Posted: Dec 18, 2011 4:15 am
by jlowder
Thommo wrote:
jlowder wrote:
Thommo wrote:If you regularly buy lunch at McDonalds 6 days a week (for example), then Pr(E|H) can easily be > 0.5 regardless of any predictions of H.


Not necessarily. I'm using the epistemic interpretation of probability, not the frequency interpretation. If I regularly buy lunch at McDonald's 6 days a week, that could only be relevant IF:

1. We expand the expression Pr(E|H) to Pr(E|H&B), where B represents our background knowledge.
2. We include in B the fact that I regularly buy lunch at McDonald's 6 days a week.

If we do that, then Pr(E|H&B) > 0.5. Of course, H will be explanatorily irrelevant, since it will B, not H, that will make it possible for this value to be > 0.5. In fact, H will be irrelevant precisely because Pr(E|H&B) = Pr(E|B).


Yes indeed, and it's also true if we replace the 0.5 with any 0 < x < 1, which is rather why I was wondering where the 0.5 comes into it!


This is an excellent question. I may be using terminology in a slightly non-standard way here--I'm not sure--but I make a distinction between H's relevance and whether H predicts E. H is relevant to E just in case Pr(E|H&) does not equal Pr(E|B). H predicts E just in case H predicts that E is more likely than not, i.e., Pr(E|H&B) > 0.5. If H is relevant to E but does not predict E, there are other possibilities:

* H mystifies E; that is, H predicts that not-E or ~E is more likely than not i.e., Pr(E|H&B) < 0.5.
* And note that if H1 and H2 are mutually exclusive but NOT exhaustive alternatives, then E can be evidence favoring H1 over H2 even if both H1 and H2 "mystify" E. This would be the case if the ratio of Pr(E|H1) to Pr(E|H2) is greater than one, i.e., [Pr(E|H1) / Pr(E|H2)] > 1. (I have to admit this is subtle and it took me a while to wrap my mind around this one.)

Thommo wrote:Of course, if you don't like the frequentist example, feel free to consider any other method for estimating the probability of your eating lunch at McDonald's.

To put it another way, it seems rather odd to suggest it is a prediction of the theory of heliocentrism (say) that if I toss a coin twice in a row I won't get two heads, though it surely has P > 0.5 on the assumption that heliocentrism is true (again the theory is irrelevant as the probability is independent).


Hmmm... I may have oversimplified in an earlier post. In this example, I would say that the ratio of Pr(2 heads in a row | heliocentrism) to Pr(not 2 heads in a row | heliocentrism) is 1, which is just a fancy way of saying that heliocentrism is irrelevant.