Let $(\Omega, \mathcal F, P)$ be a probability space, and let $\mathcal G \subseteq \mathcal F$ be a sub-$\sigma$-algebra of $\mathcal F$ and $X : \Omega \to \mathbb R$ a random variable. Then the conditional expection of $X$ conditioned on $\mathcal G$ is defined to be the a.e. unique random variable $Y$ such that

(i) $Y$ is $\mathcal G$-measurable, and

(ii) for each $A \in \mathcal G$ we have $$ E[X\cdot 1_A] = E[Y\cdot 1_A]. $$ And it is denoted as $Y := E[X | \mathcal G]$. For a random variable $Z$ the conditional expection conditioned on $Z$ is defined as $E[X | Z] := E[X | \sigma(Z)]$.

These two notions are equivalent, so now my question. Given a condtional expectation $E[X | \mathcal G]$ w.r.t. some sub-$\sigma$-algebra $\mathcal G$, how to find a random variable $Z$ such that $$ E[X | Z] = E[X | \mathcal G] \quad \mbox{a.e.}? $$


1 Answer 1


If you want a real-valued random variables $Z$, then $\mathcal G$ must be countably generated. (At least up to null sets.)

Suppose $\mathcal G$ is countably generated. We want to find a random variable $Z \colon \Omega \to \mathbb R$ such that $\sigma(Z) = \mathcal G$. Say $\mathcal G$ is generated by $A_1, A_2, \dots$. How about $$ Z = \sum_{k=1}^\infty 3^{-k} \mathbb 1_{A_k} $$ (here $\mathbb 1_A$ is the indicator function of $A$).

On the other hand, if $\mathcal G = \sigma(Z)$, then $\mathcal G$ is generated by the countable family $E_r = \{Z>r\}$, where $r$ ranges over the rationals.

  • $\begingroup$ Could the conclusion be drawn that the two notions of conditional expectations are not equivalent because you can not always translate from $E[X|\mathcal G]$ to $E[X|Z]$? $\endgroup$
    – StefanH
    Jan 23, 2015 at 21:37
  • 3
    $\begingroup$ The two notions are equivalent in standard Borel space, which includes all (or almost all) situations of interest in probability theory. But, in the abstract, the two notions are not equivalent. $\endgroup$ Jan 23, 2015 at 21:41

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