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Is this actually true for the Bayesian model? Throw in a parameter for whether or not Bill and Lorena were trying to have children until they had a boy and a gi

Is this actually true for the Bayesian model? Throw in a parameter for whether or not Bill and Lorena were trying to have children until they had a boy and a girl. Now your answer will depend quite heavily on your prior!

btilly 11y ago
This is entirely true. The posterior odds are computed ENTIRELY from the odds of the observed events under the prior beliefs. There is NO WAY in which might-have-beens and didn't-happens enter in. Therefore the posterior probabilities cannot depend on the knowledge of what they would have done if something different had happened. Of course frequentist statistics are heavily affected by what would have happened if something different had happened.
- thinkmoore 11y ago
  No, even as a Bayesian I could use either of two models to understand my data: In world one, I use bayesian update on the model P(# boys | Bias) = Mutltinomial(n = 7, p = Bias) In world two, I use bayesian update on the model P(# children | Bias) = Geometric(p = Bias) Might-have-beens and didn't happens do play in, in my choice of model. I should choose the one that I believe, and if I'm not certain, I should use an even more complicated model that incorporates my beliefs about what models might be appropriate.
  - btilly 11y ago
    Not if you follow Bayes' theorem. If you start with a prior distribution of beliefs about the likelihood of various ratios of boy vs girl births, the posterior distribution only depends on the observed outcomes. And the posterior distribution is exactly given by Bayes' theorem. One possible source of confusion for you is that Bayesian ideas have been a source of inspiration for a lot of ad hoc techniques (eg naive Bayes) which do NOT follow Bayesian rules of inference. The reason is that exact inference in Bayes nets is NP-hard. So you're used to hearing "Bayesian" applied to things that have nothing to do with Bayes' formula.
    - thinkmoore 11y ago
      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

btilly 11y ago

This is entirely true. The posterior odds are computed ENTIRELY from the odds of the observed events under the prior beliefs. There is NO WAY in which might-have-beens and didn't-happens enter in. Therefore the posterior probabilities cannot depend on the knowledge of what they would have done if something different had happened. Of course frequentist statistics are heavily affected by what would have happened if something different had happened.

thinkmoore 11y ago
No, even as a Bayesian I could use either of two models to understand my data: In world one, I use bayesian update on the model P(# boys | Bias) = Mutltinomial(n = 7, p = Bias) In world two, I use bayesian update on the model P(# children | Bias) = Geometric(p = Bias) Might-have-beens and didn't happens do play in, in my choice of model. I should choose the one that I believe, and if I'm not certain, I should use an even more complicated model that incorporates my beliefs about what models might be appropriate.
- btilly 11y ago
  Not if you follow Bayes' theorem. If you start with a prior distribution of beliefs about the likelihood of various ratios of boy vs girl births, the posterior distribution only depends on the observed outcomes. And the posterior distribution is exactly given by Bayes' theorem. One possible source of confusion for you is that Bayesian ideas have been a source of inspiration for a lot of ad hoc techniques (eg naive Bayes) which do NOT follow Bayesian rules of inference. The reason is that exact inference in Bayes nets is NP-hard. So you're used to hearing "Bayesian" applied to things that have nothing to do with Bayes' formula.
  - thinkmoore 11y ago
    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