The floor on averaging correlated returns

Question

You average more and more returns, each with standard deviation $\sigma = 8$ , every pair correlated at $\rho = 0.25$ .

As the number of returns grows without bound, what value does the standard deviation of the average approach?

Accepted Answer

With independent data the spread of the average would head to zero as you add more. With correlated data it does not: a shared component survives no matter how many you average, putting a floor under the precision. Start from the correlated-average variance: 	ext{Var}(\bar{X}) = \frac{\sigma^2}{n}\Big[1 + (n-1)ho\Big] = \frac{\sigma^2}{n} + \frac{n-1}{n}ho\sigma^2. As n 	o \infty, the first term vanishes and the fraction (n-1)/n 	o 1, leaving 	ext{Var}(\bar{X}) 	o ho\sigma^2, \qquad 	ext{SD}(\bar{X}) 	o \sigma\sqrt{ho}. Plug in \sigma = 8, ho = 0.25: \sigma\sqrt{ho} = 8 	i…

The floor on averaging correlated returns

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