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When data are sampled from a lognormal distribution with a reasonably large geometric standard deviation, the distribution is asymmetrical and the arithmetic mean will be distinct from (and larger than) the geometric mean.

  • The arithmetic mean can be thought of as the center of gravity of the distribution. Parkin and Robinson suggest it makes sense to use the arithmetic mean "for situations where an estimate of the total mass
    of a variable (either a chemical constituent or a process) is required" (1).

  • The geometric mean is equivalent to the median (for the entire distribution; not for any particular sample). It is the center point, in the sense that half the values are higher, and half are lower.

I've always thought that the geometric mean is a more sensible (and standard) way to describe the distribution.

Of course, there is no need to choose, and reporting both means makes sense in some situations. But if you want to report only one mean, what is the best way to choose (and explain your choice)?


Addendum. Here is the key section from page 222 of Parkin and Robinson (using N not for population size, but for the element nitrogen): enter image description here

1.Parkin, T. B. & Robinson, J. A. Analysis of lognormal data. Adv Soil Sci 20, 193–235 (1992).
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    $\begingroup$ If you're interested in learning about the population mean of the original variable, it would make sense to report an estimate of that specific quantity (though you won't necessarily use the sample mean to estimate it) rather than of the mean of its log. But in any case I don't really understand the common default to only consider a single measure of location (or, rather, scale in the case of a lognormal), when two or three measures may be considerably more informative while taking up only a few more characters. That's not intended as criticism of your post; it's the publishing environment $\endgroup$
    – Glen_b
    Oct 22, 2023 at 1:33
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    $\begingroup$ I agree strongly on the interest and importance of the geometric mean. An intriguing educational question is that the geometric mean seems often to be considered too advanced for introductory texts but perhaps also too elementary or obvious for more advanced texts. The neglect in introductory texts may often arise because, or be encouraged by, reluctance to use logarithms in front of students (or even scientists etc.). $\endgroup$
    – Nick Cox
    Oct 28, 2023 at 15:17
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    $\begingroup$ Working in industry, I generally would just report the median and note something like "the mean is less useful here, because of the lopsided distribution". I don't fancy explaining geometric means or log-normal distributions to stakeholders. $\endgroup$
    – Eoin
    Oct 29, 2023 at 10:35
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    $\begingroup$ @NickCox, did anyone take a geometric mean of more than two numbers before 1800? I think no, which suggests that the concept is advanced after all. As benchmarks, logs were common with Napier's book of 1614, and the geometric mean was familiar for Cauchy in 1821, but this paper (cited in Cajori's History of Mathematical Notations as an early example of nth-root notation) suggests that even in 1788 people only thought of the geometric mean as between two quantities: babel.hathitrust.org/cgi/pt?id=hvd.hxin1i&seq=403 $\endgroup$
    – Matt F.
    Oct 30, 2023 at 16:01
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    $\begingroup$ The answers are interesting, but the question more or less answers itself, if vacuously: it all depends on exactly what you are trying to do. I'd add a trivial but homely example. The distribution of my expenditure is markedly skew with occasional enormous items (house, car, whatever) and numerous minor items (coffee, lunch, whatever). But the total -- as implied by the arithmetic mean -- is undoubtedly the key quantity -- yet it is not contradictory to say that otherwise the geometric mean is more interesting to me than the arithmetic mean. $\endgroup$
    – Nick Cox
    Oct 31, 2023 at 12:38

5 Answers 5

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If data comes from a lognormal distribution, then yes, the geometric mean is the single best summary statistic.

The key is the hypothesis. If the post had asked about choosing between normal and lognormal models, then the arithmetic mean might be more useful. Instead, the post asked about "data sampled from a lognormal distribution" -- i.e. a distribution already assumed or chosen to be lognormal. So whatever reasoning went into that assumption or choice probably would also suggest using the geometric mean.

