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I have recently come across below result.

Proposition 12.4. Let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}$ and $X, Y$ two random variables such that $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable. Let $\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$ be Borel-measurable such that $\mathbb{E}[|\varphi(X, Y)|] < \infty$. Then $$ \mathbb{E}[ \varphi(X, Y) | \mathcal{G}] = \psi(Y) \quad \text { a.s.} \quad \text{where} \quad \psi(y) := \mathbb{E}[ \varphi(X, y)]. $$

A direct consequence is that $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is $\sigma(Y)$-measurable.

Can we prove that $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is $\sigma(Y)$-measurable without knowing $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}] = \psi(Y)$?

Update It seems I was wrong in claiming "$\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is $\sigma(Y)$-measurable".

  • Let $Z$ be an integrable random variable. Any version of $\mathbb E[Z | \mathcal G]$ must be $\mathcal G$-measurable in the standard sense.

  • Assume $U$ is a random variable such that $\mathbb E[U1_B] = \mathbb E[Z 1_B]$ for all $B \in \mathcal G$. Then $U$ is equal to every version of $\mathbb E[Z | \mathcal G]$ almost everywhere, but $U$ is not necessarily a version of $\mathbb E[Z | \mathcal G]$.

Analyst
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  • I do not think so, excepted if you already know some properties of conditional independence... which are proved in the same way. – Christophe Leuridan Jan 15 '23 at 14:02
  • @ChristopheLeuridan It seems from this thread that I was wrong in claiming $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is $\sigma(Y)$-measurable. Could you confirm this? – Analyst Jan 15 '23 at 14:09
  • Remember that the conditional expectation is defined up to a negligible set. So $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is $\sigma(Y)$-measurable in the sense that it is almost surely equal to a $\sigma(Y)$-measurable random variable, namely $\psi(Y)$. – Will Jan 15 '23 at 14:34
  • @Will IMHO, if we allow being $\sigma(Y)$-measurable in "the sense" you mentioned, it would cause some confusion. For example, let's consider $\mathbb E[Z | \mathcal G]$ for some random variable $Z$. If $U$ is a random variable such that $\mathbb E[U1_B] = \mathbb E[Z 1_B]$ for all $B \in \mathcal G$. Then $U = \mathbb E[Z | \mathcal G]$ a.s. and thus $U$ is $\mathcal G$-measurable in "the sense" you mentioned, but it's not necessarily $\mathcal G$-measurable in the standard sense. – Analyst Jan 15 '23 at 14:53
  • Usually in the definition of the conditional expected value we add that $U$ is $\mathcal G$-measurable. Otherwise $U=Z$ satisfies the condition you mentioned and you would deduce that any random variable is (up to almost sure equality) $\mathcal G$-measurable, which is of course wrong. – Will Jan 15 '23 at 15:03
  • And again, conditional expected value is defined up to a negligible set. There is no such thing as THE conditional expected value. Strictly speaking you must speak of A conditional expected value. With that in mind, speaking of measurability of a conditional expected value can only be meaningful if you allow flexibility up to a negligible set. – Will Jan 15 '23 at 15:05
  • @Will That's exactly what I wanted to say. I meant being $\mathcal G$-measurable in the sense used in the definition of conditional expectation is not the same as being $\mathcal G$-measurable in the sense mentioned in your third comment. – Analyst Jan 15 '23 at 15:05
  • @Will Any version of $\mathbb E[Z | \mathcal G]$ must be $\mathcal G$-measurable in the standard sense. Assume a random variable $U$ is such that $\mathcal G$-measurable in the sense mentioned in your third comment, and that $\mathbb E[U1_B] = \mathbb E[Z 1_B]$ for all $B \in \mathcal G$. Then $U$ is equal to every version of $\mathbb E[Z | \mathcal G]$ almost everywhere, but $U$ is not necessarily a version of $\mathbb E[Z | \mathcal G]$. [...] – Analyst Jan 15 '23 at 15:17
  • [...] My final goal is to verify that $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ is not necessarily $\sigma(Y)$-measurable in the standard sense. I'm sorry for being stubborn. I just want to make sure I understand related definitions/concepts correctly... – Analyst Jan 15 '23 at 15:17
  • If by $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ you mean one of the almost surely equal $\mathcal G$- random variables which satisfy the definition of the conditional expectation, then no it is not necessarily $\sigma(Y)$-measurable. But at least one of them is $\sigma(Y)$-measurable. – Will Jan 15 '23 at 15:30
  • @Analyst $E[\varphi(X,Y)|\mathcal{G}$ is an equivalence class of random variables for the almost sure equality. In this equivalence class, there is at least one measurable $\mathcal{G}$ random variable. – Christophe Leuridan Jan 15 '23 at 15:31
  • @Will Thank you so much for your verification! This is exactly what I'm looking for. Could you post your comment as an answer? In my usage, I say $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ satisfies a property if and only every version of $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ satisfies that property. – Analyst Jan 15 '23 at 15:39
  • @ChristopheLeuridan by $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$, I meant a random variable $Z:\Omega\to \mathbb R$ that satisfies two conditions of conditional expectation. Of course, being $\mathcal G$-measurable does not make sense for an equivalence class. – Analyst Jan 15 '23 at 15:44
  • My comments did not really answer your question whether it is possible to deduce the measurability without the so called freezing lemma – Will Jan 15 '23 at 18:15
  • @Will You have already confirmed that there are possibly versions of $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}]$ that are not $\sigma(Y)$-measurable in the standard sense... – Analyst Jan 15 '23 at 19:03
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    Indeed a possible answer could be "We cannot because the version you chose for the conditional expectation might not be $\sigma(Y)$-measurable". But many people would believe I am playing the fool, as they would like an answer to the much more interesting question: "how can we prove that there exists a $\sigma(Y)$-measurable version of the conditional expectation without using the freezing lemma?". Do you see my point? – Will Jan 15 '23 at 20:11
  • @Will I got your point. Anyway, I very appreciate your comments. Thank you so much for your help! – Analyst Jan 15 '23 at 20:12

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