a = (bc) / [bc + d(1-c)]
Solve for (a-c) as a function of (b-d).
You may recognize the expression above as Bayes' Theorem, where: a = P(A|B) b = P(B|A) c = P(A) d = P(B|not A) 1 - c = P(not A)
I'd like to know the function describing the relationship between the difference between the posterior and prior probability as a function of the difference between two components of the likelihood ratio.
Given $$ a = \frac{bc}{bc + d(1-c)},$$ we can write it as $$a = \frac{bc}{c(b-d)+d}$$ so that $$a - c = \frac{bc}{c(b-d)+d} - c = \frac{bc - c^2(b-d) - cd}{c(b-d)+d} = \frac{(c - c^2)(b-d)}{c(b-d)+d} = \frac{c(1 - c)(b-d)}{c(b-d)+d}.$$ I'm not sure if it could be simplified any more than this much.