Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

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

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
Perfect! 3ks... –  Eastsun Sep 13 '12 at 12:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.