a question from McGill about The Bayesian Choice

I received an email from a group of McGill students working on Bayesian statistics and using The Bayesian Choice (although the exercise pictured below is not in the book, the closest being exercise 1.53 inspired from Raiffa and Shlaiffer, 1961, and exercise 5.10 as mentioned in the email):

There was a question that some of us cannot seem to decide what is the correct answer. Here are the issues,

Some people believe that the answer to both is ½, while others believe it is 1. The reasoning for ½ is that since Beta is a continuous distribution, we never could have θ exactly equal to ½. Thus regardless of α, the probability that θ=½ in that case is 0. Hence it is ½. I found a related stack exchange question that seems to indicate this as well.

The other side is that by Markov property and mean of Beta(a,a), as α goes to infinity , we will approach ½ with probability 1. And hence the limit as α goes to infinity for both (a) and (b) is 1. I think this also could make sense in another context, as if you use the Bayes factor representation. This is similar I believe to the questions in the Bayesian Choice, 5.10, and 5.11.

As it happens, the answer is ½ in the first case (a) because π(H⁰) is ½ regardless of α and 1 in the second case (b) because the evidence against H⁰ goes to zero as α goes to zero (watch out!), along with the mass of the prior on any compact of (0,1) since Γ(2α)/Γ(α)². (The limit does not correspond to a proper prior and hence is somewhat meaningless.) However, when α goes to infinity, the evidence against H⁰ goes to infinity and the posterior probability of ½ goes to zero, despite the prior under the alternative being more and more concentrated around ½!

2 Responses to “a question from McGill about The Bayesian Choice”

  1. Georges Henry Says:

    On n’y voit rien. Deja, avec cet affreux fond noir du blog…Mais la photo du tableau est pire. Si je comprends bien, tu as besoin de la limite de la loi $Beta(a,a)$ si $a$ tend vers l’infini. Pourquoi ne pas dire que si $X\sim Beta(a,a) alors $X\sim Y/(Y+Z)$ ou $Y$ et $Z$ sont iid de loi gamma de parametre de forme $a?$ Car alors la loi des grands nombres montre que $Beta(a,a)\rightarrow \delta{1/2}.$ Amicalement.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: