# Defining the conditional distribution of $Z$ as $E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$

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.

• 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 $C_b(M)$. Dec 29, 2019 at 19:40
• 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. Dec 29, 2019 at 19:42

$$\newcommand{\M}{\mathcal M}$$ $$\newcommand{\G}{\mathcal G}$$ $$\newcommand{\F}{\mathcal F}$$ $$\newcommand{\P}{\mathsf P}$$ $$\newcommand{\E}{\mathsf E}$$ 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 $$\Omega\ni\omega\mapsto\mu_Z(\omega,\cdot)$$.