# Bayes Rule clarification

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

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

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.