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I've run through a bunch of searches, especially here on SO, but I simply couldn't find something that answers a question that has been on my mind lately. How was the geometric mean derived? What is the intuition behind it. Most simply use the final equation as a justification for its existence.

The $n$-th root of the product of all elements equation really doesn't do it justice.

Could someone elaborate on how and why is it interesting?

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The phrase "why it works" doesn't make sense. Do you mean "why is it interesting"? –  KCd Apr 29 '12 at 21:00
@KCd Yes, that is acceptable. I just wish some intuition about it, how it was derived etc. –  Rebuq Apr 29 '12 at 21:03
It is sounds terrible to ask how the concept was derived. Terminology aside, suppose you make a\$500 investment over $n$ years with annual growth rates of $a_1\%$, $a_2\%,\dots,a_n\%$. This is the same as a constant growth rate per year, over $n$ years, of what percent? You need to solve $(1+a_1/100)(1+a_2/100)\cdots(1+a_n/100) = (1+r/100)^n$, so $1+r/100$ is the geometric mean of $1+a_1/100, 1+a_2/100,\dots,1+a_n/100$. (The initial investment could have been anything and the answer for $r$ would come out the same; the use of 500 as the principal doesn't affect the choice for $r$.) –  KCd Apr 29 '12 at 21:05
For example, if your initial investment had growth rates over 3 years of 10\%, 20\%, and -30\%, this does not mean you come out even; in fact you end up with a net loss. The arithmetic-geometric mean inequality shows that if your investment grows over $n$ years at rates of $a_1\%, a_2\%, \dots,a_n\%$ and $a_1 + a_2 + \cdots + a_n = 0$, you will wind up with less than the principal unless every $a_i$ is $0$. This is a concrete application of the arithmetic-geometric mean inequality where the result to many seems counterintuitive at first (being misled by sum of growth rates being $0$). –  KCd Apr 29 '12 at 21:10
What's wrong with the geometric definition of the geometric mean? A rectangle with side lengths $a$, $b$ has the same area as the square of side length $\sqrt{ab}$; a shoebox with side lengths $a$, $b$, $c$ has the same volume as the cube with side length $\sqrt[3]{abc}$; likewise for higher dimensions. In general, a mean is simply a single value that can replace each member of a list of values yet achieve the same result. Just as the arithmetic mean of values preserves their sum, the geometric mean of values preserves their product. (The harmonic mean ... well ... look it up.) –  Blue Apr 30 '12 at 0:36
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3 Answers

up vote 5 down vote accepted

If I may join on the festivities, here are my two cents!

I personally take great pleasure in finding the intuition behind mathematical concepts (if there is some to find, read on), as well as the historical background that drove the development of a concept, so this might get lengthy. I hope it will provide you the insight you need to use the concepts without guilt.

First of all, it seems to me that you have an issue with the general formalization of mathematical concepts/constructs and the expected intuition behind them (inferrred from KCd's comment). You're not the only one, this is generated by lousy school systems which portray many mathematical concepts like they were pulled out of someone's behind. And this bugs people, some turn to malicious attempts of disproving mathematics ( to a point of bitter hate) and some invest the time to investigate the backdrop and find out whether that's truly the case.

It most certainly is not. Let me address the problem in logical chunks relating to your questions.

Mathematics and intuition

Even the most simple concepts within mathematics have a profound background and logical underpinnings. A very good definition of mathematics is that it is a response to problems. Mathematics is a response to problems. Any problem. As people have ventured through their lives, they've found patterns. They have found ways of strictly expressing these patterns. They have found ways of describing relationships and inferring others based on the same tools they've been developing all these years for describing nature.

As those tools became more refined, people understood the value of mathematics. It perhaps began as a nature-describing tool, but the concepts are so broad that one can actually develop consistent mathematical constructs which have no known basis in reality. This notion that one can actually develop a mathematical tool which perhaps won't find an application for hundreds of years (and even then get used in new, unforeseen ways) is the magic of the consistency and rigour that is mathematics.

So, you can have problems which have a known basis in reality or those which are problems within mathematics, logical expansion on the basic (or not so basic) constructs that have already been developed.

And in the other scenario, realworld intuition is not always available as much as one would like. In such scenarios, you are trusting your current tools and logical thinking to extrapolate a new concept or generalize an existing one. That way, we got hypercubes, $n$-dimensional objects which cannot be imagined by puny humans who are constricted to our $3$-dimensional space. That's where our intuition lacks, which is solely based on first-hand experience and is, by fact, just the tip of an iceberg that is the Universe.

