# Variance of annual return based on variance of monthly return

I'm trying to understand the whole variance/std error thing of a time series of financial returns, and I think I'm stuck. I have a series of monthly stock return data (let's call it $X$), which has expected value 1.00795, and variance 0.000228 (std. dev is 0.01512). I'm trying to calculate the worst case of the annual return (let's say expected value minus twice the standard error). Which way is the best way to do it?

A. Calculate it for a single month ( $\mu_X-2\cdot \sigma_X=0.977$ ), and multiply it by itself 12 times (=0.7630).

B. Assuming the months are independent, define $Y=X\cdot X\cdot ...\cdot X$ 12 times, find it's expected value $E[Y]=(E[X])^{12}$) and variance $\operatorname{var}[Y]=(\operatorname{var}[X]+(E[X])^2)^{12} - ((E[X]^2)^{12}$. The standard dev in this case is 0.0572, and the expected value minus twice the std. dev is 0.9853.

C. Multiply the monthly std. dev with $\sqrt{12}$ to get the annual one. use it to find the worst case annual value ($\mu - 2\cdot \sigma$). It comes out as 0.9949.

Which one is correct? What is the right way to calculate the expected annual value minus twice the std. dev if you know these properties only for the monthly data? (In general - if $Y=X\cdot X\cdot ...\cdot X$ 12 times and $\mu_X$, $\sigma_X$ are known, what is $\mu_Y-2\cdot \sigma_Y$ ?)

If you define the proportional return as $\Delta P/P = (P_{t+1}-P_t)/P_t$, where $P$ is the price, it's not uncommon with daily returns to simply multiply the proportional return by $250$ (number of working days in a year) and the standard deviation by $\sqrt{250}$ to annualize them. This corresponds to your case C. The point here is to rescale so that a meaningful annual figure can be reported from daily figures (but you wouldn't use this to rigorously compare metrics derived from daily against those derived from monthly). In general, you'd do all your calculations and make all your decisions at the frequency you collected your data (monthly in your case).

The theoretically correct approach is to use log returns = $\log(P_{t+1}/P_t)$ (using natural logs). The formula for the expectation of a sum of random variables can then be used correctly, because the sum of log returns is the log of the product of the returns.

Moreover, if you use log returns the Central Limit Theorem gives some theoretical justification that the log returns are normally distributed (essentially the Central Limit Theorem says that the sum of independent variables tends to a normal distribution as the number of random variables in the sum increases). This allows you to assign a probability to seeing a return less than $\mu - 2\sigma$ (the probability is given by the cumulative distribution function for the normal distribution: $\Phi(-2) \simeq 0.023)$. If the log returns are normally distributed, then we say that the returns are lognormally distributed -- this is one of the assumptions used deriving the famous Black Scholes option pricing formula.

One thing to note is that when a proportional return is small, then the proportional return is approximately equal to log returns. The reason for this is that the Taylor series for the natural logarithm is given by $\log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + \ldots$, and when the proportional return $x$ is small you can ignore terms with $x^2$, $x^3$, etc. This approximation gives a bit more comfort to those who choose to work with proportional returns and multiply the mean by $n$ and the standard deviation by $\sqrt{n}$!

You ought to be able to find further information on the web. E.g., I tried searching for "log returns" to refresh my memory, and the first hit seemed pretty good.

What you have put in case A is wrong. In the rest of your post you use the facts that (i) the expectation of a sum of random variables is the sum of their expectations, and (ii) the variance of a sum of independent random variables is the sum of their variances. From (ii), it follows that the standard deviation of $n$ independent identically distributed random variables with standard deviation $\sigma$ is $\sqrt{n}\sigma$. But in case A you have multiplied both the mean $\mu_X$ and the standard deviation $\sigma_X$ by $n$, whereas the mean needs to be multiplied by $n$ and the standard deviation by $\sqrt{n}$.

A subtle but important point, as noted by in @whuber's comment, is that rule (ii) requires correlation, which in the case of time series means no serial correlation (usually true but worth checking). The requirement for independence holds in both the proportional and log returns case.

(I haven't seen case B, the product of random variables, before. I don't think this approach is commonly used. I haven't looked in detail at your calculations, but your numbers look about right, and the formula can be found on wikipedia. In my opinion this approach seems a lot more complicated than either the approximation involved in using proportional returns or the theoretically sound approach of using log returns. And, compared to using log returns, what can you say about the distribution of Y? How can you assign probabilities to your worst case return, for example?)

• +1 Using logs is the key. It may be worth noting the implicit assumption in both the question and this answer that the monthly returns exhibit no appreciable serial correlation. (In my experience that's a reasonable assumption for most financial time series, but it's always worth checking.) – whuber Jul 12 '13 at 20:48
• Thank you on the log-return suggestion! I will look it up. However - regarding the rest of your answer - in my post I actually calculated P_t+1/P_t (and not [Pt+1-Pt /Pt]), so the expected value 1.00795 actually means return of 0.795%. That's why I multiplied the monthly values and not added them. (So the annual value in A is actually the monthly "worst-case" value to the power of 12). I would be glad to know if now you think differently on A or B, given that my question refers to a product of random variables and not their sum. Again, thanks a lot. – lyosef Jul 12 '13 at 21:16
• @NightMaster769 Sorry, I should have referred more directly to your post. I realized you were multiplying in order to compound the returns correctly but I didn't state that explicitly. This is after all why you were rightly concerned about using the formulas for adding random variables. Nevertheless, A just compounds the "2 standard deviation monthly bad return" over 12 months. It doesn't give you "2 standard deviation annual bad return". Regarding B, your approach seems sound but is complicated compared to log returns, and begs the question "What is the distribution of Y?". – TooTone Jul 12 '13 at 21:43
• @whuber Thanks I've added your point about serial correlation. – TooTone Jul 12 '13 at 21:54