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Does anyone have a nice example where the solution is the geometric mean? There are nice examples where the solution is the harmonic mean; see, e.g., Arithmetic mean vs Harmonic mean An example similar to that (not rooted in geometry) would be great.

max_zorn
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  • Have a look at https://www.cut-the-knot.org/pythagoras/GeometricMean.shtml – Jack D'Aurizio Jan 06 '18 at 18:30
  • Thanks Jack, I am looking for something algebraic though (see my link). Nonmathematicians couldn't care less about geometric constructions, but the average speed example is nice, algebraic and counter-intuitive. – max_zorn Jan 06 '18 at 18:33
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    What about something along the following lines? You have a deposit of money, subject to an unknown compound interest. One year ago your deposit was $2000$ dollars, one year from now your deposit will be $8000$ dollars. What is the amount of your deposit now? – Jack D'Aurizio Jan 06 '18 at 18:37
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    Something similar can be designed by using a radioactive decay in place of a compound interest. The key point still is that the geometric mean is the pullback of the arithmetic mean through an exponential map. – Jack D'Aurizio Jan 06 '18 at 18:38

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Suppose you are given $n$ independent samples $X_1, \dots X_n$ from a normal distribution with known variance $\sigma^2$ but unknown mean $\mu$. Then the maximum likelihood estimate for $\mu$ is given by the arithmetic mean $\frac{X_1 + \dots + X_n}{n}$ of the samples. This is a common example of how arithmetic means are relevant to probability and statistics.

Now suppose you are instead given $n$ independent samples from a lognormal distribution given by the exponential of a normal distribution with known variance $\sigma^2$ but unknown mean $\mu$. Then the maximum likelihood estimate for $\exp(\mu)$ is given by the geometric mean of the samples. This is essentially the same fact as the previous but exponentiated.

In the same way that the central limit theorem leads to approximately normally distributed quantities if they are given as the sum of many small independent contributions, it also leads to approximately lognormally distributed quantities if they are given as the product of many small independent contributions. See Wikipedia for examples.

Qiaochu Yuan
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  • Very nice, thanks! I am still looking for something I can explain to nonmathematicians (see my comment above). Thanks again. – max_zorn Jan 06 '18 at 21:28
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    @max_zorn: the way I would phrase it to a nonmathematician is the following. Suppose you want to find the average of some data points in a data set as a way of roughly summarizing it. It often makes sense to take the arithmetic mean as this average, e.g. if you wanted to summarize the average height among your friends. But the arithmetic mean is sometimes a bad summary, e.g. if you wanted to summarize the average income among your friends; it heavily weights large outliers. In many situations the geometric mean offers a more sensible summary; said another way, many numbers should be... – Qiaochu Yuan Jan 06 '18 at 21:30
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    ...measured on a log scale. – Qiaochu Yuan Jan 06 '18 at 21:30
  • Very nice Qiaochu, thanks again! However, if you follow my link in the original post, you will come to the speed example. This is very compelling, you discover the harmonic mean as something very natural. Plus I can explain it to anyone with high school background. I look for a counterpart for the geometric mean. – max_zorn Jan 06 '18 at 22:05
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    If you don't allow geometric examples then I'm not aware of such a thing. Note that an important difference between the geometric mean and both the arithmetic and harmonic means has to do with units. For example, as an intermediate step in the computation of the geometric mean you compute the product $x_1 \dots x_n$. If each of the $x_i$ have some units $u$ then this product has units $u^n$. By contrast, when computing the arithmetic mean and the harmonic mean you compute either $x_1 + \dots x_n$ which has units $u$ or $\frac{1}{x_1} + \dots + \frac{1}{x_n}$ which has units $u^{-1}$. – Qiaochu Yuan Jan 06 '18 at 22:18
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    It's usually easy to interpret $u^{-1}$ but it can be hard to interpret $u^n$. A related point is that the arithmetic and harmonic mean are related by the map $x \mapsto \frac{1}{x}$, which sends things with units $u$ to things with units $u^{-1}$, but the arithmetic and geometric mean are related by the map $x \mapsto \exp(x)$ which cannot operate on unital quantities at all. – Qiaochu Yuan Jan 06 '18 at 22:21
  • Hi Qiaochu, why is the unit observation not an issue for geometric constructions where the mean originally comes from? – max_zorn Jan 06 '18 at 23:48
  • Good point. In that case the relevant quantities have units of length and we know how to interpret lengths raised to powers as volumes etc. so that's not so bad. – Qiaochu Yuan Jan 06 '18 at 23:56
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    @max_zorn: A similar thing, but with estimates instead of data points: if you want to estimate the size of something and can find a rough lower bound and a rough upper bound, then usually their geometric mean is a much better guess than their arithmetic mean. See here, for example. – Hans Lundmark Jan 07 '18 at 09:46
  • Hans, very nice - thanks a bunch! – max_zorn Jan 07 '18 at 19:29