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Let $(\Omega, \mathcal{F}, \mathbf{P})$ be a probability space, $(\mathcal{X}, \mathcal{B})$ a measurable space, and $X : \Omega \to \mathcal{X}$ a random element of $\mathcal{X}$. Also, let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{F}$.


Question. How unique are regular conditional distributions of $X$ given $\mathcal{G}$?


A regular conditional distribution of $X$ given $\mathcal{G}$ is a function $P : \Omega \times \mathcal{B} \to [0, 1]$ such that the following properties hold.

  1. For all $\omega \in \Omega$, the map $B \mapsto P(\omega, B)$ from $\mathcal{B}$ into $[0, 1]$ is a probability measure on $(\mathcal{X}, \mathcal{B})$.
  2. For all $B \in \mathcal{B}$, the map $\omega \mapsto P(\omega, B)$ from $\Omega$ into $[0, 1]$ is $(\mathcal{G}, \mathcal{B}_{[0, 1]})$-measurable (where $\mathcal{B}_{[0, 1]}$ denotes the Borel $\sigma$-algebra of $[0, 1]$).
  3. For all $B \in \mathcal{B}$ and all $G \in \mathcal{G}$, we have $$ \mathbf{P}(\{X \in B\} \cap G) = \int_G P(\cdot, B) \, d\mathbf{P}. $$

(Items 2. and 3. just say that, for each $B \in \mathcal{B}$, the random variable $P(\cdot, B)$ is a version of the conditional probability $\mathbf{P}(X \in B\mid \mathcal{G})$.)

Suppose $P$ and $Q$ are two regular conditional distributions of $X$ given $\mathcal{G}$.

On the one hand, it is not necessarily true that $P(\omega, B) = Q(\omega, B)$ for all $\omega \in \Omega$ and $B \in \mathcal{B}$. For example, for any $\mathbf{P}$-null set $N \in \mathcal{F}$ and any probability measure $\mu$ on $(\mathcal{X}, \mathcal{B})$, we can define $P^\prime : \Omega \times \mathcal{B} \to [0, 1]$ by $$ P^\prime(\omega, B) = \begin{cases} P(\omega, B), & \text{if $\omega \notin N$,} \\ \mu(B), & \text{if $\omega \in N$.} \end{cases} $$ Then $P^\prime$ is another regular conditional distribution of $X$ given $\mathcal{G}$, but it might hold that $P(\omega, B) \neq P^\prime(\omega, B)$ for some $\omega \in \Omega$ and $B \in \mathcal{B}$.

On the other hand, suppose $B \in \mathcal{B}$ is fixed. Then we have $$ \int_G P(\cdot, B) \, d\mathbf{P} = \mathbf{P}(\{X \in B\} \cap G) = \int_G Q(\cdot, B) \, d\mathbf{P} $$ for every $G \in \mathcal{G}$. Since $P(\cdot, B)$ and $Q(\cdot, B)$ are $\mathcal{G}$-measurable, this implies that there exists a $\mathcal{P}$-null set $N \in \mathcal{F}$ such that $P(\omega, B) = Q(\omega, B)$ for all $\omega \in \Omega \setminus N$. However, this null set depends on $B$, so we can't a priori conclude that there exists a $\mathbf{P}$-null set $N^\prime \in \mathcal{F}$ such that $P(\omega, B) = Q(\omega, B)$ for all $\omega \in \Omega \setminus N^\prime$ and all $B \in \mathcal{B}$.


More Precise Question. Suppose $P$ and $Q$ are two regular conditional distributions of $X$ given $\mathcal{G}$. Does there always exist a $\mathbf{P}$-null set $N \in \mathcal{F}$ such that $$ P(\omega, B) = Q(\omega, B) $$ for all $\omega \in \Omega \setminus N$ and all $B \in \mathcal{B}$?


I think I remember reading that this is true somewhere, but I can't find a proof. I'm fine with assuming that any measurable spaces in question are standard Borel, if needed.

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This is true if $\mathcal{B}$ is countably generated. Specifically, $$ P(\omega,A)=Q(\omega,A) \quad\text{a.s.} \tag{1}\label{1} $$ for all $A\in \mathcal{A}$ (a countable algebra that generates $\mathcal{B}$). Therefore, there exists a $\mathbf{P}$-null set $N$ s.t. $\eqref{1}$ holds for all $A\in\mathcal{A}$ and all $\omega\in \Omega\setminus N$. Now extrapolate this result to $\mathcal{B}$.

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  • $\begingroup$ Perfect, thank you! The monotone class theorem does the trick to extend to $\mathcal{B}$ $\endgroup$ – Artem Mavrin Jan 29 at 16:52

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