What is the intuition or proof behind the conditional Bayes' theorem?

Question

In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:

$$\Pr(B_i\mid A \cap C) = \dfrac{\Pr(B_i\mid C) \cdot \Pr(A\mid B_i \cap C)}{\sum (\Pr(B_i\mid C)\cdot \Pr(A\mid B_i \cap C))}$$

What is the intuition or mathematical proof behind this?

Answer 1

It's directly equivalent to: $$\begin{align} \Pr(B_i\mid A) & = \frac{\Pr(B_i \cap A)}{\Pr(A)} \\[1ex] &= \frac{\Pr(B_i)\cdot \Pr(A\mid B_i)}{\sum_k \Pr(B_k)\cdot\Pr(A\mid B_k)} \end{align}$$ But with conditioning when given $C$: $$\begin{align} \Pr(B_i\mid A\cap C) & = \frac{\Pr(B_i \cap A\mid C)}{\Pr(A\mid C)} \\[1ex] &= \frac{\Pr(B_i \mid C)\cdot\Pr(A\mid B_i\cap C)}{\sum_k \Pr(B_k\mid C)\cdot\Pr(A\mid B_k\cap C)} \end{align}$$

How do you know equation 3 is true?

$\begin{align} \Pr(B_i\mid A\cap C) & = \frac{\Pr(B_i\cap A \cap C)}{\Pr(A\cap C)} & \text{by definition} \\[1ex] & = \frac{\Pr(B_i\cap A\mid C)\cdot \Pr(C)}{\Pr(A\mid C)\cdot\Pr(C)} & \text{by the same} \\[1ex] & = \frac{\Pr(B_i\cap A\mid C)}{\Pr(A\mid C)} & \text{by cancelation} \end{align}$

Also, can you explain how you derived the denominator in equation 4?

By the Law of Total Probability. If the set of $\{B_k: k\in \{1..n\}\}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.

$$\begin{align}\Pr(A) \quad & = \Pr(B_1)\cdot\Pr(A\mid B_1) + \cdots + \Pr(B_k)\cdot\Pr(A\mid B_k) + \cdots \Pr(B_n)\cdot\Pr(A\mid B_n) \\ & = \sum_{k=1}^n \Pr(B_k)\cdot\Pr(A\mid B_k) \end{align}$$

I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).

Yes. Thuswise: $$\begin{align} \Pr(A\mid C) \quad & = \frac{\Pr(A\cap C)}{\Pr(C)} & \text{by conditional probability} \\[1ex] & = \frac{\sum_k \Pr(A\cap C\cap B_k)}{\Pr(C)} & \text{by total probability} \\[1ex] & = \frac{\sum_k \Pr(B_k\cap C)\cdot\Pr(A\mid C\cap B_k)}{\Pr(C)} & \text{by conditional probability} \\[1ex] & = \sum_k \frac{\Pr(B_k\cap C)}{\Pr(C)}\cdot\Pr(A\mid C\cap B_k) & \text{by rearrangement} \\[1ex] & = \sum_k \Pr(B_k\mid C)\cdot\Pr(A\mid C\cap B_k) & \text{by conditional probability} \end{align}$$