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My simple understanding of arithmetic mean is that the arithmetic mean represents a "central point" (central tendency) where all others numbers converge as the accumulative linear distance from the left equals the accumulative linear distance from the right.

While I know (memorize) the definition of geometric mean, I cannot "interpret/visualize" it (maybe I should not?) in a way analogous to that of the arithmetic mean. I suspect that it may have to do with the fundamental concept of "root", which I fail to grasp, (I did try to read up on the concept of root but have yet found anything beyond its definition and application) but still I cannot find a satisfying answer how taking an nth root would yield a "mean" that reasonably represents the group.

Is there an easier way to "understand" geometric mean? (I tried to look for a similar question in this forum but have not found one so far).

B Chen
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    $$ \begin{align} A & = \big( \text{arithmetic mean of } a_1,\ldots,a_n\big) \ {} \ & \underbrace{A+\cdots+A}\text{$n$ terms}{} = a_1+\cdots+a_n \{} \ B & = \big( \text{geometric mean of } a_1,\ldots,a_n \big) \ {} \ & \underbrace{A\times\cdots\times A}\text{$n$ factors} {} = a_1\times\cdots\times a_n \end{align} $$ – Michael Hardy Jul 09 '21 at 01:39
  • @achillehui : Why do you say "scaling + translation" instead of just "scaling"? – Michael Hardy Jul 09 '21 at 01:42
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    If you define the "average" growth rate $\ \rho\ $ of some quantity achieved by a sequence $\ \rho_1,\rho_2,\dots,\rho_n\ $ of growth rates occurring in each successive year over a period of $\ n\ $ years to be that constant rate groth rate that will achieve the same total growth over the given period, then $\ 1+\rho\ $ is the geometric mean of $\ 1+\rho_1,1+\rho_2,\dots,1+\rho_n\ $. – lonza leggiera Jul 09 '21 at 01:58

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Ever since I was in about eighth grade I have assumed the reason it's called "geometric" is that it's about things having the same shape, but I've never heard anyone other than myself say so.

$$ \begin{array}{l} \overbrace{\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad}}^\text{The length of this is 9.} \\ \underbrace{\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\vphantom{\sum_{\displaystyle\bigotimes}}}_\text{The length of this is 12.} \\[12pt] \overbrace{\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}}^\text{The length of this is 12.} \\ \underbrace{\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\vphantom{\sum_{\displaystyle\bigotimes}}}_\text{The length of this is 16.} \end{array} $$ In the first pair of lines, the second line exceeds the first in length by an amount that is one-third the length of the first.

In the second pair of lines, the second line exceeds the first in length by an amount that is one-third the length of the first.

Thus both pairs of lines have the same shape.

So $12$ is the geometric mean of $9$ and $16.$

See also my comment under the question. If a group of people in a room take all their money out of their pockets and divide it equally among them, then the amount that each gets is the arithmetic mean of the amounts they had before that.

Michael Hardy
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I think of the geometric mean as a log-scale version of the arithmetic mean. That requires a little explanation.

The arithmetic mean of $x_1, \ldots, x_n$ is $\mathrm{AM}(x) = \frac{x_1 + \cdots + x_n}{n}$. Set $y_i = b^{x_i}$ for a fixed base $b>1$. Then the geometric mean of $y_1, \ldots, y_n$ is \begin{align*} \mathrm{GM}(y) &= (y_1 \cdots y_n)^{1/n} = (b^{x_1 + \cdots + x_n})^{1/n} = b^\left(\frac{x_1 + \cdots + x_n}{n}\right) \\ &= b^{\mathrm{AM}(x)}. \end{align*}

As an example, suppose we had 3 bank accounts with $y=\$100, \$1000, \$10000$ in them. The arithmetic mean is basically only going to pick up the \$10000 amount and will be basically $\mathrm{AM}(y) \approx \$10000/3$. However, the geometric mean effectively thinks of these as $\$10^2, \$10^3, \$10^4$ where $x=2, 3, 4$ with $\mathrm{AM}(x) = 3$, so the geometric mean of the amounts in the bank is $\mathrm{GM}(y) = \$10^3 = \$1000$.

Lots of data is really of the form $b^x$. For instance, this is basically the insight behind Benford's law. Consequently the geometric mean is sometimes more appropriate than the arithmetic mean.

Joshua P. Swanson
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As Jordan Ellenberg nicely shows in his book Shape, the geometric mean of $a_1$ and $a_2$ is the length of the side of a square having the same area as a rectangle with side $a_1$ and $a_2$. This generalizes to the geometric mean of more numbers as well. For more numbers/higher dimensions, the geometric mean is the length of the side of the regular shape having the same multi-dimensional volume.

nosuchthingasmagic
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