So we have arithmetic mean (AM), geometric mean (GM) and harmonic mean (HM). Their mathematical formulation is also well known along with their associated stereotypical examples (e.g., Harmonic mean and it's application to 'speed' related problems).

However, a question that has always intrigued me is "how do I decide which mean is the most appropriate to use in a given context?" There must be at least some rule of thumb to help understand the applicability and yet the most common answer I've come across is: "It depends" (but on what?).

This may seem to be a rather trivial question but even high-school texts failed to explain this -- they only provide mathematical definitions!

I prefer an English explanation over a mathematical one -- simple test would be "would your mom/child understand it?"

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    $\begingroup$ This perhaps oversimplifies but I've always used range and observations. If range is the same = AM (compare scores 0-100, to 0-100), if range is different but observation is the same = GM (compare scores 1-5, to 0-10), if range is same but observations are different = HM (speed of a car at different obs, heights of two ladders, other "rates"). $\endgroup$ Commented Feb 19, 2012 at 20:25
  • $\begingroup$ > "It depends" (but on what?) It depends on the data processing algorithm. $\endgroup$
    – Macson
    Commented Oct 20, 2017 at 11:11
  • 1
    $\begingroup$ It's not just a choice of which mean to use. It's also a choice of what set of summary statistics to describe the population or process of interest. One shouldn't think that all that's necessary is a single number to describe something of maybe great complexity. $\endgroup$
    – JimB
    Commented May 11, 2018 at 14:44

5 Answers 5


This answer may have a slightly more mathematical bent than you were looking for.

The important thing to recognize is that all of these means are simply the arithmetic mean in disguise.

The important characteristic in identifying which (if any!) of the three common means (arithmetic, geometric or harmonic) is the "right" mean is to find the "additive structure" in the question at hand.

In other words suppose we're given some abstract quantities $x_1, x_2,\ldots,x_n$, which I will call "measurements", somewhat abusing this term below for the sake of consistency. Each of these three means can be obtained by (1) transforming each $x_i$ into some $y_i$, (2) taking the arithmetic mean and then (3) transforming back to the original scale of measurement.

Arithmetic mean: Obviously, we use the "identity" transformation: $y_i = x_i$. So, steps (1) and (3) are trivial (nothing is done) and $\bar x_{\mathrm{AM}} = \bar y$.

Geometric mean: Here the additive structure is on the logarithms of the original observations. So, we take $y_i = \log x_i$ and then to get the GM in step (3) we convert back via the inverse function of the $\log$, i.e., $\bar x_{\mathrm{GM}} = \exp(\bar{y})$.

Harmonic mean: Here the additive structure is on the reciprocals of our observations. So, $y_i = 1/x_i$, whence $\bar x_{\mathrm{HM}} = 1/\bar{y}$.

In physical problems, these often arise through the following process: We have some quantity $w$ that remains fixed in relation to our measurements $x_1,\ldots,x_n$ and some other quantities, say $z_1,\ldots,z_n$. Now, we play the following game: Keep $w$ and $z_1+\cdots+z_n$ constant and try to find some $\bar x$ such that if we replace each of our individual observations $x_i$ by $\bar x$, then the "total" relationship is still conserved.

The distance–velocity–time example appears to be popular, so let's use it.

Constant distance, varying times

Consider a fixed distance traveled $d$. Now suppose we travel this distance $n$ different times at speeds $v_1,\ldots,v_n$, taking times $t_1,\ldots,t_n$. We now play our game. Suppose we wanted to replace our individual velocities with some fixed velocity $\bar v$ such that the total time remains constant. Note that we have $$ d - v_i t_i = 0 \>, $$ so that $\sum_i (d - v_i t_i) = 0$. We want this total relationship (total time and total distance traveled) conserved when we replace each of the $v_i$ by $\bar v$ in our game. Hence, $$ n d - \bar v \sum_i t_i = 0 \>, $$ and since each $t_i = d / v_i$, we get that $$ \bar v = \frac{n}{\frac{1}{v_1}+\cdots+\frac{1}{v_n}} = \bar v_{\mathrm{HM}} \>. $$

Note that the "additive structure" here is with respect to the individual times, and our measurements are inversely related to them, hence the harmonic mean applies.

Varying distances, constant time

Now, let's change the situation. Suppose that for $n$ instances we travel a fixed time $t$ at velocities $v_1,\ldots,v_n$ over distances $d_1,\ldots,d_n$. Now, we want the total distance conserved. We have $$ d_i - v_i t = 0 \>, $$ and the total system is conserved if $\sum_i (d_i - v_i t) = 0$. Playing our game again, we seek a $\bar v$ such that $$ \sum_i (d_i - \bar v t) = 0 \>, $$ but, since $d_i = v_i t$, we get that $$ \bar v = \frac{1}{n} \sum_i v_i = \bar v_{\mathrm{AM}} \>. $$

Here the additive structure we are trying to maintain is proportional to the measurements we have, so the arithmetic mean applies.

