Annualizing Volatility Across Sampling Frequencies

# Will volatility smoothing effects exist for returns driven by geometric brownian motion?


# Will volatility smoothing effects exist for returns driven by geometric brownian motion?












Say I randomly simulate a one-year pathway of 252 prices, where the underlying price model is driven by geometric brownian motion.

where $t = (1 / 252)$, $mu = 5$% and annual $st.dev = 10%$%.

My understanding is that since the returns will be normally distributed, I can calculate the annual volatility by taking the standard deviation of the daily returns and scaling it by $√(252)$, and this should equal exactly 10%.

My question is: say I take the same vector of prices but now only measures prices at weekly intervals, and only calculate the weekly return instead of the daily return. Will the standard deviation of the weekly returns, scaled by $√(52)$, still equal 10%? Or will it be lower due to a smoothing effect from the lower frequency? If so, how can I prove that mathematically?

## Answer by David Addison (score 1, accepted)

https://quant.stackexchange.com/a/32764

Just a quibble, but you say that you "can calculate the annual volatility by taking the standard deviation of the daily returns and scaling it by $\sqrt{252}$, and this should equal exactly 10%." I think it should read that the "annualized standard deviation will tend towards 10%".

The "tends" is actually important, thus "central tendency". The central limit theorem (CLT) states that as sample size $N$ increases from a random sampling of i.i.d. variables, with sample mean ($\mu_s$) and variance ($\sigma^2_s$), that sampling will converge with the population mean ($\mu_p$) and variance ($\sigma^2_p$). This idea is classically demonstrated through the Lindeberg–Lévy CLT:

$\sqrt{n}((\frac{1}{n}\sum_{i=1}^nX_i)-\mu)\to $ distribution $N(0,\sigma^2)$

where:

$\sqrt{n}$ is sample size

$X_i = \{X_1,X_2...\}$ is a sequence of i.i.d. variables with with an expected value equal to $\mu_p$ and expected variance equal $\sigma_p^2$.

$N(0,\sigma^2)$ is the cumulative distribution function

So, actually, your daily sampling will give you the smoother estimate of your actual variation and will converge to "$\mu_p=5\%$ and annual $\sigma_p=10\%$" quicker than your weekly sampling. In either case, the expected values of sample mean and variance are the same; weekly sampling should not result in smoothing if your underlying distribution is, indeed, normal. Note: the "root time" rule works because the variance scales linearly with respect to time.

If, however, the underlying stochastic process is mean-reverting -- as may exist in stock prices -- your intuition about smoothing of longer time periods actually will bear out (i.e., annualized daily variance will over-state actual annual variance).