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We have the following definition for a conditional expectation:

Let $X: \Omega \rightarrow \mathbb{R}$ be a randomn variable on ($\Omega, \Sigma, \mathbb{P})$. Let $F \subset \Sigma$ be a sub $\sigma-$ algebra, then $Z$ is called a conditional expectation of $X $ under $F$, if $Z$ is $F$ measurable and $\mathbb{E}(X \chi_C) = \mathbb{E}(Z \chi_C)$ for all $C \in F$. What I don't understand here is: What is wrong with $Z=X$? In this case $X$ is $F$ measurable as $X$ is $\Sigma$-measurable and we certainly have $\mathbb{E}(X \chi_C) = \mathbb{E}(X \chi_C)$, so somehow I don't trust this definition. Did I copy anything wrong from the blackboard?

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  • $\begingroup$ If $F\subset\Sigma$, the fact that $X$ is $\Sigma$-measurable does not imply that $X$ is $F$-measurable. $\endgroup$ – Did Jun 5 '14 at 19:38
  • $\begingroup$ ah, but otherwise this definition is correct? $\endgroup$ – user66906 Jun 5 '14 at 19:40
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    $\begingroup$ Everything before "What I don't understand" is correct, except that one should assume that $X$ is integrable. $\endgroup$ – Did Jun 5 '14 at 19:41
  • $\begingroup$ thank you, is there a general equation that defines $Z$? I mean okay, apparently this definition defines $Z$ uniquely, but it does not tell me how I can actually determine $Z$. $\endgroup$ – user66906 Jun 5 '14 at 20:00
  • $\begingroup$ No, there is only this definition--and some special cases such as X F-measurable or X independent of F. $\endgroup$ – Did Jun 5 '14 at 20:09

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