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I've been reading the first section Furstenberg's Noncommuting Random Products and I am confused with how he is defining conditional distribution.

Here he is considering a group $G$ acting on a space $M$. For a $M$-valued random variable $Z$, he defies the distribution $E^{*}[Z]$ of $Z$ as the functional on $C_{b}(M)$ (bounded continuous real valued functions on $M$)

$$E^{*}[Z](f)=E[f(Z)]$$

I interpret this as integrating $f$ with respect to the distribution of $Z$.

He then considers $Z$ to be a random variable on some $\Omega$ with $\sigma$-algebra $\mathcal{F}$, and defines the conditional distribution $$E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$$

and states that $E^{*}{[Z|\mathcal{F}]}$ is itself a random variable with values in the space of probability measures on $M$.

I'm confused because $E[f(Z)|\mathcal{F}]$ looks to be the conditional expectation of $f(Z)$ with respect to $\mathcal{F}$ which is itself a random variable on $\Omega$. I'm also not seeing how $E^{*}[Z|\mathcal{F}]$ is itself a random variable.

Pardon my ignorance.

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    $\begingroup$ Here is a sketchy answer. As you correctly guessed, $E^*[Z]$ corresponds to what one uses to call the “law”, or “distribution”, of $Z$. $E^*[Z|\mathcal{F}]$ is the conditional law of $Z$, which is a law-valued random variable. Morally, for a given $\omega \in \Omega$, assume that you already have all the information on $\mathcal{F}$ (at point $\omega$): then it remains some uncertainty on $\omega$, and the value of $E^*[Z|\mathcal{F}]$ at $\omega$ is the law of $Z$ when you consider that remaining uncertainty. So, $E^*[Z|\mathcal{F}]$ maps $\Omega$ to the dual space of $C_b(M)$. $\endgroup$
    – Rémi Peyre
    Dec 29, 2019 at 19:40
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    $\begingroup$ PS. Your question is not research-level and hence should rather be asked on math.stackexchange. Which is why I did not write a full-detail answer. $\endgroup$
    – Rémi Peyre
    Dec 29, 2019 at 19:42

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$\newcommand{\M}{\mathcal M}$ $\newcommand{\G}{\mathcal G}$ $\newcommand{\F}{\mathcal F}$ $\newcommand{\P}{\mathsf P}$ $\newcommand{\E}{\mathsf E}$ Suppose that $M$ is a Polish (i.e., complete separable metrizable) space with the Borel sigma-algebra $\M$ over it. Let $Z$ be an $M$-valued random variable (r.v.) defined on a probability space $(\Omega,\G,\P)$. Let $\F$ be a sub-sigma-algebra of $\G$. The key here is that then there exists a so-called regular conditional probability distribution (Theorems 1.13 and 1.17, and Remark 1.7 on pp. 8--9) $\mu_Z\colon \Omega\times\M\to[0,1]$ such that

  1. for each $\omega\in\Omega$, the map $\M\ni B\mapsto \mu_{Z;\omega}(B):=\mu_Z(\omega,B)$ is a probability measure and

  2. for each $B\in\M$, $\P(Z\in B|\F)=\mu_Z(\cdot,B)$ almost surely (a.s.).

So, for each $\omega\in\Omega$ and each $f\in C_b(M)$, we can introduce $\mu_Z(f)(\omega):=\int_M f\,d\mu_{Z;\omega}$; then it is easy to see that $$\mu_Z(f)=\E(f(Z)|\F)=E^*[Z|\F](f). $$ a.s. So, we can identify $E^*[Z|\F]$ with $\mu_Z$, which can in turn be identified with the random probability measure $\Omega\ni\omega\mapsto\mu_Z(\omega,\cdot)$.

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  • $\begingroup$ Thank you for the helpful answer and reference! $\endgroup$
    – user135520
    Dec 30, 2019 at 4:10

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