Here are two examples:

  1. A lognormal distribution might model weather or disasters such as water flows, rainfall, and hurricanes, or earthquakes, fire damages, and insurance claims. Then the sampled events can differ by orders of magnitude. So the arithmetic mean will be roughly $1/n$ times the largest case, which is not an especially useful number. By contrast, the geometric mean will estimate the median of the population, and be a reasonable guide to a common size for the events.

  2. A lognormal distribution might model repeated financial processes, such as changes in stock prices, currency rates, or crypto values. Then the arithmetic mean does not scale nicely, so it is not an especially useful number. By contrast, reporting that the geometric mean is an increase of (e.g.) $1\%$ in a month or year or day is useful, because the powers $1.01^t$ will estimate the geometric means of samples over other time scales.

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    $\begingroup$ What you say makes lots of sense to me, and how I would have answered until reading the paper/review by Parkin doi: 10.1007/978-1-4612-2930-8_4 They prefer to use the arithmetic mean from clear lognormal distributions, when the values are concentrations or masses, saying that the center of gravity is what matters most. I don't really understand that argument, which is why I posted the question here. $\endgroup$ Oct 30, 2023 at 16:53
  • $\begingroup$ Parkin URL: link.springer.com/chapter/10.1007/978-1-4612-2930-8_4. (paywalled) $\endgroup$ Oct 30, 2023 at 16:56
  • $\begingroup$ Sorry. That is a long paper. I should have been more specific. I added a screenshot the the relevant page to the original question. $\endgroup$ Oct 30, 2023 at 22:34
  • $\begingroup$ I responded to the quote from the paper in a comment on the post $\endgroup$
    – Matt F.
    Oct 31, 2023 at 13:32
  • $\begingroup$ I would argue that for some of the variables that you mention in your first example, that percentiles rather than the mean or median are more likely to be used when the objective is to build something strong enough to withstand more extreme events. Again, which and how many summary statistics are needed depend on the objective. $\endgroup$
    – JimB
    Nov 5, 2023 at 23:58
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As the definition of the lognormal distribution suggests, if $Z\sim\mathcal{N}(\mu,\sigma^2)$ then $X=\exp(Z)$ follows the $\mathcal{logN}(\mu,\sigma^2)$ distribution. I think that the answer to "which mean to report?" first needs to go through "which part of the data are you interested in?". That is, are you interested in the measurements themselves ($X$) or in the underlying normal variable ($Z$)?

If you have some measurements and you are interested in the underlying variable (for example, your measure some power ratio and would like to learn something regarding this data in dB), then I'd report the geometric mean. Not only you inform the reader about $X$, but it also follows that $\sqrt[n]{\prod_{i=1}^{n}X_i}=\exp(\frac{1}{n}\sum_{i=1}^{n}Z_i)$. That is, one quantity tells two tales.

However, If you are interested in the measurements themselves - that is, you have some size which traditionally follows the lognormal distribution, such as the concentration of elements and their radioactivity in the Earth’s crust (example taken from here), then I'd report the arithmetic mean as it is the most common central metric.

PS. As far as I know, for a unimodal distribution the center of gravity lies in the mode, not the mean. While for $Z\sim\mathcal{N}(\mu,\sigma^2)$ we know that $\mu$ is not only the mean but also the median and the mode, this isn't the case for $X=\exp(Z)$. Its mean is defined as $\exp(\mu+0.5\sigma^2)$, its median is $\exp(\mu)$ and its mode is $\exp(\mu-\sigma^2)$.

