How to convert annualized Standard Deviation of innovation of AR1 process to monthly? I want to calculate the standard deviation of innovation of AR1 process from its annual value to its corresponding monthly value. Say, I have
$x_t=\rho x_{t-1} +\epsilon_t$ where $\epsilon \sim N(0,\sigma^2)$ 
 and that  x is annual data. Now to compute the monthly counterpart of this process, I need to know the monthly values for $\rho$ and $\sigma$ to approximate the process. I know how to get $\rho$, but I am not sure how to get the other one, i.e $\sigma$. I searched quickly online, finance people scale the number by root 12, but I am not sure if that is correct.
Any help on this is much appreciated.
 A: Let $y_i$ be the monthly process, where $i = 1, 2, ...$ are the months. Let $n=12$ to convert months to years, so that $y_{n t} = x_t$.
Suppose ${y_i}$ is AR(1), with $y_i = \alpha y_{i-1} + \eta_i$, where $\eta \sim N(0,\gamma^2)$. Then, 
$$ y_{i} = \alpha^n y_{i-n} + \sum_{j=0}^{n-1} \alpha^{j} \eta_{i-j} $$
Then we can express $\epsilon_t$ as,
$$\epsilon_t = \sum_{j=0}^{n-1} \alpha^{j} \eta_{nt-j}$$
${\eta_i}$ are iid $N(0,\gamma^2)$, so $\epsilon_t$ is normally distributed with mean zero and variance,
$$ Var(\epsilon_t) = \sum_{j=0}^{n-1} \alpha^{2j} \gamma^2 $$
Since $Var(\epsilon_t) = \sigma^2$, then we have that the variance of the monthly innovations is,
$$\gamma^2 = \frac{\sigma^2}{\sum_{j=0}^{n-1} \alpha^{2j}}$$
Update:
Assuming you estimate $\hat\rho$ and $\hat\sigma^2$ from the annual AR(1), you can get $\hat\alpha = \hat\rho^{1/n}$, and then compute the estimate of the monthly innovation variance as,
$$\hat\gamma^2 = \frac{\hat\sigma^2}{\sum_{j=0}^{n-1}\hat\alpha^{2j}}$$.
Note that if we have a random walk for the annual process, $x_t = x_{t-1} + \epsilon_t$, then we also have a random walk for the monthly process, and so $\hat \alpha = 1$, and $\hat\gamma^2 = \hat\sigma^2 / n$. So the monthly vol becomes the annual vol scaled by square root of 12.
