In the Wikipedia article for conditional expectation, conditional probability is defined in terms of conditional expectation.

  1. Given a sub sigma algebra of the one on a probability space.

    Given a probability space $(\Omega, \mathcal{F}, P)$, a conditional probability $P(A \mid \mathcal{B})$ of a measurable subset $A \in \mathcal{F}$ given a sub sigma algebra $\mathcal{B}$ of $\mathcal{F}$, is defined as the conditional expectation $E(A \mid \mathcal{B})$ of indicator function $i_A$ of $A$ given $\mathcal{B}$, i.e. $$ P(A \mid \mathcal{B}): = E(A \mid \mathcal{B}), \forall A \in \mathcal{F}.$$

    So actually the conditional probability $P(\cdot \mid \mathcal{B})$ is a mapping $: \Omega \times \mathcal{F} \rightarrow \mathbb{R}$.

    A conditional probability $P(\cdot \mid \mathcal{B})$ is called regular if $P(\cdot|\mathcal{B})(\omega), \forall \omega \in \Omega$ is also a probability measure.


    what are some necessary and/or sufficient conditions for a conditional probability $P(\cdot \mid \mathcal{B})$ to be regular?

  2. Given a r.v. on a probability space.

    Suppose $(\Omega, \mathcal{F}, P)$ is a probability space and $(U, \mathcal{\Sigma})$ is a measurable space. There seem to be two ways of defining the conditional expectation $E(X\mid Y)$ of a r.v. $X: \Omega \rightarrow \mathbb{R}$ given another r.v. $Y: \Omega \rightarrow U$, either as a $\sigma(Y)$-measurable mapping $: \Omega \rightarrow \mathbb{R}$, or as a $\Sigma$-measurable mapping $: U \rightarrow \mathbb{R}$, as in my previous post.

    If one let $X$ to be the indicator function $1_A$ for some $A \in \mathcal{F}$, one can similarly define $E(1_A \mid Y)$ to be conditional probability of $A$ given $Y$, denoted as $P(A\mid Y)$. Therefore $P(\cdot \mid Y)$ is a mapping $: \Omega \times \mathcal{F} \rightarrow \mathbb{R}$ or a mapping $: U \times \mathcal{F} \rightarrow \mathbb{R}$.


    (1). What are some necessary and/or sufficient conditions for $P(\cdot \mid Y)$ to be regular, i.e. to be a mapping $: \Omega \rightarrow \{ \text{probability measures on }(\Omega, \mathcal{F}) \}$ or a mapping $: U \rightarrow \{ \text{probability measures on }(\Omega, \mathcal{F}) \}$?

    (2). Under what kinds of conditions, will $P(X \mid Y)$ defined as above be equal to the ratio $\frac{P(X, Y)}{P(Y)}$, the definition used in elementary probability?

Thanks and regards! References (links or books) will also be appreciated!


Conditional probabilities do not give a unique function on the sample space. Since conditional expectations are only defined up to a measure zero set and one has to make an uncountable number of selections, the essential problem is whether one can "glue" them together in coherent way, so that you can actually calculate conditional probabilities by integrating the function. There are several notions of regular conditional probabilities and this paper by Faden gives necessary and sufficient conditions for some of them. For the particular version you mentioned, little is known about necessary conditions. The strongest results on the existence of regular conditional probabilities can be found in this paper by Pachl, but he only requires them to be measurable with respect to the completion of the measure. The machinery he uses is rather sophisticated, his method is based on using a lifting that he then shows (under some condition, compactness) to give a countably additive probability. The most extensive resource on conditional probabilities is probably the book Conditional Measures and Applications by M.M. Rao. The book is not recommended for its readability. Your question is addressed in chapter 3 in a comprehensive manner.


For question 1, I don't know if necessary and sufficient conditions exist but here are two sufficient conditions I know :

  • the space is a standard probability space, look here,
  • the space is a Radon space, look here.

I think that there is another one but I can't remember its exact form, except that I know that the space should be Polish and the sigma-algebra countably generated, and maybe another condition.


  • $\begingroup$ Thanks! Some other references on this topic? $\endgroup$
    – Tim
    Feb 25 '11 at 8:57

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.