Following the excellent answer here, it is stated that
Connection: Let $P^X(\cdot\mid\cdot)$ be a regular conditional probability of $P$ given $X$. Then for any $A \in \mathcal{F}$ we have $$ {\rm E}[1_A\mid X]=\varphi(X), $$ where $\varphi(x) = P^X(A\mid x)$. In short we write ${\rm E}[1_A\mid X]=P^X(A\mid X)$.
I'm trying to prove this for myself, but I'd appreciate some help. We need to show two things:
- $P^X(A\mid X)$ is $\sigma(X)$-measurable
- $\rm{E}[1_C P^X(A\mid X)] = \rm{E}[1_C 1_A]$ for all $C \in \sigma(X)$.
Since $P^X(\cdot\mid\cdot)$ is a regular conditional probability, we know that for fixed $A \in \mathcal{F}$, the mapping $x \mapsto P^X(A\mid x)$ is $(\mathcal{B}(\mathbb{R}),\mathcal{B}(\mathbb{R}))$-measurable. Hence the composition $P^X(A\mid \cdot) \circ X: \Omega \to \mathbb{R}$ is a random variable for fixed $A \in \mathcal{F}$, which the author writes as $P^X(A\mid \cdot) \circ X = P^X(A|X)$. Thus $P^X(A\mid X)$ is $\sigma(X)$-measurable, and we have 1.
For 2, recall that for all $A \in \mathcal{F}$ and $B \in \mathcal{B}(\mathbb{R})$ the conditional probability is defined to satisfy $$ \int_B P^X(A\mid x) \,d\mu_X(x) = P(A \cap \{X \in B\}), $$ where $\mu_X$ is the distribution measure of $X$ on $\mathcal{B}(\mathbb{R})$. This is the part I'm having trouble with. Here's my attempt: \begin{align*} \rm{E}[1_C P^X(A\mid X)] & = \int_{\Omega} 1_C(\omega)P^X(A\mid X)(\omega) dP(\omega) \\ & = \int_{\mathbb{R}} 1_{\{1\}}(x) P^X(A\mid x)d\mu_X(x) &\qquad& (2) \\ & = \int_{\{1\}}P^X(A\mid x)d\mu_X(x) \\ & = P(A \cap \{X = 1\}) \\ & = \int_\Omega 1_A(\omega) 1_{\{X = 1\}}(\omega) dP(\omega) \\ & = \int_\Omega 1_A(\omega) 1_C(\omega) dP(\omega) &\qquad& (6) \\ & = \rm{E}[1_A 1_C]. \end{align*}
I'm almost certain lines (2) and (6) aren't true, but I'm not sure how else to write the indicator random variable when transitioning to the distribution measure in (2). For (6), I don't see why $\{X = 1\} = C$, either, but it's my best shot at it right now :/
