# Volatility estimation based on a 60 days range

In Hutchinson et al: A Nonparametric Approach to Pricing and Hedging Derivative Securities Via Learning Network (1994) paper (link), to estimate $$\sigma$$ for the Black-Scholes formula, it says (p. 881):

I'm not sure to understand. If $$s$$ is the standard deviation of the 60 last daily returns, it's the daily volatility based on a sample of 60 days. Why don't we multiply by $$\sqrt{252}$$ to have the annualized volatility ? I don't understand why he divides by $$\sqrt{60}$$.

This is indeed very strange, and is probably a typo in the paper.

It would be correct if $$s^2$$ is the sum of squares of the last 60 days returns, and $$s$$ is the square root of that. Then the division by $$\sqrt{60}$$ would give the daily vol. But if $$s$$ is the standard deviation, as they claim, then we would be doing the division twice and that would be wrong.

So I believe $$s$$ is not what they claim. Any other ideas?

• Agree that this is probably a mistake. There is good reason to look at 60-62 days in that it captures a whole quarter (and so avoids possible quarter-end effects). However, the math posted is indeed just wrong. Aug 6 '20 at 23:37
• If $s^{2}$ is the sum of square, and we take the square root and then we divide by $\sqrt{60}$, to have the daily vol, do we implicitly consider that the mean of the return is 0 ? Aug 7 '20 at 13:15
• Yes, that is right, and it is common to assume average stock returns are zero in computing volatility (esp. on a short term basis. 60 days is not enough to estimate the mean return properly). Aug 7 '20 at 13:36

Volatility over N periods is roughly proportional to sqrt(n). Using the annual number will give a different result that will make the monthly appear lower.

Because volatility is ergodic it doenst just increase linearly over time.

• i'm not sure to understand the link with my question Aug 7 '20 at 13:18