statisticalrethinkin..

2.6. PRACTICE 57implied by the equal abundance of both species. So prior to observing the birth, we have Pr(A) = 0.5.So:Pr(A|twins 1 ) = (0.1)(0.5)0.15= 5 15 = 1 3 .So the posterior probability of species A, aer observing twins, falls to 1/3, from a prior probability of1/2. is also implies a posterior probability of 2/3 that our panda is species B, since we are assumingonly two possible species. ese are small world probabilities that trust the assumptions.2.6.14. Update that panda. Continuing on from the previous problem, suppose the samepanda mother has a second birth and that it is not twins, but a singleton infant. Computethe posterior probability that this panda is species A.Solution. ere are a few ways to arrive at the answer. e easiest is perhaps to recall that Bayes’theorem accumulates evidence, using Bayesian updating. So we can take the posterior probabilitiesfrom the previous problem and use them as prior probabilities in this problem. is implies Pr(A) =1/3. Now we can ignore the first observation, the twins, and concern ourselves with only the latestobservation, the singleton birth. e previous observation is embodied in the prior, so there’s no needto account for it again.e formula is:Pr(A|singleton) =Pr(singleton|A) Pr(A)Pr(singleton)We already have the prior, Pr(A). e other pieces are straightforward:Pr(singleton|A) = 1 − 0.1 = 0.9Pr(singleton) = Pr(singleton|A) Pr(A) + Pr(singleton|B) Pr(B)Combining everything, we get:= (0.9) 1 3 + (0.8) 2 3 = 5 6Pr(A|singleton) = (0.9) 1 356= 9 25 = 0.36is is a modest increase in posterior probability of species A, an increase from about 0.33 to 0.36.e other way to proceed is to go back to the original prior, Pr(A) = 0.5, before observed anybirths. en you can treat both observations (twins, singleton) as data and update the original prior.I’m going to start abbreviating “twins” as T and “singleton” as S. e formula:Pr(A|T, S) =Pr(T, S|A) Pr(A)Pr(T, S)Let’s start with the average likelihood, Pr(T, S), because it will force us to define the likelihoods anyway.Pr(T, S) = Pr(T, S|A) Pr(A) + Pr(T, S|B) Pr(B)I’ll go slowly through this, so I don’t lose anyone along the way. e first likelihood is just the probabilitya species A mother has twins and then a singleton:Pr(T, S|A) = (0.1)(0.9) = 0.09And the second likelihood is similar, but for species B:Pr(T, S|B) = (0.2)(0.8) = 0.16