Equal volume cube

Suppose we have constructed an $n$-dimensional box with a given volume $V$ and our measurements are the side-lengths of the box. Then $$ V = x_1 \cdot x_2 \cdots x_n \>, $$ and suppose we wanted to construct an $n$-dimensional (hyper)cube with the same volume. That is, we want to replace our individual side-lengths $x_i$ by a common side-length $\bar x$. Then $$ V = \bar x \cdot \bar x \cdots \bar x = \bar x^n \>. $$

This easily indicates that we should take $\bar x = (x_i \cdots x_n)^{1/n} = \bar x_{\mathrm{GM}}$.

Note that the additive structure is in the logarithms, that is, $\log V = \sum_i \log x_i$ and we are trying to conserve the left-hand quantity.

New means from old

As an exercise, think about what the "natural" mean is in the situation where you let both the distances and times vary in the first example. That is, we have distances $d_i$, velocities $v_i$ and times $t_i$. We want to conserve the total distance and time traveled and find a constant $\bar v$ to achieve this.

Exercise: What is the "natural" mean in this situation?

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    $\begingroup$ +1 This is a great answer. However, I think it's incomplete in an important way: in many cases the right mean to use is determined by the question we are trying to answer rather than by any mathematical structure in the data. A good example of this occurs in environmental risk assessment: regulatory authorities want to estimate a population's total exposure to contaminants over time. This requires an appropriately weighted arithmetic mean, even though environmental concentration data usually have a multiplicative structure. The geometric mean would be the wrong estimator or estimand. $\endgroup$
    – whuber
    Commented Feb 20, 2012 at 15:03
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    $\begingroup$ @whuber: (+1) This is an excellent comment. On my path to constructing an answer, I took a decidedly nonstatistical fork, so I'm glad you mentioned this. It's a topic worthy of a complete answer (hint). $\endgroup$
    – cardinal
    Commented Feb 20, 2012 at 17:20
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    $\begingroup$ @whuber: It also brings up the fact (perhaps unintentionally), that statistical analysis can oftentimes be subject to the oversight of domain experts (or, perhaps in your example, even nonexperts), who want to estimate something meaningful to their domain but almost wholely unnatural statistically. The issue I've run into there in the past is that they sometimes want to also dictate the way that statistical estimation is carried out! :) $\endgroup$
    – cardinal
    Commented Feb 20, 2012 at 17:20
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    $\begingroup$ @whuber: It'd be very much appreciated if you could add that viewpoint to the answer too, with some elaboration. Honestly, your explanations are one of the best I've seen on Stats.SE! $\endgroup$
    – PhD
    Commented Feb 20, 2012 at 18:05
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    $\begingroup$ The usual great comment from @whuber. Sometimes (perhaps often!) the right mean to use is none; rather, the question often needs to be broadened to "what measure of central tendency should I use?". $\endgroup$
    – Peter Flom
    Commented Feb 20, 2012 at 22:41

Expanding on @Brandon 's excellent comment (which I think should be promoted to answer):

The geometric mean should be used when you are interested in multiplicative differences. Brandon notes that geometric mean should be used when the ranges are different. This is usually correct. The reason is that we want to equalize the ranges. For example, suppose college applicants are rated on SAT score (0 to 800), grade point average in HS (0 to 4) and extracurricular activities (1 to 10). If a college wanted to average these and equalize the ranges (that is, weight increases in each quality relative to the range) then geometric mean would be the way to go.

But this isn't always true when we have scales with different ranges. If we were comparing income in different countries (including poor and rich ones), we would probably not want the geometric mean, but the arithmetic mean (or, more likely, the median or perhaps a trimmed mean).

The only use I've seen for harmonic mean is that of comparing rates. As an example: If you drive from New York to Boston at 40 MPH, and return at 60 MPH, then your overall average is not the arithmetic mean of 50 MPH, but the harmonic mean.

AM = $(40 + 60)/2 = 50$ HM = $2/(1/40 + 1/60) = 48$

to check that this is right for this simple example, imagine it is 120 miles from NYC to Boston. Then the drive there takes 3 hours, the drive home takes 2 hours, the total is 5 hours, and the distance is 240 miles. $240/5 = 48$

  • 6
    $\begingroup$ Why would your SAT/GPA/extracurricular example use a geometric mean rather than a weighted or scaled arithmetic mean? Why should an SAT or GPA of zero mean that the other two values become irrelevant (as a geometric mean would imply)? And what if (say) extracurricular-activities tend to cluster in a much narrower band than its theoretical range? It seems like it would make more sense to take an arithmetic mean of percentiles (or other adjusted values) than a geometric mean of raw values. $\endgroup$
    – ruakh
    Commented Feb 20, 2012 at 2:08
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    $\begingroup$ @ruakh Interesting. The 0 issue doesn't really matter in this case, as SAT and GPA can't really be 0 (SAT = 0 is almost impossible, and GPA of 0 would not graduate). I think an arithmetic mean of percentiles will be close to the geometric mean in its conclusions (even though not in the actual numbers). $\endgroup$
    – Peter Flom
    Commented Feb 20, 2012 at 12:04

I'll try to boil it down to 3-4 rules of thumb and provide some more examples of the Pythagorean means.

