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I would like to show or see a proof that the conditional expected value $ E[X|Y] $ is a function of $Y$ for arbitrary random variables $X,Y$ (especially not discrete or continuous).

I think for discrete random variables this is easy to see however I cannot seem to wrap my head around the general case. I have read that the Doob-Dynkin lemma should do the trick however I do not see how.

Any help would be very much appreciated.

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    $\begingroup$ What is your starting point definition of $E[X|Y]$? $\endgroup$
    – Michael
    Commented Apr 24 at 12:09
  • $\begingroup$ We have only defined it for discrete random variables in lecture so far as $ E[X|Y] = \sum_{y} E[X|Y=y] 1_{Y=y} $ $\endgroup$
    – user007
    Commented Apr 24 at 12:11
  • $\begingroup$ This has been answered many times. $\endgroup$
    – geetha290krm
    Commented Apr 24 at 12:13
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    $\begingroup$ So the general measure theory version of your question is "Why is a function that is $\sigma(Y)$-measurable a (measurable) function of $Y$?" The non-measure theory version of your question may just want to define $E[X|Y]$ by $g(Y)$ where $g(y)=E[X|Y=y]$. $\endgroup$
    – Michael
    Commented Apr 24 at 12:17
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    $\begingroup$ Oh I see why the Doob-Dynkin lemma solves the problem. Thanks @Michael ! Since the random variable is $ \sigma (Y)$ measurable it has to be a function of $Y$. Am I correct? $\endgroup$
    – user007
    Commented Apr 24 at 12:21

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