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?

  • $\begingroup$ This follows from equation (3.1). $\endgroup$
    – Xi'an
    Apr 10, 2016 at 9:02
  • $\begingroup$ How does it follow? Is it just a standard derivative? $\endgroup$ Apr 12, 2016 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 $\phi$ ? $\endgroup$ Apr 12, 2016 at 23:25

1 Answer 1


$$ P(c_i=c|\vec{c_{-i}},y_i,\phi) =\frac{p(\vec{c_i},y_i,\phi)}{p(\vec{c_{-i}},y_i,\phi)} = \frac{p(y_i|\vec{c_{i}},\phi)p(\vec{c_{i}},\phi)}{p(y_i|\vec{c_{-i}},\phi)p(\vec{c_{-i}},\phi)} \\ = \frac{p(y_i|c_i,\phi)p(c_i|\vec{c_{-i},\phi})p(\vec{c_{-i}},\phi)}{p(y_i|\vec{c_{-i}},\phi)p(\vec{c_{-i}},\phi)} \\= \frac{p(y_i|c_i,\phi)p(c_i|\vec{c_{-i}},\phi)}{p(y_i)} $$

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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