Chain Rule for Conditional Probability?

Question

So while the most basic form of the product rule for probability is $P(A \cap B) = P(A) P(B|A)$, I heard that for any events $A, B, C,$ the following also holds:

$$P(A \cap B | C) = P(A|C) P(B| A \cap C). $$

I've been trying to derive this formula and/or find the general form of this for $n$ events, but so far haven't had any success. Could someone help me see why

$$P(A \cap B | C) = P(A|C) P(B| A \cap C)$$

is true (how we get from $P(A \cap B) = P(A) P(B|A)$ to this) and if there's a more general formula for this?

Answer 1

The short form is, all real world probabilities are this way. The "given $C$" aspect is just a way of saying, ceteris paribus or "all other things being equal."

Imagine a Venn diagram with all three circles. Now restrict your attention to the circle $C$. Now imagine that that is the entire set universe.

$$P(A \cap B) = P(A|B)\cdot P(B)$$ is actually the definition of $P(A|B)$. So you can say that the formula you want to prove is also true by definition.

Answer 2

It is very simple

$$P(AB|C)=\frac{P(ABC)}{P(C)}=\frac{P(AC)P(B|AC)}{P(C)}=\frac{P(AC)}{P(C)}\cdot P(B|AC)=P(A|C)P(B|AC)$$

Answer 3

The general form of this equation has no name that I remember, now, but here it is:

For events $A_1, \ldots, A_n$ with non-zero probability we have $$\mathbb P\left(\bigcap_{i=1}^nA_i\right)=\mathbb P(A_1)\cdot\prod_{i=2}^n\mathbb P\left(A_i|\bigcap_{k=1}^{i-1}A_k\right)$$

or equivalently $$\mathbb P\left(\bigcap_{i=2}^nA_i|A_1\right)=\prod_{i=2}^n\mathbb P\left(A_i|\bigcap_{k=1}^{i-1}A_k\right).$$

Of course, this is also true for any other numeration of the events.

The case $n=3$ is proved by introducing a $1$ $$\mathbb P(A\cap B\cap C)=\frac{\mathbb P(A\cap B\cap C)}{\mathbb P(B\cap C)}\cdot \frac{\mathbb P(B\cap C)}{\mathbb P(C)}\cdot\mathbb P(C)$$

You can prove your result in the original question by deviding both sides of the equation by $\mathbb P(C)$.

The strategy to prove the general result is then just induction.