H2 Mathematics syllabus 9758 includes the product-moment correlation coefficient as a measure of fit of a linear model, and its interpretation near −1, 0 and 1. This guide focuses on the meaning of r≈0 and the danger of inferring independence from linear correlation alone. It is not a hypothesis test for correlation; the syllabus explicitly excludes correlation and regression hypothesis tests. All datasets below are fictional mathematics examples.
Quick answer: r measures linear association, not every possible relationship
For paired observations (xᵢ,yᵢ), Pearson’s sample correlation coefficient is r=Σ(xᵢ−x̄)(yᵢ−ȳ)/√[Σ(xᵢ−x̄)²Σ(yᵢ−ȳ)²], provided both variables vary. A value near +1 indicates a strong positive linear trend, a value near −1 a strong negative linear trend, and a value near zero indicates little linear association in that dataset. It does not establish that the variables are statistically independent or unrelated in every nonlinear way.
- r near +1: the points tend to follow a rising straight-line pattern.
- r near −1: the points tend to follow a falling straight-line pattern.
- r near zero: a straight-line pattern may be weak or absent, but a curved relationship may still exist.
- Independence: knowing one variable does not change the distribution of the other; this is a stronger condition.
- Correlation requires variation: if all x-values or all y-values are identical, Pearson r is undefined, not zero.
- Plot before interpreting: a scatter diagram can reveal U shapes, outliers, clusters or other structures that one coefficient cannot capture.
A complete counterexample: X takes −1, 0, 1 and Y=X²
Let X take the values −1, 0, 1 with equal probability 1/3, and set Y=X². The corresponding possible pairs are (−1,1), (0,0), (1,1). Their means are E(X)=0 and E(Y)=2/3. Since XY=X³, symmetry gives E(XY)=0. Therefore the covariance is E(XY)−E(X)E(Y)=0−0(2/3)=0.
Both variables have nonzero variance. Var(X)=E(X²)=2/3; Var(Y)=E(Y²)−[E(Y)]²=2/3−4/9=2/9. Thus their population correlation is zero, because covariance is zero and the standard deviations are positive. Nonetheless the variables are dependent: if X=0, Y=0 with probability one, whereas without knowing X, P(Y=0)=1/3. Those two probabilities are not equal.
The example shows exactly what r=0 can fail to see: a U-shaped relationship. As X moves from −1 to 0, Y falls; as X moves from 0 to 1, Y rises. A straight-line trend cannot describe both halves with a single signed gradient. The positive and negative co-movements cancel in covariance, even though the function Y=X² is completely deterministic.
A four-point scatter example with exact r=0
Consider four fictional pairs: (−2,4), (−1,1), (1,1), (2,4). The x-mean is zero, the y-mean is 2.5, and the product of centred deviations sums to zero. Specifically, Σxy=(−2)(4)+(−1)(1)+(1)(1)+(2)(4)=−8−1+1+8=0, while Σx=0. Therefore the numerator of Pearson r is zero. Both x and y vary, so r=0 is defined and correct.
But every point lies on y=x². If you know x you know y exactly, while the best flat linear description misses the upward curve. A scatter diagram drawn with x horizontally and y vertically would show two high points at either end and two low points near the centre. This geometry is the crucial independent check against interpreting r=0 as “there is no relationship”.
Why independence is a stronger statement
For random variables, independence means that probabilities factor appropriately for their possible events or outcomes. In a simple discrete example it requires P(X=x,Y=y)=P(X=x)P(Y=y) for all relevant pairs. When Y=X² and X takes three values, that condition fails: Y=0 is certain if X=0, but not otherwise. The relationship is deterministic and therefore dependent except for degenerate special cases.
Independent random variables with finite variances have covariance zero, but the reverse is not generally true. The mathematical implication has one direction in general: independence ⇒ zero covariance ⇒ zero correlation when standard deviations are positive. Zero correlation alone is insufficient to conclude independence. This lesson deliberately avoids advanced special joint-distribution exceptions, which are not necessary to interpret H2 scatter diagrams correctly.
A genuinely independent comparison with r=0
Now take X and Y independently, each equal to −1 or +1 with probability one-half. The four joint pairs (−1,−1), (−1,+1), (+1,−1), (+1,+1) are equally likely. The means are zero and E(XY)=0, so the correlation is zero, and in this case independence truly holds because the joint probabilities factor. The same correlation value r=0 therefore occurs in both an independent model and the dependent U-shaped model.
