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):

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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?

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  • $\begingroup$ 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. $\endgroup$ – kurtosis Aug 6 at 23:37
  • $\begingroup$ 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 ? $\endgroup$ – TmSmth Aug 7 at 13:15
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    $\begingroup$ 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). $\endgroup$ – noob2 Aug 7 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.

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  • $\begingroup$ i'm not sure to understand the link with my question $\endgroup$ – TmSmth Aug 7 at 13:18

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