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I'd like to check my understanding on the following, and I'd appreciate a thorough combing with the rigor comb. I'm following the notation in the excellent answer given here. In the end, I'd like to string together three different integral representations for $\rm{E}[Y \mid X = x]$.

Let $(\Omega,\mathcal{F},P)$ be a probability space and let $X: \Omega \to S \subset \mathbb{R}$, $Y: \Omega \to T \subset \mathbb{R}$ be random variables.

Let $P^X(\cdot \mid \cdot): \mathcal{F} \times S \to \mathbb{R}$ be the regular conditional probability of $P$ given $X$. Then $$ \rm{E}[Y \mid X=x] = \int_\Omega Y(\omega)\, P^X(d\omega | x). \qquad (1) $$ Let $P_{Y \mid X}(\cdot \mid \cdot): \mathcal{B}(T) \times S \to \mathbb{R}$ be the regular conditional distribution of $Y$ given $X$. Then $$ \int_\Omega Y(\omega)\, P^X(\rm{d}\omega | x) = \int_T y\, P_{Y \mid X}(\rm{d}y | x). \qquad (2) $$ Finally, recall for fixed $x \in S$ that the mapping $B \mapsto P_{Y \mid X}(B \mid x)$ is a probability measure on $(T,\mathcal{B}(T))$ for $B \in \mathcal{B}$ by definition of a regular conditional distribution. Let $\lambda$ be the Lebesgue measure on $(T,\mathcal{B}(T))$. Then for fixed $x \in S$, since $P_{Y \mid X}(B|x)$ and $\lambda$ are $\sigma$-finite on $(T,\mathcal{B}(T))$, if $P_{Y \mid X}(\cdot \mid x) << \lambda$ then there exists a Radon-Nikodym derivative denoted by $h(\cdot \mid x): T \to \mathbb{R}$ such that $$ \int_T y\, P_{Y \mid X}(\rm{d}y | x) = \int_T y\, h(y|x)\, \lambda(\rm{d}y). \qquad (3) $$ Assuming (1) - (3) are correct, we get $$ \rm{E}[Y \mid X=x] = \int_\Omega Y(\omega)\, P^X(d\omega | x) = \int_T y\, P_{Y \mid X}(\rm{d}y | x) = \int_T y\, h(y|x)\, \lambda(\rm{d}y). $$

Does everything seem correct? Any suggestions for improving this exposition?

Update: Added absolute continuity condition $P_{Y \mid X}(\cdot \mid x) << \lambda$ on $(T, \mathcal{B}(T))$ as required by the Radon-Nikodym theorem (thanks @Did). I don't think this is guaranteed, though, unless maybe $Y$ is a continuous random variable (?) This means the distribution of $Y$ is absolutely continuous w.r.t $\lambda$, but I'm not sure it would imply $P_{Y \mid X}(\cdot \mid x) << \lambda$,

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  • $\begingroup$ Re the introduction of $h$: why two measures being sigma-finite on the same measurable space should imply the existence of a Radon-Nykodym derivative? $\endgroup$ – Did Jul 17 '15 at 14:27
  • $\begingroup$ @Did Thanks, added absolute continuity and updated my question. $\endgroup$ – bcf Jul 17 '15 at 14:47
  • $\begingroup$ Indeed absolute continuity is not guaranteed, this is even a rather unnatural hypothesis. Example: $X=Y$ almost surely. $\endgroup$ – Did Jul 17 '15 at 15:13

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