Apply the CLT after estimating the means of two variables from sample data.
The idea
When the sample means of two variables X and Y come from the same n independent pairs of observations, the CLT applies jointly. For large n the pair of sample means is approximately bivariate normal, centred on the true means, with the population covariance matrix divided by n. So each mean has a variance equal to that variable's variance divided by n, and the two means have covariance σXY ÷ n. Positive correlation makes the difference of the means more precise, as its variance shows: Variance = (σX² + σY² − 2σXY) ÷ n. For the sum of the means, the covariance enters with a plus sign: Variance = (σX² + σY² + 2σXY) ÷ n. The usual mistake is to treat the two sample means as independent and ignore the covariance term or its sign.
Formulas
Variance of the difference between two sample means from paired data
Var(x̄ − ȳ) = (σX² + σY² − 2 × σXY) ÷ n
- σX
- the standard deviation of X
- σY
- the standard deviation of Y
- σXY
- the covariance of X and Y
- n
- the number of pairs
The means come from the same n independent pairs. For the sum of the means the covariance enters with a plus sign; ignoring it treats the means as independent.
A practice question
The CLT is applied to the sample means of two random variables X and Y estimated from the same n independent pairs of observations. For large n, the pair (x-bar, y-bar) is approximately:
- bivariate normal, with covariance matrix equal to the population covariance matrix divided by the square root of n.Why not: The covariances of means scale with 1/n; only the standard deviations scale with 1/sqrt(n).
- bivariate normal, centred on (mu_X, mu_Y), with covariance matrix equal to the population covariance matrix of (X, Y) divided by n. - correct
- two independent normals, whatever the correlation between X and Y.Why not: Sample means inherit the dependence between X and Y; their covariance is sigma_XY/n.
- bivariate normal only if the original variables X and Y are jointly normal.Why not: The joint CLT does not require normality of the data; finite second moments are enough.
The multivariate CLT: sqrt(n) x ((x-bar, y-bar) - (mu_X, mu_Y)) tends to a bivariate normal with the covariance matrix of (X, Y). In finite samples the means have variances sigma²/n and covariance sigma_XY/n.
A harder one
Two funds have return variances of 4 and 9 (%²) and a correlation of 0.5. From n = 100 independent pairs of observations, what is the standard error of the difference of the sample means (x-bar minus y-bar), closest to?
- 0.361%Why not: Ignores the covariance, treating the sample means as independent.
- 0.436%Why not: Adds twice the covariance instead of subtracting it, as for a sum rather than a difference.
- 0.070%Why not: Reports the variance of the difference, not its standard deviation.
- 0.265% - correct
Covariance = 0.5 x 2 x 3 = 3.0. Var(x-bar - y-bar) = (4 + 9 - 2 x 3.0)/100 = 0.070. Standard error = 0.265%.
The hardest tier
Two assets have equal return variances and a correlation of 0.8. Relative to the standard error of the difference of their sample means that would apply if the samples were independent, the correct standard error (using the joint CLT on paired observations) is what multiple?
- 0.20Why not: Reports the ratio of variances, not of standard errors; the square root is missing.
- 0.45 - correct
- 0.60Why not: Uses sqrt(1 - rho²), which belongs to residual variance in regression, not to the variance of a difference.
- 1.34Why not: Has the sign of the covariance term wrong, as for a sum of means.
Paired variance = 2 sigma² (1 - rho)/n, independent variance = 2 sigma²/n. The ratio of standard errors is sqrt(1 - 0.8) = 0.45. Positive correlation makes the difference of means more precise.
In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.
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