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The arithmetic mean of a set of numbers minimizes the squared error. The geometric mean minimizes the squared log difference $\left[\log(x)-\log(\overline{x})\right]^2.$ The median minimizes the mean absolute error. What does the harmonic mean minimize?

MachineLearner
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Gavin
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  • You can get the proper font and spacing for $\log$ using \log. For operators that don't have a command of their own, you can use \operatorname{name}. – joriki Apr 23 '20 at 15:23

1 Answers1

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Since the geometric mean yields the arithmetic mean of the logarithms and minimizes the squared difference of the logarithms, it stands to reason that the harmonic mean, which yields the arithmetic mean of the reciprocals, minimizes the squared difference of the reciprocals:

$$ \sum_i\left(\frac1{x_i}-\frac1{\overline x}\right)^2\to\min $$

yields

$$ \sum_i\left(\frac1{x_i}-\frac1{\overline x}\right)=0 $$

and thus

$$ \frac1{\overline x}=\frac1n\sum_i\frac1{x_i}\;. $$

joriki
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