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Describe how covariance and correlation of two random variables relate to each other and to whether the variables are independent.

FRM® Part I · Quantitative Analysis · Multivariate Random Variables · FRM1.QA.4.e

The idea

Correlation is covariance rescaled by the two standard deviations: Corr = Cov(X, Y) ÷ (σX × σY). This removes the units and forces the value into the range from −1 to +1, so correlations can be compared across pairs while raw covariances cannot. A correlation of +1 or −1 means an exact linear relationship, and the covariance can never exceed σX × σY in absolute value. If X and Y are independent, E(XY) = E(X) × E(Y), so both covariance and correlation are zero. The reverse is false. Zero covariance only rules out a linear relationship, and variables can be strongly dependent in a nonlinear way, for example when Y moves with X² and X is symmetric around zero. Exams often test exactly this one-way implication.

Formulas

Correlation from covariance

Corr(X, Y) = Cov(X, Y) ÷ (σX × σY)

Cov(X, Y)
the covariance of X and Y
σX
the standard deviation of X
σY
the standard deviation of Y

Always between −1 and +1. A result outside that range means the inputs are inconsistent.

A practice question

From the question bank, with the reason each wrong option is wrong.

How does the correlation coefficient of X and Y relate to their covariance?

  1. Correlation equals the covariance divided by the product of the two standard deviations, so it is unit-free and lies between -1 and +1. - correct
  2. Correlation equals the covariance divided by the product of the two variances, so it lies between -1 and +1.Why not: Divides by the variances instead of the standard deviations; the result would not be bounded by 1 or unit-free.
  3. Correlation equals the covariance multiplied by the product of the two standard deviations, so it has the same units as covariance.Why not: Multiplies instead of divides, which makes the units worse and removes the normalisation.
  4. Correlation equals the covariance divided by the sum of the two standard deviations, so it lies between 0 and 1.Why not: Divides by the sum of the standard deviations; the normalisation uses their product and the range includes negative values.

Corr(X, Y) = Cov(X, Y) / (sigma_X sigma_Y). Standardising by the two standard deviations removes the units and bounds the result between -1 and +1 (by the Cauchy-Schwarz inequality).

A harder one

Two risk factors have Var(X) = 25, Var(Y) = 16 and Cov(X, Y) = 10. What is their correlation?

  1. 0.025Why not: Divides the covariance by the product of the variances (400) instead of the product of the standard deviations (20).
  2. 0.25Why not: Reports the squared correlation (R²), not the correlation.
  3. 0.50 - correct
  4. 0.40Why not: Divides the covariance by Var(X) only; this is the regression slope of Y on X, not the correlation.

sigma_X = 5 and sigma_Y = 4, so Corr = 10 / (5 x 4) = 0.50.

The hardest tier

X takes values -1, 0, 1 and Y takes values 0, 1. The joint pmf is: P(-1,0) = 0.05, P(-1,1) = 0.25, P(0,0) = 0.40, P(0,1) = 0, P(1,0) = 0.05, P(1,1) = 0.25. Which conclusion is correct?

  1. Cov(X, Y) = 0, so X and Y are independent.Why not: Treats zero covariance as proof of independence; the joint cells must factor into the product of the marginals, and here they do not.
  2. Cov(X, Y) = 0 but X and Y are not independent, because P(X = -1, Y = 1) differs from P(X = -1) P(Y = 1). - correct
  3. Cov(X, Y) is positive, because Y = 1 occurs whenever X is not 0.Why not: Ignores the symmetry: the positive and negative values of X pair with Y = 1 equally often, so the products cancel.
  4. Cov(X, Y) is negative, because Y is larger when X is more extreme.Why not: Confuses dependence on the size of X with linear co-movement; the covariance only sees linear association and the two sides cancel.

E[X] = 0 by symmetry and E[XY] = (-1)(0.25) + (1)(0.25) = 0, so Cov = 0. But P(X = -1) P(Y = 1) = 0.30 x 0.50 = 0.15, not the joint value 0.25. Zero covariance does not imply independence.

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