Why to downweight Precision in nominator of F-beta when I actually want to upweight Precision?

Question

F-beta score's formula calculates like this:

$$ F_{\beta} = (1+ \beta^2) \frac{PR}{\beta^2P + R} $$

However, according to some sources, in case I want to add more emphasis to Precision I should use beta < 1, and complementary, in case I want to add less emphasis to Precision than Recall, I should use beta > 1. This somehow seems me contrary to the purpose; why to downweight beta in the denominator when I actually want to assign more weight to Precision in the caluculation?

*Bonus question: either way the answer is for the question above, is there any particular formula to define beta if I want to assign different weights to Precision and Recall? Maybe in proportion to the cost difference in false negatives and false positives? Or simply just use beta=0.5 and beta=2 as a rule of thumb?

Answer 1

By setting $\beta$ to $0$ we are effectively getting "just Precision" that is because the Recall multipliers cancel each other out and we are left with Precision only. By using a lower $\beta$ we do not down-weight Precision in any way, we are up-weighting as we allow it to dominate the fraction's final value through the numerator.

If we already know the cost of our FP/FP we can directly use them. $\beta$ itself reflects our trade-off between Recall and Precision in the sense of $\beta=\frac{\text{Recall}}{\text{Precision}}$; therefore the values of $0.5$ and $2$ merely reflect that we hypotheises that: "we value Precision twice as much as Recall" (for $\beta=0.5$) or "we value Recall twice as much as Precision" (for $\beta=2$). Obviously if we value them equally $\beta=1$ and we get our standard $F_1$ score.

Sasaki (2007) The truth of the F-measure presents this discussion very nicely in a formal manner where it grounds it firmly on van Rijsbergen's original 1979 work on Information Retrieval.