Sampling Concentration Parameter of DP according to Escobar and West

Question

I am reading Escobar&West paper and in particular am interested in their Gibbs sampler for the concentration parameter of Dirichlet Process (eq 13, eq 14). Given this,

$$p(\alpha,\eta|k)\propto p(\alpha)\alpha^{k-1}(\alpha+n)\eta^\alpha(1-\eta)^{n-1}$$

If we replace $p(\alpha)$, with it's $Gamma(a,b)$ distribution replacement, we get,

$$p(\alpha,\eta|k)\propto \large(\frac{1}{\Gamma(a)b^a}\alpha^{a-1}e^{\frac{-\alpha}{b}}\large)\alpha^{k-1}(\alpha+n)\eta^\alpha(1-\eta)^{n-1}$$

which can be re-written as,

$$p(\alpha,\eta|k)\propto \frac{1}{\Gamma(a)b^a}e^{\frac{-\alpha}{b}}\alpha^{a+k-2}(\alpha+n)e^{\log\eta^\alpha}(1-\eta)^{n-1}$$

and can be simplified again as,

$$p(\alpha,\eta|k)\propto \frac{1}{\Gamma(a)b^a}\alpha^{a+k-2}(\alpha+n)e^{(-\frac{\alpha}{b}+\alpha\log\eta)}(1-\eta)^{n-1}$$

and to convert this to $p(\alpha|\eta,k)$, we must take the integral,

$$p(\alpha|\eta,k)\propto \frac{1}{\Gamma(a)b^a}\alpha^{a+k-2}(\alpha+n)\int e^{(-\frac{\alpha}{b}+\alpha\log\eta)}(1-\eta)^{n-1}d\eta$$

but I don't understand how from that we can get to the following relation mentioned in the paper,

$$p(\alpha|\eta,k)\propto \alpha^{a+k-2}(\alpha+n)e^{\alpha(-b+\log(\eta))}$$

Assuming that they dropped redundant terms that do not depend on $\alpha$, for example, $\frac{1}{\Gamma(a)b^a}$ and $(1-\eta)^{n-1}$, still it is not clear how they computed the integral. Also, the power of $e$ must be ${\alpha(-\frac{1}{b}+\log(\eta))}$ unless I am skipping a step.

Answer 1

You are making an error when taking the integral. Indeed,

$$p(\alpha|\eta,k) \neq p(\alpha|k) = \int p(\alpha, \eta|k) d\eta.$$

Instead of integrating, you ought to use Bayes' rule;

$$ p(\alpha|\eta, k) \propto p(\alpha, \eta|k) $$

where the proportionality constant $1/p(\eta|k)$ does not need to be computed since it is constant with respect to $\alpha$. Indeed, we can remove more factors which do not depend on $\alpha$ in your expression of $p(\alpha, \eta|k)$, which yields an answer similar to the one the authors arrive at. (There is a small difference which leads me to suspect that there is a transcription error somewhere.)