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    $\begingroup$ @spatzie thanks. I don't understand the end of the second paragraph. "Two tales"? I also would prefer stronger statements about pros/cons of geo and arithmetic means. "interested in the underlying value" vs "interest in the measurements themselves" isn't very clear to me. $\endgroup$ Oct 30, 2023 at 15:13
  • $\begingroup$ Two tales: It tells you something about both $X$ and $Z$ $\endgroup$
    – Spätzle
    Oct 30, 2023 at 15:14
  • $\begingroup$ @spatzie. I don't think your equation in paragraph 2 is right. Please check. Beyond that, I don't understand the point. What two concepts does the geometric mean covey/ $\endgroup$ Oct 30, 2023 at 15:16
  • $\begingroup$ The arithmetic/geometric means are different tools, serving different purposes. When we need to pick which tool do use use, we first need to consider what is the purpose of our investigation. One does not simply take some data and spills all the central/dispersion metric available, right? What are these measurements? What do we want to know about the data collected? If we report some quantities, what does the addressee want to know? These are the questions you should ask yourself before discussing general pros/cons of different tools. $\endgroup$
    – Spätzle
    Oct 30, 2023 at 15:20
  • $\begingroup$ Thanks for pointing out the mistake in par.2, I've corrected it $\endgroup$
    – Spätzle
    Oct 30, 2023 at 15:23
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I think that the standard deviation (or standard error) will help you decide. The variation within a log-normal distribution is asymmetric and so the conventionally calculated standard deviation will overstate the variation on the low end and understate it on the high end. That leads to confidence intervals that do not behave well.

In an example that will be familiar to you the confidence interval for a drug EC50 might go down to near or, sometimes, below zero. As that would imply the statistical possibility of super-duper potency, or more than infinite, that is misleading. (Unless you are a homeopathy practitioner!) The confidence interval for the logEC50 behaves correctly.

Given that, I prefer by far the geometric mean, and would call it the mean log EC50.

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    $\begingroup$ Hi @Michael Lew. I think you miss the point of the question. Yes, lognormal distributions (with a sufficiently high GeoSD) are asymmetrical. But Parker and Robinson prefer using the arithmetic mean when the distribution is of mass or concentration because it is the center-of-gravity of the distribution. BUT.. there is a better way to compute the arithmetic mean than add them up and divide by n. And there are better equations to compute the CI of the arithmetic mean (that are based on sampling from lognormal, not normal). I agree with you, of course, on reporting log(EC50) not EC50. $\endgroup$ Oct 28, 2023 at 20:44
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[I'm going to post an answer to my own question, in hopes of getting comments and a better answer. ]

Imagine your goal is to compare a variable like pollutant concentration or rainfall in various locations. At each general location, you collect measurements and find that over several time intervals. The distribution among those values is lognormal. You don't care about the typical value (median) in each location. What you really want to compare is the total amount of rain or pollutant in the locations. You can't know that, but the arithmetic mean is proportional to the total. Therefore comparing the arithmetic means is more sensible than comparing the geometric means.

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If you only report one mean, you should choose the one with the most practical relevance. In situations where an estimate of the total mass matters, it is the arithmetic mean.

The mean is dominated by extreme values which is a desirable quality.

Situations where an estimate of the total mass matters (and not most common values) are situations where the total mass has most relevance, e.g. influence on the actions.

You are interested to know how extreme magnitudes of an earthquake or changes in a stock price, or rainfall can be because you have to prepare for that and before you prepare for most common magnitudes.

EDIT: So this answer can be seen as an extension to Spätzle's answer who says that it depends on the question you are trying to answer.

Matt F. Is right that the geometric mean is a good summary for the lognormal distribution because it fits the multplicative nature of the data generating process.

However practical implications might sometimes be more relevant which is why you want to show the arithmetic mean.

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  • $\begingroup$ Will you clarify in which of your examples you want to know the total mass vs the extreme magnitudes vs the most common magnitudes? I think you have different examples in mind for each of those, and that would be helpful, but I can't quite make out which examples you intend for which statistics. $\endgroup$
    – Matt F.
    Oct 31, 2023 at 19:06
  • $\begingroup$ Examples where the total mass matters are - rainfall, e.g. when you build infrastructure that should last, changes in stock prices when you are interested to estimate a risk. It really depends on what question you are trying to answer. My answer can be seen as an extension of your answer and Spätzle's answer. $\endgroup$
    – ChrisL
    Nov 1, 2023 at 9:24

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