Derivation of M-step for pLSA

I was looking at section 6 of these notes and trying to understand the derivation of the M-step at the top of page 10. I understood the derivation for the model without background, but I do not understand where the $(1-P(Z_{d,w}=\theta_B \mid d,w))$ terms come from. Intuitively, I understand that they are down-weighting words that are mostly background and not topic-related. But, how do you get the expressions in the M-step from the derivatives?

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Firstly, thanks for referring to my note. I think the first answer is correct. I'll try to improve the note. –  sviatoslav hong Oct 12 '12 at 16:37

This derivation may be more understandable than the one in the paper.

The log likelihood of the background model's joint distribution is the following:

$$\log p(w,d,Z)=\sum_{d,w}n(d,w)\mathbb{I}\left(Z_{w,d}=\theta_{B}\right)\log\left(\lambda_{B}p(w|\theta_{B})\right)+\sum_{d,w,z}n(d,w)\mathbb{I}\left(Z_{w,d}=z\right)\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\log\left(\left(1-\lambda_{B}\right)p(w|z)p(z|d)p(d)\right)$$

Then take the expectation to get a bound on the marginal log likelihood

$$\log p(w,d) \ge \sum_{d,w} n(d,w)E\left[\mathbb{I}\left(Z_{w,d}=\theta_{B}\right)\right]\log\left(\lambda_{B}p(w|\theta_{B})\right)+\sum_{d,w,z} n(d,w) E\left[\mathbb{I}\left(Z_{w,d}=z\right)\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\right]\log\left(\left(1-\lambda_{B}\right)p(w|z)p(z|d)p(d)\right)$$

Then take the derivative with respect to the bound to maximize ($\beta$ is a Lagrange multiplier enforcing the normalization constraint $\sum_w p(w|z)=1$):

$$\frac{\partial H}{\partial p(w|z)}=\frac{\sum_d n(d,w)E\left[\mathbb{I}\left(Z_{w,d}=z\right)\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\right]}{p(w|z)}-\beta=0$$

$$p(w|z)\propto \sum_d n(d,w)E\left[\mathbb{I}\left(Z_{w,d}=z\right)\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\right]$$

The expectation above ends up being

$$E\left[\mathbb{I}\left(Z_{w,d}=z\right)\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\right]=E\left[\mathbb{I}\left(Z_{w,d}=z\right)\right] E\left[\mathbb{I}\left(Z_{w,d}\neq\theta_{B}\right)\right]=P\left(Z_{w,d}=z\right |w,d)P\left(Z_{w,d}\neq\theta_{B}|w,d\right)=P\left(Z_{w,d}=z\right|w,d) (1-P\left(Z_{w,d}=\theta_{B}|w,d\right))$$

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I don't understand the $P(Z_{d,w}=z|Z_{d,w}\neq \theta_B)=1-P(Z_{d,w}=\theta_B)$. Shouldn't the right hand side have a $z$ in it somewhere? Otherwise, all possible values of $Z_{d,w}$ that are not $\theta_B$ have the same probability for a particular $d,w$ (i.e. the topic probabilities, $P(Z_{d,w}=z|Z_{d,w}\neq \theta_B)$, seem to be independent of the topic). –  ccb Oct 12 '12 at 19:16
(multiple comments since too long) Perhaps, the notation is overloaded. Maybe, $P(Z_{d,w}=z) = P(Z_{d,w}=z, Z_{d,w}\neq \theta_B)$, at first. –  ccb Oct 12 '12 at 19:36
This gives $P(Z_{d,w}=z, Z_{d,w}\neq \theta_B) = P(Z_{d,w}\neq \theta_B)P(Z_{d,w}=z|Z_{d,w}\neq \theta_B) = (1-P(Z_{d,w}= \theta_B))P(Z_{d,w}=z|Z_{d,w}\neq \theta_B)$. –  ccb Oct 12 '12 at 19:36
But, now (overloading notation - this is the notational shortcut you mentioned) $P(Z_{d,w}=z|Z_{d,w}\neq \theta_B) = P(Z_{d,w}=z)$. Plugging this in to the expression above, would give $(1-P(Z_{d,w}= \theta_B))P(Z_{d,w}=z|Z_{d,w}\neq \theta_B) = (1-P(Z_{d,w}= \theta_B))P(Z_{d,w}=z)$. –  ccb Oct 12 '12 at 19:36
What all of this would mean is that the $(1-P(Z_{d,w}= \theta_B))P(Z_{d,w}=z)$ portion of the M-step updates is trying to represent $P(Z_{d,w}=z, Z_{d,w}\neq \theta_B)$, which might be interpreted as the expectation of two sets of hidden variables. Am I understanding any of this correctly? It might be easier to understand if there was a separate hidden variable, say $B_{d,w}$, to indicate background. –  ccb Oct 12 '12 at 19:36