2 ms·

He-he, the sides of inequality in my formula are always equal :) (c&d/c)/(d/total) > (c&d/d)/(c/total) c&d*total/c*d > c&d*total/d*c

• avodonosov 3y ago
  Looking at it further, it's indeed not so easy. If we have a statistics produced by some casual relationship, e.g. function takeMeasurement() { // c (the cause) itself has probability 0.5 // d (the dependent) in absence of c has probability 0.5 // but the presence of c increases it to 0.7 const c = Math.random(1.0) < 0.5; const d = Math.random(1.0) < (c ? 0.7 : 0.5); return {c, d}; } one can calculate an equally plausible hypothesis of causality in opposite direction, which would produce an equal statistics: P(c) = 0.5 P(d|!c) = 0.5 P(d|c) = 0.7 P(d) = ? P(c|!d) = ? P(c|d) = ? ------------------------------- P(d) = P(c)*P(d|c) + P(!c)*P(d|!c) = = P(c)*P(d|c) + (1 - P(c))*P(d|!c) = = 0.5*0.7 + 0.5*0.5 = 0.35 + 0.25 = = 0.6 P(c|!d) = P(c) * P(!d|c) / P(!d) = = P(c) * (1 - P(d|c)) / (1 - P(d)) = = 0.5 * 0.3 / 0.4 = = 0.374(9) P(c|d) = P(c) * P(d|c) / P(d) = = 0.5 * 0.7 / 0.6 = 0.35 / 0.6 = = 0.58(3) ------------------------------- Bayes formula: P(a|b) = P(a) * P(b|a) / P(b) a resulting function: function takeMeasurement2() { const d = Math.random(1.0) < 0.6; const c = Math.random(1.0) < (d ? 0.58333 : 0.374999); return {c, d}; }
  • avodonosov 3y ago
    Is it really hopeless? Maybe not? For example, a boundary case when d only happens in case of c? Are the hypotheses "c causes d" and "d causes c" still equally plausable?