4 ms·

Isn't it possible to distinguish casuality between A and B from correlation by simply considering set sizes? !A & B A & !B A & B !A & !B A

• pseudocomposer 3y ago
  Could you explain which distributions you think are correlation and which are causation?
  • avodonosov 3y ago
    I hoped someone who knows statistics will just tell us, instead of me stretching my brain. The famous phrase "correlation does not mean casuation" is not equal to "correlation is indistingushable from casuation", but that's how people seem to often treat it. Ideally, the paper abstract should be clear about what can or can not be concluded. In the title they say "subsequent" as if suggesting casuality, but in the abstract they say "associated". Confusing for a layman such as me. My intuition of how to distinguish correlation from casuation: If cannabis and depression were symmetrically connected, then incease of chances for cannabis use between depressed was equal to the increase of chances for depression between cannabis users. If cannabis increases chances for depression more that depression increases chances for cannabis, than we conclude cannabis tends to cause depression. Formula: percentage of depressed between canabis users / percentage of depressed in general population > persentage of cannabis users between depressed / persentage of cannabis users in general population. Something like that.
    • avodonosov 3y ago
      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?
• jitl 3y ago
  no?
• PheonixPharts 3y ago
  No because you can never know for sure that there isn't a latent factor C which is the real causal factor and A is merely correlating with it (if we presume causality to exist at all, which is a pretty standard assumption). We can compose DAGs that control for the factors we do know about, but it's impossible to exhaust or even know all possible latent processes that are impacting the outcome.
  • avodonosov 3y ago
    Ok. But can statistics disprove casuality? E.g. to show that A does not cause B, while B (or a latent factor B correlates with) causes A? When people say that the study can equally be interpretted as cannabis disorder causes depression and as depression causes cannabis disorder, can't statistics show that these two hypothesys are not equal?
    • avodonosov 3y ago
      No so easy: https://news.ycombinator.com/item?id=37633222 https://news.ycombinator.com/item?id=37633222