# Volatility of investment: Joint-Volatility of mutually independent variables

I´ve been trying to compute a volatility of invesment with currency hedging and I have a question. Let's take this example. We have our money in a fond copying the S&P500 index, which has 16% volatility, we also know that the current volatility of a dollar toward our currency is 5%. We want to know the volatility of the whole invesment.

Can I compute as following? If so, what is the reason for adding the two deviations instead of mulitplying them considering the volalitity of an index and a currency are mutualy independent.

$$\sigma=\sqrt{(16^2)+(5^2)}$$

• I'm voting to close this question as off-topic because this is clearly a question about basic statistics. (Perhaps a migration to stats.SE is in order?) – Giskard Apr 30 '16 at 18:22
• A lot of clarifications needed: What I understand from what you write is that you have made an investment $W$ denominated in foreign currency (USD) , and you want to know what will be its volatility (vague term - since you use percentage I guess you mean the standard deviation over mean value ratio "coefficient of variation"), in your local currency. But then you are looking at the entity $W\cdot S$ where $S$ is the "local currency per unit of USD" exchange rate. Independent they may (assumed to) be as random variables, but you have a product of ind. rv's, not a sum. – Alecos Papadopoulos May 4 '16 at 11:09
• It does not make sense to multiply them: suppose one had volatility 0, and the other 1. What would be the total volatility? 0? You can also see it in logs, log(ab)= log(a)+log(b). So that var(log(ab))=var(log(a))+var(log(b)) if hey are independent – Fix.B. May 11 '16 at 3:21

The variances of the sum of independent random variables are additive. Therefore, if X and Y are independent, Var(X+Y)= Var(X) + Var(Y). The standard deviation is defined as $\sigma_z = \sqrt{Var(z)}$. Therefore, if X and Y are independent, $\sigma_{X+Y} = \sqrt{Var(X+Y)} = \sqrt{Var(X) + Var(Y)}$. This gives the result above.