Why Log-Return Volatility Scales with the Square Root of Time
Summary
The document explains the standard time-scaling rule for continuously compounded stock returns. Under a Brownian-motion model with independent return increments, the variance accumulated over a horizon is the sum of the variances of its increments. Variance therefore grows in proportion to time, while standard deviation grows with the square root of time. Applying that rule to the stated annual volatility gives the four-year result described in the source.
The discussion distinguishes stock price from log return and sketches how a geometric Brownian price process leads to additive log returns. It also notes that the model is a conventional default when a problem supplies no other assumptions, not a claim that real stock returns truly follow Brownian motion. Independence and the model assumptions matter; dependence, changing volatility, or other departures could invalidate the simple scaling relationship.
Key ideas
- Independent log-return increments have variances that add across periods.
- Under the Brownian-motion assumption, return variance is proportional to the time horizon.
- Standard deviation consequently scales with the square root of elapsed time.
- The scaling rule depends on the model and independence assumptions, which may not hold in actual markets.
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Full text
# How to calculate standard deviation of continuously compounded four-year stock returns?
# How to calculate standard deviation of continuously compounded four-year stock returns?
Currently I am preparing for quant interview and I encounter the following question in Heard on the street.
> Question: If the standard deviation of continuously compounded annual stock returns is $10\%,$ what is the standard deviation of continuously compounded four-year stock returns?
Solution:
> Assuming continuously compounded returns follow an arithmetic Brownian motion, variance of returns grows linearly with the compounding period. This is because consecutive returns in a random walk are independent, and the variance of a sum of independent random variables is just the sum of variances. This means that the four-year $\sigma^2$ equals four times the one-year $\sigma^2.$ It follows that the four-year $\sigma$ is two times the one-year $\sigma.$ The answer is therefore $20\%.$
I have a few doubts on the solution.
- Why can we assume that the returns follow an arithmetic Brownian motion (ABM)? I think ABM satisfies the SDE $$dS_t = \mu dt+\sigma dW_t$$ where $S_t$ is stock returns and $W_t$ is Brownian motion.
- For second bolded sentence, how does it explain that the variance of returns grows linearly with the compounding period?
## Answer by Hydraxize (score 3)
https://quant.stackexchange.com/a/49234
Classic asset price model in the continuous-time limit using a Wiener process notation can be written as $$ dS_t=\mu S_tdt+\sigma S_t dX $$ where $S_t$ is the stock price (not the stock return) and $dX$ is an independent random variable with normal distribution. If we eliminate the drift ($\mu = 0$) and only focus on randomness as asked in your question we can simplify by $$ dS_t=\sigma S_t dX\\ \frac{dS_t}{S_t}=\sigma dX_t $$ Note the left side is now the stock return and is equivalent to your first equation. Let's do some math now:
\begin{eqnarray*} \frac{dS_t}{S_t}&=&\frac{S_t}{S_{t-1}}-1\\ \frac{S_t}{S_{t-1}}&=&1+\sigma dX_t\\ \ln\left(\frac{S_t}{S_{t-1}}\right) &=&\ln\left(1+\sigma dX_t\right)\\ V_t-V_{t-1} &=&\ln\left(1+\sigma dX_t\right) \text{ with }V_t=\ln S_t\\ V_t &=&V_{t-1}+\ln\left(1+\sigma dX_t\right) \end{eqnarray*} The following is the most important part and why the question assumes continuous-time. When $dX_t \rightarrow 0$ the last equation becomes:
$$ V_t \approx V_{t-1}+\sigma dX_t $$ This can be rewrite as $$ V_t = V_{t-1}+x_t \text{ where } x_t\sim \mathcal{N}(0,\sigma^2) $$ We can now have a relation between $V_0$ and $V_T$ at a certain time $T$ $$ V_T = V_0+\sum_{i=1}^T{x_i}\\ \text{Var}\left(V_T - V_0\right) = \text{Var}\left(\sum_{i=1}^T{x_i}\right) $$
Now you use your second bolded sentence. This random variable is independent from each other, i.e. $Cov(x_t,x_{t-1})=0$ so \begin{eqnarray*} \text{Var}\left(V_T - V_0\right)&=&\sum_{i=1}^T{\text{Var}\left(x_i\right)}\\ &=&T\sigma^2 \end{eqnarray*} Now $V_T - V_0$ is nothing else than $\ln\left(S_T/S_0\right)$ which is the log return of the stock $S_t$ over a period $T$ and $\text{Var}\left(V_T - V_0\right)$ is the variance of the log return over that same period or $\sigma_T^2$. We can now write: $$ \sigma_T^2=T\sigma^2\\ \sigma_T=\sqrt{T}\sigma $$ We now have the famous equation of time scaling volatility. Put in application to your question: $$ \sigma_4=10\%\sqrt{4}\\ =20\% $$
## Answer by Chris (score 2)
https://quant.stackexchange.com/a/49222
The first is something of a theoretical question. It's widely held/assumed that stocks follow a BM process, it appears as though the author is setting the table for the subsequent statement.
The second is an artifact of applying Ito's lemma...the $dW_tdt$ and $dtdt$ terms both equal 0, hence fall out, leaving only $dW_t^2$ = dt. Thus, the variance scales with the change in time (ie, linearly). A bit more detail here.
## Answer by Drew (score 1)
https://quant.stackexchange.com/a/49224
For the first question, it is the standard assumption to make for stock returns if no other information is given. That's not to say it's a great assumption, but there it is clearly the only one that can be justified in this context.
For the second part, independence of returns tells you that investment for T years has cumulative variance $T \sigma^2$ (when each year's investment has variance $\sigma^2$). Thus, the variance grows linearly in T. (The standard deviation grows at rate $\sqrt T$.)Shown 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.