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