I apologize if this is not a smart question, but it seems contradictory that the standard deviations of two price series show that series A is riskier, but when I plot the annualized volatilities (using the log returns), series B seems riskier. Is it normal? In the volatility plot below, the pink series is the one with the lowest standard deviation for its prices (30.00 vs 47.4 for the blue one) enter image description here


1 Answer 1


It's not really a contradiction. Computing the standard deviation of the returns rather than the prices discards information about the scale of the prices.

Take a price series $X_t$ and define the returns:

$$r_t = \log\left(\frac{X_t}{X_{t-1}}\right)$$

Then, any rescaling $Y_t = a X_t$ has the same returns, for any $a \neq 0$:

$$\log\left(\frac{aX_t}{aX_{t-1}}\right) = \log\left(\frac{X_t}{X_{t-1}}\right) = r_t$$

The standard deviation of the returns of $X_t$ and of $Y_t$ are therefore identical.

However, if you compare the standard deviation of the prices, if $a > 1$ then $Y_t$ will have a greater standard deviation, and if $a < 1$, it will be smaller.

More generally, for any two price series, if the first has smaller standard deviation for returns, the second could have larger, smaller, or the same standard deviation for prices, depending on the relative scale of prices.

In a financial context, the variance of the returns is typically the one that is considered, for a number of reasons.

  • $\begingroup$ Oh yes! thank you! the standard deviation of returns is indeed higher for the series that appears to have a higher volatility. $\endgroup$
    – Narjems
    Commented Dec 29, 2021 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.