Suppose I have a sequence of monthly returns of a stock, $r_1,r_2,\ldots$. Suppose further that this is an i.i.d. sequence with with finite second moments.

In every paper, report, lecture note etc. the annualized volatility of the return of this stock is given as $\sigma(r_1)\sqrt{12}$.

On the other hand, if I annualize the monthly returns first, that is if I consider $(1+r_1)^{12}-1,(1+r_2)^{12}-1,\ldots$, then I get $\sigma((1+r_1)^{12}-1) \approx 12\sigma(r_1)$ since $(1+x)^{12}-1 \approx 12x$.

My question is what is wrong with what I am doing? Is it only a matter of convention that people use the first formula to report annualized volatility?

  • $\begingroup$ @noob2 Since I assumed the returns to be an i.i.d. sequence, $\sigma(r_1) = \sigma(r_2) = \ldots$ $\endgroup$ – Calculon Oct 12 '16 at 15:27

It makes no difference whether you work with annualized numbers or not.

If you work with monthly logarithmic returns $\{r_1,r_2,\cdots,r_{12}\}$ then the return for the year is $R=r_1+r_2+\cdots+r_{12}$. Assuming only that the returns are i.i.d and the standard deviation $\sigma$ exists, then the standard deviation of the annual return $R$ is $\sqrt {12} \sigma$.

If you prefer to work with annualized returns, then you are looking at $\{12 r_1,12 r_2,\cdots,12r_{12}\}$. The return for the full year is $\frac{12r_1+12r_2+\cdots+12r_{12}}{12}$ which is the identical expression as before and its volatility is again $\sqrt {12} \sigma$.


Actually what you are referring as a conventions comes from an assumption that the returns are driven by a normal distribution. If you consider a stochastic variable (time-series of a random variable) that is normally distributed you can demonstrate that the variance of the distribution grows linearly with time. It means that if you go from a 1 day return time-series to a two-days return time series, the variance of the later is going to be twice as the former.

$\sigma^2_{annual} = 12 \sigma^2_{month} $

However, the standard deviation (which is simplest estimation of the volatility) of the later is going to be:

$\sigma_{annual} = \sqrt{12} \sigma_{month}$

The definition of standard deviation, as being the square-root of the variance, is what makes financial industry, as a whole, to consider the square root of the time (compared to the daily volatility, for instance) as the correct way to convert to an annual volatility.

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    $\begingroup$ I understand that the underlying assumption dictates the square-root formula. But this is still a convention, right? I mean people could have reported annualized volatility like I do from the beginning and it would have been fine I guess. It is just in a different unit it seems. $\endgroup$ – Calculon Oct 12 '16 at 15:44
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    $\begingroup$ @guga Actually this has nothing to do with the normal distribution but everything to do with the iid assumption. $\endgroup$ – Quantuple Oct 12 '16 at 16:02
  • $\begingroup$ yes, you are right. "....iid assumption is important in the classical form of the central limit theorem, which states that the probability distribution of the sum (or average) of i.i.d. variables with finite variance approaches a normal distribution." If you compoung, or sum, those iid distributions over time you end up getting a normal distribution anyway. $\endgroup$ – Guga Oct 12 '16 at 16:05
  • $\begingroup$ No that's not the point. $\endgroup$ – Quantuple Oct 12 '16 at 22:27

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