5
$\begingroup$

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.

Improve this question
$\endgroup$
7
  • $\begingroup$ 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 ♦
    Commented 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 ♦
    Commented 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
    Commented May 21, 2015 at 20:59
  • $\begingroup$ @user777 Thank you very much for editing it :) $\endgroup$
    – yvetterowe
    Commented 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 ♦
    Commented May 21, 2015 at 22:01

1 Answer 1

Reset to default
-3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ 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
    Commented May 21, 2015 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.