Bayes Theorem - one event dependent on two

Question

I do have problems to put up the Bayes Theorem for one event depending on two: $\mathbb{P}(q|z,x)$. Two events depending on one was still ok, \begin{equation*} \mathbb{P}(x,q|z) = \frac{\mathbb{P}(z|x,q) }{\mathbb{P}(z))} \mathbb{P}(x|z) \mathbb{P}(q) \end{equation*}.

Can somebody please help me with that?

Answer 1

Here $x$, $q$ and $z$ are just labels for events. You can just reassigns them so you have: $$ \mathbb{P}(z,x|q) = \frac{\mathbb{P}(q|z,x) }{\mathbb{P}(q)} \mathbb{P}(z|x) \mathbb{P}(x) $$ Now rearrange.

You can also derive all of these relations from first principals by going back to looking at $\mathbb{P}(z,x,q)$ (the order of $z,x,q$ is arbitary and they can be reordered at will). Then: $$ \mathbb{P}(z,x,q)=\mathbb{P}(z,x|q)\mathbb{P}(q)=\mathbb{P}(q|z,x)\mathbb{P}(z,x)=\mathbb{P}(q|z,x)\mathbb{P}(z|x)\mathbb{P}(x) $$

Answer 2

The theorem is no different in this case: $$P(q | z, x) = \frac{P(z, x | q)P(q)}{P(z, x)}$$ You can further break down $P(z, x)$ as $$P(z, x) = P(z | x)P(x) = P(x | z)P(z)$$