tag:blogger.com,1999:blog-15453937.post6882884158865664687..comments2015-03-18T02:41:05.630-07:00Comments on The Lippard Blog: Skeptics and Bayesian epistemologyJim Lippardhttps://plus.google.com/117399630461557245031noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-15453937.post-64902070872594657832011-12-01T11:40:23.775-07:002011-12-01T11:40:23.775-07:00My description was pretty poor, and there's re...My description was pretty poor, and there's really a whole cluster of problems around conditionalization and the treatment of evidence in Bayesian epistemology. What I called the problem of new evidence is just the objection to simple conditionalization viewing the acquisition of new evidence as coming to a position of certainty about that evidence, which the Stanford Encyclopedia of Philosophy discusses under the name "the problem of uncertain evidence." The problem of old evidence is a different issue and is due to Clark Glymour. See sections 6.2 A & B in <a href="http://plato.stanford.edu/entries/epistemology-bayesian/" rel="nofollow">that article</a>. You might also find John Pollock's <a href="http://oscarhome.soc-sci.arizona.edu/ftp/PAPERS/Problems%20for%20Bayesian%20Epistemology.pdf" rel="nofollow">"Problems for Bayesian Epistemology"</a> (PDF) of interest, especially pp. 14-15.Jim Lippardhttp://www.blogger.com/profile/16826768452963498005noreply@blogger.comtag:blogger.com,1999:blog-15453937.post-61346878029764203182011-11-22T13:33:01.561-07:002011-11-22T13:33:01.561-07:00I don't understand why you would say that &quo...I don't understand why you would say that "a Bayesian approach considers new evidence to have a probability of 1". This means that the denominator of Bayes theorem is always 1, which is nonsense!<br /><br />The total evidence is the marginal probability of E, or p(E) = p(E|H)p(H) + p(E|-H)p(-H), and this is rarely equal to 1!<br /><br />For instance, take two mutually exclusive and exhaustive hypotheses and set the prior of each to 0.5. Assume that one of them predicts the data perfectly, p(E|H)=1 and that the other predicts it poorly, p(E|-H)=0. So p(E)= .5*1+.5*0 = .5, a value far from 1! The so-called 'problem of old evidence' is a non-problem for exactly the same reason.Freethoughtfulhttp://www.blogger.com/profile/12512248958676351959noreply@blogger.com