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> [P(H|E) : P(¬H|E)] = [P(H) : P(¬H)] × [P(E|H) : P(E|¬H)]

Hmm, you can do even better by formatting it as:

  p( H|E)   p( H)   p(E| H)
  ─────── = ───── * ───────
  p(¬H|E)   p(¬H)   p(E|¬H)
which is vertically symmetric about H/¬H, or possibly:

  ⎡p( H|E)⎤ = ⎡p( H)⎤ [*] ⎡p(E| H)⎤
  ⎣p(¬H|E)⎦   ⎣p(¬H)⎦     ⎣p(E|¬H)⎦
(where `[*]` is elementwise mutiplication over a vector).


What I wanted to say is, the :-separated blocks are not fractions, they are two-component vectors up to a multiple (aka coordinates on [a part of] the projective real line, for those who like that sort of thing), and the “two” here is completely incidental.

For example, suppose that, in the Monty Hall paradox, you chose door one and observed Monty open door three. Then the posterior odds for the car being behind each door are calculated as

    prior        =  1  : 1 : 1
  × likelihoods  = 1/2 : 1 : 0
  = posterior    = 1/2 : 1 : 0
  =                 1  : 2 : 0
(Exercise: prove this formulation of the Bayes theorem for >2 hypotheses.)


> the :-separated blocks are not fractions, they are two-component vectors up to a multiple

Projective coordinates are fractions, or more precisely the ℤ⁺²-projective vector space[1] (over the positive integers) is ℚ⁺, the positive rational numbers[0]. You generally want to use ℝ instead of ℤ for obvious reasons, but that's not really "not fractions" in any useful sense.

But writing it as vectors does have the benefit of making the generalization from ℝ² to ℝⁿ (n≥2) somewhere between obvious and trivial, whereas fractions are at-least-implicitly specific to the ℝ² case.

0: give or take some issues with 0 and ∞ that we don't care about because 0 and 1 are not probabilities

1: I'm sure someone can out-pedant me on the terminology here, but the point is it's two-dimensional, which (2) is the fewest dimensions this works non-degenerately for (leaving one degree of freedom, versus zero for the ℝ¹ case, and negative one (aka doesn't work at all) for ℝ⁰)


> the ℤ⁺²-projective vector space

Semi-erratum: that's "(ℤ⁺)²" (aka "ℤ⁺×ℤ⁺"), not "ℤ⁽⁺²⁾".




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