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Suppose the average monthly return is $\mu$, the monthly standard deviation is $\sigma$ and denote the autocorrelation of monthly returns by

$corr(r_i,r_{i+h}) = \rho(h)$

Prove that, when $\sigma$ is small,

$\displaystyle \sigma_{year} = \sigma(1+\mu)^{11}\sqrt{12+2\sum_{i=1}^{11}(12-i)\rho(i)}$

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up vote 1 down vote accepted

To compute the yearly return, you have to compound the monthly returns:

$$1+r_{year}=\prod_{i=1}^{12}(1+r_i)=\prod_{i=1}^{12}(1+\mu+\sigma \epsilon_i)\approx(1+\mu)^{12}+\sigma (1+\mu)^{11}\Big(\sum_{i=1}^{12}\epsilon_i\Big)\; .$$

So the variance of that is

$$\sigma_{year}^2=\mbox{Var}(1+r_{year})=\Big(\sigma (1+\mu)^{11}\Big)^2\mbox{Var}\Big(\sum_{i=1}^{12}\epsilon_i\Big) \; .$$

The variance of a sum of random variables can be computed as

$$\mbox{Var}\Big(\sum_{i=1}^{12}\epsilon_i\Big) = \sum_{i=1}^{12}\mbox{Var}(\epsilon_i)+2\sum_{i<j}\mbox{Cov}(\epsilon_i,\epsilon_j)=12+2\sum_{i<j}\mbox{corr}(r_i,r_j)=12+2\sum_{i<j}\rho(|j-i|)=12+2\sum_{i=1}^{11}(12-i)\rho(i) \; .$$

Combining all results we indeed get the sought after formula.

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Perfect! 3ks... – Eastsun Sep 13 '12 at 12:05

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