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 $M$)


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.

  • 1
    $\begingroup$ 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 $C_b(M)$. $\endgroup$ Dec 29, 2019 at 19:40
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 29, 2019 at 19:42

1 Answer 1


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.