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Suppose you have a prior that is a mixture of two beta distributions, like so: prior = (0.5)dbeta(a1,b1) + (1-0.5)dbeta(a2,b2). Furthermore, suppose you just recently ran a binomial experiment with 'x' successes out of 'n' trials and want to calculate the posterior probability for theta, the true probability of success, using Bayesian methods. I know that the posterior should have the form: posterior = (p)dbeta(a1+x,b1+n-x) + (1-p)dbeta(a2+x,b2+n-x) where 'p' represents some mixing weight between 0 and 1. Here's where I get confused: I know that, usually, posterior = (prior)(likelihood). How do you get from the general equation for posterior to this new equation for posterior when the prior is a mixture of 2 beta distributions? Why is the equation not just: posterior = (0.5)(prior1)(likelihood) + (0.5)(prior2)(likelihood)? And finally, how do you solve for p, the posterior mixing weights?

Thanks in advance!

PennyPrograms
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  • In this hierarchical scheme, the reason that the weights do not stay the same is that one of the priors in the mixture usually gives a higher likelihood for the observations than the other does, and you need to reweight to reflect this evidence. – Henry Mar 17 '20 at 23:40
  • I thought the likelihood function just produced the likelihood of your observed results (say x=12, n=20) occurring for supplied theta parameters? So the most 'likely' theta value would be 0.6 (since 12/20=0.6). And then the prior contains your previous beliefs about theta. Thus the posterior is (prob that theta = 'z')(likelihood of new results given that theta='z'). Are the two not independent of one another? Also, do you know the general method for solving for 'p' in a case like this? – PennyPrograms Mar 18 '20 at 00:42
  • You can find some exposition at a Cross Validated question or in some lecture notes – Henry Mar 18 '20 at 08:09
  • Thanks so much! – PennyPrograms Mar 18 '20 at 17:12

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