statisticalrethinkin..

58 2. SMALL WORLDS AND LARGE WORLDSe priors are both 0.5, so all together:Pr(T, S) = (0.09)(0.5) + (0.16)(0.5) = 1 8 = 0.125We already have the likelihood needed for the numerator, so we can go to the final answer now:Pr(A|T, S) = (0.09)(0.5) = 0.360.125Unsurprisingly, the same answer we got the other way.2.6.15. A panda for all data. A common boast of Bayesian statisticians is that Bayesianinference makes it easy to use all of the data, even if the data are of different types.So suppose now that a veterinarian comes along who has a new genetic test that sheclaims can identify the species of our mother panda. But the test, like all tests, is imperfect.is is the information you have about the test.• e probability it correctly identifies a species A panda is 0.8.• e probability it correctly identifies a species B panda is 0.65.e vet administers the test to your panda and tells you that the test is positive for species A.First ignore your previous information from the births and compute the posterior probabilitythat your panda is species A. en redo your calculation, now using the birth data as well.Solution. First, ignoring the births. is is what we know about the test:Pr(test A|A) = 0.8Pr(test A|B) = 1 − 0.65 = 0.35We use the original prior, Pr(A) = 0.5. Plugging everything into Bayes’ theorem:Pr(A|test A) =(0.8)(0.5)(0.8)(0.5) + (0.35)(0.5) ≈ 0.7So the test has increased the confidence in species A from 0.5 to 0.7.Now to use the birth information as well. e easiest way to do this is just to begin with theposterior from the previous problem. at posterior already used the birth data, so if we adopt it asour prior, we automatically use the birth data. And so our prior becomes Pr(A) = 0.36. en theapproach is just the same as just above:Pr(test A|A) Pr(A)Pr(A|test A) =Pr(test A)(0.8)(0.36)Pr(A|test A) =(0.36)(0.8) + (1 − 0.36)(0.35) ≈ 0.56And since this posterior uses all of the data, the two births and the genetic test, it would be honest tolabel it as Pr(A|test A, twins, singleton) ≈ 0.56.