This comparison proves why a correlation coefficient cannot, by itself, answer an independence question. To decide independence, we need information about the joint distribution or the underlying data-generating mechanism, not simply a linear summary. When the problem is a scatter diagram, an appropriate description is “little evidence of a linear association”, with any visible nonlinear pattern explained separately.
Correlation does not prove causation either
A strong numerical correlation does not by itself prove that one variable causes the other. They could move together because of a third variable, selection effects, or a shared external influence. A value near zero does not prove they have no causal relationship either; a causal effect could be nonlinear and cancel in the linear statistic. These are limitations of interpreting associations from observational data.
In a school answer, keep the inference within the information given. If the scatter plot curves upward, say so. If r≈0, say that a linear relationship is weak, not that no dependence is possible. If a regression line is used, explain how well the straight-line model represents the visible points and whether interpolation or extrapolation is being attempted.
Outliers and the single-number trap
Pearson correlation can be sensitive to extreme points. A dataset with a clear main trend but one influential observation may have an unexpected correlation coefficient. Conversely, mixtures of separate clusters can produce a coefficient that conceals important structure. A single reported r value does not tell you where the points lie or whether the observations form groups. The scatter diagram is essential context.
The H2 syllabus explicitly includes using a scatter diagram to judge whether a plausible linear relationship exists, then interpreting r as a measure of linear fit. That is not a decorative plotting exercise. It is a check that the numerical summary describes the type of relationship you are trying to model.
Why r is undefined for constant data
Suppose every y-value equals 4, regardless of x. The variation Σ(y−ȳ)² is zero, so the denominator of Pearson r is zero. The correlation coefficient is undefined, not r=0. A flat constant y sequence cannot be assigned a Pearson correlation by simply claiming that there is no linear slope. The mathematical formula has a domain condition, just as a reciprocal or logarithm does.
This is a useful boundary check when a student invents a dataset to illustrate zero correlation. To demonstrate a defined r=0, ensure both x and y vary. The symmetric U-shaped examples below use two different positive x magnitudes so the squared y-values differ, giving nonzero variation in both coordinates.
One-minute diagnostic
Show four pairs (−2,4), (−1,1), (1,1), (2,4). Ask for the shape of the scatter plot. It is U-shaped. Then calculate Σxy=0 with mean x=0 and conclude Pearson r=0. Finally ask whether y is independent of x. It is not, because every point satisfies y=x². A student who can say “zero linear correlation, not zero relationship” has captured the essential distinction.
Thirty-six original symmetric scatter datasets
Each clinic uses four fictional paired observations symmetrically arranged around an x-centre h, with y determined by a quadratic rule. The computed Pearson r is exactly zero, yet the output still depends deterministically on the input. The values vary by centre, scale and vertical shift. The point is to calculate a zero linear summary correctly without erasing the visible nonlinear structure.
1. Four points around x=-4, two different heights
Try the scatter data. Consider the original fictional paired points (-5,-4), (-3,-4), (-6,-1), (-2,-1). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -4, but the y-values are not all equal.
Exact calculation. The means are x̄=-4 and ȳ=-2.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 10 for x and 9 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-4))²+(-5). Therefore knowing x determines y in this dataset. The horizontal symmetry makes positive and negative centred products cancel. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
2. Four points around x=-3, two different heights
Try the scatter data. Consider the original fictional paired points (-5,2), (-1,2), (-7,14), (1,14). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -3, but the y-values are not all equal.
Exact calculation. The means are x̄=-3 and ȳ=8. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 40 for x and 144 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-3))²+(-2). Therefore knowing x determines y in this dataset. The scatter diagram is U-shaped, so a straight-line model can miss the relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
3. Four points around x=-2, two different heights
Try the scatter data. Consider the original fictional paired points (-5,10), (1,10), (-8,37), (4,37). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -2, but the y-values are not all equal.
Exact calculation. The means are x̄=-2 and ȳ=23.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 90 for x and 729 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-2))²+(1). Therefore knowing x determines y in this dataset. A correlation of zero is well-defined here because both variables have nonzero spread. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
4. Four points around x=-1, two different heights
Try the scatter data. Consider the original fictional paired points (-5,20), (3,20), (-9,68), (7,68). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -1, but the y-values are not all equal.
