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.

  • $\begingroup$ It likely depends on what you intend to do with the CV you have computed. Could you elaborate a little on that? $\endgroup$
    – whuber
    Commented Sep 11, 2018 at 22:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Sep 12, 2018 at 16:15
  • $\begingroup$ @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. $\endgroup$ Commented Sep 12, 2018 at 16:43
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Sep 12, 2018 at 16:46
  • $\begingroup$ 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? $\endgroup$ Commented Sep 12, 2018 at 16:51

1 Answer 1


One nice property of the CV is that it does not change if you scale all the data by a constant factor. The SD of the log-transformed data shares this property. Your proposed measure (SD/mean of the log-transformed data) does not share this property. Lewontin (1966) may help elucidate some of these issues.

  • $\begingroup$ The link is inaccessible. Can you provide another one? $\endgroup$ Commented Sep 12, 2018 at 15:29
  • $\begingroup$ The link is an excellent one: accessible, short, and to the point. I had no problems accessing it online, either. It proposes using the S.D. of the logarithms as a (unitless) measure of relative variability for purposes of testing relative variability between groups, while retaining the CV of the original data as a descriptive statistic that (roughly) corresponds to the S.D. of the logs. $\endgroup$
    – whuber
    Commented Sep 12, 2018 at 17:39
  • 2
    $\begingroup$ The full cite is: Lewontin RC (1966) On the Measurement of Relative Variability. Systematic Biology 15(2):141–142. The DOI is here. $\endgroup$ Commented Sep 12, 2018 at 21:44
  • $\begingroup$ filogenetica.org/bibliografia/morfometria/estadistica/… $\endgroup$
    – flies
    Commented Aug 30, 2023 at 20:55

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.