Can anyone help me in an intuitive manner how to simplify the following conditional probabilities and expectations - $E(Y|X)|X$ and $E(E(Y|X)|X)$ -? I am trying to use their literal interpretation (the expectation of $Y$ conditional on $X$ conditinalized on $X$) to see how to move forward, but that has failed me so far. any tips welcome.
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$\begingroup$ @User1885516 - please see derivation on files.nyu.edu/cds2083/public/docs/methods/lie_notes.pdf $\endgroup$– user148019May 5, 2014 at 15:08
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1 Answer
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First, the entity $E(Y|X)|X$ does not exist (and the phrase "the expectation of $Y$ conditional on $X$ conditinalized on $X$" is pure mumbo-jumbo).
Second, $E(E(Y|X)|X) = E(Y|X)$ by definition, because the random variable $E(Y|X)$ is $σ(X)$-measurable and because $E(Z|X) = Z$ for every $σ(X)$-measurable (integrable) random variable $Z$.
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$\begingroup$ apologies - on the first question - i misread the line - the original line stated E(E(Y−E(Y|X)|X)), and hence i made a mistakes with regards to the brackets. that said - can you expand more what σ(X)-measurable implies (or send a link)? unfortunately - my knowledge falls short here $\endgroup$ May 5, 2014 at 15:04
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$\begingroup$ This is all over the place, really. Williams' book Probability with martingales, Rosenthal's probability books, anywhere... Which book are you following yourself? $\endgroup$– DidMay 5, 2014 at 15:29
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$\begingroup$ And my answer implies that
E(E(Y−E(Y|X)|X))=0, and even thatE(Y−E(Y|X)|X)=0 a.s., do you see this? $\endgroup$– DidMay 5, 2014 at 15:30