Sometimes traders approximate realized volatility to compare it to the annualized implied volatilities in options by multiplying the 1-day daily return (as a substitute for the daily volatility since the average return is assumed to be zero) by 16 since volatility is proportional to the square root of time and if there are 252 trading days in a year:


$\sqrt{252}=15.8745 \approx 16$

so, $16*\sigma_{daily}\approx\sigma_{annualized}$

The absolute value of the percent return on a given day is used in place of daily volatility to get a rough approximation of the realized volatility for that day. However, in "Volatility Trading" by Euan Sinclair he claims we should use 20 as the multiplier instead of 16. He writes the difference "is due to confusing the square root of average squared returns with daily returns". He arrives at this multiplier by the following equations:

$E[| R_{t} |]=\sqrt{2/\pi}\sigma$ (equation 1)

$\sigma=19.896(\frac{1}{N}\sum_{t=1}^{N}| R_t |)$ (equation 2)

so, $average move = 0.04986\sigma S \approx \frac{\sigma S}{20}$ (equation 3)

How is he coming to equation 1 and 2? Why isn't the first approach correct? Please explain in more detail.

  • $\begingroup$ Probably naive, while how is equation 3 derived? My understanding $averagemove = E[|R_t|] S$ thus the denominator 19.896 rather than 20.05 (1/0.04986)? $\endgroup$ Commented Jul 19, 2022 at 2:26
  • $\begingroup$ Can anyone explain me what is S in that formula? $\endgroup$ Commented Aug 1, 2022 at 20:01

1 Answer 1


Let's first discuss how author arrived at equation 1 and 2. It is based on the fact that if $X \sim N(0, \sigma^2)$, then $ Y = |X|$ follows half-normal distribution. In your question, $X$ is daily return (ie $R_t$). It can be shown that $$E(|X|) = \sqrt{\frac{2}{\pi}} \sigma \tag{equation 1}$$ this provides $$\sigma = E(|X|) \cdot\sqrt{\frac{\pi}{2}} $$ For sample data $$\hat{\sigma} = 1/N\sum_{I =1}^{N} |X_i| \cdot \sqrt{\frac{\pi}{2}} $$ where $\hat{\sigma}$ is estimator of daily volatility. Assuming 252 days in a year, this provides (assuming $X_i's$ are i.i.d.) $$\sigma_{annual} = \sqrt{252} \hat{\sigma} = 19.896 \cdot 1/N\sum_{I =1}^{N} |X_i| \tag{equation 2} $$

You also asked, which approach is more correct. For asymptotic large $N$ both approach are equivalent and will provide same estimator. Under the first approach volatility is defined as average of square root of return, whereas under the second approach volatility is simply average of absolute return. Notice, $$1/N \sum_{I=1}^{N} |X_i| \leq \sqrt{1/N \sum_{I=1}^{N} X_i^2}$$ I think this provide nice relationship between standard deviation and absolute standard deviation.

  • 1
    $\begingroup$ Note that $19.896=15.8745 \sqrt{\pi/2}$ $\endgroup$
    – Brian B
    Commented Aug 5, 2021 at 16:51
  • 2
    $\begingroup$ Further addition as the question popped up in the feed again; the very last line ("MAD of a Gaussian is about $\sqrt{2/\pi}$ = 80% of its SD") was also mentioned in this brief experimental note of Goldstein and Taleb, where they observed that practicioners tend to think about MAD but use it like SD (thus underestimating vols). $\endgroup$
    – KevinT
    Commented Aug 6, 2021 at 10:43
  • $\begingroup$ Is X a simple daily return, or a log return? $\endgroup$ Commented Aug 1, 2022 at 20:04
  • $\begingroup$ @DenisGorodetskiy You can approximate date return using log return. $\endgroup$
    – Neeraj
    Commented Aug 3, 2022 at 10:42

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