Bayes rule with multiple conditions I am wondering how I would apply Bayes rule to expand an expression with multiple variables on either side of the conditioning bar.
In another forum post, for example, I read that you could expand $P(a,z \mid b)$ using Bayes rule like this
(see Summing over conditional probabilities):
$$P( a,z \mid b) = P(a \mid z,b) P(z \mid b)$$
However, directly using Bayes rule to expand $P(a,z \mid b)$ doesn't seem to be the right way to start out:
$$P(a,z\mid b) = { P(b\mid a,z)P(a,z) \over P(a)}$$
 A: FYI, using the chain rule, P(a,z,b) = P(a|z,b)*P(z|b)*P(b)
A: Note that you didn't apply Bayes' Rule correctly; Bayes' Rule says that:

$$P(X|Y)={P(Y|X)P(X) \over P(Y)}$$

so your denominator should have actually been $P(b)$.
Instead, I will use the definition of conditional probability and multiplication rule (which together imply Bayes' Rule):

\begin{array}{cc}
P(X|Y) =\dfrac{P(X,Y)}{P(Y)} & (1)\\
P(X)P(Y|X) =P(X,Y)=P(Y)P(X|Y) & (2)
\end{array}

Thus, observe that:
$$ \begin{array}{r@{}ll}
P\big( (a,z) \mid b \big) &= \dfrac{P(a,z,b)}{P(b)} & \text{by (1), where } X=a,z \text{ and } Y=b\\
&= \dfrac{P(z,b)P\big(a \mid (z,b) \big)}{P(b)} &\text{by (2), where } X=a \text{ and } Y=z,b\\
&= \dfrac{P(b)P(z \mid b)P\big(a \mid (z,b) \big)}{P(b)} &\text{by (2), where } X=z \text{ and } Y=b\\
&= P(z \mid b)P\big(a \mid (z,b) \big)  \\
&= P\big(a \mid (z,b) \big) P(z \mid b) \\
\end{array} $$
as desired.
A: Since the op mentioned expand an expression with multiple variables on EITHER side of the conditioning bar., here is my attempt to derive P(a|z,b) using Bayesian rule, assuming that A is a discrete variable:
$$\begin{align}
P(A=a1|z,b) &= \frac{P(a_1, z, b)}{P(z,b)} \\
         &= \frac{P(a_1,b|z) P(z)}{P(b|z)P(z)} \\
         &= \frac{P(a_1,b|z)}{P(b|z)} \\
         &= \frac{P(b|a_1,z)P(a_1|z)}{P(b|z)} \\
         &=  \frac{P(b|a_1,z)P(a_1|z)}{P(b|a_1,z)P(a_1|z) + P(b|a_2,z)P(a_2|z) + ...}
\end{align}$$
Compare this with the "single variable" version where
$$\begin{align}
P(A=a_1|b) &= \frac{P(a_1,b)}{P(b)} \\
         &= \frac{P(b|a_1)P(a_1)}{P(b)} \\
         &=  \frac{P(b|a_1)P(a_1)}
                  {P(b|a_1)P(a_1) + P(b|a_2)P(a_2) + ..}
\end{align}$$
Can we think of Z as a extra "condition" that's added to both the numerator and denominator?
