The Pearson correlation coefficient is covariance divided by the product of two standard deviations. It measures the direction and strength of linear association on a scale from −1 to 1. In portfolio work it lets you separate co-movement from the size and units of each asset's fluctuations.
“Pearson” specifies which correlation is being measured. It uses numerical values and their deviations from their means; it is not a rank correlation or a complete description of dependence.
The population quantity
For random variables with finite, strictly positive variances, define
where , , and are the population standard deviations. Thus Pearson correlation is the expected product of the two centred, standardised variables.
The Cauchy–Schwarz inequality gives . At , almost surely for some ; at , the slope is negative. These are exact affine relationships, not statements that the raw values must be identical or opposite. If either variable is constant, its variance is zero and correlation is undefined, rather than zero. Infinite-variance models also fall outside this definition.
Four return scenarios, fully specified
Suppose a hypothetical one-period model gives two assets the following four joint scenarios, each with probability . Returns are expressed in percentage points: a value of 1 means a 1% simple return.
| Scenario | (pp) | (pp) | (pp) | (pp) | Product of deviations (pp²) |
|---|---|---|---|---|---|
| A | 1 | 2 | −1.5 | −0.5 | 0.75 |
| B | 2 | 1 | −0.5 | −1.5 | 0.75 |
| C | 3 | 4 | 0.5 | 1.5 | 0.75 |
| D | 4 | 3 | 1.5 | 0.5 | 0.75 |
Both means are 2.5 pp. Weighting by the stated probabilities gives
Consequently,
Every scenario places the two assets on the same side of their respective means. Yet correlation is below 1: their centred fluctuations are not proportional. Positive correlation does not require a perfect straight line, and a coefficient of 0.6 does not mean “the returns agree 60% of the time.”
Original finite scenario model, not observed market data. The relevant origin for co-movement is the pair of means. All four attained outcomes are filled markers.
For a fixed equal-weight portfolio, the four returns are 1.5, 1.5, 3.5 and 3.5 pp. Its population variance is 1 pp², compared with 1.25 pp² for either asset. The same result follows from portfolio variance: . Correlation enters risk through covariance; the weights and individual volatilities still matter. Even perfect negative correlation produces a zero-variance combination only when the exposures offset the centred fluctuations exactly.
A sample uses the same centring, but answers another question
If those four rows instead represent four observed, aligned holding periods, their sample correlation is
Here the centred cross-sum is 3 and both square-sums are 5, giving . Using the usual divisor, sample covariance is 1 pp² and each sample variance is pp². Using consistently gives 0.75 and 1.25 instead. Either common divisor cancels in the correlation ratio; mixing conventions does not.
The matching answer does not turn a short history into a known population law. In the scenario model, probabilities define the population coefficient exactly. In the empirical reading, summarises four rows and may estimate an unknown under an appropriate sampling model. Alignment, horizon, selection and serial dependence affect what that estimate can support.
There is also a geometric reading: is the cosine of the angle between the two mean-centred observation vectors. Omitting centring gives the raw cosine for these rows, which is a different quantity. Adding a constant to either series leaves Pearson correlation unchanged; converting pp to decimal returns also leaves it unchanged. Multiplying one series by a negative number reverses its sign. Correlation is symmetric in and is not a regression slope.
What zero leaves unresolved
Take a separate dimensionless model: is equally likely to be −1, 0 or 1, and . Then , , and . Both variances are positive, so their Pearson correlation is zero. Nevertheless, observing determines exactly.
Independence implies zero correlation when finite positive variances exist; the converse needs additional structure. It does hold for a jointly Gaussian pair. Having two individually normal marginal distributions is insufficient, and a normality assumption is not required merely to define or calculate Pearson correlation.
Inspect a scatter plot alongside the coefficient. Curvature, an influential extreme observation or a change of regime can matter more than the single summary. Neither a large sample coefficient nor a positive population coefficient establishes causality, stable future diversification or protection in joint tail events. NIST's scatter-plot guidance illustrates why nonlinear patterns and outliers need their own inspection.
For the expectation and covariance calculations, Expectation, Variance and Covariance is available in full with a free signed-in account. Covariance and Correlation develops the population identities and perfect-correlation condition; the full lesson requires membership, with a public preview.
Sources
- Dimitri P. Bertsekas and John N. Tsitsiklis, Introduction to Probability, 2nd edition (Athena Scientific, 2008), §4.2, pp. 218–222 — covariance, correlation, affine equality cases and variance of sums.
- NIST Dataplot: Correlation — the centred sample formula and its interpretation as linear association.
- MIT OpenCourseWare, 1.010 Uncertainty in Engineering, Fall 2008, Brief Notes 8: Selected Distribution Models, multivariate normal properties, p. 5 — independence of uncorrelated components under joint normality.
- NIST/SEMATECH e-Handbook: Scatter Plot — linear and nonlinear patterns, outliers and the distinction between association and causality.