Conditional version of the total probability theorem (Ch 1 Problem 28 Intro to Probability)

This is Problem 1.28 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.

Let $$C_1,...,C_n$$ be disjoint events that form a partition of the state space. Let also A and B be events such that $$P(A \cap C_i) \gt 0$$ for all i. Show that

$$P(A | B) = \sum_{i=1}^n P(C_i | B) P(A | B \cap C_i)$$

The solution says

$$P(A \cap B) = \sum_{i=1}^n P((A \cap B) \cap C_i)$$

and converts that to $$P((A \cap B) \cap C_i) = P(B) P(C_i | B) P(A | B \cap C_i)$$ and uses the conditional probability formula to combine them into the proposed formula.

Ah, got it. This is the second time I've typed a question and it has triggered the answer. Let me know if this is right: they are replacing A $$\cap$$ B and turning it into one event which they put, collectively, into the total probability theorem.

I'll still post this in case others are working through a similar problem.

• I suppose they assume $P(B) > 0$? You are absolutely right. They first consider $A \cap B$ as one event. Next, they do the same for $B \cap C_i$ to get $P(A \cap B \cap C_i) = P(B \cap C_i)P(A|B \cap C_i)$ using the conditional probability formula. A final application of the same formula then gives the desired result. The well-known identity $(A \cap B) \cap C$ plays a more prominent role than one first would have thought. – Furrer Oct 12 '18 at 23:01