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Our goal is to calculate P(pb > 0.5 | "Six girls and one boy"), where pb is the probability of having a boy. (Ignoring that we have already assumed p is fixed),

Our goal is to calculate P(pb > 0.5 | "Six girls and one boy"), where pb is the probability of having a boy. (Ignoring that we have already assumed p is fixed), by applying Bayes' Theorem, we have: P(pb > 0.5 | "Six girls and one boy") = (P("Six girls and one boy" | pb > 0.5) * P(pb > 0.5)) / P("Six girls and one boy") Applying Bayes theorem thus requires us to calculate P("Six girls and one boy" | pb > 0.5). How do you suggest that we do this? Why is your answer the unique correct solution?

btilly 11y ago
Well first you have to start with a prior distribution of beliefs, which that is not. And to reduce confusion I'll switch back to the actual genders (6 boys then a girl). Suppose our prior distribution of beliefs is 0.5 that the probability is exactly 1/2, versus 0.5 that the probability of a boy is some value P which is equally likely to be any value from 0 to 1. In the first case, the probability of 6 boys and 1 girl is 0.5^7 = 1/2^7. In the second case the probability of 6 boys and 1 girl is P^6(1-P) = P^6 - P^7. The integral from 0 to 1 of P^6 - P^7 is 1/6-1/7 = 1/42. Each case also has a priori odds of 1/2 of holding true. After observing 6 boys and 1 girl, the first case now has probability (0.5/2^7)/(0.5/2^7 + 0.5/42) = 1/(1 + 64/21) = 21/129 = 0.162790697674419. The second case now has probability 1 - this, which is 0.837209302325581. Furthermore if the second case is true, P is no longer uniformly distributed. In fact its density is now proportional to P^6-P^7. So the posterior distribution is now going to be: With probability 21/129, exactly 0.5. And otherwise any value P from 0 to 1 with a probability density of 108/129*(P^6-P^7)/42. Given this prior and this set of observations, any other answer is wrong. Given a different prior you would get a different posterior, but as long as the prior gives a constant probability of male/female, the only fact that matters is how many boys and girls there are. The order of births can only start to matter if you start with a prior that gives different probabilities of different genders based on prior events. Even then it is hard to come up with a realistic scenario in which the plans of the parents would make an order of magnitude difference in the posterior distribution.
- thinkmoore 11y ago
  But if they are having children until they have one of each, the probability of the different observations does depend on the prior events!
  - btilly 11y ago
    The absolute probability of the observation is irrelevant. Only the RELATIVE probabilities of said observation under the different possible theories which are part of the prior. If the set of prior theories does not include anything that depends on birth order, then birth order and the experimental design are irrelevant to the posterior conclusions.
    - thinkmoore 11y ago
      Right, but this is exactly my point: the correct answer depends on the set of prior theories, which is exactly what the frequentist scenarios you consider are varying.
      - btilly 11y ago
        If you are arguing THAT point, then you shouldn't have disagreed with me anywhere! We have 3 types of facts. 1. What is the set of prior theories? Probability theory says this should matter. Bayesians are explicit about its involvement. Frequentists ignore it. 2. Observed data. Probability theory says that this should matter. Everyone takes this into account. 3. Experimental design for what would have happened had something different than the observed actually happened. This matters a lot to frequentist approaches and does not matter at all to Bayes' theorem. Bayesian approaches generally do not care about it at all. The difference between scenario 1 and scenario 2 is a fact of type 3, the conditions under which Bill and Lorena would have stopped having children. Unless you believe that Bill and Lorena's desire for one gender has a material impact on the probability of boys vs girls, this fact is irrelevant to any calculation of posterior probabilities. And is irrelevant in classical Bayesian approaches. Yet, despite being irrelevant, it matters a lot for frequentist approaches.

btilly 11y ago

Well first you have to start with a prior distribution of beliefs, which that is not. And to reduce confusion I'll switch back to the actual genders (6 boys then a girl). Suppose our prior distribution of beliefs is 0.5 that the probability is exactly 1/2, versus 0.5 that the probability of a boy is some value P which is equally likely to be any value from 0 to 1. In the first case, the probability of 6 boys and 1 girl is 0.5^7 = 1/2^7. In the second case the probability of 6 boys and 1 girl is P^6(1-P) = P^6 - P^7. The integral from 0 to 1 of P^6 - P^7 is 1/6-1/7 = 1/42. Each case also has a priori odds of 1/2 of holding true. After observing 6 boys and 1 girl, the first case now has probability (0.5/2^7)/(0.5/2^7 + 0.5/42) = 1/(1 + 64/21) = 21/129 = 0.162790697674419. The second case now has probability 1 - this, which is 0.837209302325581. Furthermore if the second case is true, P is no longer uniformly distributed. In fact its density is now proportional to P^6-P^7. So the posterior distribution is now going to be: With probability 21/129, exactly 0.5. And otherwise any value P from 0 to 1 with a probability density of 108/129*(P^6-P^7)/42. Given this prior and this set of observations, any other answer is wrong. Given a different prior you would get a different posterior, but as long as the prior gives a constant probability of male/female, the only fact that matters is how many boys and girls there are. The order of births can only start to matter if you start with a prior that gives different probabilities of different genders based on prior events. Even then it is hard to come up with a realistic scenario in which the plans of the parents would make an order of magnitude difference in the posterior distribution.

thinkmoore 11y ago
But if they are having children until they have one of each, the probability of the different observations does depend on the prior events!
- btilly 11y ago
  The absolute probability of the observation is irrelevant. Only the RELATIVE probabilities of said observation under the different possible theories which are part of the prior. If the set of prior theories does not include anything that depends on birth order, then birth order and the experimental design are irrelevant to the posterior conclusions.
  - thinkmoore 11y ago
    Right, but this is exactly my point: the correct answer depends on the set of prior theories, which is exactly what the frequentist scenarios you consider are varying.
    - btilly 11y ago
      If you are arguing THAT point, then you shouldn't have disagreed with me anywhere! We have 3 types of facts. 1. What is the set of prior theories? Probability theory says this should matter. Bayesians are explicit about its involvement. Frequentists ignore it. 2. Observed data. Probability theory says that this should matter. Everyone takes this into account. 3. Experimental design for what would have happened had something different than the observed actually happened. This matters a lot to frequentist approaches and does not matter at all to Bayes' theorem. Bayesian approaches generally do not care about it at all. The difference between scenario 1 and scenario 2 is a fact of type 3, the conditions under which Bill and Lorena would have stopped having children. Unless you believe that Bill and Lorena's desire for one gender has a material impact on the probability of boys vs girls, this fact is irrelevant to any calculation of posterior probabilities. And is irrelevant in classical Bayesian approaches. Yet, despite being irrelevant, it matters a lot for frequentist approaches.