D
dannyp123
Member
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!
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!