# Coefficient of variation (CV) of log-transformed data

I understand that with log-transformed data, the coefficient of variation (CV) on the original scale is equal to sqrt(exp(sigma^2)-1), where sigma is the standard deviation of log-transformed data.

But is there anything inherently wrong with simply calculating CV on log scale as sigma/xbar, where xbar is the mean of the log-transformed data?

For instance, would this calculation of CV on log-scale not really represent what is thought of as a coefficient of variation?

EDIT to explain my intended use of CV

My intended use is to report descriptive statistics for two sets of data:

1. price data for homes in different cities in Europe and;
2. price level indexes of homes for different cities in Europe using London as a 'base', i.e. (price home in city x/price home in London) x 100.

Because city prices and indexes generally, but not always, follow a log-normal distribution I decided to perform a log transformation to better visualize the distance of each city price or each city price level index from the center of each respective distribution.

• It likely depends on what you intend to do with the CV you have computed. Could you elaborate a little on that?
– whuber
Sep 11 '18 at 22:05
• A CV makes sense, if at all, only for positive data. Your logarithms are not necessarily positive. For instance, the logarithms of many of your relative prices (especially if you did not arbitrarily multiply the ratios by 100) would be negative. Your initial sentences suggest that what you really want to ask about is whether you could use $\sqrt{\exp(s^2)-1},$ where $s$ is the sample standard deviation of the logarithms, in a fashion similar to the CV of the original prices.
– whuber
Sep 12 '18 at 16:15
• @whuber. What about my 1st intended use, when my data are all positive. Would calculation of CV on log-scale still represent what is thought of as a coefficient of variation? Also, for my second case, I may change the base to the geometric mean of prices across all cities. This will ensure all positive data. Sep 12 '18 at 16:43
• My point is that even when all data are positive, the logarithms may have any sign. Any procedure whose applicability depends on an arbitrary choice of base is arbitrary, and so scarcely is worth considering.
– whuber
Sep 12 '18 at 16:46
• So, to be clear I have a tradeoff between using CV and transforming the data to logs. So that I either: a) use CV as a measure of dispersion on the original (not log-transformed) data or b) transform the data to logs but don't use CV as a measure of dispersion. Is this correct? Sep 12 '18 at 16:51