2
$\begingroup$

On page 12 of this tutorial, it states

$$P(\pi \mid \mathbf{L}; \gamma_{\pi1}, \gamma_{\pi0}) = P(\mathbf{L} \mid \pi) P(\pi \mid \gamma_{\pi1}, \gamma_{\pi0})$$

I'm having some trouble seeing why this equality holds from Bayes' rule.

For context (if important):

$\gamma_{\pi1}, \gamma_{\pi0}$ are fixed parameters of the Beta distribution.

$\pi$ is drawn from the $Beta(\gamma_{\pi1}, \gamma_{\pi0})$.

And $\mathbf{L}$ is a 2-d vector drawn from the $Bernoulli(\pi)$.

From my understanding of Bayes' rule, I get something like

$$P(\pi \mid \mathbf{L}; \gamma_{\pi1}, \gamma_{\pi0}) = \frac{P(\mathbf{L} \mid \pi, \gamma_{\pi1}, \gamma_{\pi0}) P(\pi \mid \gamma_{\pi1}, \gamma_{\pi0})}{P(\mathbf{L \mid \gamma_{\pi1}, \gamma_{\pi0}})}$$

$\endgroup$
2
$\begingroup$

Yes, your answer is correct although you can drop $\gamma_{\pi_0}$ and $\gamma_{\pi_1}$ from the condition in $P(L \mid \pi, \gamma_{\pi_0}, \gamma_{\pi_1})$ because $L$ is conditionally independent of these hyper parameters given $\pi$. However, even though their expression is off by a factor of $P(L \mid \gamma_{\pi_0}, \gamma_{\pi_1})$ this doesn't really change the argument because they would only need to replace "$=$" with "$\propto$" in the first statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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