The spread of an average of mildly correlated readings

Question

You average n = 50 readings, each with standard deviation \sigma = 4. Every pair is correlated with common correlation \rho = 0.1. What is the standard deviation of the average?

Accepted Answer

Because the readings share a common component, the plain \sigma/\sqrt{n} rule overstates how much the averaging helps. Use the correlated-average formula: 	ext{Var}(\bar{X}) = \frac{\sigma^2}{n}\Big[1 + (n-1)ho\Big]. Plug in \sigma^2 = 16, n = 50, ho = 0.1: 	ext{Var}(\bar{X}) = \frac{16}{50}\Big[1 + 49 	imes 0.1\Big] = 0.32 	imes [1 + 4.9] = 0.32 	imes 5.9 = 1.888. So the standard deviation of the average is \sqrt{1.888} \approx 1.37. The independent case would give \sigma/\sqrt{n} = 4/\sqrt{50} \approx 0.57. Even a modest correlation of 0.1 more than doubles the true spread of the a…

The spread of an average of mildly correlated readings

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