tag:blogger.com,1999:blog-715339341803133734.post7679936510795950877..comments2023-08-18T04:14:50.151-05:00Comments on Maximum Entropy: Ockham's RazorTom Campbell-Rickettshttp://www.blogger.com/profile/07387943617652130729noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-715339341803133734.post-83131342059023455842012-09-17T22:10:58.288-05:002012-09-17T22:10:58.288-05:00Thanks, you are perfectly correct, of course. The ...Thanks, you are perfectly correct, of course. The two numbers must add up to 1. I just checked my spreadsheet with the calculation on it, and it is indeed 0.034 - I updated the text accordingly.<br /><br />Congratulations, you passed the test, and your golden key will arrive in the post shortly. You are well on your way to becoming a fully fledged member of the Bayesian Conspiracy! (Keep it very quiet, though.)<br /><br />Thanks again. Tom Campbell-Rickettshttps://www.blogger.com/profile/07387943617652130729noreply@blogger.comtag:blogger.com,1999:blog-715339341803133734.post-41308000495233347472012-09-17T13:32:29.783-05:002012-09-17T13:32:29.783-05:00PS: Great article nonetheless, thanks!PS: Great article nonetheless, thanks!Pnoreply@blogger.comtag:blogger.com,1999:blog-715339341803133734.post-65452877109599135632012-09-17T13:26:41.462-05:002012-09-17T13:26:41.462-05:00For P(N|a,I) I got 0.034, rather than 0.044. Howev...For P(N|a,I) I got 0.034, rather than 0.044. However, my value of B = 28.6 is correct according to the cited paper. Also, P(E|mu,I)+P(N|mu,I)=0.966+0.034=1.00. Possible typo?<br /><br />I like the use of the Einstein controversy as an application example, although it does not add much complexity to the statistics -- it seems merely to provide a singular predicted value to work with.Petorrnoreply@blogger.com