(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An urn has 6 balls: black and white. Initially, the number of white balls W is distributed uniformly in {0,1,...,6}. Two balls are taken from the urn: one white and one black.

Find the distribution of W before the experiment.

3. The attempt at a solution

So, initially, if p is the proportion of white balls, then f_{p}(p) = 1/71{p=0,1/6,...,1}. This is the prior distribution of the parameter p.

Given the samplex=(B,W), if I consider each element of the sample a Bernoulli variable of parameter p, then the likelihood function would be L(p|B,W) = p*(1-p).

(Now, this would rule out the possibility of p being 0 or 1. I don't know how that translates into the posterior distribution of p...)

Then, the posterior distribution of p would be:

[tex]\pi (p|B,W) = \frac{{\frac{1}{7}p(1 - p)1\left\{ {p = 0,1/6,...,1} \right\}}}{{\sum\limits_p {\frac{1}{7}p(1 - p)1\left\{ {p = 0,1/6,...,1} \right\}} }}[/tex]

My problem is that the marginal distribution of the sample (which I think is what the problem asks for) yields 5/361{p=0,1/6,...,1}.

This is impossible! since the sample shows at least one white (p can't be 0) and one black (p can't be 1). Besides, that should be a uniform distribution over {p=0,1/6,...,1}, and with the 5/36 that is impossible (5/36*7 = 35/36 != 1).

What am I doing wrong?

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# Homework Help: Bayesian inference

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