Why Annualized Volatility Scales with the Square Root of Time
Summary
The question compares two ways of relating monthly and annual return volatility. It assigns each month one twelfth of that year’s annual return, so all twelve monthly observations within a year are identical. Under that assumption, monthly volatility is one twelfth of the volatility across annual returns. This appears to conflict with the familiar square-root-of-time rule for scaling volatility.
The response identifies the key distinction: the constructed monthly returns are perfectly correlated within each year, whereas square-root-of-time scaling assumes independent increments of a stochastic process. With independent returns of equal variance, aggregation over multiple periods increases standard deviation with the square root of the number of periods. The example therefore does not show a contradiction; it uses a dependence structure that violates the assumption behind the usual scaling rule. It is a conceptual explanation, not an empirical estimate, and the scaling rule may also require adjustment when returns are dependent or other model assumptions fail.
Key ideas
- Dividing each annual return equally across its months makes the monthly observations within a year identical.
- That construction yields monthly volatility equal to annual volatility divided by twelve.
- Square-root-of-time volatility scaling relies on independent increments with equal variance.
- Perfect correlation across monthly returns changes how their aggregate variance scales.
- Volatility scaling assumptions should be checked when returns are dependent.
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Full text
# Monthly returns annualized vs annual returns
# Monthly returns annualized vs annual returns
Lets say that I have a stock with annual returns, $a_i $ for year $i\in \left\{1,...n\right\}$ and monthly returns $m_{i,j}$ for month $j\in \left\{1,...12\right\}$. Lets define monthly returns to be equal each month within a year so $m_{i,j} = \frac{a_i}{12}$.
Let $\bar{a}$ and $\bar{m}$ denote the means of yearly and monthly returns respectively so $$\bar{a}=\frac{1}{n}\sum_{i=1}^na_i\quad\text{and}\quad{}\bar{m}=\frac{1}{12n}\sum_{i=1}^n\sum_{j=1}^{12} m_{i,j}=\frac{1}{12n}\sum_{i=1}^n\sum_{j=1}^{12}\frac{a_i}{12}=\frac{1}{12n}\sum_{i=1}^na_i=\frac{\bar{a}}{12}$$
Then the monthly volatlity of returns is: $$\begin{align*} \sigma_{m} &= \sqrt{\frac{1}{12n}\sum_{i=1}^n\sum_{j=1}^{12} \left(m_{i,j}-\bar{m} \right)^2}\\ &= \sqrt{\frac{1}{12n}\sum_{i=1}^n\sum_{j=1}^{12} \left(\frac{a_i}{12}-\frac{\bar{a}}{12}\right)^2 }\\ &= \sqrt{\frac{1}{12n}\left(\frac{1}{144}\right)\sum_{i=1}^n\sum_{j=1}^{12} \left(a_i-\bar{a}\right)^2 }\\ &= \sqrt{\frac{1}{12n}\left(\frac{1}{144}\right) \sum_{i=1}^n12\left(a_i-\bar{a}\right)^2 }\\ &= \sqrt{\frac{1}{n}\left(\frac{1}{144}\right) \sum_{i=1}^n\left(a_i-\bar{a}\right)^2 }\\ &= \frac{1}{12}\sqrt{\frac{1}{n} \sum_{i=1}^n\left(a_i-\bar{a}\right)^2 }\\ &= \frac{\sigma_a}{12} \end{align*}$$
So in other words $$\sigma_a = 12\sigma_m$$. This seems to contradict https://en.wikipedia.org/wiki/Volatility_(finance) where they say generalized volatility in $T$ time periods is $\sigma_T = \sigma \sqrt{T}$. What am I doing wrong?
## Answer by Randor (score 1)
https://quant.stackexchange.com/a/32486
i think you are assuming 100% correlation between returns, whereas for a stochastic process, increments are independentShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.