Variance of the average of two correlated readings

Question

Two sensor readings X and Y each have variance 8, with covariance \mathrm{Cov}(X, Y) = 2. You report their average, M = \frac{X + Y}{2}. What is the variance of the average?

Accepted Answer

Handle this in two moves: first the variance of the sum X + Y, then the effect of dividing by 2. Variance of the sum. With a covariance present, \mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\,\mathrm{Cov}(X, Y) = 8 + 8 + 2(2) = 20. Divide by two. Dividing a quantity by 2 divides its variance by 2^2 = 4 (the multiplier gets squared): \mathrm{Var}(M) = \frac{1}{4}\,\mathrm{Var}(X + Y) = \frac{20}{4} = 5. The standard deviation is \sqrt{5} \approx 2.24, noticeably smaller than a single reading's spread of \sqrt{8} \approx 2.83. Averaging reduces noise, though the positive covarianc…

Variance of the average of two correlated readings

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