Where does this conditional probability law come from?

Question

I was trying to follow a computation done in my class notes, and was having difficulty seeing the inspiration for a part of the manipulation in a question regarding probability.

I did some Googling, found another set of notes online which gave this rule:

$$\mathbb{P}(A \cap B |C) = \mathbb{P}(A|C)\cdot\mathbb{P}(B|A\cap C) $$

This cleared up my confusion with regards the manipulation in my original notes, but now I don't actually understand what that formula is based on. It doesn't look like a variant of Bayes' Theorem as far as I can see, and it extends beyond my knowledge of conditional probability.

Can someone explain how one might derive this rule from basic probability laws? I imagine it's actually fairly simple but I can't find a source online that explains it rather than just stating without proof or explanation.

Answer 1

It is known as the chain rule. A justification can be seen, assuming of course $\Pr[C] > 0$ and $\Pr[A\cap C] > 0$, as $$\begin{align} \Pr[A\cap B \mid C ] &= \frac{\Pr[A\cap B \cap C ]}{\Pr[C]} = \frac{\Pr[A\cap B \cap C ]}{\Pr[A\cap C]}\cdot \frac{\Pr[A \cap C ]}{\Pr[C]} \\ &= \Pr[B\mid A\cap C ]\cdot \Pr[A \mid C ] \end{align}$$ where the first and last equalities are by definition of conditional probabilities.

Answer 2

In general $\mathbb{P}\left(A\mid B\right)$ is defined by the equality $$\mathbb{P}\left(B\right)\mathbb{P}\left(A\mid B\right)=\mathbb{P}\left(A\cap B\right)$$ If $\mathbb{P}\left(B\right)\neq0$ then this comes to the same as $\mathbb{P}\left(A\mid B\right)=\frac{\mathbb{P}\left(A\cap B\right)}{\mathbb{P}\left(B\right)}$.

Multiply both sides in the equation mentioned in your question by $\mathbb{P}\left(C\right)$.

The LHS is: $\mathbb{P}\left(C\right)\mathbb{P}\left(A\cap B\mid C\right)=\mathbb{P}\left(A\cap B\cap C\right)$

The RHS is: $\mathbb{P}\left(C\right)\mathbb{P}\left(A\mid C\right)\mathbb{P}\left(B\mid A\cap C\right)=\mathbb{P}\left(A\cap C\right)\mathbb{P}\left(B\mid A\cap C\right)=\mathbb{P}\left(A\cap B\cap C\right)$

So LHS$=$RHS implying that the equation is correct if $\mathbb{P}\left(C\right)\neq0$.

Answer 3

Conditional probabilities: for the probability measure $\mathbb{P}$ (appropriately defined now, we don't like zero divisions), $$ \mathbb{P}\left(B \,|\, A\right){}={}\frac{\mathbb{P}\left(B \,\cap \, A\right)}{\mathbb{P}\left(A\right)} $$

Choose the probability measure $\mathbb{P}\left( \dot\,\,|\,\, C\right)$ and apply it to this definition above, replacing $\mathbb{P}\left(\,\dot\, \right)$ as stated, to get the result.