On another hand - for quite a while, we actually thought that space is flat (Euclidean space), but it's actually quite curved. That's where our intuition betrays us. Euclidean space is still valid in our heads and in mathematics and a very good approximation of the true state of things, but has no real basis in reality. At a big enough scale, every line experiences curvature. See what happened? Nature played a trick on us with our own intuition which is quite limited. By applying the basic notions, we were able to infer the true nature. And very likely, nature has again tricked us. But every time we uncover one of its tricks, we learn quite a bit about the Universe we inhabit.

By generalizing the notions of intuition we have with strict logical thinking, mathematics gives us an insight in things that sometimes seem counter-intuitive, but which are nonetheless true. Or a subset of the truth which we continually try to expand by bootstrapping ourselves into the vortex of discovery.

Hopefully, now you've got a better idea of mathematics as a problem responder (both intuitive/every day and something that is beyond us but still very much based in the common notions we have about the problem or idea at hand). Also, you should have a grasp around intuition (or more commonly known as "making sense"), for what it's good and, in turn, for what it's considered a damping factor in our exploration of the Universe and, in turn, the Universe of Mathematics (and a broad Universe it is). There is no why, we just have how. Philosophers try to work out the why, we just analyze various phenomena - be it natural or mathematical ( a superset of natural ).


Intentionally capitalized for purposes of drama.

That is a mouthful, isn't it? Now, let's apply some of the staff from our discussion a moment ago. Mathematics solve problems. Sometimes they're evidently represented in nature. Sometimes they're represented in nature, but we can't see it due to our limitations. And sometimes it's just expanding into the perfect universe of mathematics which simply cannot exist in the relative chaos of our Universe.

Luckily, the geometric mean is a problem of the first kind. It's an evident problem with an relatively evident solution. Let's take a step back and take a look at the arithmetic mean, from which you should be able to understand the need for multiple means and how they can be formulated (simply by trying to answer problems - and that will be left to you to analyze and comprehend):

$$A = \frac{1}{k}\sum_{i=1}^{k}x_i$$

It is astonishing how many people use this, without having the slightest idea of what it is. Of what it represents. You often hear "Add all the elements together and divide by the number of elements - that's the average."

Ask that person what is this average, where does this relationship stem from - you'll probably get a wall of silence - or even worse - more explanations which use the thing they're actually trying to explain.

So, let's explore this. Say we have an array of a few rather random (accidentally odd) numbers:

$$R = [3, 5, 11]$$

They could be anything. Money, pure amounts, grades in school - irrelevant. Now, the problem:

How do we find out a number $n$ which relates to all inputs as close as possible?

What does it mean to lie as close as possible? That means that the absolute difference between the input and a supposed value $n$ is as low as possible. Let's formalize such a statement by trying to estimate how $n=6$ relates to all of our values:

$$\Delta x_1 = | 3 - 6 |= 3 $$

$$\Delta x_2 = | 5 - 6 |= 1 $$

$$\Delta x_3 = | 11 - 6 |= 5 $$


Now, we can see the how far $n$ is from every input. We want minimal numbers here, as small as the inputs permit (because they modulate the final output). We could spend all day searching for the smallest possible $\Delta$-difference array which is given as a function of $n$, or we could do something clever. We want the a number whose distance from every input is as small as possible. Well, we could add up all the differences

$$ d(n) = \sum_{i=1}^{k}( | x_i - n | )$$

Where $k$ is the number of inputs. In the previous example, that gives us a total "miss for $n=6$" of $d(6) = 9$. We want $d(n)$ to be as small as possible which simply means to be maximally close to all of the inputs.

You could track the possible solution as the lowest point which the function $d(n)$ reaches. And this is great, that's the value that represents the central tendency, which considers inputs... Unfortunately, because of the nature of the abs ($||$), there isn't a unique solution in most cases ( a minimum might drag itself on a interval, not really useful). Unfortunate (find out more on your own). How can we remedy this?

