Currently, I am confused about the calculation of realised daily volatility. Assume I have daily returns, for example, FTSE, then I need to estimate the daily realised volatility. I read some materials and I get the idea:

$v = 100 * \sqrt{\frac{252}{n}\sum_{i = 1}^n R_t^2}$,

however, some other materials explain it should be $\frac{252}{n-1}$ in the equation, which one is right?

Then it comes to why $v$ in the equation could be treated as the realised volatility. Here is my understanding, if we want to estimate the realised volatility on day $N$, we use the standard deviation of returns $R_{N - n}, R_{N - n + 1}, \dots, R_N $ multiply $\sqrt{252}$ as an approximation, if that true? If it is true, then how to decide the size of $n$, $n= 10, 20 \dots$.

Besides this method, any other method possible? Of course, except the estimation from GARCH family model.



2 Answers 2


NN Taleb has some discussion of this in his book Dynamic Hedging. You'll find a lot of criticism of the book out in the aether, and there are certainly a good number of typos, but it is probably the least academic and most experience-based resource out there, and certainly worth considering. Augen is another big experience-based proponent of volatility (e.g. "The Volatility Edge in Options Trading").

Putting the two together, with what you have given: your equation is simply calculating the simple moving average of the root-mean-square deviation (RMSD) from zero of returns, and expressing it as annualized volatility (based on the assumption of a Gaussian distribution). Firstly, it's not clear if you're using simple returns or log-returns. Both Taleb and Augen urge the use of log returns. Augen uses the RMSD from the average, Taleb advocates just using zero as you have. Augen uses the simple average as you have, Taleb also suggests an exponential moving average.

All of the above are variations on the theme of time-averaging. In contrast, Taleb also suggests the Garman-Klass estimator, which uses only a single day's common price details: Open, High, Low, Close: $$ \sigma_{GK} = \left\{\frac{1}{2}\left[\ln\left(\frac{H}{L}\right)\right]^2 - (2\ln2 - 1)\left[\ln\left(\frac{C}{O}\right)\right]^2\right\} $$ Note that this equation is different to his; it's the one I use that I believe corrects the typos/mistakes.

  • $\begingroup$ These estimators are only consistent for GBMS. and they aren't tradable, while you can trade the classical estimator. It doesnt make sense to use them. $\endgroup$
    – Drew
    May 18, 2015 at 21:07
  • $\begingroup$ Thanks, @Drew. But the situation for me is that I can not get the high frequency data.:-( $\endgroup$
    – Fly_back
    May 18, 2015 at 21:15
  • $\begingroup$ @Drew, The moving average estimates do seem to correspond better to the Black-Scholes estimates of implied volatility, if that's what you mean, but we have found trade-able predictive value in comparisons of the GK and historical estimates, akin to Taleb's discussion of Parkinson estimate (which I didn't mention) vs historical estimates. $\endgroup$
    – GoneAsync
    May 18, 2015 at 22:47
  • $\begingroup$ that doesnt make sense. i mean an estimate you can replicate as a stochastic integral, not replicate roughly. Example delta hedging a parabola, gives you the realized quadratic variation $\endgroup$
    – Drew
    May 19, 2015 at 15:05
  • 1
    $\begingroup$ Not sure -- the difference may be in our semantics of "tradable". The statement may not hold under theory, but we're operating empirically in our operation; by "tradable" I simply mean "makes money". $\endgroup$
    – GoneAsync
    May 20, 2015 at 1:25

The factor $\frac{n}{n-1}$ (Bessel's correction) is used when estimating sample variance. This is because using $n$ in the denominator yields a biased estimator of the variance. That being said, if one assumes that the mean is $0$ (not an unusual assumption), then one doesn't lose a degree of freedom in estimating sample mean, so Bessel's correction isn't necessary.

Also, while the correction yields an unbiased estimator of variance, it does not yield an unbiased estimator of standard deviation (volatility), so the point is a bit moot; the difference is small anyway.

As for the second question, the realized volatility is estimated over a period of time. So, if $n = 10$, you're estimating RV over a ten day period. This is not the same as estimating the RV on a single day. Multiplying it by $\sqrt{252}$ is simply to transform the estimated RV to an annualized RV.

As far as I know, your first equation is the most common way of estimating RV. Though I do remember Tauchen was working on using Laplace transforms on high-frequency data, but I'm not sure that's what you're looking for.

  • $\begingroup$ Thanks @ocstl. So if $n = 10$, it is just a ten days' realised volatility and it is impossible to estimate the daily volatility with only the daily returns. it that true? well, what I need is daily volatility, so I need 5 min or 1 hour returns to estimate the daily volatility? $\endgroup$
    – Fly_back
    May 15, 2015 at 18:09
  • $\begingroup$ If you use $n = 1$, that would yield the day's realized volatility (assuming of course a mean of $0$, since otherwise $\sigma = 0$. Expect a time series to be quite jumpy. And I'm not sure it's entirely useful. It really depends on what you're planning to use it for. $\endgroup$
    – ocstl
    May 15, 2015 at 18:57
  • $\begingroup$ if $n = 1$, then the return is the RV as well. :-(. I am using it to compare the simulated conditional variance of GARCH model. In other words, I want to get absolute error between RV and simulated ones and then get a conclusion about which GARCH model is better to describe the volatility. $\endgroup$
    – Fly_back
    May 15, 2015 at 19:10
  • $\begingroup$ Indeed it is, but is it a problem? If you're trying to compare how best the models fit the data, use an appropriate loss function to estimate your parameters, and you're done. As for intraday vol estimation, I believe Andersen & Bollerslev have a number of good papers on this. $\endgroup$
    – ocstl
    May 15, 2015 at 21:04

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