Deriving mathematical model of pLSA After knowing how LSA works, I went on continue reading on pLSA but couldn't really make sense of the mathematical formula. This is what I get from wikipedia (other academic papers/tutorial show similar form)
\begin{align}
P(w,d) & = \sum_{c} P(c) P(d|c) P(w|c)\\
& = P(d) \sum_{c} P(c|d) P(w|c)\\
\end{align}
I gave up trying to derive it, and found this instead
\begin{align}
P(c|d) & = \frac{P(d|c)P(c)}{P(d)}\\
P(c|d)P(d) & = P(d|c)P(c)\\
P(w|c)P(c|d)P(d) & = P(w|c)P(d|c)P(c)\\
P(d) \sum_{c} P(w|c)P(c|d) & = \sum_{c} P(w|c)P(d|c)P(c)
\end{align}
How does the summation appear at the last line? I am currently reading through some tutorial on Bayesian Inferencing (learnt basic probability rules and Bayesian theorem before but can't really see them being useful enough here).
 A: I am assuming you want to derive:
\begin{align*} 
P(w,d) = \sum_{c} P(c) P(d|c) P(w|c)
&= P(d) \sum_{c} P(c|d) P(w|c)
\end{align*}
Further, this is similar to Probabilistic latent semantic indexing (cf. Blei, Jordan, and Ng (2003) Latent Dirichlet Allocation. JMLR section 4.3). PLSI posits that a document label $d$ and a word $w$ are conditionally independent given an unobserved topic $z$.
If this is true, your formula is a simple consequence of Bayes theorem. Here are the steps:
\begin{align*}
P(w, d) &= \displaystyle \sum_z P(w, z, d)\\
 & = \displaystyle \sum_z P(w, d | z) p(z)\\
 &= \displaystyle \sum_z P(w | z) p(d|z) p(z),
\end{align*}
where factorization into products is because of conditional independence. 
Now use Bayes theorem again to get
\begin{align}
\displaystyle \sum_z P(w | z) p(d|z) p(z) &= \displaystyle \sum_z P(w | z) p(z,d)\\
&= \displaystyle \sum_z P(w | z) p(z|d)p(d)\\
&= p(d)\displaystyle \sum_z P(w | z) p(z|d)
\end{align}
A: The line $P(c|d)P(c) = P(d|c)P(c)$ (your eq 2) should be $P(c|d)P(d) = P(d|c)P(c)$. 
I'm not sure why you don't think Bayes theorem and basic probability rules are useful:
Eq 1 is Bayes theorem (ie recognizing that $P(d|c)P(c) = P(c,d)$ and plugging in to the definition of conditional probability)
Eq 2 follows immediately from eq 1
Eq 3 is just eq 2 multiplied through by $P(w|c)$. 
Since eq 3 holds for all $c$ the sums are equal. Then since $w$ is independent of $d$ given $c$ (an assumption from the model), $P(w|c)P(c|d) = P(w|c, d)P(c|d) = P(w,c|d)$ and so $\sum_c\ P(w,c|d) = P(w|d)$ by the law of total probability, giving you $P(w|d)P(d)$. 
Finally, $P(w|d)P(d)=P(w,d)$ from the definition of conditional probability.
So basic probability is in fact both necessary and sufficient for the derivation!
