Why we don't use weighted arithmetic mean instead of harmonic mean?

Question

I wonder what is an intrinsic value of using harmonic mean (for instance to calculate F-measures), as opposed to weighted arithmetic mean in combining precision and recall? I am thinking that weighted arithmetic average could play the role of harmonic mean, or am I missing something?

Answer 1

In general, harmonic means are preferred when one is trying to average rates, instead of whole numbers. In the case of an F1-measure, a harmonic mean will penalize very small precisions or recalls whereas the unweighted arithmetic mean won't. Imagine averaging 100% and 0%: Arithmetic mean is 50% and Harmonic mean is 0%. The harmonic mean requires that both precision and recall be high.

In addition, when the precision and recall are close together, the harmonic mean will be close to the arithmetic mean. Example: the harmonic mean of 95% and 90% is 92.4% compared to the arithmetic mean of 92.5%.

Whether this is a desirable property is probably dependent on your use case, but typically it's considered good.

Finally, note that, as @whuber stated in the comments, the harmonic mean is indeed a weighted arithmetic mean.

Answer 2

The harmonic mean may be a handy substitute to the arithmetic mean when the latter has no expectation or no variance. It may indeed be the case that $\mathbb{E}[X]$ does not exist or is infinite, while $\mathbb{E}[1/X]$ exists. For instance, the Pareto distribution with density$$f(x)=\dfrac{\alpha x_0^{\alpha}}{x^{\alpha+1}}\mathbb{I}_{x\ge x_0}$$has no finite expectation when $\alpha\le 1$, which implies that the arithmetic mean has an infinite expectation, while$$\mathbb{E}[1/X]=\int_{x_0}^\infty \dfrac{\alpha x_0^{\alpha}}{x^{\alpha+2}}\,\text{d}x=\dfrac{\alpha x_0^{\alpha}}{(\alpha+1) x_0^{\alpha+1}}=\dfrac{\alpha}{(\alpha+1) x_0}$$which implies that the harmonic mean has a finite expectation.

Conversely, there are distributions for which the harmonic mean has no expectation, as for instance the Beta $\mathcal{B}e(\alpha,\beta)$ distribution when $\alpha\le1$. And many more for which it has no variance.

There is also a link with Monte Carlo approximations to integrals, and particularly normalising constants, based on the Bayesian posterior identity$$\mathbb{E}\left[\dfrac{\varphi(\theta)}{\pi(\theta)L(\theta|x)}\Big| x\right]=\dfrac{1}{m(x)}$$where $\varphi(\cdot)$ is any density, $\pi(\cdot)$ is the prior, $L(\cdot|x)$ the likelihood, and $m(\cdot)$ the marginal, as discussed on that other question on X validated, where I comment on the dangers of using what Radford Neal (U Toronto) calls the worst Monte Carlo estimator ever. (I also wrote several entries on my blog on that topic.)