Hi, Can someone give me some advice of Q13.7 Ch13? Part (ii) - as n1,n2,n3,n4,n5 toys have been found I had assumed that this was a neg bin type 1 distribution with k=1,2,3,4,5 respectively. Ie: prob(first success is n1) x prob(second success is n2) x ... x prob(5th success is n5). This gives a posterior beta dist with alpha = 16 and beta = SUM(ni) - 14. (i think!) However the answer is actually given as alpha = 6 and beta = SUM(ni) - 4. Because, i ASSUME, it is treating each individual ni as geometric distribution probabilities, ie. each n1, n2,n3,n4,n5 are individual "first successes". Why is this? Or rather, how can I know the question wants me to do this rather than the neg bin type 1 approach? Any guidance on this would be appreciated, as it seems quite tricky to discern between the two. many thanks in advance!
I believe you get the same answer using negative binomial - although I think you are double counting. We can't take our "experiments" as selecting n1 then n1+n2 boxes then n1+n2+n3 etc. You are double (or more than double!) counting the early observed data. I think you get the same answer by using negative binomial on our one "experiment" with k=5 on the overall data set.
