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 $P(A\cap G)$?

  • $\begingroup$ 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 $P(G)=0$. $\endgroup$ Aug 7 at 22:33
  • $\begingroup$ What book is this from? $\endgroup$
    – littleO
    Aug 7 at 22:38
  • 1
    $\begingroup$ @littleO en.wikipedia.org/wiki/Conditional_probability_distribution But I was reading Billingsley Probability and Measure $\endgroup$ Aug 7 at 22:38

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Aug 7 at 22:41
  • $\begingroup$ Yes it does (up to sets of measure $0$). $\endgroup$ Aug 7 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.