# Annualization of discrete returns

There is a well known approach to annualize volatility of log-returns for a given frequency. Let $$P(t)$$ a price process and define a log return $$r_l(t)$$ as $$r_l(t) = \ln \left( \frac{P(t)}{P(t-1)} \right).$$

An aggregate return over $$n$$ periods is \begin{aligned} r_l^A(t) &= \ln \left( \frac{P(t)}{P(t-n)} \right) \\ &= \sum_{t-n+1}^t r_l(i). \end{aligned}

Let $$\sigma_l^2$$ denote variance of log-returns. Then, variance of an aggregate return is \begin{aligned} Var[r_l^A] &= Var \left[ \sum_{i=t-n+1}^t r_l(i) \right] \\ &= \left( n + 2\sum^{t-1}_{i=1} (n-i) \rho(i) \right) \times \sigma^2, \end{aligned} where $$\rho(i)$$ is an autocorrelation of order $$i$$.

If observed series has autocorrelations zero for all lags than this simplifies to $$Var[r_l^A] = n \times \sigma^2.$$

The formula is, however, not valid for discrete returns. One could argue that it is a good approximation for higher frequencies since log-returns are close to discrete ones, but how would annualize volatility of discrete returns for, say, weekly or monthly data?

## Normal log-returns case

For a case when log-returns are normally distributed, we can derive a closed form solution for a variance of aggregated dicrete returns. In the below I assume that all autocorrelations are zero to simplify formulas and that $$E[r_l(t)] = \mu$$.

(Aggregated) Discrete return is related to a log-return via $$r_d^A(t) = \exp (r_l^A(t)) - 1,$$ which means that $$r_d^A(t)$$ is log-normally distributed with variance $$Var[r_d^A(t)] = \exp \left( n Var[r_l^A(t)] - 1 \right) \times \exp \left( 2n\mu + nVar[r_l^A(t)] \right).$$

This is a monotonically increasing function in a mean of log-returns, $$\mu$$, as well as in their variance, $$Var[r_l^A(t)]$$. Below are two figures that illustrate that the divergence between the two volatilities (took square root of the above variances) increase as $$\mu$$ and $$Var[r_l^A(t)]$$ increase (number of aggregation periods is $$12$$ for monthly data).

• Hi: This is not a complete answer by any means but I can provide an outline. What you provided assumes that log returns are normally distributed with correlation $\rho_k$ and variance $\sigma$.. There should be a formula somewhere that converts this to what the discrete return would be. It's lognormal with mean $\mu_d$ and var $\sigma^2_d$ but I forget what they are. Also the correlations $\rho_k$ will change. But, once you have all those things, it should be okay to use the same relationsip that you have except put in those things instead of $\sigma$ and $\rho_k$. Sep 19 '21 at 13:32
• NOTE: I didn't read the link that you provided as carefully as I should have so you should confirm that the formula for $\sigma_{a}$ does not depend on the normality assumption. I don't think it does but my comment above obviously depends on that being the case. Sep 19 '21 at 13:36
• Hi @markleeds, the formulas above are independent of a distribution. I edited the question to reflect that. However, assuming normality for log-returns allows us to study the difference between aggregated volatilities, I added analysis on that as well. Sep 19 '21 at 18:31
• Hi tosik: I still think you need to do the following: Assuming that an rv, $log(X)$ is normal with mean $\mu$ and variance $\sigma^2$ and autocorrelation $\rho_k$, then what is the distribution of $exp(log(X) = X$. In other words, $X$ is lognormal and you need its parameters given that you have the parameters of the normal distribution of $log(X)$. If you get that distribution, then its straightforward to just plug the variance and the correlation into the formula that you already have written out for $Var(r^{A}_{l})$. Does that make sense ? It must be around but I'm not sure where it is. Sep 20 '21 at 2:16

• @tosik: the issue comes down to the following: Assume one has a random variable, $log(X)$ that is normally distributed with autocorrelation $\rho_k$. Then, if one takes $exp(log(X) = X$, what is the resulting autocorrelation of $X$. If that answer can be found, you're done. Unfortunately, I don't know the answer and can't find it on the net . If it can be derived, I don't know how to derive it either. Sep 21 '21 at 18:19
• @tosik yes the i.i.d. assumption is quite restrictive however the solution provided by the author can still be considered superior to the usual approach of of multiplying by e.g. $252$ or $\sqrt{252}$ Dec 16 '21 at 0:08