# Conditional probability measure theoretic definition

Let $$(\Omega, \mathcal{F}, P)$$ be a probability space, $$\mathcal{G} \subseteq \mathcal{F}$$ a $$\sigma$$-field in $$\mathcal{F}$$. Given $$A \in \mathcal{F}$$, the Radon-Nikodym theorem implies that there is $${ }^{[3]}$$ a $$\mathcal{G}$$-measurable random variable $$P(A \mid \mathcal{G}): \Omega \rightarrow \mathbb{R}$$, called the conditional probability, such that $$\int_{G} P(A \mid \mathcal{G})(\omega) d P(\omega)=P(A \cap G)$$ for every $$G \in \mathcal{G}$$, and such a random variable is uniquely defined up to sets of probability zero. A conditional probability is called regular if $$\mathrm{P}(\cdot \mid \mathcal{B})(\omega)$$ is a probability measure on $$(\Omega, \mathcal{F})$$ for all $$\omega \in \Omega$$ a.e.

I am used to the non-measure theoretic definition of conditional probability defined for events where $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$. Why is this function not defined like

$$\frac{1}{P(G)}\int_{G} P(A \mid \mathcal{G})(\omega) d P(\omega)=\frac{P(A \cap G)}{P(G)}$$

which would make it in line with the definition for events? I understand that $$P(G)=0$$ is an issue, but what is the motivation of defining this representation for $$P(A\cap G)$$?

• Note that the thing being defined is the random variable $P(A|\mathcal{G})$ and clearly, if a random variable fits into the one definition it fits into the other, except in the case where $P(G)=0$. Aug 7 at 22:33
• What book is this from? Aug 7 at 22:38
• @littleO en.wikipedia.org/wiki/Conditional_probability_distribution But I was reading Billingsley Probability and Measure Aug 7 at 22:38

## 1 Answer

In the case where $$\mathcal G$$ is the $$\sigma$$-algebra generated by event $$G$$, $$P(A \mid \mathcal G)$$ is (up to sets of measure $$0$$) $$P(A|G)$$ when $$G$$ is true, $$P(A|G^c)$$ when $$G$$ is false. The equation then says $$P(A \cap G) = P(G) P(A | G)$$ and $$P(A \cap G^c) = P(G^c) P(A | G^c)$$, which fits with your non-measure theoretic definition.

• Does this mean that $P(A|\mathcal{G})$ takes on exactly two values? And if I had events $B_1,\dots,B_n$ partitioning $\Omega$, then $P(A|\sigma(B_1,\dots,B_n))$ would take on $n$ values where $P(A|\sigma(B_1,\dots,B_n)) (\omega)=P(A|B_i)$? Is $P(A|\mathcal{G})$ doing the same type of "smoothing" that conditional expectation is doing? Aug 7 at 22:41
• Yes it does (up to sets of measure $0$). Aug 7 at 23:17