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Suppose $X_t$ is a collection of random variables from a measure space $(\Omega, \mathcal{F}, P)$ to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ (to keep things simple), where $t \in \mathbb{R}$. Moreover let $\tau$ denote a random variable from $(\Omega, P)$ with values in $\mathbb{R}$. If $\mathcal{G}$ is a sub $\sigma$-algebra of $\mathcal{F}$ and these two conditions hold: 1) $X_t$ is independent of $\mathcal{G} \ \forall t\in \mathbb{R}$ 2) $\tau$ is $\mathcal{G}$-measurable; is it true that \begin{equation} \mathbb{E}[X_{\tau} \mid \mathcal{G}] = g(\tau) \end{equation} where $g(s)= \mathbb{E}[X_s]$? If $\tau$ takes only finitely many values (say $\tau(\omega) \in \{t_1,...,t_n\} \ \forall \omega \in \Omega$) the statement holds since we can write $X_{\tau}$ as

\begin{equation} X_{\tau} = \sum_{i=1}^{n} \mathbb{1}_{\{\tau = t_i\}}X_{t_i} \end{equation} and the statement becomes a simple application of the very well known fact for measurable functions $\phi(X,Y)$ where $X$ and $Y$ are random variables independent from $\mathcal{G}$ and $\mathcal{G}$-measurable respectively. What about the general case? Can I use some approximation argument based on the finite valued case I have already shown?

Edit: Thanks for the answers. As it has been shown we have at least to suppose the measurability of $X_{\tau}$, so what can be said in this case?

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  • $\begingroup$ How do you ensure measurability of $X_\tau$? $\endgroup$ Commented Mar 11 at 21:23
  • $\begingroup$ @unwissen Given the assumption of independence, a measurable version can be constructed using the product measure and product spaces (assuming both processes exist separately). $\endgroup$ Commented Mar 11 at 21:43
  • $\begingroup$ @IosifPinelis Ah, you’re right that as stated in the book i cannot apply the freezing lemma (just checked). I will delete my answer until (and if) i find a fix. $\endgroup$ Commented Mar 11 at 21:58
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented Mar 14 at 1:10
  • $\begingroup$ I think you should accept the answer of Iosif and maybe ask a new question with the assumption that $(t, \omega) \mapsto X_t(\omega)$ be jointly measurable. Then $X_\tau$ should be measurable and the map $t \mapsto \mathbb{E}[X_t]$ too, if you add integrability conditions (also for $X_\tau$) to avoid further trivial counterexamples. $\endgroup$ Commented Mar 14 at 10:16

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$\newcommand\G{\mathcal G}\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}\newcommand\R{\mathbb R}\newcommand\Om{\Omega}\newcommand\om{\omega}$

This will not hold in general, for the simple reason that $X_\tau$ may be non-measurable.

Indeed, let $(\Om,\F,P)=(\R,\B(\R),N(0,1))$ and let $\G=\F$. Let $f\colon\R\to\R$ be a non-Borel-measurable function.

For each $t\in\R$ and all $\om\in\Om$, let $$X_t(\om):=f(t),$$ so that $X_t$ is a constant random variable (r.v.) and hence independent of any $\sigma$-algebra over $\Om$, so that your condition 1) holds. (One may also note that here $g=f$.)

Let $\tau(\om):=\om$ for all $\om\in\Om$, so that your condition 2) holds as well.

Letting now $Y:=X_\tau$, we have $Y(\om)=X_{\tau(\om)}(\om)=X_\om(\om)=f(\om)$ for all $\om\in\Om$, so that $X_\tau=Y=f$, so that the function $X_\tau\colon\Om\to\R$ is not measurable wrt the $\sigma$-algebras $\F$ and $\B(\R)$ over $\Om$ and $\R$, respectively. $\quad\Box$

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