Exact calculation. The means are x̄=-1 and ȳ=44. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 160 for x and 2304 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-1))²+(4). Therefore knowing x determines y in this dataset. Knowing x determines y exactly under the quadratic rule; the variables are dependent. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
5. Four points around x=0, two different heights
Try the scatter data. Consider the original fictional paired points (-5,21), (5,21), (-6,32), (6,32). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 0, but the y-values are not all equal.
Exact calculation. The means are x̄=0 and ȳ=26.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 122 for x and 121 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(0))²+(-4). Therefore knowing x determines y in this dataset. Adding a constant to every x or y value does not create linear correlation in this symmetric dataset. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
6. Four points around x=1, two different heights
Try the scatter data. Consider the original fictional paired points (0,0), (2,0), (-2,8), (4,8). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 1, but the y-values are not all equal.
Exact calculation. The means are x̄=1 and ȳ=4. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 20 for x and 64 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(1))²+(-1). Therefore knowing x determines y in this dataset. The two high outer points and two low inner points create a visible curved pattern. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
7. Four points around x=2, two different heights
Try the scatter data. Consider the original fictional paired points (0,6), (4,6), (-3,27), (7,27). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 2, but the y-values are not all equal.
Exact calculation. The means are x̄=2 and ȳ=16.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 58 for x and 441 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(2))²+(2). Therefore knowing x determines y in this dataset. Pearson r summarises linear co-movement, not an arbitrary nonlinear relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
8. Four points around x=3, two different heights
Try the scatter data. Consider the original fictional paired points (0,14), (6,14), (-4,54), (10,54). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 3, but the y-values are not all equal.
Exact calculation. The means are x̄=3 and ȳ=34. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 116 for x and 1600 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(3))²+(5). Therefore knowing x determines y in this dataset. This is not a claim that every real dataset with r≈0 has a U shape; it is a counterexample to a false general implication. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
9. Four points around x=4, two different heights
Try the scatter data. Consider the original fictional paired points (0,13), (8,13), (-1,22), (9,22). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 4, but the y-values are not all equal.
Exact calculation. The means are x̄=4 and ȳ=17.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 82 for x and 81 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(4))²+(-3). Therefore knowing x determines y in this dataset. A student should inspect the scatter plot rather than interpreting a coefficient without geometry. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
10. Four points around x=-4, two different heights
Try the scatter data. Consider the original fictional paired points (-9,25), (1,25), (-11,49), (3,49). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -4, but the y-values are not all equal.
Exact calculation. The means are x̄=-4 and ȳ=37. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 148 for x and 576 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-4))²+(0). Therefore knowing x determines y in this dataset. It is possible for two datasets with the same r=0 to have very different underlying relationships. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
11. Four points around x=-3, two different heights
Try the scatter data. Consider the original fictional paired points (-4,4), (-2,4), (-7,19), (1,19). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -3, but the y-values are not all equal.
Exact calculation. The means are x̄=-3 and ȳ=11.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 34 for x and 225 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-3))²+(3). Therefore knowing x determines y in this dataset. The horizontal symmetry makes positive and negative centred products cancel. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
12. Four points around x=-2, two different heights
Try the scatter data. Consider the original fictional paired points (-4,-1), (0,-1), (-8,31), (4,31). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -2, but the y-values are not all equal.
Exact calculation. The means are x̄=-2 and ȳ=15. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 80 for x and 1024 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-2))²+(-5). Therefore knowing x determines y in this dataset. The scatter diagram is U-shaped, so a straight-line model can miss the relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
13. Four points around x=-1, two different heights
Try the scatter data. Consider the original fictional paired points (-4,7), (2,7), (-5,14), (3,14). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -1, but the y-values are not all equal.
Exact calculation. The means are x̄=-1 and ȳ=10.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 50 for x and 49 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-1))²+(-2). Therefore knowing x determines y in this dataset. A correlation of zero is well-defined here because both variables have nonzero spread. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
14. Four points around x=0, two different heights
Try the scatter data. Consider the original fictional paired points (-4,17), (4,17), (-6,37), (6,37). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 0, but the y-values are not all equal.
