# 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.

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?

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: $$\sigma_4=10\%\sqrt{4}\\ =20\%$$

• Why can we assume that $\mu=0$? – Idonknow Oct 17 '19 at 9:45
• Because the problem only focus on the volatility. There is no randomness in the drift, $\text{Var}(\mu) = 0$ – Hydraxize Oct 17 '19 at 9:54
• So continuously compounding implies that $\mu = 0?$ – Idonknow Oct 17 '19 at 12:41
• No, $\mu$ is a constant hence $\text{Var}(\mu)=0$. I took $\mu=0$ to simplify the equations but you could have taken any value it would have changed the result of the time scaling equation. The continuously compounding assumption is used when I do the approximation $\text{ln}(1+x)\approx x$ – Hydraxize Oct 17 '19 at 16:59

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.

• For my first question, why can't we assume that the stock price follow a geometric brownian motion satisfying $$dS_t = S_t \mu dt + S_t\sigma dW_t?$$ – Idonknow Oct 17 '19 at 9:47
• As it relates to variance, you could. Conceptually, the drift term won't add anything to the variance. Formally, it will fall out for the reason I mentioned above. – Chris Oct 17 '19 at 15:17

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 $$\sqrt T$$.)

• For my first question, whenever I encounter similar kind of question, I can assume that the stock returns follow brownian motion? – Idonknow Oct 17 '19 at 9:47
• Not sure about that one. I think you need to read between the lines: if you need some assumption on the distribution in order to answer the question, and you are not told what to assume, this is the first thing that should come to mind. In this case, it is sufficient that yearly returns are independent and identically distributed -- somewhat weaker than Brownian motion, and not really a bad assumption. In this case, if the return has variance $\sigma^2$ in each year, and each year's return is i.i.d., $T$ years give cumulative variance $T\sigma^2$. – Drew Oct 19 '19 at 16:06
• BTW, when I say "not really a bad assumption", and I am speaking specifically of yearly returns. I.i.d. daily returns is not nearly so realistic. (If you have to make an assumption to answer the question, however, you do what you can!) – Drew Oct 19 '19 at 16:10