Harmonic Mean

Equal-distance average speed, F1-score, parallel resistance, and HM ≤ GM ≤ AM.

To see the pattern behind the harmonic mean, write two positive numbers as

𝑎 = \frac{𝑐}{𝑥}, 𝑏 = \frac{𝑐}{𝑦}, 𝑐, 𝑥, 𝑦 > 0 .

Same numerator, different denominators: both fractions have $𝑐$ on top, while $𝑥$ and $𝑦$ are the denominators. Any positive $𝑎, 𝑏$ can be written this way: choose $𝑐 > 0$ , then set $𝑥 = \frac{𝑐}{𝑎}$ and $𝑦 = \frac{𝑐}{𝑏}$ .

Keep the shared numerator and average the denominators:

𝐻 = \frac{𝑐}{\frac{𝑥 + 𝑦}{2}} = \frac{2 𝑐}{𝑥 + 𝑦} .

Since $\frac{1}{𝑎} = \frac{𝑥}{𝑐}$ and $\frac{1}{𝑏} = \frac{𝑦}{𝑐}$ , this is exactly the reciprocal of the arithmetic mean of the reciprocals:

𝐻 = \frac{1}{\frac{\frac{1}{𝑎} + \frac{1}{𝑏}}{2}} = \frac{2 𝑎 𝑏}{𝑎 + 𝑏} .

Driving there and back

Drive a distance $𝑑$ at speed $𝑎$ , then return the same distance at speed $𝑏$ . Same numerator, different denominators: the two speeds are $\frac{𝑑}{𝑡_{1}}$ and $\frac{𝑑}{𝑡_{2}}$ . Both legs cover the same distance $𝑑$ , but take different times. Average speed is total distance divided by total time:

𝑣_{avg} = \frac{2 𝑑}{\frac{𝑑}{𝑎} + \frac{𝑑}{𝑏}} = \frac{2 𝑎 𝑏}{𝑎 + 𝑏} = 𝐻 .

For example, drive 60 km out at 30 km/h and 60 km back at 60 km/h. The legs take 2 hours and 1 hour, so the average speed is $\frac{120}{3} = 40$ km/h, not $\frac{30 + 60}{2} = 45$ km/h. You spend more time at the slower speed. The arithmetic mean would apply to equal times, not equal distances.

The upside-down trick. Arithmetic averaging works when the rates refer to equal amounts of their denominator: km/h for equal times. Here the equal quantity is distance, so flip the rates to h/km (time per distance), average arithmetically, then flip back. At 30 and 60 km/h, each kilometre takes 2 and 1 minutes respectively. Their arithmetic mean is 1.5 min/km; turning that upside down gives $\frac{60}{1.5} = 40$ km/h. The harmonic mean is just an arithmetic mean upside down: reciprocate, average, reciprocate.

F1-score

Precision and recall measure a classifier’s positive predictions:

𝑃 = \frac{TP}{TP + FP}, 𝑅 = \frac{TP}{TP + FN},

where TP, FP, and FN count true positives, false positives, and false negatives. Same numerator, different denominators: both ratios count the same true positives. Precision divides by all predicted positives; recall divides by all actual positives.

Their harmonic mean is the F1-score:

𝐹_{1} = \frac{2}{\frac{1}{𝑃} + \frac{1}{𝑅}} = \frac{2 𝑃 𝑅}{𝑃 + 𝑅} = \frac{2 TP}{2 TP + FP + FN} .

For example, with 30 true positives, 10 false positives, and 30 false negatives, $𝑃 = \frac{30}{40} = 0.75$ and $𝑅 = \frac{30}{60} = 0.5$ . Thus $𝐹_{1} = \frac{60}{100} = 0.6$ , below the arithmetic mean $0.625$ : a high precision cannot fully make up for a low recall, or vice versa.

The shared numerator makes the upside-down trick explicit: average $\frac{\frac{1}{𝑃} + \frac{1}{𝑅}}{2} = \frac{(TP + FP) + (TP + FN)}{2 TP}$ , then reciprocate. F1 gives precision and recall equal weight. This is a choice of metric, not a rule that every pair of ratios with a shared numerator must be averaged harmonically.