Well, we used the $||$ to denote that we want positive values, because negative distances don't make sense here. What else can give us positive values, preserve scale and relieve our problem of a non-unique solution? Raise the equations to the power of $2$!

$$ d(n) = \sum_{i=1}^{k}(x_i - n )^2$$

Well, that was quick. What have we gained? We now have our distances squared and summed, but positive. And guess what, we now have a unique peak value (the point where the parabola switches its direction.) How do we calculate that beautiful point? Well, it's that slim little point when the slope of the quadratic equation is equal to $0$. And how do we get that point?

$$\frac{d}{dn}d(n) = 0$$

So, let's first expand our $d(n)$ before deriving and extrapolating the $n$ at $d'(n) = 0$, here are the steps:

$$ d(n) = kn^2 - 2n\sum_{i=1}^{k}(x_i) + \sum_{i=1}^{k}(x_i^2) $$


$$ \frac{d}{dn}d(n) = 2kn - 2\sum_{i=1}^{k}(x_i) $$

And the winner is:

$$ 2kn - 2\sum_{i=1}^{k}(x_i) = 0 $$

$$ n = \frac{2\sum_{i=1}^{k}(x_i)}{2k} $$

$$ n = \frac{\sum_{i=1}^{k}(x_i)}{k} $$

$$ n = \frac{1}{k}\sum_{i=1}^{k}(x_i) = A $$

Could it be? Oh, yes. That's the elusive, proper definition of an arithmetic mean. Nobody woke up one day and simply wrote out the final equation. It was an important issue to resolve, to find what is the value around which all inputs tend to be accumulating.

And that's the whole point of it, when you try to reduce the sum of accumulated distances by respecting all the inputs included, you will find that the value drops around the most populated areas of the number line. Also, when you have only two values, that's simply resolves down to the middle between two points on the numberline. The more inputs you have, the more precisely can you define the central tendency.

And the geometric mean...

A lot of people here have given you a great amount of information which connect the geometric with the arithmetic mean. The actual extrapolation of the final equation most people refer to as the geometric mean has been developed just like arithmetic mean -

To solve a problem. And here the problem was, among many, how to analyze normalized values ($[0,1]$ or percentages of a total) because some comparison data might vary numerically and therefore the arithmetic mean would assign it much more weight than it's actually worth. And also, problems of exponential growth as were described by a few people here.

Hope it helps and if you find yourself desiring more information, do not hesitate to ask/comment. I'd write a lot more, but I'm getting dangerously high in word count. Now, off to bed.

The Geometric Mean - 2nd Ed.

A tale of the geometric mean - Mysterium Xarxes - Necromancia Mathematica

Per request of the commenting audience, I acknowledge the need to expand and complete the answer into the realm of the geometric mean in more detail, carrying with us the torch of discovery that unveiled the mysterious arithmetic mean and offered some background for independent research for other means. For completeness and a definite answer to sate the thirst of those interested - here goes:

We have greatly covered some of the background concerning the arithmetic mean which answers the question we posed as: "What single number best describes a set of numbers, a number which therefore lies as close as possible to all inputs?"

We interpreted it as the least squared sum of the errors (geometrically, a point on a number line which is as the bare minimum distance from all inputs, naturally moving to the highest concentration of inputs). The derived expression is as follows (from the previous discussion):

$$A = \frac{1}{k}\sum_{i=1}^{k}x_i$$

Verbosely, it adds up all the elements and divides by the number of elements. You have a total sum of all the inputs and you divide it in $k$ equal parts. The number that you get from the former definition is one which when added to itself $k$ times gives the original sum (which is simply inferred by the definition of basic arithmetic operations).

So basically, it replaces the varying elements with a constant term which preserves the sum. And this is the average by the very definition of the concept. If every day is the same, it's considered average. If every day is legendary then every day is, again, average, there is no oscillation. It's the same.

And this is facilitated simply by having a total sum of all the inputs and dividing that sum in $k$ equal parts. This is also why when you take the average between two points, it's the middle point. The total length is divided into two equal parts, defining a point on the number line directly inbetween separating the two partitions.

So, in simplest terms, arithmetic mean replaces all the inputs with one constant and requires it to add up to the original total sum. In contrast, geometric mean answers the question: "If all inputs were, again, the same value, what would be that value to multiply up to the original product." It is expressed as:

$$G = \sqrt[n]{\prod_{i=1}^{n}x_i} = (\prod_{i=1}^{n}x_i)^{\frac{1}{n}} $$

Looks wild? It's just simple algebra which drives a simple concept, as I hope you'll see further below.

I hope everyone now sees why I emphasized on understanding the arithmetic mean first, since the concepts are related in terms of goals, providing an average value which needs to be employed in the calculation that must satisfy some properties which are specific to its problem.

So, what is the problem of the geometric mean? Relative values. When each of the inputs or elements of the set is defined in terms of the previous. That is at the heart of percentages, stating how much of the total you have.

