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Geometric CV seems to be an appropriate measure for log-normalized data. Also, according to the discussions mentioned in theorems 1 and 2 of here, it seems to me that CV and geometric CV are both calculating the same quantity $\frac{\sqrt{Var(X)}}{E(x)}$, but since one performs the calculations on the raw data and the other performs them on the log-transformed data, CV is computed by $\frac{\sigma}{\mu}$ but geometric CV is simplified to $\sqrt{e^{\sigma^2}-1}$ (and $\mu$ and $\sigma$ are on raw data in the first formula but $\sigma$ is on the log-transformed data in the second). Thus, if the assumption of log-normality is almost true, I expect to obtain roughly similar quantities by them.

The problem is that in the 2000 values that I have, it seems that the log-normality holds (See the image below), but CV is obtained as 22.209 (calculated with "=STDEV(B:B)/AVERAGE(B:B)" in LibreOffice) while geometric CV is obtained as 435821.347 (STDEV=5.096; calculated with "=SQRT(EXP(STDEVP(F:F)^2)-1)" on the log-transformed data), which tremendously differ (Data are available here). What's the problem? Is any of my explanations wrong?

The histogram of the log-transformed data

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    $\begingroup$ There may be multiple problems. First, assuming your plot shows counts of logarithms of values, it doesn't look normal. Second, it's hard for us to tell why you might have obtained such different values unless you show how you performed the calculation--after all, you likely just made some arithmetic error. Third, your expectation concerning the relationship between the two forms of CV may hold for theoretical distributions, but it's unlikely to be even remotely correct for data unless both CVs are small. Please edit your post to clarify what you did. $\endgroup$ – whuber Jun 30 '18 at 15:21
  • $\begingroup$ Obviously the distribution is not a theoretical normal distribution. However, it's almost normal. Such little difference with normal shouldn't probably make such big difference in the values. I showed how I did. I reported the formulas. I even put a link to the spreadsheet file. All the logs performed are natural logarithms. Means and standard deviations are calculated based on standard Libreoffice functions (STDEVP, STDEV, AVERAGE). However, if you still want more details I add the formulas to the question. $\endgroup$ – Shayan Jun 30 '18 at 15:43
  • $\begingroup$ About "Third, your expectation concerning the relationship between the two forms of CV may hold for theoretical distributions, but it's unlikely to be even remotely correct for data unless both CVs are small", from where do you say that? My first link proves the formulas. Why do you say "unless both CVs are small"? Moreover, such huge difference may not be explainable by minor points in the assumptions. $\endgroup$ – Shayan Jun 30 '18 at 15:47
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    $\begingroup$ @Shayan Not implying that this is what is wrong but the data is clearly non-normal and it is unclear why you say this shouldn’t make a difference. Why, exactly, do you think the violation of the normality assumption would not make a difference? If the data is not log-normal (an assumption which has huge implications about tail behavior) there is no reason to expect the formula you have based on the log-normal to be close to correct. $\endgroup$ – guy Jun 30 '18 at 16:00
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    $\begingroup$ You would be amazed how large numbers can become when you exponentiate them. Bear in mind, too, that the formulas you rely on do not apply to data--ever. Your large CV belies some extremely large values in the dataset, indicating that most calculations you care to do will be extraordinarily sensitive to a few--or just one--of the numbers in the data, which is why it's not reasonable to expect that a formula for theoretical distributions would continue to hold. $\endgroup$ – whuber Jun 30 '18 at 17:30

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