The reciprocal derivation assumes $𝑃, 𝑅 > 0$ . If either is zero, F1 is zero whenever $2 TP + FP + FN > 0$ ; if all three counts are zero, the score is undefined unless a convention is specified.

Parallel resistance

Two positive resistances $𝑅_{1}$ and $𝑅_{2}$ connected in parallel have the same voltage $𝑉 > 0$ across them. By Ohm’s law,

𝑅_{1} = \frac{𝑉}{𝐼_{1}}, 𝑅_{2} = \frac{𝑉}{𝐼_{2}} .

Same numerator, different denominators: both ratios have the same voltage $𝑉$ on top, but the branch currents $𝐼_{1}$ and $𝐼_{2}$ differ.

The currents add, so the equivalent resistance is

𝑅_{eq} = \frac{𝑉}{𝐼_{1} + 𝐼_{2}} = \frac{1}{\frac{1}{𝑅_{1}} + \frac{1}{𝑅_{2}}} = \frac{𝑅_{1} 𝑅_{2}}{𝑅_{1} + 𝑅_{2}} .

The equivalent resistance is half the harmonic mean, not the harmonic mean itself:

𝐻 = \frac{𝑉}{\frac{𝐼_{1} + 𝐼_{2}}{2}} = 2 𝑅_{eq} .

Dividing the shared voltage by the average branch current gives $𝐻$ ; dividing it by the total current gives $𝑅_{eq}$ .

For example, put 6 Ω and 3 Ω resistors in parallel across 12 V. The branch currents are 2 A and 4 A, so $𝑅_{eq} = \frac{12}{6} = 2$ Ω, while their harmonic mean is $\frac{12}{3} = 4$ Ω.

Harmonic, geometric, and arithmetic means

For $𝑎, 𝑏 > 0$ , let $𝐺 = \sqrt{𝑎 𝑏}$ and $𝐴 = \frac{𝑎 + 𝑏}{2}$ . Then

𝐻 \leq 𝐺 \leq 𝐴 .

Short proof of AM–GM. Since a square is nonnegative,

{(\sqrt{𝑎} - \sqrt{𝑏})}^{2} \geq 0 \Rightarrow 𝑎 + 𝑏 \geq 2 \sqrt{𝑎 𝑏} \Rightarrow 𝐴 \geq 𝐺 .

Also, $𝐻 = \frac{𝐺^{2}}{𝐴} \leq 𝐺$ , giving the full chain. Equality in either inequality holds exactly when $𝑎 = 𝑏$ .

Geometric proof (Thales and similar triangles). Draw a semicircle with diameter $𝑃 𝑄 = 𝑎 + 𝑏$ . Choose $𝐷$ on the diameter with $𝑃 𝐷 = 𝑎$ and $𝐷 𝑄 = 𝑏$ , and let the perpendicular through $𝐷$ meet the semicircle at $𝑇$ .

Let $𝑂$ be the centre. By Thales’ theorem, $∠ 𝑃 𝑇 𝑄 = 90 °$ . The right triangles $𝑃 𝐷 𝑇$ and $𝑇 𝐷 𝑄$ are similar, so

\frac{𝑃 𝐷}{𝐷 𝑇} = \frac{𝐷 𝑇}{𝐷 𝑄} \Rightarrow 𝐷 𝑇 = \sqrt{𝑎 𝑏} .

This height cannot exceed the semicircle’s radius $𝑂 𝑇 = \frac{𝑎 + 𝑏}{2}$ , hence $\sqrt{𝑎 𝑏} \leq \frac{𝑎 + 𝑏}{2}$ . Equality holds exactly when $𝐷$ is the centre, that is, $𝑎 = 𝑏$ .

Harmonic Mean

Driving there and back

F1-score

Parallel resistance

Harmonic, geometric, and arithmetic means

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