How to calculate standard deviation of continuously compounded four-year stock returns?

Question

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.

1. 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.
2. For second bolded sentence, how does it explain that the variance of returns grows linearly with the compounding period?

Answer 1

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 2

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 3

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 $\sqrt T$.)