The percentage, $\%$ - that's a unit. It's defined as:

$$\% = \frac{1}{100}$$

$\%$ is a hundreth. Standing alone it means you've got one hundredth of the base unit $1$. Doesn't make much sense on it's own. It needs to be defined in terms of a definite initial value. Then you can say that you've got $20\%$ of it, which amounts to $20/100$ or $2/10$ or $1/5$ of the total value.

Imagine now a problem that involves relative growth, which is quite common. In the first year, you gain 30%, second - 40%, third - 50%. The naive approach would be to calculate the arithmetic mean to return the average change over the 3 years - returning a value of $40\%$.

But here you're making an assumption that for each year you use the original total multiplied by respectively $1.3$, $1.4$, $1.5$. Let's assume that the initial value is $v_0$:

$$\frac{1.3v_0 + 1.4v_0 + 1.5v_0}{3} = A$$

You see how the actual value is the same for all elements? What actually happens is this ($v_0$ is constant, we can factor it out):

$$v_0\frac{1.3 + 1.4+ 1.5}{3} = 1.4v = v_0 + 0.4v_0 $$

And this simply isn't right. Since we've expressed our values in terms of percentages of the initial value, the elements are dependent on one another, after the first year, the $v$ is no longer $v_o$. And this breaks down.

We need to understand the motivation for the geometric mean, like we did with the arithmetic mean. When considering the geometric mean, most get fooled by its name. To someone new to the concept, such a term might be intimidating.

So, we've talked about relative growth, expressing values as percentages which depend upon an undisclosed value (when evaluating the geometric term, we are agnostic of the value because we don't need to know it). If they indeed depend on one another and must be expressed in terms of the previous, we can express it as a product whose very definition lends itself to the problem:

$$v_0*1.3*1.4*1.5 = v_0x^n$$

We are, essentially looking for a value which, when multiplied by itself $3$ times (the number of percentages per respective years), gives the same final value as if we simply repeatedly multiplied each year with the respective $1.3$, $1.4$ and $1.5$. Now, since we don't care about the $v_0$ and we are trying to express the values in relative notions of percentages, we can just get rid of it:

$$1.3*1.4*1.5 = x^3$$

Let's be even more explicit:

$$1.3*1.4*1.5 = x*x*x$$

I hope this drives the notion home. One, average $x$ that does the same job as every other element on the left side. Now, we've defined the concept of a root as the number which when multiplied by itself $n$ times gives the value of which we're taking the root in the first place. So, take the $n$-th root of both sides, where in this case $n=3$:

$$\sqrt[3]{1.3 * 1.4 * 1.5} = \sqrt[3]{x^3}$$

What is the number which multiplied by itself three times gives $x^3$? Well, that's $x$, right? And that gives the following expression:

$$x = \sqrt[3]{1.3 * 1.4 * 1.5} = ~1.3976$$

That totals down to an average growth $39.76\%$ over the period of three years. See how it is just a little bit smaller than the arithmetic mean? That's actually quite a useful property called the arithmetic-geometric inequality which states that it is either less or equal to the arithmetic mean. The actual differences can be quite drastic, especially when dealing with ratios, relative values, growth etc.

And this simply generalizes to the first expression:

$$G = \sqrt[n]{x_1x_2 \ldots x_n}= \sqrt[n]{\prod_{i=1}^{n}x_i}$$

As the arithmetic mean protects the total sum, the geometric mean protects the total product (and solves our problem of dependent values). We mentioned that the arithmetic mean can be visualized as something that divides a number line in $n$ equal parts, which when added together give the original total.

EQUAL, you say?

Indeed. You could imagine having four sides of a rectangle (that means that the two values are the same). If you add them up, that is the perimeter of the rectangle. Now, divide it in four equal $a$ parts. Woah, what's that? Add them together to make sure we have the sum:

$$a + a + a + a = 4a = S$$

That seems awfully like the circumference of a square. And that's exactly what it is. It has the same circumference as the rectangle, but just one value is needed to describe the same thing.

Now, onto the geometric mean. If you had, as Day Late Don described, a rectangle with sides $a$ and $b$, their product is the area (as per agreement when the multiplication was defined, that's the geometric interpretation). So, which number when multiplied by itself $n = 2$ times gives the same area as the rectangle $ab$? That's right, $G = \sqrt{ab}$.

Do you see what we have here? G is the side of a square which when multiplied with itself gives the same area as the more complicated product of two different values. We have found a central value that does the job of all other values. As the arithmetic mean preserves the imaginary circumference, the geometric mean preserves the imaginary area. Respectively, that's protecting the sum. And protecting the area.