The relationship between the 3 means is HM < GM < AM for non-negative data with some variation. They will be equal if and only if there's no variation at all in sample data.

For data in levels, use the AM. Prices are a good example. For ratios, use the GM. Investment returns, relative prices, and the UN's Human Development Index are all examples. HM is appropriate when dealing with rates. Here's a non-automotive example courtesy of David Giles:

For instance, consider data on "hours worked per week" (a rate). Suppose that we have four people (sample observations), each of whom work a total of 2,000 hours. However, they work for different numbers of hours per week, as follows:

Person      Total Hours       Hours per Week          Weeks Taken
1                  2,000                  40                   50
2                  2,000                  45                   44.4444
3                  2,000                  35                   57.142857
4                  2,000                  50                   40

Total:           8,000                                       191.587297

The Arithmetic Mean of the values in the third column is AM = 42.5 hours per week. However, notice what this value implies. Dividing the total number of weeks worked by the sample members (8,000) by this average value yields a value of 188.2353 as the total number of weeks worked by all four people.

Now look at the last column in the table above. In fact the correct value for the total number of weeks worked by sample members is 191.5873 weeks. If we compute the Harmonic Mean for the values for Hours per Week in the third column of the table we get HM = 41.75642 hours (< AM), and dividing this number into the 8,000 hours gives us the correct result of 191.5873 for the total number of weeks worked. Here is a case where the Harmonic Mean provides the appropriate measure for the sample average.

David also discusses the weighted version of the 3 means, which come up in price indices used to measure inflation.

I often find it hard to figure out if something is a rate or a ratio. Returns on an investment are usually treated as ratios when calculating means, but they are also a rate since they are usually denominated in "% per unit of time." I think a useful distinction is that ratios are usually unitless, so returns are ratios because \$ of current value over $ invested has the dollars signs cancel. Rates have different units in the numerator and the denominator.

Thus if you wanted to summarize the Big Mac Index for Northern European countries, you would use the equally weighted HM, because it is a rate. Divided by the number countries, the HM would tell you how much currency you would need to afford a BM under the constraint that you had to have the same amount of each currency.

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    $\begingroup$ A couple of years late, but did you ever find an answer to your question re: "If you wanted to summarize the Big Mac Index for Northern European countries, would you use the GM?" ? $\endgroup$ Commented Jan 26, 2015 at 20:29
  • 2
    $\begingroup$ @StatsScared Nope, but that would make a nice question! $\endgroup$
    – dimitriy
    Commented Jan 26, 2015 at 21:00

A possible answer to your question ("how do I decide which mean is the most appropriate to use in a given context?") is the definition of mean as given by the Italian mathematician Oscar Chisini.

How to Compute a Mean? The Chisini Approach and Its Applications is a paper with a more detailed explanation and some examples (mean travelling speed and others). Citation: R Graziani, P Veronese (2009). How to compute a mean? The Chisini approach and its applications. The American Statistician 63(1), pp. 33-36.

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    $\begingroup$ It might be ideal if you could add a few lines about Chisini's definition here in case the link goes dead, &/or to help readers know if they want to click the link to pursue the ideas further. $\endgroup$ Commented Jun 2, 2014 at 19:54

I think a simple way to answer the question would be:

  1. If the mathematical structure is xy = k (an inverse relationship between variables) and you're looking for an average, then you need to use the harmonic mean--which amounts to a weighted arithmetic mean--consider

Harmonic average = 2ab/(a+b) = a(b/a+b) + b(a/(a+b)

For example: dollar cost averaging falls into this category because the amount of money you're investing (A) stays fixed, but the price per share (P) and number of shares (N) vary (A = PN). In fact, if you think of an arithmetic average as a number equally centered between two numbers, the harmonic average is also a number equally centered between two numbers but (and this is nice) the "center" is where the percentages (ratios) are equal. That is: (x - a)/a = (b -x)/b, where x is the harmonic average.

  1. If the mathematical structure is a direct variation y = kx, you use the arithmetic mean--which is what the harmonic mean reduces to in this case.
  • 2
    $\begingroup$ I think you need to check your brackets match in your harmonic average equation - note that you can use Latex markup for your maths typesetting by surrounding it with dollar signs, e.g. $x$ produces $x$. For fractions note that \frac{a}{b} produces $\frac{a}{b}$. See our editing help for more information. $\endgroup$
    – Silverfish
    Commented Jun 20, 2016 at 23:17
  • $\begingroup$ Let's say you want to ensemble average the probabilities of several different models. In that case does it ever make sense to use geometric or harmonic mean? $\endgroup$
    – thecity2
    Commented Mar 20, 2018 at 18:04

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