((P(H|E) > P(H)) if and only if (P(H) > P(H|~E))) and
((P(H|E) = P(H)) if and only if (P(H) = P(H|~E)))
So, if some statement is evidence of a hypothesis, its negation must be evidence against. And if some statement’s truth value is independent of a hypothesis, then so is that statements negation.
This is implied by the expectation of posterior probabilities version. Since P(E) + P(~E) = 1, that means that P(H|E) and P(H|~E) are either equal, or one is greater than P(H) and one is less than. If they were both less than P(H), then P(H|E)P(E)+P(H|~E)P(~E) would have a lesser value than the largest conditional probability in that formula; suppose P(H|E) is the greater one, then P(H|E)P(E)+P(H|~E)P(~E) < P(H|E) and P(H|E) < P(H), so P(H|E)P(E)+P(H|~E)P(~E) ≠ P(H). If they are both larger than P(H), then P(H|E)P(E)+P(H|~E)P(~E) must be larger than the smallest conditional probability in that formula; suppose that P(H|E) is the smaller one, then we have P(H|E)P(E)+P(H|~E)P(~E) > P(H|E), and P(H|E) > P(H), so P(H) ≠ P(H|E)P(E)+P(H|~E)P(~E). And if both posterior probabilities are equal, then P(H|E)P(E)+P(H|~E)P(~E) = P(H|E), and both posteriors must eqaul the prior.
Q.e.d.
I think that the formula that expresses the prior as the average of the posterior probability weighted by the probabilities of observing that evidence and not observing that evidence, is a great way to express the point of this article. But it might not be trivial for everyone to get:
((P(H|E) > P(H)) if and only if (P(H) > P(H|~E))) and
((P(H|E) = P(H)) if and only if (P(H) = P(H|~E)))
from
P(H) = P(H|E)P(E) + P(H|~E)P(~E)
That something is evidence in favor if and only if its negation is evidence against, and that some result is independent of some hypothesis if and only if not observing that result is independent of that hypothesis, are the take home messages of this post as far as i can tell. The law that “P(H) = P(H|E)P(E) + P(H|~E)P(~E)” says more than that, it also tells you how to get P(H|~E) from P(H|E), P(H) and P(E). But adding the boolean statement and its proof from the weighted average statement to the post, or at least to a comment on this post, not even necessarily using the boolean symbols or formalisms, might help a lot of students that come across this long after algebra class. I know it would have helped me.
Wouldn’t the rule be something more like:
((P(H|E) > P(H)) if and only if (P(H) > P(H|~E))) and ((P(H|E) = P(H)) if and only if (P(H) = P(H|~E)))
So, if some statement is evidence of a hypothesis, its negation must be evidence against. And if some statement’s truth value is independent of a hypothesis, then so is that statements negation.
This is implied by the expectation of posterior probabilities version. Since P(E) + P(~E) = 1, that means that P(H|E) and P(H|~E) are either equal, or one is greater than P(H) and one is less than. If they were both less than P(H), then P(H|E)P(E)+P(H|~E)P(~E) would have a lesser value than the largest conditional probability in that formula; suppose P(H|E) is the greater one, then P(H|E)P(E)+P(H|~E)P(~E) < P(H|E) and P(H|E) < P(H), so P(H|E)P(E)+P(H|~E)P(~E) ≠ P(H). If they are both larger than P(H), then P(H|E)P(E)+P(H|~E)P(~E) must be larger than the smallest conditional probability in that formula; suppose that P(H|E) is the smaller one, then we have P(H|E)P(E)+P(H|~E)P(~E) > P(H|E), and P(H|E) > P(H), so P(H) ≠ P(H|E)P(E)+P(H|~E)P(~E). And if both posterior probabilities are equal, then P(H|E)P(E)+P(H|~E)P(~E) = P(H|E), and both posteriors must eqaul the prior. Q.e.d.
I think that the formula that expresses the prior as the average of the posterior probability weighted by the probabilities of observing that evidence and not observing that evidence, is a great way to express the point of this article. But it might not be trivial for everyone to get:
from
That something is evidence in favor if and only if its negation is evidence against, and that some result is independent of some hypothesis if and only if not observing that result is independent of that hypothesis, are the take home messages of this post as far as i can tell. The law that “P(H) = P(H|E)P(E) + P(H|~E)P(~E)” says more than that, it also tells you how to get P(H|~E) from P(H|E), P(H) and P(E). But adding the boolean statement and its proof from the weighted average statement to the post, or at least to a comment on this post, not even necessarily using the boolean symbols or formalisms, might help a lot of students that come across this long after algebra class. I know it would have helped me.