Exact calculation. The means are x̄=0 and ȳ=27. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 104 for x and 400 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(0))²+(1). Therefore knowing x determines y in this dataset. Knowing x determines y exactly under the quadratic rule; the variables are dependent. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
15. Four points around x=1, two different heights
Try the scatter data. Consider the original fictional paired points (-4,29), (6,29), (-7,68), (9,68). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 1, but the y-values are not all equal.
Exact calculation. The means are x̄=1 and ȳ=48.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 178 for x and 1521 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(1))²+(4). Therefore knowing x determines y in this dataset. Adding a constant to every x or y value does not create linear correlation in this symmetric dataset. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
16. Four points around x=2, two different heights
Try the scatter data. Consider the original fictional paired points (1,-3), (3,-3), (-3,21), (7,21). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 2, but the y-values are not all equal.
Exact calculation. The means are x̄=2 and ȳ=9. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 52 for x and 576 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(2))²+(-4). Therefore knowing x determines y in this dataset. The two high outer points and two low inner points create a visible curved pattern. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
17. Four points around x=3, two different heights
Try the scatter data. Consider the original fictional paired points (1,3), (5,3), (0,8), (6,8). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 3, but the y-values are not all equal.
Exact calculation. The means are x̄=3 and ȳ=5.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 26 for x and 25 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(3))²+(-1). Therefore knowing x determines y in this dataset. Pearson r summarises linear co-movement, not an arbitrary nonlinear relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
18. Four points around x=4, two different heights
Try the scatter data. Consider the original fictional paired points (1,11), (7,11), (-1,27), (9,27). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 4, but the y-values are not all equal.
Exact calculation. The means are x̄=4 and ȳ=19. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 68 for x and 256 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(4))²+(2). Therefore knowing x determines y in this dataset. This is not a claim that every real dataset with r≈0 has a U shape; it is a counterexample to a false general implication. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
19. Four points around x=-4, two different heights
Try the scatter data. Consider the original fictional paired points (-8,21), (0,21), (-11,54), (3,54). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -4, but the y-values are not all equal.
Exact calculation. The means are x̄=-4 and ȳ=37.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 130 for x and 1089 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-4))²+(5). Therefore knowing x determines y in this dataset. A student should inspect the scatter plot rather than interpreting a coefficient without geometry. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
20. Four points around x=-3, two different heights
Try the scatter data. Consider the original fictional paired points (-8,22), (2,22), (-12,78), (6,78). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -3, but the y-values are not all equal.
Exact calculation. The means are x̄=-3 and ȳ=50. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 212 for x and 3136 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-3))²+(-3). Therefore knowing x determines y in this dataset. It is possible for two datasets with the same r=0 to have very different underlying relationships. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
21. Four points around x=-2, two different heights
Try the scatter data. Consider the original fictional paired points (-3,1), (-1,1), (-4,4), (0,4). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -2, but the y-values are not all equal.
Exact calculation. The means are x̄=-2 and ȳ=2.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 10 for x and 9 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-2))²+(0). Therefore knowing x determines y in this dataset. The horizontal symmetry makes positive and negative centred products cancel. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
22. Four points around x=-1, two different heights
Try the scatter data. Consider the original fictional paired points (-3,7), (1,7), (-5,19), (3,19). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -1, but the y-values are not all equal.
Exact calculation. The means are x̄=-1 and ȳ=13. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 40 for x and 144 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-1))²+(3). Therefore knowing x determines y in this dataset. The scatter diagram is U-shaped, so a straight-line model can miss the relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
23. Four points around x=0, two different heights
Try the scatter data. Consider the original fictional paired points (-3,4), (3,4), (-6,31), (6,31). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 0, but the y-values are not all equal.
Exact calculation. The means are x̄=0 and ȳ=17.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 90 for x and 729 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(0))²+(-5). Therefore knowing x determines y in this dataset. A correlation of zero is well-defined here because both variables have nonzero spread. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
24. Four points around x=1, two different heights
Try the scatter data. Consider the original fictional paired points (-3,14), (5,14), (-7,62), (9,62). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 1, but the y-values are not all equal.
Exact calculation. The means are x̄=1 and ȳ=38. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 160 for x and 2304 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(1))²+(-2). Therefore knowing x determines y in this dataset. Knowing x determines y exactly under the quadratic rule; the variables are dependent. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
25. Four points around x=2, two different heights
Try the scatter data. Consider the original fictional paired points (-3,26), (7,26), (-4,37), (8,37). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 2, but the y-values are not all equal.
