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I am working on a project where I need to do workload balancing, and to see how well my approach is working I want to compute both the average amount of time between requests for a workload and the average amount of time for a requested workload to be obtained.

Here is what is bothering me. My data consists of a list of time intervals (e.g., 0.4 seconds, 1.3 seconds, etc.). From this data, I can compute either the average rate at which an event is taking place by taking the reciprocal of each time interval to turn it into an instantaneous rate and averaging over that, or I can compute the average period by directly averaging over the time interval data. Doing the first is equivalent to computing the harmonic mean of the data, and doing the second involves computing the arithmetic mean of the data, so the two analyses get different results.

What confuses me is that, although the two means get different results, a rate is equal to the inverse of a period, and so the two results seems like they should be related via reciprocation, even though they are not.

Another option that I have is to use the geometric mean which will obtain consistent results for the rate and the period (as reciprocation commutes with taking the root), but my limited understanding says that one should only use the geometric mean in cases where you are taking the mean of something like growth rates.

So in short, it is not clear to me which of these means --- harmonic, arithmetic, geometric, or something entirely different --- should be used in this situation. Any insight would be appreciated.

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    $\begingroup$ Maybe this will help: stats.stackexchange.com/questions/23117/… $\endgroup$
    – dimitriy
    Feb 14, 2013 at 6:58
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    $\begingroup$ Note that $E(1/Y) \neq 1/E(Y)$; so, the things the arithmetic and harmonic means 'estimate' are different. Your prime consideration should probably be 'what expectation do you want to know about - the rate or the interval?' $\endgroup$
    – Glen_b
    Feb 14, 2013 at 8:10
  • $\begingroup$ Dimitriy, thanks, your link went to a detailed discussion about how to choose what estimate you want based on the quantity being conserved, which was exactly the kind of answer I was looking for. :-) (I can't believe that I didn't see that link in my previous searches...) $\endgroup$
    – Gregory Crosswhite
    Feb 14, 2013 at 10:31

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Geometric mean is useful either when (1) the distribution of observations is highly right-skewed and you expect that measurement error is greater on that right end of big values than on the left end of small values; or (2) the observations (values) are multiplicative for you rather than additive: you prefer to see their collective effect as their product, not their sum.

The choice between arithmetic and harmonic mean is usually less obvious and is partly a matter of philosophy. They say that you take harmonic mean instead of arithmetic one when the characteristic which varies hides in the denominator - not the numerator - of the value. It's easy to say - not so easy to do. For example, sprint running, a sportsman got it 100 meters for 10 sec. His speed was 100/10=10 m/s. You want to average speeds of several sportsmen. If the effective feature that varies for you lies in the numerator - the distance ran in a second - you'd choose arithmetic mean of the speed values. But if the effective feature that varies for you lies in the denominator - the time elapsed in a meter - you'd choose harmonic mean of the speeds. The problem here is that the decision between the two alternatives in this example is quite arbitrary due to vague "philosophy" of such terms as "speed" or "rate".

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