Amazing, huh? By trying to figure out a completely different problem without even contemplating the applications within geometry, we have acquired a new useful tool should we ever need it in our investigations.

Mr. Ayman Hourieh has answered the correspondence between the arithmetic mean and the geometric mean when we try to log both sides. The product reduces to a sum because of the nature of logarithms ($\log{(ab)} = \log{a} + \log{b}$. Should any questions arise, again, don't hesitate to ask.

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Wow, that's quite a bit of text! I love the approach to deriving the arithmetic mean, is that the way it was originally defined? A lot of useful info! –  Fractal Resurgence Apr 30 '12 at 3:27
@Fractal: that's basically the concept of regression/least squares. Here it is applied to the problem "what is the one number that 'best' approximates this bunch of numbers", for some mathematically precise definition of "best" (in this case the least squares criterion). –  J. M. Apr 30 '12 at 3:29
Of course, when Gauss took the least squares route, he made assumptions like the errors in the data having a normal distribution, but that looks to be a tale for another time... –  J. M. Apr 30 '12 at 3:30
Sir, you just made my day. Great answer. –  Pedro Tamaroff Apr 30 '12 at 3:45
How does this address the question at all? There is a long discussion on the arithmetic mean, but when it comes to the geometric mean nothing is said. –  KCd Apr 30 '12 at 4:41
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Here is one motivation for the geometric mean: By taking the logarithm of both sides of the geometric mean definition, you'll find that the logarithm of the geometric mean is the mean of logarithms of values:

\begin{align*} G &= \left(\prod_{i=1}^{n}a_{i}\right)^\frac{1}{n} \\ \Rightarrow \log{G} &= \frac{1}{n}\sum_{i=1}^{n}\log{a_{i}} \end{align*}

When the numerical ranges of values are too large, you may want to use logarithms of values, and hence the geometric mean.

For example, let's say you're studying a value that grows exponentially over time (like human population, compound interest, etc). The geometric mean makes more sense when studying this value, as illustrated by this example in Wikipedia.

As for why the geometric mean works better for values that grow exponentially: When a value grows exponentially, its logarithm grows linearly. Therefore, linear approximation via the geometric mean works, because it's based on the logarithms of values.

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What's an example where you'd want to do this (i.e., use logarithms of values simply because the numerical range is too large)? –  KCd Apr 29 '12 at 21:11
@KCd Please see the update in my answer and the link to an example in Wikipedia. –  Ayman Hourieh Apr 29 '12 at 21:16
Question, why does the arithmetic mean over-state the growth in comparison to the geometric mean? –  Rebuq Apr 29 '12 at 21:20
@Rebuq See my latest addition to the answer. –  Ayman Hourieh Apr 29 '12 at 21:31
And as a point of interest, there are at least two other means which are $f^{-1}(\textrm{arithmetic mean}(f(x_i)))$, namely the harmonic mean ($f(x) = x^{-1}$) and the RMS ($f(x) = x^2$). –  Peter Taylor Apr 29 '12 at 22:01
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One important motivation for the geometric mean is in sequential percentage increases.

If you have a sequence of percentage increases, and want to know the average percentage increase over the period, you must use the geometric mean.

The arithmetic mean (or other forms of average) will not cut it.

So, this has important applications in financial mathematics.

If the average percentage increase is $\bar{p}$% per year, then we expect an amount to increase by a factor of $(1+\bar{p}/100)$ per year.

So, the total increase ought to be $(1+\bar{p}/100)^n$, if we want the average to make sense.

If we know the individual increases are $p_1, p_2, ... p_n$, then the total increase will be $(1+p_1/100)(1+p_2/100)...(1+p_n/100)$

So, the motivation for the n-th root should be clear - if we have the individual increases, we can access the total increase, and then we'll need to take the n-th root to correctly calculate the average increase.

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Again, the most important question - why doesn't the arithmetic mean cut it? –  Rebuq Apr 29 '12 at 21:27
I have updated my answer in this respect. The geometric mean is necessary because increases multiply up, rather than add up. –  Ronald Apr 29 '12 at 21:30
@Rebuq: I illustrated in my comment to your question why the arithmetic mean is not the correct tool to use: even if the sum of the growth rates averages to 0, that does not mean your investment returns to its original value, but in fact there will be a net loss. –  KCd Apr 30 '12 at 1:03
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