tag:blogger.com,1999:blog-8288194986820249216.post3016759073102139554..comments2024-03-22T05:09:17.789-07:00Comments on Abstract Heresies: A Slightly Less Simple AnswerJoe Marshallhttp://www.blogger.com/profile/03233353484280456977noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-8288194986820249216.post-84233013317139734502008-08-07T19:09:00.000-07:002008-08-07T19:09:00.000-07:00The point was just that, if you want to know what ...The point was just that, if you want to know what the probability of getting two heads on a coin with p(head)=mu, it's just mu*mu.<BR/><BR/>If you're not certain of the probability and, in fact, have a distribution in mu, then you have to integrate over that distribution, giving something like integral(mu*mu*p(mu) dmu).<BR/><BR/>Here, we know p(mu)=Beta(mu, 30, 13), so we can just do that integral. I find it somewhat amazing (and reassuring) that this gives the same answer as just multiplying the averages. (i.e., in this case, E[mu|29 heads]*E[mu|30 heads].<BR/><BR/>I'm not much of a Bayesian myself, but I've recently been working through some of this logic, so this caught my eye.Johann Hibschmanhttps://www.blogger.com/profile/18302360204130445141noreply@blogger.comtag:blogger.com,1999:blog-8288194986820249216.post-68799482273465766892008-08-07T09:44:00.000-07:002008-08-07T09:44:00.000-07:00Not late at all.I'm quite new to this Bayesian pro...Not late at all.<BR/><BR/>I'm quite new to this Bayesian probability stuff and I'm still unsure of my grasp of it at the more abstract level. I sort of understand your direct method (it basically unifies the first two equations) and I know that dealing with Beta priors is so common that it becomes second nature to experienced Bayesians. But I keep having to go back and relate the problem to a Bernoulli Urn situation and re-derive the equations from that because that way I feel I'm on familiar ground and I trust the results.Joe Marshallhttps://www.blogger.com/profile/03233353484280456977noreply@blogger.comtag:blogger.com,1999:blog-8288194986820249216.post-5896092257805939752008-08-06T19:17:00.000-07:002008-08-06T19:17:00.000-07:00This is a little late, but you can get the same re...This is a little late, but you can get the same result as the two-step inference by the (arguably) more direct method of integrating the distribution function.<BR/><BR/>Using R, this would just be "integrate(function (x) {x*x*dbeta(x,30,13)}, 0, 1)", using a Beta(1, 1) prior.Johann Hibschmanhttps://www.blogger.com/profile/18302360204130445141noreply@blogger.com