# Multiplication rule and regular conditional probability

I've been studying the conditions of existence of the regular conditional probability and have a question about it. Let's $(\Omega, \mathcal{B}, P)$ be a product probability space, and let's say the regular conditional probability exists :

\begin{align} P_{\times}(A\times B) = \int_B P(A|\omega)dP_0(\omega) \end{align}

with $P_0$ being the marginal of $P_{\times}$.

I was wondering how to derive from this the well known multiplication rule for measurable sets : \begin{align} P(A\cap B) = P(A|B)P(B) \end{align}

those two results seems pretty general to me and are somewhat related (I think). I just want to wrap my head around the two equations. If we assume the sets to be independents I managed to arrive to the same result $P(A\cap B) = P(A)P(B)$. So how theses equations relate ?

• This is asking to deduce that avery prime greater than 3 is odd from Fermat last theorem. What you call "multiplication rule for measurable sets" is usually taken as the definition of P(A|B) when P(B) is positive. – Did Jan 28 '15 at 10:56
• @Did I'm not sure that's a fair comparison. It seems that the OP already has a definition of conditional probability in terms of "more basic" notions of the space, and seeks to prove the equation usually quoted as the definition in more axiomatic treatments of probability spaces that ignore the details of the measure itself. – Mario Carneiro Jan 28 '15 at 13:13
• @Did Thank you for your input. So if we have $(\Omega_i, \mathcal{B}_i, P_i)$ two measure spaces and we build a product space where $P_i$ are the marginals of the two measures, and if we know (build) a conditional measure $P_{1|2}$, can we use independently the two equations ? I mean I have been using the second one all the time and got confused when I introduced myself to the regular conditional probability concept. That's why I needed somehow to reconciliate the two. – user149705 Jan 28 '15 at 15:53
• @MarioCarneiro What could be more basic than to define P(A|B) as P(A|B)=P(A∩B)/P(B) when P(B) is positive? – Did Jan 28 '15 at 16:44
• @user149705 I am afraid I fail to understand your last comment. – Did Jan 28 '15 at 20:27

Assume you have $(\Omega, \mathcal{B},\Bbb{P}_0)$ be a probability space where you define $X,Y$ two random variables. Let $\sigma(Y)$ be the sigma-algebra generated by $Y$. Assume that a regular conditional probability for $\Bbb{P}_0$ given $\sigma(Y)$ exists, that is assume there is a function $K_Y: \mathcal{B}\times \Omega \to [0,1]$ with the following properties:

1) $A \mapsto K_Y (A,\omega)$ is a probability for every $\omega \in \Omega$

2) $\omega \mapsto K_Y(A,\omega)$ is $\sigma(Y)$- measurable for every $A \in \mathcal{B}$

3) $K_Y(A,\omega) = \Bbb{E}[1_A \vert \sigma(Y)]$

Now note that

$$\Bbb{P}_0(X \in A, Y \in B) = \Bbb{E}[\Bbb{E}[1_{\{X \in A\}} 1_{\{Y \in B\}} \vert \sigma(Y)]] = \Bbb{E}[1_{\{Y \in B\}}\Bbb{E}[1_{\{X \in A\}} \vert \sigma(Y)]] = \Bbb{E}[1_{\{Y \in B\}}K(\{X \in A\},\omega)] = \int_B K_Y(\{X \in A\},\omega) \, d\Bbb{P}_0$$

So now calculate

$$\Bbb{P}_0(X \in A \vert Y \in B) = \frac{\Bbb{P}_0(X \in A, Y \in B)}{\Bbb{P}_0( Y \in B)} = \int_B K_Y(\{X \in A\},\omega) \, \frac{d\Bbb{P}_0}{\Bbb{P}_0( Y \in B)}$$

The probability of $[A] = \{X \in A\}$ given $[B] = \{Y \in B\}$ is therefore the mean of the conditional probabilities of $[A]$ given $\omega \in [B]$.

Note finally that $K(A,\omega)$ gives you a more general notion of conditional probability, once it allows you to speak of the conditional probability of an event given an outcome $\omega$ (even if $\Bbb{P}_0(\{\omega\}) = 0$)

• Thank you. This is somewhat the answer I was looking for. Both definition of conditional probabilities ( the mean and the more general one ) have their own requirements to exist. I understand both are also defined in their own way and related by your equation. – user149705 Jul 22 '15 at 15:19