# Regular Conditional Probability given $X$

In Williams' "Probability with Martingales" he defines a regular conditional probability given $\mathcal{G} \subset \mathcal{F}$ as the map $$\mathbf{P}(\cdot, \cdot) : \Omega \times \mathcal{F} \to [0,1)$$ such that \begin{align*} &1.\quad \forall F \in \mathcal{F}, \omega \mapsto \mathbf{P}(\omega,F) \text{ is a version of } \mathbf{P}(F \mid \mathcal{G}),\\ &2.\quad\text{ for almost every } \omega \in \Omega, F \mapsto \mathbf{P}(\omega,F) \text{ is a probability measure on } (\Omega,\mathcal{F}). \end{align*} Property 1 tells us that for any $F \in \mathcal{F}$ and any $G \in \mathcal{G}$, \begin{align*} \int_G \mathbf{P}(\omega, F) \, \mathrm{d}P(\omega) & = \int_G \mathbf{P}(F \mid \mathcal{G})(\omega) \, \mathrm{d}P(\omega) \\ & = \int_G E(\mathbb{I}_F \mid \mathcal{G})(\omega) \, \mathrm{d}P(\omega) \\ & = \int_G \mathbb{I}_F(\omega) \, \mathrm{d}P(\omega) \\ & = P(F \cap G). \end{align*}

Fair enough, but I'm trying to get a handle on the regular conditional probability given a random variable $X: \Omega \to \mathbb{R}$. It seems like we should define it as the map $$\mathbf{P}(\cdot, \cdot) : \mathbb{R} \times \mathcal{F} \to [0,1)$$ such that \begin{align*} &1.\quad \forall F \in \mathcal{F}, x \mapsto \mathbf{P}(x,F) \text{ is a version of } \mathbf{P}(F \mid \sigma(X)),\\ &2.\quad\text{ for almost every } x \in \mathbb{R}, F \mapsto \mathbf{P}(x,F) \text{ is a probability measure on } (\Omega, \mathcal{F}). \end{align*} However, I'm not sure how to get the "defining equation" as above, but here's my shot. Let $F \in \mathcal{F}$ and $C \in \sigma(X) \subset \mathcal{F}$. Note $C = X^{-1}(B)$ for some $B \in \mathcal{B}$. Letting $\Lambda_X$ be the distribution of $X$, by property 1 we get \begin{align*} \int_B \mathbf{P}(x,F) \, \mathrm{d}\Lambda_X(x) & = \int_C \mathbf{P}(X(\omega), F) \, \mathrm{d} P(\omega) \\ & = \int_C \mathbf{P}(F \mid \sigma(X))(\omega) \, \mathrm{d} P(\omega) \\ & = \int_C E(\mathbb{I}_F \mid \sigma(X))(\omega) \, \mathrm{d} P(\omega) \\ & = \int_C \mathbb{I}_F(\omega) \, \mathrm{d} P(\omega) \\ & = P(F \cap \{X \in B\}). \end{align*} Does this seem correct?

Update Based on Conrado Costa's answer, I think the proper definition for the regular conditional probability given $X$ should be as follows. Below, let $X: (\Omega, \mathcal{F}) \to (E, \mathcal{E})$.

Assume the regular conditional probability given $\sigma(X)$ exists. That is, there exists a function $\mathbf{P}^2 : \Omega \times \mathcal{F} \to [0,1]$ such that

1. $\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F)$ is a version of $\mathbf{P}(F \mid \sigma({X}))$,
2. $\forall \omega \in \Omega$, $F \mapsto \mathbf{P}^2(\omega, F)$ is a probability measure on $(\Omega, \mathcal{F})$.

In particular, property 1 tells us that $\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F)$ is $(\sigma(X), \mathcal{B}[0,1])$-measurable. Then, by Theorem 18 listed below in Conrado Costa's answer, there exists a function $\mathbf{P}^1: E \times \mathcal{F} \to [0,1]$ that is $(\mathcal{E}, \mathcal{B}[0,1])$-measurable such that $\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F) = \mathbf{P}^1(X(\omega),F)$.

So, for any $F \in \mathcal{F}$ and any $C \in \sigma(X)$, by the change of variables theorem we get \begin{align*} \int_{X(C)} \mathbf{P}^1(x,F) \, d\Lambda_X(x) & = \int_C \mathbf{P}^2(\omega,F) \, dP(\omega) \\ & = \int_C \mathbf{P}(F \mid \sigma(X))(\omega) \, dP(\omega) \\ & = \int_C E(\mathbb{I}_F \mid \sigma(X))(\omega) \, dP(\omega) \\ & = \int_C \mathbb{I}_F(\omega) \, dP(\omega) \\ & = P(F \cap \{X \in B\}). \end{align*}

This is a subtle question.

you wrote: \begin{align*} &1.\quad \forall F \in \mathcal{F}, x \mapsto \mathbf{P}(x,F) \text{ is a version of } \mathbf{P}(F \mid \sigma(X)),\\ &2.\quad\text{ for almost every } x \in \mathbb{R}, F \mapsto \mathbf{P}(x,F) \text{ is a probability measure on } (\Omega, \mathcal{F}). \end{align*}

for 1 It is important to note that the first $\mathbf{P}$ the one atributed to $\mathbf{P}(x,F)$ is a function from $\Bbb{R} \times \mathcal{F}$ into $[0,1]$, the second $\mathbf{P}$ (atributed to $\mathbf{P}(F \mid \sigma(X))$ is a function from $\Omega \times \mathcal{F}$ into $[0,1]$.

So Let's give them different names, call the first one $\mathbf{P}^1(x,F)$ and the second one $\mathbf{P}^2(\omega, F)$

The second one is readily defined from your previous consideration as you already know how to deal with regular conditional probability distributions.

The first one needs the following theorem: (taken from Meyers Probability and potentials, page 10)

It tells us that $\mathbf{P}^2(\omega,F) = P^1(X(\omega),F)$ Use this $\mathbf{P}^1(x,F)$ as your object and then you can safely say that

\begin{align*} \int_B \mathbf{P}^1(x,F) \, \mathrm{d}\Lambda_X(x) & = \int_C \mathbf{P}^1(X(\omega), F) \, \mathrm{d} P(\omega) = \int_C\mathbf{P}^2(\omega, F) \, \mathrm{d} P(\omega) \; (\ldots)\\ \end{align*}

• Thanks for the great explanation. Do you have a good definition for the regular conditional probability given $X$, then? – bcf Sep 15 '15 at 23:54
• :) The best definition I know is on the Book of Karatzas Brownian motion and Stochastoc Calculus on page 307. Should you need more detail just ask. – Conrado Costa Sep 15 '15 at 23:57
• I updated by question based on your answer, would you mind seeing if my understanding is correct? – bcf Sep 20 '15 at 15:37
• I.e., $\mathbf{P}^1$ is the $h$ in Theorem 18, which is guaranteed to exist as soon as $\mathbf{P}^2$ (which would be $g$ in the theorem) exists? – bcf Sep 20 '15 at 15:46
• Your understanding is correct. – Conrado Costa Sep 20 '15 at 19:26