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can someone give me some pointers on deriving p(alpha|rest) on page 4 of (pdf warning)? I'm having trouble separating it out from beta. This isn't homework, but it seems substantially similar, so I'll mark it as such.

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what exactly are you asking for? – Nana Jun 24 '11 at 6:00
Post the question here because future readers will understand nothing if the PDF is removed. – Emre Jun 24 '11 at 15:57

You have an expression $$p(y,\theta,\alpha,\beta)= \prod_{i=1}^{N} {n_i \choose y_i} \theta_i^{y_i} (1-\theta_i)^{n_i-y_i} \prod_{i=1}^{N} \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\theta_i^{\alpha} (1-\theta_i)^{\beta} p(\alpha,\beta).$$

The proportionate expression for $p(\alpha | y,\theta,\beta)$ treats $y$, $\theta$ and $\beta$ as known constants, and so removes any multiplicative term which does not depend on $\alpha$, including all of the lefthand product and part of the righthand product to give $$p(\alpha | y,\theta,\beta) \propto \prod_{i=1}^{N} \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)}\theta_i^{\alpha} p(\alpha,\beta).$$ The next stage is to notice for example that $\prod_{i=1}^N k = k^N$ and similarly for the terms in the product not depending on $i$. That gives the stated result of $$p(\alpha | y,\theta,\beta) \propto \left[\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)}\right]^N \prod_{i=1}^{N} \theta_i^{\alpha} p(\alpha,\beta).$$ [This "key observation" also applies to the conditional expressions for $\theta_i$ and $\beta$.]

If you want to turn the proportionate expression into an exact expression, you then need to divide by the integral over $\alpha$ of the proportionate expression.

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