Exact calculation. The means are x̄=2 and ȳ=31.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 122 for x and 121 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(2))²+(1). Therefore knowing x determines y in this dataset. Adding a constant to every x or y value does not create linear correlation in this symmetric dataset. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
26. Four points around x=3, two different heights
Try the scatter data. Consider the original fictional paired points (2,5), (4,5), (0,13), (6,13). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 3, but the y-values are not all equal.
Exact calculation. The means are x̄=3 and ȳ=9. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 20 for x and 64 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(3))²+(4). Therefore knowing x determines y in this dataset. The two high outer points and two low inner points create a visible curved pattern. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
27. Four points around x=4, two different heights
Try the scatter data. Consider the original fictional paired points (2,0), (6,0), (-1,21), (9,21). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 4, but the y-values are not all equal.
Exact calculation. The means are x̄=4 and ȳ=10.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 58 for x and 441 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(4))²+(-4). Therefore knowing x determines y in this dataset. Pearson r summarises linear co-movement, not an arbitrary nonlinear relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
28. Four points around x=-4, two different heights
Try the scatter data. Consider the original fictional paired points (-7,8), (-1,8), (-11,48), (3,48). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -4, but the y-values are not all equal.
Exact calculation. The means are x̄=-4 and ȳ=28. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 116 for x and 1600 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-4))²+(-1). Therefore knowing x determines y in this dataset. This is not a claim that every real dataset with r≈0 has a U shape; it is a counterexample to a false general implication. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
29. Four points around x=-3, two different heights
Try the scatter data. Consider the original fictional paired points (-7,18), (1,18), (-8,27), (2,27). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -3, but the y-values are not all equal.
Exact calculation. The means are x̄=-3 and ȳ=22.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 82 for x and 81 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-3))²+(2). Therefore knowing x determines y in this dataset. A student should inspect the scatter plot rather than interpreting a coefficient without geometry. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
30. Four points around x=-2, two different heights
Try the scatter data. Consider the original fictional paired points (-7,30), (3,30), (-9,54), (5,54). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -2, but the y-values are not all equal.
Exact calculation. The means are x̄=-2 and ȳ=42. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 148 for x and 576 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-2))²+(5). Therefore knowing x determines y in this dataset. It is possible for two datasets with the same r=0 to have very different underlying relationships. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
31. Four points around x=-1, two different heights
Try the scatter data. Consider the original fictional paired points (-2,-2), (0,-2), (-5,13), (3,13). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around -1, but the y-values are not all equal.
Exact calculation. The means are x̄=-1 and ȳ=5.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 34 for x and 225 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(-1))²+(-3). Therefore knowing x determines y in this dataset. The horizontal symmetry makes positive and negative centred products cancel. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
32. Four points around x=0, two different heights
Try the scatter data. Consider the original fictional paired points (-2,4), (2,4), (-6,36), (6,36). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 0, but the y-values are not all equal.
Exact calculation. The means are x̄=0 and ȳ=20. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 80 for x and 1024 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(0))²+(0). Therefore knowing x determines y in this dataset. The scatter diagram is U-shaped, so a straight-line model can miss the relationship. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
33. Four points around x=1, two different heights
Try the scatter data. Consider the original fictional paired points (-2,12), (4,12), (-3,19), (5,19). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 1, but the y-values are not all equal.
Exact calculation. The means are x̄=1 and ȳ=15.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 50 for x and 49 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(1))²+(3). Therefore knowing x determines y in this dataset. A correlation of zero is well-defined here because both variables have nonzero spread. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
34. Four points around x=2, two different heights
Try the scatter data. Consider the original fictional paired points (-2,11), (6,11), (-4,31), (8,31). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 2, but the y-values are not all equal.
Exact calculation. The means are x̄=2 and ȳ=21. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 104 for x and 400 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(2))²+(-5). Therefore knowing x determines y in this dataset. Knowing x determines y exactly under the quadratic rule; the variables are dependent. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
35. Four points around x=3, two different heights
Try the scatter data. Consider the original fictional paired points (-2,23), (8,23), (-5,62), (11,62). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 3, but the y-values are not all equal.
