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I have series A that is daily levels of a stock index, from which I calculate daily returns, then calculate the std dev of the daily returns, and multiply by (250)^0.5 to get annual std dev of returns for the series = sd(A).

I have a separate series B that shows daily values of yields for a bond index. If I want to compare the volatility between both series, is it correct to compare the coefficient of variation of B = cv(B) with sd(A) and make a statement of the form 'series A is sd(A)/cv(B) more volatile than series B'?

Or, are std dev and coefficient of variation inconsistent measures that cannot be compared?

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The population coefficient of variation (CV) is $\sigma/\mu,$ where $\sigma$ is the population standard deviation and $\mu$ is the population mean. [Perhaps see Wikipedia for definition and examples of useful and improper applications.]

One commonly used estimate of the population CV uses the sample standard deviation and mean $S/\bar X$ and, for small sample sizes $n.$ the adjusted value $(1 + \frac 4 n)S/\bar X.$ Appropriate uses are for positive interval data (height, weight, etc.).

The CV has no units, so it is the same for a group of stock prices, whether they are measured in dollars or yen. [There is a sense in which ants are of more variable weights than elephants, which is captured by the CV. Various species of ants can vary in weight by an order of magnitude or more, but the same can't be said for elephants.]

So it is usually a mistake to make comparisons between standard deviations (which have units) and CVs (which do not). If you are using an F-test to compare variances, as suggested in the answer by @josef_joestarr, you should make sure both sample variances have the same units.

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why not just do an equal variance test ? (F-test)

https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm

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