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I run in to this formula when reading a tutorial:

$$ \begin{align} P(\pi|\mathbf L;\gamma_{\pi1}, \gamma_{\pi0}) & =P(\mathbf L|\pi)P(\pi|\gamma_{\pi1},\gamma_{\pi0})\tag{28} \\ &\propto [\pi^{C_1}(1-\pi)^{C_0}][\pi^{\gamma_{\pi1}-1}(1-\pi)^{\gamma_{\pi0}-1}]\tag{29}\\ &\propto\pi^{C_1+\gamma_{\pi1}-1}(1-\pi)^{C_0+\gamma_{\pi0}-1}\tag{30} \end{align} $$

I was just wondering for formula (28), if we get the left side from the right side, why the left side of equation is $P(\pi \mid \mathbf L;\gamma_{\pi1}, \gamma_{\pi2})$ instead of $P(\mathbf L \mid \gamma_{\pi1}, \gamma_{\pi2})$ (according to the chain rule)?

Edit: This is the link of the tutorial. Basically, this is to derive a Gibbs sampler for Naive Bayes model and Figure 4 is the plate representation of its graphical model. $\pi \sim Beta(\gamma_{\pi1}, \gamma_{\pi2})$, $L \sim Bernoulli(\pi)$. $C_{0}$ and $C_{1}$ denote all documents with negative label and all documents with positive label respectively.

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  • $\begingroup$ I've edited your question to use Mathjax in place of the image. Please review that it still says what you intend. A tutorial for mathjax can be found here. meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax ♦
    May 21, 2015 at 14:23
  • $\begingroup$ Could you provide a little more context? Exactly how does this tutorial define or describe the key objects in your question, including $\pi$, $\mathbf{L}$, and the $C_i$? $\endgroup$
    – whuber ♦
    May 21, 2015 at 17:08
  • $\begingroup$ @whuber Thanks for letting me know. This is the link of the tutorial umiacs.umd.edu/~resnik/pubs/LAMP-TR-153.pdf You can find the graphical model in Figure 4. Basically, this is to derive a Gibbs sampler for Naive Bayes model. $\pi \sim Beta(\gamma_{\pi1}, \gamma_{\pi2})$, $L \sim Bernoulli(\pi)$ $\endgroup$
    – yvetterowe
    May 21, 2015 at 20:59
  • $\begingroup$ @user777 Thank you very much for editing it :) $\endgroup$
    – yvetterowe
    May 21, 2015 at 21:03
  • $\begingroup$ That description of $\mathbf{L}$ seems inconsistent with its use in the equations. It appears instead that $\mathbf{L}$ might be synonymous with the ordered pair $(C_0,C_1)$ of counts of failures and successes, respectively, in $C_0+C_1$ independent Bernoulli$(\pi)$ trials. I suppose if $C_0+C_1=1$ then the description of $\mathbf{L}$ as Bernoulli would be accurate, but I cannot determine that because I still don't know what $C_0$ and $C_1$ might be. It would be strange to use two variables in that way to represent a single 0/1 outcome. $\endgroup$
    – whuber ♦
    May 21, 2015 at 22:01

1 Answer 1

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This is a Bayes theorem application, as in: $$P(A|B)\sim P(B|A)P(A)$$

The semicolon separates the parameters, in your case it's $\gamma_{\pi1}, \gamma_{\pi0}$. So the Equation 28 has in the right side the probability distribution of $\pi$ given the parameters $\gamma_{\pi1}, \gamma_{\pi0}$ denoted as: $P(\pi|\gamma_{\pi1},\gamma_{\pi0})$. So, you can map $A=L$, $B=\pi$ to see how the Bayes theorem is applied here.

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    $\begingroup$ While I agree that you answer is correct, I think additional explanation is needed. In particular, I think you'd need to explain where the gammas are (implicitly) in the first term on the right, and why they're not written there. [I can imagine a reader could be likely to try to associate them either with A or with B in your answer, and become confused.] $\endgroup$
    – Glen_b
    May 21, 2015 at 14:34

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