# Annualized rolling volatility? [closed]

I have 600 days of closing prices of a stock. I want to calculate the annualized volatility for 6 day window. How do i do that?

If I calculate the std dev of the first 6 days, i get, say 1%. This is the daily volatility of the first 6 days. To annualize it, should i just multiply 1% with sqrt(252)? Or, is this 1% the daily volatility of 6 days? In that case i should multiply with sqrt(252/6)? Is this number 1% the daily volatility for 1 day, or is it the dayily volatility for 6 days?

I dont get it, I am missing a parameter. The period should also be involved in the calculation, right?

So to annualize 6 day volatility, i multiply with sqrt(252/6)? And when do i multiply with sqrt(252)?

UPDATE1: Ami44 writes that the correct procedure to annualize a 6 day window, is to multiply with sqrt (252/6). See Converting 30day annualized vol to 2day annualized vol

UPDATE2: in the answer below, ForeignVolatility says that I should multiply with sqrt (252). This is contradictory to "UPDATE1" above. So I am confused. Should I multiply with sqrt (252) or sqrt(252/6)? And, if I have a 30 day window, should I still multiply with sqrt (252), or should I use sqrt(252/30)? Great confusion. Some say Ba, and other say Bu.

• Ami44 writes: quant.stackexchange.com/questions/46765/… that the correct procedure is to get the annualized volatility for this 6 day window, by multiplying with sqrt (252/6) May 12 at 21:41

We work in annual units because $$T=1$$ means one year. This means that the time units must be converted to portions of a year. For example, in the case of daily observations, $$\Delta t = 1 / 252$$. Hence, in your example, we multiply by $$\sqrt{252}$$ because it's assumed that the variance is measured daily.
More generally, say you have $$n$$ returns observed with frequency $$\Delta t$$ arbitrary. Assume they are i.i.d. and follow a distribution $$N(0,\sigma^2\Delta t)$$. You then have $$\mathbb E\left[\frac 1n \sum_{t=1}^n r_t^2\right] = \frac 1n \sum_{t=1}^n\mathbb E\left[r_t^2\right] = \frac 1n \sum_{t=1}^n\sigma^2\Delta t = \sigma^2\Delta t.$$ What you described, the sdt dev of the first 6 days, corresponds to the square-root to an estimation of the LHS with $$n=6$$. The value you're looking for is $$\sigma^2$$. Hence, you need to multiply by $$1/\sqrt{\Delta t} = 1/(1/\sqrt{252}) = \sqrt{252}$$.
• @OrvarKorvar the number you multiply by depends on the frequency of the sampling ($\Delta t)$, not on the number of observations ($n$). This is because, by computing std dev, you are already taking the number of observations out of the equation. May 13 at 15:21