Latent Dirichlet Allocation Derivation I am exploring different derivations for the the LDA and was a bit surprised about a step I found in the following paper : https://cxwangyi.files.wordpress.com/2012/01/llt.pdf
My question is about the transition between step 2.19 -> 2.20 which goes as follow : 
$$
p(W|Z,\beta) = \prod_{k=1}^{K} \left ( \frac{1}{B(\beta )} \int \prod_{v=1}^{V} \phi_{k,v}^{\psi_{k,v} + \beta_{v}-1} d\phi_k \right )
$$
to
$$
p(W|Z,\beta) = \prod_{k=1}^{K} \frac{B(\psi_k + \beta)}{B(\beta )}
$$
The precise question is : If we integrate out the right part of the equation it should sum to 1 and shouldn't be left with the Beta numerator... Or should we ?
Thx in advance for any details.
Keep.
 A: Better revise your Bayesian conjugacy properties. 
When you integrate out the merged terms of the binomial and the B Beta distribution you get a new Beta distribution summing up the corresponding terms like the second LaTex line you posted which is perfectly correct... The Dirichlet works just the same way as it is based on a multinomial and no more on the stated binomial. BTW It's where the term "collapsed" comes from.
Hope it helps [=
Regards.
A: Here is the proof for that.
First, note that:
$$ 
\int_{\phi_k} \frac{\Gamma(\sum_{v = 1}^{V} \psi_{k,v} + \beta_{v})}{\prod_{v=1}^{V}\Gamma (\psi_{k,v} + \beta_{v})} \prod_{v=1}^{V}  \phi_{k,v}^{\psi_{k,v} + \beta_{v}-1} d\phi_k = 1 \label{sum1}\tag{#}
$$
Now consider eq. 2.19:
$$p(W|Z,\beta) = \prod_{k=1}^{K} \left ( \frac{1}{B(\beta )} \int \prod_{v=1}^{V} \phi_{k,v}^{\psi_{k,v} + \beta_{v}-1} d\phi_k \right )$$
Multiply eq 2.19 by 1:
$$p(W|Z,\beta) = \prod_{k=1}^{K} \left ( \frac{1}{B(\beta )} \frac {\prod_{v=1}^{V}\Gamma (\psi_{k,v} + \beta_{v})} {\Gamma(\sum_{v = 1}^{V} \psi_{k,v} + \beta_{v})}  \int \frac{\Gamma(\sum_{v = 1}^{V} \psi_{k,v} + \beta_{v})} {\prod_{v=1}^{V}\Gamma (\psi_{k,v} + \beta_{v})} \prod_{v=1}^{V} \phi_{k,v}^{\psi_{k,v} + \beta_{v}-1} d\phi_k \right )$$
Now using eq \ref{sum1} from here and eq 2.5 from the article  https://cxwangyi.files.wordpress.com/2012/01/llt.pdf we have the result:
$$p(W|Z,\beta) = \prod_{k=1}^{K} \frac{B(\psi_k + \beta)}{B(\beta )}$$