Exact calculation. The means are x̄=3 and ȳ=42.5. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 178 for x and 1521 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(3))²+(-2). Therefore knowing x determines y in this dataset. Adding a constant to every x or y value does not create linear correlation in this symmetric dataset. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
36. Four points around x=4, two different heights
Try the scatter data. Consider the original fictional paired points (3,2), (5,2), (-1,26), (9,26). Find the Pearson correlation coefficient and describe the visible relationship before making any claim about independence. The x-values have been deliberately placed symmetrically around 4, but the y-values are not all equal.
Exact calculation. The means are x̄=4 and ȳ=14. The centred cross-product sum Σ(x−x̄)(y−ȳ) is 0, because the two positive and negative x deviations at each common y-value cancel. The sums of squared deviations are 52 for x and 576 for y, both positive, so the denominator of Pearson r is nonzero. Therefore r=0 exactly.
Interpretation and challenge. Each point obeys the deterministic nonlinear rule y=(x−(4))²+(1). Therefore knowing x determines y in this dataset. The two high outer points and two low inner points create a visible curved pattern. Draw the points and explain why the left half slopes down while the right half slopes up. The cancellation justifies “no linear correlation”, not “no relationship” or “statistical independence”.
Eight short checks with answers
- If X is uniform on {−1,0,1} and Y=X², then E(X)=0, E(Y)=2/3 and Cov(X,Y)=0.
- In that model Var(X)=2/3 and Var(Y)=2/9, so correlation is zero and defined.
- The variables are dependent because P(Y=0|X=0)=1 whereas P(Y=0)=1/3.
- The pairs (−2,4),(−1,1),(1,1),(2,4) have sample Pearson r=0.
- All four pairs still satisfy y=x², a deterministic curved relationship.
- If y is constant for all observations, Pearson r is undefined rather than zero.
- Independence implies zero covariance when finite variances exist, but zero covariance does not generally imply independence.
- Correlation alone does not prove causation or rule out nonlinear patterns.
A five-day learning sequence
Day one: plot several scatter diagrams and identify increasing straight-line, decreasing straight-line and curved shapes. Day two: calculate x and y means and the centred cross-product sum for symmetric four-point data. Day three: compare dependent U-shaped examples with independently generated discrete pairs having the same zero correlation. Day four: test boundary cases where one variable is constant and correlation is undefined. Day five: interpret r values in short written conclusions while refusing to infer causation or independence from the coefficient alone.
In a three-learner small-group tutorial, one student can sketch the points, another can calculate Pearson r and the third can challenge the interpretation. Rotate the roles. The strongest answer explains both why the linear covariance cancels and why the variables remain related, rather than simply presenting a zero from the calculator.
Frequently asked questions
Does r=0 mean the variables are independent?
No. Zero Pearson correlation means no linear covariance; nonlinear dependence can still exist.
Can y=x² give zero correlation with x?
Yes. For a symmetric distribution of x around zero with suitable nonzero variation in x², the positive and negative linear contributions cancel.
Does independent always mean zero covariance?
For independent variables with finite second moments, yes. The converse is not generally valid.
Can r be zero for a U-shaped scatter plot?
Yes. A symmetric U shape can cancel linear co-movement even when the curved structure is obvious.
What if one variable never changes?
Pearson correlation is undefined because one of the standard deviations is zero.
Does r=1 prove one variable causes the other?
No. Even a perfect linear association does not by itself establish a causal mechanism.
Does H2 Mathematics assess correlation hypothesis tests?
No. The published 2027 syllabus excludes correlation and regression hypothesis tests, while including interpreting Pearson r and scatter diagrams.
Why look at the scatter plot if r is known?
The plot reveals curves, clusters and outliers that cannot be described adequately by one linear summary statistic.
Continue with the JC2 H2 series
- Composite Functions: Hidden Domain After a Square Root
- Implicit Differentiation: Tangent and Normal to a Circle
- Without Replacement: Dependent Draws
- Correlation r = 0 Does Not Prove Independence: current guide
Related subject-owner content includes Bukit Timah H2 Correlation and Linear Regression, Which Regression Line Predicts Correctly?, and the Bukit Timah JC2 tuition hub. The immutable Clementi small-group Mathematics tutorials reference illustrates a different-level teaching format.
Official syllabus: SEAB H2 Mathematics 9758, 2027 examination, Correlation and Linear Regression §6.6. Every dataset above is an original fictional mathematical construction, not an examination paper or observed tuition result. Interpret r as a measure of linear relationship, not as proof of independence or causation.