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I'm reading Markov Chain Sampling Methods for Dirichlet Process Mixture Models by Radford M. Neal. Equation (3.6) states that

$$ \text{If } c=c_{j} \text{ for some } j\neq i: P\left(c_{i}=c\;|\;c_{-i}, y_{i}, \boldsymbol{\phi}\right) = b\frac{n_{-i,c}}{n-1+\alpha}F\left(y_{i},\phi_{c}\right)$$

$$P\left(c_{i}\neq c_{j} \text{ for all } j\neq i\;|\; c_{-i},y_{i},\boldsymbol{\phi}\right)=b\frac{\alpha}{n-1+\alpha}\int F\left(y_{i},\phi\right)\,dG_{0}\left(\phi\right)$$

It doesn't give any derivation. How can I compute this by hand?

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  • $\begingroup$ This follows from equation (3.1). $\endgroup$ – Xi'an Apr 10 '16 at 9:02
  • $\begingroup$ How does it follow? Is it just a standard derivative? $\endgroup$ – Gabriel Fair Apr 12 '16 at 23:12
  • $\begingroup$ @Xi'an i know it uses equation (3.1) but how? Why does one case have the integral whereas the other doesn't? And why is there an integral in the first place? Isn't that marginalizing $\phi$ ? $\endgroup$ – Daeyoung Lim Apr 12 '16 at 23:25
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$$ P(c_i=c|\vec{c_{-i}},y_i,\phi) =\frac{p(\vec{c_i},y_i,\phi)}{p(\vec{c_{-i}},y_i,\phi)} = \frac{p(y_i|\vec{c_{i}},\phi)p(\vec{c_{i}},\phi)}{p(y_i|\vec{c_{-i}},\phi)p(\vec{c_{-i}},\phi)} \\ = \frac{p(y_i|c_i,\phi)p(c_i|\vec{c_{-i},\phi})p(\vec{c_{-i}},\phi)}{p(y_i|\vec{c_{-i}},\phi)p(\vec{c_{-i}},\phi)} \\= \frac{p(y_i|c_i,\phi)p(c_i|\vec{c_{-i}},\phi)}{p(y_i)} $$

Here, $p(c_i|\vec{c_{-i}},\phi)=\frac{n_{-i,c}}{n-1+\alpha}$.

When $c_i$ is a existing one, then:

$$ p(y_i|c_i,\phi)=F(y_i,\phi_c) $$

When $c_i$ is a new cluster, then:

$$ p(y_i|c_i,\phi)=\int p(y_i|\phi_c,c_i)p(\phi_c|\phi,c_i)d\phi_c $$ Since $dG_0=p(\phi_c|\phi)$, we can conclude that: $$ p(y_i|c_i,\phi)=\int F(y_i,\phi_c)dG_0 $$

This is what I thought about the derivation. I am not quite understand that if it is correct. Did you solve this problem? @Daeyoung Lim

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