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:
$$P(a,z\mid b) = { P(b\mid a,z)P(a,z) \over P(a)}$$
$\endgroup$ 33 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$FYI, using the chain rule, P(a,z,b) = P(a|z,b)*P(z|b)*P(b)
$\endgroup$ 2 $\begingroup$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?
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